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Assume that the experimenter has chosen a beta density to describe the state of his knowledge about \( x \) before the experiment. Then he gives the drug to \( n \) subjects and records the number \( i \) of successes. The number \( i \) is a discrete random variable, so we may conveniently describe the set of possible outcomes of this experiment by referring to the ordered pair \( \left( {x, i}\right) \).
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We let \( m\left( {i \mid x}\right) \) denote the probability that we observe \( i \) successes given the value of \( x \) . By our assumptions, \( m\left( {i \mid x}\right) \) is the binomial distribution with probability \( x \) for success:\n\[ m\left( {i \mid x}\right) = b\left( {n, x, i}\right) = \left( \begin{matrix} n \\ i \end{matrix}\right) {x}^{i}{\left( 1 - x\right) }^{j}, \]\nwhere \( j = n - i \).\n\nIf \( x \) is chosen at random from \( \left\lbrack {0,1}\right\rbrack \) with a beta density \( B\left( {\alpha ,\beta, x}\right) \), then the density function for the outcome of the pair \( \left( {x, i}\right) \) is\n\[ f\left( {x, i}\right) = m\left( {i \mid x}\right) B\left( {\alpha ,\beta, x}\right) \]\n\[ = \left( \begin{matrix} n \\ i \end{matrix}\right) {x}^{i}{\left( 1 - x\right) }^{j}\frac{1}{B\left( {\alpha ,\beta }\right) }{x}^{\alpha - 1}{\left( 1 - x\right) }^{\beta - 1} \]\n\[ = \left( \begin{matrix} n \\ i \end{matrix}\right) \frac{1}{B\left( {\alpha ,\beta }\right) }{x}^{\alpha + i - 1}{\left( 1 - x\right) }^{\beta + j - 1}. \]\n\nNow let \( m\left( i\right) \) be the probability that we observe \( i \) successes not knowing the value of \( x \) . Then\n\[ m\left( i\right) = {\int }_{0}^{1}m\left( {i \mid x}\right) B\left( {\alpha ,\beta, x}\right) {dx} \]\n\[ = \left( \begin{matrix} n \\ i \end{matrix}\right) \frac{1}{B\left( {\alpha ,\beta }\right) }{\int }_{0}^{1}{x}^{\alpha + i - 1}{\left( 1 - x\right) }^{\beta + j - 1}{dx} \]\n\[ = \left( \begin{matrix} n \\ i \end{matrix}\right) \frac{B\left( {\alpha + i,\beta + j}\right) }{B\left( {\alpha ,\beta }\right) }. \]\n\nHence, the probability density \( f\left( {x \mid i}\right) \) for \( x \), given that \( i \) successes were observed, is\n\[ f\left( {x \mid i}\right) = \frac{f\left( {x, i}\right) }{m\left( i\right) } \]\n\[ = \frac{{x}^{\alpha + i - 1}{\left( 1 - x\right) }^{\beta + j - 1}}{B\left( {\alpha + i,\beta + j}\right) } \]\n(4.5)\n\nthat is, \( f\left( {x \mid i}\right) \) is another beta density. This says that if we observe \( i \) successes and \( j \) failures in \( n \) subjects, then the new density for the probability that the drug is effective is again a beta density but with parameters \( \alpha + i,\beta + j \) .
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Yes
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Example 4.24 (Two-armed bandit problem) You are in a casino and confronted by two slot machines. Each machine pays off either 1 dollar or nothing. The probability that the first machine pays off a dollar is \( x \) and that the second machine pays off a dollar is \( y \) . We assume that \( x \) and \( y \) are random numbers chosen independently from the interval \( \left\lbrack {0,1}\right\rbrack \) and unknown to you. You are permitted to make a series of ten plays, each time choosing one machine or the other. How should you choose to maximize the number of times that you win?
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One strategy that sounds reasonable is to calculate, at every stage, the probability that each machine will pay off and choose the machine with the higher probability. Let \( \operatorname{win}\left( i\right) \), for \( i = 1 \) or 2, be the number of times that you have won on the \( i \) th machine. Similarly, let lose \( \left( i\right) \) be the number of times you have lost on the \( i \) th machine. Then, from Example 4.23, the probability \( p\left( i\right) \) that you win if you choose the \( i \) th machine is \[ p\left( i\right) = \frac{\operatorname{win}\left( i\right) + 1}{\operatorname{win}\left( i\right) + \operatorname{lose}\left( i\right) + 2}. \] Thus, if \( p\left( 1\right) > p\left( 2\right) \) you would play machine 1 and otherwise you would play machine 2. We have written a program TwoArm to simulate this experiment. In the program, the user specifies the initial values for \( x \) and \( y \) (but these are unknown to the experimenter). The program calculates at each stage the two conditional densities for \( x \) and \( y \), given the outcomes of the previous trials, and then computes \( p\left( i\right) \), for \( i = 1,2 \) . It then chooses the machine with the highest value for the probability of winning for the next play. The program prints the machine chosen on each play and the outcome of this play. It also plots the new densities for \( x \) (solid line) and \( y \) (dotted line), showing only the current densities. We have run the program for ten plays for the case \( x = {.6} \) and \( y = {.7} \) . The result is shown in Figure 4.10.
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Yes
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Consider a family with two children. Given that one of the children is a boy, what is the probability that both children are boys?
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One way to approach this problem is to say that the other child is equally likely to be a boy or a girl, so the probability that both children are boys is \( 1/2 \) . The \
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No
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Example 4.26 Mr. Smith is the father of two. We meet him walking along the street with a young boy whom he proudly introduces as his son. What is the probability that Mr. Smith's other child is also a boy?
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As usual we have to make some additional assumptions. For example, we will assume that if \( \mathrm{{Mr}} \) . Smith has a boy and a girl, he is equally likely to choose either one to accompany him on his walk. In Figure 4.13 we show the tree analysis of this problem and we see that \( 1/2 \) is, indeed, the correct answer.
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No
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Two envelopes each contain a certain amount of money. One envelope is given to Ali and the other to Baba and they are told that one envelope contains twice as much money as the other. However, neither knows who has the larger prize. Before anyone has opened their envelope, Ali is asked if she would like to trade her envelope with Baba. She reasons as follows: Assume that the amount in my envelope is \( x \) . If I switch, I will end up with \( x/2 \) with probability \( 1/2 \), and \( {2x} \) with probability \( 1/2 \) . If I were given the opportunity to play this game many times, and if I were to switch each time, I would, on average, get\n\n\[ \n\frac{1}{2}\frac{x}{2} + \frac{1}{2}{2x} = \frac{5}{4}x.\n\]\n\nThis is greater than my average winnings if I didn't switch.
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Of course, Baba is presented with the same opportunity and reasons in the same way to conclude that he too would like to switch. So they switch and each thinks that his/her net worth just went up by \( {25}\% \) .\n\nSince neither has yet opened any envelope, this process can be repeated and so again they switch. Now they are back with their original envelopes and yet they think that their fortune has increased \( {25}\% \) twice. By this reasoning, they could convince themselves that by repeatedly switching the envelopes, they could become arbitrarily wealthy. Clearly, something is wrong with the above reasoning, but where is the mistake?\n\nOne of the tricks of making paradoxes is to make them slightly more difficult than is necessary to further befuddle us. As John Finn has suggested, in this paradox we could just have well started with a simpler problem. Suppose Ali and Baba know that I am going to give then either an envelope with \( \$ 5 \) or one with \( \$ {10} \) and I am going to toss a coin to decide which to give to Ali, and then give the other to Baba. Then Ali can argue that Baba has \( {2x} \) with probability \( 1/2 \) and \( x/2 \) with probability \( 1/2 \) . This leads Ali to the same conclusion as before. But now it is clear that this is nonsense, since if Ali has the envelope containing \( \$ 5 \), Baba cannot possibly have half of this, namely \$2.50, since that was not even one of the choices. Similarly, if Ali has \( \$ {10} \), Baba cannot have twice as much, namely \( \$ {20} \) . In fact, in this simpler problem the possibly outcomes are given by the tree diagram in Figure 4.14. From the diagram, it is clear that neither is made better off by switching.\n\nIn the above example, Ali's reasoning is incorrect because he infers that if the amount in his envelope is \( x \), then the probability that his envelope contains the\n\n\n\nFigure 4.14: John Finn's version of Example 4.28.\n\nsmaller amount is \( 1/2 \), and the probability that her envelope contains the larger amount is also \( 1/2 \) . In fact, these conditional probabilities depend upon the distribution of the amounts that are placed in the envelopes.
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Yes
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Example 4.29 Suppose that we have two envelopes in front of us, and that one envelope contains twice the amount of money as the other (both amounts are positive integers). We are given one of the envelopes, and asked if we would like to switch.
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As above, we let \( X \) denote the smaller of the two amounts in the envelopes, and let\n\n\[ \n{p}_{x} = P\left( {X = x}\right) .\n\]\n\nWe are now in a position where we can calculate the long-term average winnings, if we switch. (This long-term average is an example of a probabilistic concept known as expectation, and will be discussed in Chapter 6.) Given that one of the two sample points has occurred, the probability that it is the point \( \left( {x, x/2}\right) \) is\n\n\[ \n\frac{{p}_{x/2}}{{p}_{x/2} + {p}_{x}}\n\]\n\nand the probability that it is the point \( \left( {x,{2x}}\right) \) is\n\n\[ \n\frac{{p}_{x}}{{p}_{x/2} + {p}_{x}}.\n\]\n\nThus, if we switch, our long-term average winnings are\n\n\[ \n\frac{{p}_{x/2}}{{p}_{x/2} + {p}_{x}}\frac{x}{2} + \frac{{p}_{x}}{{p}_{x/2} + {p}_{x}}{2x}.\n\]\n\nIf this is greater than \( x \), then it pays in the long run for us to switch. Some routine algebra shows that the above expression is greater than \( x \) if and only if\n\n\[ \n\frac{{p}_{x/2}}{{p}_{x/2} + {p}_{x}} < \frac{2}{3}.\n\]\n\n(4.6)\n\nIt is interesting to consider whether there is a distribution on the positive integers such that the inequality 4.6 is true for all even values of \( x \) . Brams and Kilgour \( {}^{22} \) give the following example.\n\nWe define \( {p}_{x} \) as follows:\n\n\[ \n{p}_{x} = \left\{ \begin{matrix} \frac{1}{3}{\left( \frac{2}{3}\right) }^{k - 1}, & \text{ if }x = {2}^{k}, \\ 0, & \text{ otherwise. } \end{matrix}\right.\n\]\n\nIt is easy to calculate (see Exercise 4) that for all relevant values of \( x \), we have\n\n\[ \n\frac{{p}_{x/2}}{{p}_{x/2} + {p}_{x}} = \frac{3}{5}\n\]\n\nwhich means that the inequality 4.6 is always true.
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Yes
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Suppose that we have two envelopes in front of us, and we are told that the envelopes contain \( X \) and \( Y \) dollars, respectively, where \( X \) and \( Y \) are different positive integers. We randomly choose one of the envelopes, and we open it, revealing \( X \), say. Is it possible to determine, with probability greater than \( 1/2 \) , whether \( X \) is the smaller of the two dollar amounts?
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Even if we have no knowledge of the joint distribution of \( X \) and \( Y \), the surprising answer is yes! Here's how to do it. Toss a fair coin until the first time that heads turns up. Let \( Z \) denote the number of tosses required plus \( 1/2 \) . If \( Z > X \), then we say that \( X \) is the smaller of the two amounts, and if \( Z < X \), then we say that \( X \) is the larger of the two amounts.\n\nFirst, if \( Z \) lies between \( X \) and \( Y \), then we are sure to be correct. Since \( X \) and \( Y \) are unequal, \( Z \) lies between them with positive probability. Second, if \( Z \) is not between \( X \) and \( Y \), then \( Z \) is either greater than both \( X \) and \( Y \), or is less than both \( X \) and \( Y \) . In either case, \( X \) is the smaller of the two amounts with probability \( 1/2 \) , by symmetry considerations (remember, we chose the envelope at random). Thus, the probability that we are correct is greater than \( 1/2 \) .
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Yes
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For example, suppose a line of customers waits for service at a counter. It is often assumed that, in each small time unit, either 0 or 1 new customers arrive at the counter. The probability that a customer arrives is \( p \) and that no customer arrives is \( q = 1 - p \) . Then the time \( T \) until the next arrival has a geometric distribution. It is natural to ask for the probability that no customer arrives in the next \( k \) time units, that is, for \( P\left( {T > k}\right) \) .
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This is given by\n\n\[ P\left( {T > k}\right) = \mathop{\sum }\limits_{{j = k + 1}}^{\infty }{q}^{j - 1}p = {q}^{k}\left( {p + {qp} + {q}^{2}p + \cdots }\right) \]\n\n\[ = {q}^{k}\text{.} \]
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Yes
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A fair coin is tossed until the second time a head turns up. The distribution for the number of tosses is \( u\left( {x,2, p}\right) \) . Thus the probability that \( x \) tosses are needed to obtain two heads is found by letting \( k = 2 \) in the above formula.
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\[ u\left( {x,2,1/2}\right) = \left( \begin{matrix} x - 1 \\ 1 \end{matrix}\right) \frac{1}{{2}^{x}}, \] for \( x = 2,3,\ldots \)
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Yes
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Example 5.3 A typesetter makes, on the average, one mistake per 1000 words. Assume that he is setting a book with 100 words to a page. Let \( {S}_{100} \) be the number of mistakes that he makes on a single page. Then the exact probability distribution for \( {S}_{100} \) would be obtained by considering \( {S}_{100} \) as a result of 100 Bernoulli trials with \( p = 1/{1000} \). The expected value of \( {S}_{100} \) is \( \lambda = {100}\left( {1/{1000}}\right) = {.1} \). The exact probability that \( {S}_{100} = j \) is \( b\left( {{100},1/{1000}, j}\right) \), and the Poisson approximation is
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\[ \frac{{e}^{-{.1}}{\left( {.1}\right) }^{j}}{j!} \]
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Yes
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Assume that you live in a district of size 10 blocks by 10 blocks so that the total district is divided into 100 small squares. How likely is it that the square in which you live will receive no hits if the total area is hit by 400 bombs?
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We assume that a particular bomb will hit your square with probability \( 1/{100} \) . Since there are 400 bombs, we can regard the number of hits that your square receives as the number of successes in a Bernoulli trials process with \( n = {400} \) and \( p = 1/{100} \) . Thus we can use the Poisson distribution with \( \lambda = {400} \cdot 1/{100} = 4 \) to approximate the probability that your square will receive \( j \) hits. This probability is \( p\left( j\right) = {e}^{-4}{4}^{j}/j \) !. The expected number of squares that receive exactly \( j \) hits is then \( {100} \cdot p\left( j\right) \) .
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Yes
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Example 5.6 It is often of interest to consider two traits, such as eye color and hair color, and to ask whether there is an association between the two traits. Two traits are associated if knowing the value of one of the traits for a given person allows us to predict the value of the other trait for that person. The stronger the association, the more accurate the predictions become. If there is no association between the traits, then we say that the traits are independent. In this example, we will use the traits of gender and political party, and we will assume that there are only two possible genders, female and male, and only two possible political parties, Democratic and Republican.
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Suppose that we have collected data concerning these traits. To test whether there is an association between the traits, we first assume that there is no association between the two traits. This gives rise to an \
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No
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Suppose that customers arrive at random times at a service station with one server, and suppose that each customer is served immediately if no one is ahead of him, but must wait his turn in line otherwise. How long should each customer expect to wait? (We define the waiting time of a customer to be the length of time between the time that he arrives and the time that he begins to be served.)
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Let us assume that the interarrival times between successive customers are given by random variables \( {X}_{1},{X}_{2},\ldots ,{X}_{n} \) that are mutually independent and identically distributed with an exponential cumulative distribution function given by\n\n\[ \n{F}_{X}\left( t\right) = 1 - {e}^{-{\lambda t}}. \n\]\n\nLet us assume, too, that the service times for successive customers are given by random variables \( {Y}_{1},{Y}_{2},\ldots ,{Y}_{n} \) that again are mutually independent and identically distributed with another exponential cumulative distribution function given by\n\n\[ \n{F}_{Y}\left( t\right) = 1 - {e}^{-{\mu t}}. \n\]\n\nThe parameters \( \lambda \) and \( \mu \) represent, respectively, the reciprocals of the average time between arrivals of customers and the average service time of the customers. Thus, for example, the larger the value of \( \lambda \), the smaller the average time between arrivals of customers. We can guess that the length of time a customer will spend in the queue depends on the relative sizes of the average interarrival time and the average service time.\n\nIt is easy to verify this conjecture by simulation. The program Queue simulates this queueing process. Let \( N\left( t\right) \) be the number of customers in the queue at time \( t \) .\n\n\n\nFigure 5.7: Queue sizes.\n\n\n\nThen we plot \( N\left( t\right) \) as a function of \( t \) for different choices of the parameters \( \lambda \) and \( \mu \) (see Figure 5.7).\n\nWe note that when \( \lambda < \mu \), then \( 1/\lambda > 1/\mu \), so the average interarrival time is greater than the average service time, i.e., customers are served more quickly, on average, than new ones arrive. Thus, in this case, it is reasonable to expect that \( N\left( t\right) \) remains small. However, if \( \lambda > \mu \) then customers arrive more quickly than they are served, and, as expected, \( N\left( t\right) \) appears to grow without limit.\n\nWe can now ask: How long will a customer have to wait in the queue for service? To examine this question, we let \( {W}_{i} \) be the length of time that the \( i \) th customer has to remain in the system (waiting in line and being served). Then we can present these data in a bar graph, using the program Queue, to give some idea of how the \( {W}_{i} \) are distributed (see Figure 5.8). (Here \( \lambda = 1 \) and \( \mu = {1.1} \).\n\nWe see that these waiting times appear to be distributed exponentially. This is always the case when \( \lambda < \mu \) . The proof of this fact is too complicated to give here, but we can verify it by simulation for different choices of \( \lambda \) and \( \mu \), as above.
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No
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Theorem 5.1 Let \( X \) be a continuous random variable, and suppose that \( \phi \left( x\right) \) is a strictly increasing function on the range of \( X \) . Define \( Y = \phi \left( X\right) \) . Suppose that \( X \) and \( Y \) have cumulative distribution functions \( {F}_{X} \) and \( {F}_{Y} \) respectively. Then these functions are related by\n\n\[ \n{F}_{Y}\left( y\right) = {F}_{X}\left( {{\phi }^{-1}\left( y\right) }\right) \n\] \n\nIf \( \phi \left( x\right) \) is strictly decreasing on the range of \( X \), then \n\n\[ \n{F}_{Y}\left( y\right) = 1 - {F}_{X}\left( {{\phi }^{-1}\left( y\right) }\right) . \n\]
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Proof. Since \( \phi \) is a strictly increasing function on the range of \( X \), the events \( \left( {X \leq {\phi }^{-1}\left( y\right) }\right) \) and \( \left( {\phi \left( X\right) \leq y}\right) \) are equal. Thus, we have\n\n\[ \n{F}_{Y}\left( y\right) = P\left( {Y \leq y}\right) \n\] \n\n\[ \n= P\left( {\phi \left( X\right) \leq y}\right) \n\] \n\n\[ \n= P\left( {X \leq {\phi }^{-1}\left( y\right) }\right) \n\] \n\n\[ \n= {F}_{X}\left( {{\phi }^{-1}\left( y\right) }\right) \text{.} \n\] \n\nIf \( \phi \left( x\right) \) is strictly decreasing on the range of \( X \), then we have\n\n\[ \n{F}_{Y}\left( y\right) = P\left( {Y \leq y}\right) \n\] \n\n\[ \n= \;P\left( {\phi \left( X\right) \leq y}\right) \n\] \n\n\[ \n= P\left( {X \geq {\phi }^{-1}\left( y\right) }\right) \n\] \n\n\[ \n= 1 - P\left( {X < {\phi }^{-1}\left( y\right) }\right) \n\] \n\n\[ \n= 1 - {F}_{X}\left( {{\phi }^{-1}\left( y\right) }\right) \text{.} \n\] \n\nThis completes the proof.
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Yes
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Corollary 5.1 Let \( X \) be a continuous random variable, and suppose that \( \phi \left( x\right) \) is a strictly increasing function on the range of \( X \) . Define \( Y = \phi \left( X\right) \) . Suppose that the density functions of \( X \) and \( Y \) are \( {f}_{X} \) and \( {f}_{Y} \), respectively. Then these functions are related by\n\n\[ \n{f}_{Y}\left( y\right) = {f}_{X}\left( {{\phi }^{-1}\left( y\right) }\right) \frac{d}{dy}{\phi }^{-1}\left( y\right) .\n\]\n\nIf \( \phi \left( x\right) \) is strictly decreasing on the range of \( X \), then\n\n\[ \n{f}_{Y}\left( y\right) = - {f}_{X}\left( {{\phi }^{-1}\left( y\right) }\right) \frac{d}{dy}{\phi }^{-1}\left( y\right) .\n\]
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Proof. This result follows from Theorem 5.1 by using the Chain Rule.
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No
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Suppose that \( X \) is a normally distributed random variable with parameters \( \mu = {10} \) and \( \sigma = 3 \) . Find the probability that \( X \) is between 4 and 16 .
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To solve this problem, we note that \( Z = \left( {X - {10}}\right) /3 \) is the standardized version of \( X \) . So, we have\n\n\[ P\left( {4 \leq X \leq {16}}\right) = P\left( {X \leq {16}}\right) - P\left( {X \leq 4}\right) \]\n\n\[ = {F}_{X}\left( {16}\right) - {F}_{X}\left( 4\right) \]\n\n\[ = {F}_{Z}\left( \frac{{16} - {10}}{3}\right) - {F}_{Z}\left( \frac{4 - {10}}{3}\right) \]\n\n\[ = {F}_{Z}\left( 2\right) - {F}_{Z}\left( {-2}\right) \text{.} \]\n\nThis last expression can be evaluated by using tabulated values of the standard normal distribution function (see 12.3); when we use this table, we find that \( {F}_{Z}\left( 2\right) = \) .9772 and \( {F}_{Z}\left( {-2}\right) = {.0228} \) . Thus, the answer is .9544 .
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Yes
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Suppose that we drop a dart on a large table top, which we consider as the \( {xy} \) -plane, and suppose that the \( x \) and \( y \) coordinates of the dart point are independent and have a normal distribution with parameters \( \mu = 0 \) and \( \sigma = 1 \) . How is the distance of the point from the origin distributed?
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This problem arises in physics when it is assumed that a moving particle in \( {R}^{n} \) has components of the velocity that are mutually independent and normally distributed and it is desired to find the density of the speed of the particle. The density in the case \( n = 3 \) is called the Maxwell density. The density in the case \( n = 2 \) (i.e. the dart board experiment described above) is called the Rayleigh density. We can simulate this case by picking independently a pair of coordinates \( \left( {x, y}\right) \), each from a normal distribution with \( \mu = 0 \) and \( \sigma = 1 \) on \( \left( {-\infty ,\infty }\right) \), calculating the distance \( r = \sqrt{{x}^{2} + {y}^{2}} \) of the point \( \left( {x, y}\right) \) from the origin, repeating this process a large number of times, and then presenting the results in a bar graph. The results are shown in Figure 5.11. We have also plotted the theoretical density \[ f\left( r\right) = r{e}^{-{r}^{2}/2}. \] This will be derived in Chapter 7; see Example 7.7.
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No
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Suppose that we have the data shown in Table 5.8 concerning grades and gender of students in a Calculus class. We can use the same sort of model in this situation as was used in Example 5.6. We imagine that we have an urn with 319 balls of two colors, say blue and red, corresponding to females and males, respectively. We now draw 93 balls, without replacement, from the urn. These balls correspond to the grade of A. We continue by drawing 123 balls, which correspond to the grade of B. When we finish, we have four sets of balls, with each ball belonging to exactly one set. (We could have stipulated that the balls were of four colors, corresponding to the four possible grades. In this case, we would draw a subset of size 152, which would correspond to the females. The balls remaining in the urn would correspond to the males. The choice does not affect the final determination of whether we should reject the hypothesis of independence of traits.)
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Instead, we will describe a single number which does a good job of measuring how far a given data set is from the expected one. To quantify how far apart the two sets of numbers are, we could sum the squares of the differences of the corresponding numbers. (We could also sum the absolute values of the differences, but we would not want to sum the differences.) Suppose that we have data in which we expect to see 10 objects of a certain type, but instead we see 18 , while in another case we expect to see 50 objects of a certain type, but instead we see 58 . Even though the two differences are about the same, the first difference is more surprising than the second, since the expected number of outcomes in the second case is quite a bit larger than the expected number in the first case. One way to correct for this is to divide the individual squares of the differences by the expected number for that box. Thus, if we label the values in the eight boxes in the first table by \( {O}_{i} \) (for observed values) and the values in the eight boxes in the second table by \( {E}_{i} \) (for expected values), then the following expression might be a reasonable one to use to measure how far the observed data is from what is expected:\n\n\[ \mathop{\sum }\limits_{{i = 1}}^{8}\frac{{\left( {O}_{i} - {E}_{i}\right) }^{2}}{{E}_{i}} \]\n\nThis expression is a random variable, which is usually denoted by the symbol \( {\chi }^{2} \) , pronounced \
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Yes
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Suppose that a mirror is mounted on a vertical axis, and is free to revolve about that axis. The axis of the mirror is 1 foot from a straight wall of infinite length. A pulse of light is shown onto the mirror, and the reflected ray hits the wall. Let \( \phi \) be the angle between the reflected ray and the line that is perpendicular to the wall and that runs through the axis of the mirror. We assume that \( \phi \) is uniformly distributed between \( - \pi /2 \) and \( \pi /2 \) . Let \( X \) represent the distance between the point on the wall that is hit by the reflected ray and the point on the wall that is closest to the axis of the mirror. We now determine the density of \( X \) .
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Let \( B \) be a fixed positive quantity. Then \( X \geq B \) if and only if \( \tan \left( \phi \right) \geq B \) , which happens if and only if \( \phi \geq \arctan \left( B\right) \) . This happens with probability\n\n\[ \frac{\pi /2 - \arctan \left( B\right) }{\pi }.\]\n\nThus, for positive \( B \), the cumulative distribution function of \( X \) is\n\n\[ F\left( B\right) = 1 - \frac{\pi /2 - \arctan \left( B\right) }{\pi }.\]\n\nTherefore, the density function for positive \( B \) is\n\n\[ f\left( B\right) = \frac{1}{\pi \left( {1 + {B}^{2}}\right) }.\]\n\nSince the physical situation is symmetric with respect to \( \phi = 0 \), it is easy to see that the above expression for the density is correct for negative values of \( B \) as well.
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Yes
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Let an experiment consist of tossing a fair coin three times. Let \( X \) denote the number of heads which appear. Then the possible values of \( X \) are \( 0,1,2 \) and 3 . The corresponding probabilities are \( 1/8,3/8,3/8 \), and \( 1/8 \) . Thus, the expected value of \( X \) equals
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\[ 0\left( \frac{1}{8}\right) + 1\left( \frac{3}{8}\right) + 2\left( \frac{3}{8}\right) + 3\left( \frac{1}{8}\right) = \frac{3}{2}. \]
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Yes
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Suppose that we toss a fair coin until a head first comes up, and let \( X \) represent the number of tosses which were made. Then the possible values of \( X \) are \( 1,2,\ldots \), and the distribution function of \( X \) is defined by\n\n\[ m\left( i\right) = \frac{1}{{2}^{i}}. \]\n\n(This is just the geometric distribution with parameter \( 1/2 \) .) Thus, we have\n\n\[ E\left( X\right) = \mathop{\sum }\limits_{{i = 1}}^{\infty }i\frac{1}{{2}^{i}} \]
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\[ = \mathop{\sum }\limits_{{i = 1}}^{\infty }\frac{1}{{2}^{i}} + \mathop{\sum }\limits_{{i = 2}}^{\infty }\frac{1}{{2}^{i}} + \cdots \]\n\n\[ = 1 + \frac{1}{2} + \frac{1}{{2}^{2}} + \cdots \]\n\n\[ = 2\text{.} \]
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Yes
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Suppose that we flip a coin until a head first appears, and if the number of tosses equals \( n \), then we are paid \( {2}^{n} \) dollars. What is the expected value of the payment?
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We let \( Y \) represent the payment. Then, \[ P\left( {Y = {2}^{n}}\right) = \frac{1}{{2}^{n}}, \] for \( n \geq 1 \) . Thus, \[ E\left( Y\right) = \mathop{\sum }\limits_{{n = 1}}^{\infty }{2}^{n}\frac{1}{{2}^{n}} \] which is a divergent sum. Thus, \( Y \) has no expectation. This example is called the St. Petersburg Paradox.
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Yes
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Let \( T \) be the time for the first success in a Bernoulli trials process. Then we take as sample space \( \Omega \) the integers \( 1,2,\ldots \) and assign the geometric distribution\n\n\[ m\left( j\right) = P\left( {T = j}\right) = {q}^{j - 1}p. \]
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Thus,\n\n\[ E\left( T\right) = 1 \cdot p + {2qp} + 3{q}^{2}p + \cdots \]\n\n\[ = p\left( {1 + {2q} + 3{q}^{2} + \cdots }\right) \text{.} \]\n\nNow if \( \left| x\right| < 1 \), then\n\n\[ 1 + x + {x}^{2} + {x}^{3} + \cdots = \frac{1}{1 - x}. \]\n\nDifferentiating this formula, we get\n\n\[ 1 + {2x} + 3{x}^{2} + \cdots = \frac{1}{{\left( 1 - x\right) }^{2}}, \]\n\nso\n\n\[ E\left( T\right) = \frac{p}{{\left( 1 - q\right) }^{2}} = \frac{p}{{p}^{2}} = \frac{1}{p}. \]\n\nIn particular, we see that if we toss a fair coin a sequence of times, the expected time until the first heads is \( 1/\left( {1/2}\right) = 2 \) . If we roll a die a sequence of times, the expected number of rolls until the first six is \( 1/\left( {1/6}\right) = 6 \) .
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Yes
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One can also interpret this number as the expected value of a random variable. To see this, let an experiment consist of choosing one of the women at random, and let \( X \) denote her height. Then the expected value of \( X \) equals 67.9.
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\[ \frac{{69} + {69} + {66} + {68} + {71} + {65} + {67} + {66} + {66} + {67} + {70} + {72}}{12} = {67.9}. \]
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Yes
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Now suppose an experiment consists of tossing a fair coin three times. Find the expected number of runs.
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To calculate \( E\left( Y\right) \) using the definition of expectation, we first must find the distribution function \( m\left( y\right) \) of \( Y \) i.e., we group together those values of \( X \) with a common value of \( Y \) and add their probabilities. In this case, we calculate that the distribution function of \( Y \) is: \( m\left( 1\right) = 1/4, m\left( 2\right) = 1/2 \), and \( m\left( 3\right) = 1/4 \) . One easily finds that \( E\left( Y\right) = 2 \) .
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Yes
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Theorem 6.1 If \( X \) is a discrete random variable with sample space \( \Omega \) and distribution function \( m\left( x\right) \), and if \( \phi : \Omega \rightarrow \mathrm{R} \) is a function, then\n\n\[ E\left( {\phi \left( X\right) }\right) = \mathop{\sum }\limits_{{x \in \Omega }}\phi \left( x\right) m\left( x\right) ,\]
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The proof of this theorem is straightforward, involving nothing more than grouping values of \( X \) with a common \( Y \) -value, as in Example 6.6.
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No
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We flip a coin and let \( X \) have the value 1 if the coin comes up heads and 0 if the coin comes up tails. Then, we roll a die and let \( Y \) denote the face that comes up. What does \( X + Y \) mean, and what is its distribution?
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This question is easily answered in this case, by considering, as we did in Chapter 4, the joint random variable \( Z = \left( {X, Y}\right) \), whose outcomes are ordered pairs of the form \( \left( {x, y}\right) \) , where \( 0 \leq x \leq 1 \) and \( 1 \leq y \leq 6 \) . The description of the experiment makes it reasonable to assume that \( X \) and \( Y \) are independent, so the distribution function of \( Z \) is uniform, with \( 1/{12} \) assigned to each outcome. Now it is an easy matter to find the set of outcomes of \( X + Y \), and its distribution function.
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No
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Theorem 6.2 Let \( X \) and \( Y \) be random variables with finite expected values. Then\n\n\[ E\left( {X + Y}\right) = E\left( X\right) + E\left( Y\right) ,\]\n\nand if \( c \) is any constant, then\n\n\[ E\left( {cX}\right) = {cE}\left( X\right) .
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Proof. Let the sample spaces of \( X \) and \( Y \) be denoted by \( {\Omega }_{X} \) and \( {\Omega }_{Y} \), and suppose that\n\n\[ {\Omega }_{X} = \left\{ {{x}_{1},{x}_{2},\ldots }\right\} \]\n\nand\n\n\[ {\Omega }_{Y} = \left\{ {{y}_{1},{y}_{2},\ldots }\right\} \]\n\nThen we can consider the random variable \( X + Y \) to be the result of applying the function \( \phi \left( {x, y}\right) = x + y \) to the joint random variable \( \left( {X, Y}\right) \) . Then, by Theorem 6.1, we have\n\n\[ E\left( {X + Y}\right) = \mathop{\sum }\limits_{j}\mathop{\sum }\limits_{k}\left( {{x}_{j} + {y}_{k}}\right) P\left( {X = {x}_{j}, Y = {y}_{k}}\right) \]\n\n\[ = \mathop{\sum }\limits_{j}\mathop{\sum }\limits_{k}{x}_{j}P\left( {X = {x}_{j}, Y = {y}_{k}}\right) + \mathop{\sum }\limits_{j}\mathop{\sum }\limits_{k}{y}_{k}P\left( {X = {x}_{j}, Y = {y}_{k}}\right) \]\n\n\[ = \mathop{\sum }\limits_{j}{x}_{j}P\left( {X = {x}_{j}}\right) + \mathop{\sum }\limits_{k}{y}_{k}P\left( {Y = {y}_{k}}\right) . \]\n\nThe last equality follows from the fact that\n\n\[ \mathop{\sum }\limits_{k}P\left( {X = {x}_{j}, Y = {y}_{k}}\right) = P\left( {X = {x}_{j}}\right) \]\n\nand\n\n\[ \mathop{\sum }\limits_{j}P\left( {X = {x}_{j}, Y = {y}_{k}}\right) = P\left( {Y = {y}_{k}}\right) . \]\n\nThus,\n\n\[ E\left( {X + Y}\right) = E\left( X\right) + E\left( Y\right) . \]\n\nIf \( c \) is any constant,\n\n\[ E\left( {cX}\right) = \mathop{\sum }\limits_{j}c{x}_{j}P\left( {X = {x}_{j}}\right) \]\n\n\[ = c\mathop{\sum }\limits_{j}{x}_{j}P\left( {X = {x}_{j}}\right) \]\n\n\[ = {cE}\left( X\right) \text{.} \]
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Yes
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Let \( Y \) be the number of fixed points in a random permutation of the set \( \{ a, b, c\} \) . To find the expected value of \( Y \)
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There are six possible outcomes of \( X \) , and we assign to each of them the probability \( 1/6 \) see Table 6.3. Then we can calculate \( E\left( Y\right) \) using Theorem 6.1, as\n\n\[ 3\left( \frac{1}{6}\right) + 1\left( \frac{1}{6}\right) + 1\left( \frac{1}{6}\right) + 0\left( \frac{1}{6}\right) + 0\left( \frac{1}{6}\right) + 1\left( \frac{1}{6}\right) = 1. \]
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Yes
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Theorem 6.3 Let \( {S}_{n} \) be the number of successes in \( n \) Bernoulli trials with probability \( p \) for success on each trial. Then the expected number of successes is \( {np} \) . That is,\n\n\[ E\left( {S}_{n}\right) = {np}. \]
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Proof. Let \( {X}_{j} \) be a random variable which has the value 1 if the \( j \) th outcome is a success and 0 if it is a failure. Then, for each \( {X}_{j} \) ,\n\n\[ E\left( {X}_{j}\right) = 0 \cdot \left( {1 - p}\right) + 1 \cdot p = p. \]\n\nSince\n\n\[ {S}_{n} = {X}_{1} + {X}_{2} + \cdots + {X}_{n}, \]\n\nand the expected value of the sum is the sum of the expected values, we have\n\n\[ E\left( {S}_{n}\right) = E\left( {X}_{1}\right) + E\left( {X}_{2}\right) + \cdots + E\left( {X}_{n}\right) \]\n\n\[ = {np}\text{.} \]
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Yes
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Theorem 6.4 If \( X \) and \( Y \) are independent random variables, then\n\n\[ E\left( {X \cdot Y}\right) = E\left( X\right) E\left( Y\right) . \]
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Proof. Suppose that\n\n\[ {\Omega }_{X} = \left\{ {{x}_{1},{x}_{2},\ldots }\right\} \]\n\nand\n\n\[ {\Omega }_{Y} = \left\{ {{y}_{1},{y}_{2},\ldots }\right\} \]\nare the sample spaces of \( X \) and \( Y \), respectively. Using Theorem 6.1, we have\n\n\[ E\left( {X \cdot Y}\right) = \mathop{\sum }\limits_{j}\mathop{\sum }\limits_{k}{x}_{j}{y}_{k}P\left( {X = {x}_{j}, Y = {y}_{k}}\right) . \]\n\nBut if \( X \) and \( Y \) are independent,\n\n\[ P\left( {X = {x}_{j}, Y = {y}_{k}}\right) = P\left( {X = {x}_{j}}\right) P\left( {Y = {y}_{k}}\right) . \]\n\nThus,\n\n\[ E\left( {X \cdot Y}\right) = \mathop{\sum }\limits_{j}\mathop{\sum }\limits_{k}{x}_{j}{y}_{k}P\left( {X = {x}_{j}}\right) P\left( {Y = {y}_{k}}\right) \]\n\n\[ = \left( {\mathop{\sum }\limits_{j}{x}_{j}P\left( {X = {x}_{j}}\right) }\right) \left( {\mathop{\sum }\limits_{k}{y}_{k}P\left( {Y = {y}_{k}}\right) }\right) \]\n\n\[ = E\left( X\right) E\left( Y\right) \text{.} \]
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Yes
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Consider a single toss of a coin. We define the random variable \( X \) to be 1 if heads turns up and 0 if tails turns up, and we set \( Y = 1 - X \) . Then \( E\left( X\right) = E\left( Y\right) = 1/2 \) . But \( X \cdot Y = 0 \) for either outcome. Hence, \( E\left( {X \cdot Y}\right) = 0 \neq \) \( E\left( X\right) E\left( Y\right) \) .
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Hence, \( E\left( {X \cdot Y}\right) = 0 \neq \) \( E\left( X\right) E\left( Y\right) \).
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Yes
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We start keeping snowfall records this year and want to find the expected number of records that will occur in the next \( n \) years. The first year is necessarily a record. The second year will be a record if the snowfall in the second year is greater than that in the first year. By symmetry, this probability is \( 1/2 \) . More generally, let \( {X}_{j} \) be 1 if the \( j \) th year is a record and 0 otherwise. To find \( E\left( {X}_{j}\right) \), we need only find the probability that the \( j \) th year is a record. But the record snowfall for the first \( j \) years is equally likely to fall in any one of these years,
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so \( E\left( {X}_{j}\right) = 1/j \) . Therefore, if \( {S}_{n} \) is the total number of records observed in the first \( n \) years,\n\n\[ E\left( {S}_{n}\right) = 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}. \]\n\nThis is the famous divergent harmonic series. It is easy to show that\n\n\[ E\left( {S}_{n}\right) \sim \log n \]\n\nas \( n \rightarrow \infty \) . A more accurate approximation to \( E\left( {S}_{n}\right) \) is given by the expression\n\n\[ \log n + \gamma + \frac{1}{2n} \]\n\nwhere \( \gamma \) denotes Euler’s constant, and is approximately equal to .5772 .
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Yes
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In the game of craps, the player makes a bet and rolls a pair of dice. If the sum of the numbers is 7 or 11 the player wins, if it is 2, 3 , or 12 the player loses. If any other number results, say \( r \), then \( r \) becomes the player’s point and he continues to roll until either \( r \) or 7 occurs. If \( r \) comes up first he wins, and if 7 comes up first he loses. The program Craps simulates playing this game a number of times.
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We have run the program for 1000 plays in which the player bets 1 dollar each time. The player's average winnings were -.006. The game of craps would seem to be only slightly unfavorable. Let us calculate the expected winnings on a single play and see if this is the case. We construct a two-stage tree measure as shown in Figure 6.1.\n\nThe first stage represents the possible sums for his first roll. The second stage represents the possible outcomes for the game if it has not ended on the first roll. In this stage we are representing the possible outcomes of a sequence of rolls required to determine the final outcome. The branch probabilities for the first stage are computed in the usual way assuming all 36 possibilites for outcomes for the pair of dice are equally likely. For the second stage we assume that the game will eventually end, and we compute the conditional probabilities for obtaining either the point or a 7. For example, assume that the player's point is 6 . Then the game will end when one of the eleven pairs, \( \\left( {1,5}\\right) ,\\left( {2,4}\\right) ,\\left( {3,3}\\right) ,\\left( {4,2}\\right) ,\\left( {5,1}\\right) ,\\left( {1,6}\\right) ,\\left( {2,5}\\right) ,\\left( {3,4}\\right) ,\\left( {4,3}\\right) \) , \( \\left( {5,2}\\right) ,\\left( {6,1}\\right) \), occurs. We assume that each of these possible pairs has the same probability. Then the player wins in the first five cases and loses in the last six. Thus the probability of winning is \( 5/{11} \) and the probability of losing is \( 6/{11} \) . From the path probabilities, we can find the probability that the player wins 1 dollar; it is \( {244}/{495} \) . The probability of losing is then \( {251}/{495} \) . Thus if \( X \) is his winning for\n\n\n\nFigure 6.1: Tree measure for craps.\n\n\n\na dollar bet,\n\n\\[ E\\left( X\\right) = 1\\left( \\frac{244}{495}\\right) + \\left( {-1}\\right) \\left( \\frac{251}{495}\\right) \\]\n\n\\[\n= - \\frac{7}{495} \\approx - {.0141}\n\\]\n\nThe game is unfavorable, but only slightly. The player’s expected gain in \( n \) plays is \( - n\\left( {.0141}\\right) \) . If \( n \) is not large, this is a small expected loss for the player. The casino makes a large number of plays and so can afford a small average gain per play and still expect a large profit.
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Yes
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In Las Vegas, a roulette wheel has 38 slots numbered 0,00,1,2, \\( \\ldots ,{36} \\) . The 0 and 00 slots are green, and half of the remaining 36 slots are red and half are black. A croupier spins the wheel and throws an ivory ball. If you bet 1 dollar on red, you win 1 dollar if the ball stops in a red slot, and otherwise you lose a dollar. We wish to calculate the expected value of your winnings, if you bet 1 dollar on red.
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Let \\( X \\) be the random variable which denotes your winnings in a 1 dollar bet on red in Las Vegas roulette. Then the distribution of \\( X \\) is given by\n\n\\[ {m}_{X} = \\left( \\begin{matrix} - 1 & 1 \\\\ {20}/{38} & {18}/{38} \\end{matrix}\\right) \\]\n\nand one can easily calculate (see Exercise 5) that\n\n\\[ E\\left( X\\right) \\approx - {.0526}. \\]
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No
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Theorem 6.5 Let \( X \) be a random variable with sample space \( \Omega \) . If \( {F}_{1},{F}_{2},\ldots ,{F}_{r} \) are events such that \( {F}_{i} \cap {F}_{j} = \varnothing \) for \( i \neq j \) and \( \Omega = { \cup }_{j}{F}_{j} \), then\n\n\[ E\left( X\right) = \mathop{\sum }\limits_{j}E\left( {X \mid {F}_{j}}\right) P\left( {F}_{j}\right) . \]
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Proof. We have\n\n\[ \mathop{\sum }\limits_{j}E\left( {X \mid {F}_{j}}\right) P\left( {F}_{j}\right) = \mathop{\sum }\limits_{j}\mathop{\sum }\limits_{k}{x}_{k}P\left( {X = {x}_{k} \mid {F}_{j}}\right) P\left( {F}_{j}\right) \]\n\n\[ = \mathop{\sum }\limits_{j}\mathop{\sum }\limits_{k}{x}_{k}P\left( {X = {x}_{k}}\right. \text{and}\left. {{F}_{j}\text{occurs}}\right) \]\n\n\[ = \mathop{\sum }\limits_{k}\mathop{\sum }\limits_{j}{x}_{k}P\left( {X = {x}_{k}}\right. \text{and}\left. {{F}_{j}\text{occurs}}\right) \]\n\n\[ = \mathop{\sum }\limits_{k}{x}_{k}P\left( {X = {x}_{k}}\right) \]\n\n\[ = E\left( X\right) \text{.} \]
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Yes
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Let \( T \) be the number of rolls in a single play of craps. We can think of a single play as a two-stage process. The first stage consists of a single roll of a pair of dice. The play is over if this roll is a \( 2,3,7 \) , 11, or 12. Otherwise, the player's point is established, and the second stage begins. This second stage consists of a sequence of rolls which ends when either the player's point or a 7 is rolled. We record the outcomes of this two-stage experiment using the random variables \( X \) and \( S \), where \( X \) denotes the first roll, and \( S \) denotes the number of rolls in the second stage of the experiment (of course, \( S \) is sometimes equal to 0 ). Note that \( T = S + 1 \) . Then by Theorem 6.5
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\[ E\left( T\right) = \mathop{\sum }\limits_{{j = 2}}^{{12}}E\left( {T \mid X = j}\right) P\left( {X = j}\right) . \] If \( j = 7,{11} \) or \( 2,3,{12} \), then \( E\left( {T \mid X = j}\right) = 1 \) . If \( j = 4,5,6,8,9 \), or 10, we can use Example 6.4 to calculate the expected value of \( S \) . In each of these cases, we continue rolling until we get either a \( j \) or a 7 . Thus, \( S \) is geometrically distributed with parameter \( p \), which depends upon \( j \) . If \( j = 4 \), for example, the value of \( p \) is \( 3/{36} + 6/{36} = 1/4 \) . Thus, in this case, the expected number of additional rolls is \( 1/p = 4 \), so \( E\left( {T \mid X = 4}\right) = 1 + 4 = 5 \) . Carrying out the corresponding calculations for the other possible values of \( j \) and using Theorem 6.5 gives \[ E\left( T\right) = 1\left( \frac{12}{36}\right) + \left( {1 + \frac{36}{3 + 6}}\right) \left( \frac{3}{36}\right) + \left( {1 + \frac{36}{4 + 6}}\right) \left( \frac{4}{36}\right) \] \[ + \left( {1 + \frac{36}{5 + 6}}\right) \left( \frac{5}{36}\right) + \left( {1 + \frac{36}{5 + 6}}\right) \left( \frac{5}{36}\right) \] \[ + \left( {1 + \frac{36}{4 + 6}}\right) \left( \frac{4}{36}\right) + \left( {1 + \frac{36}{3 + 6}}\right) \left( \frac{3}{36}\right) \] \[ = \frac{557}{165} \] \[ \approx {3.375}\ldots \].
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Yes
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Example 6.15 Let \( {S}_{1},{S}_{2},\ldots ,{S}_{n} \) be Peter’s accumulated fortune in playing heads or tails (see Example 1.4). Then\n\n\[ E\left( {{S}_{n} \mid {S}_{n - 1} = a,\ldots ,{S}_{1} = r}\right) = \frac{1}{2}\left( {a + 1}\right) + \frac{1}{2}\left( {a - 1}\right) = a. \]
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We note that Peter's expected fortune after the next play is equal to his present fortune. When this occurs, we say the game is fair. A fair game is also called a martingale. If the coin is biased and comes up heads with probability \( p \) and tails with probability \( q = 1 - p \), then\n\n\[ E\left( {{S}_{n} \mid {S}_{n - 1} = a,\ldots ,{S}_{1} = r}\right) = p\left( {a + 1}\right) + q\left( {a - 1}\right) = a + p - q. \]
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Yes
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Let us assume that a stock increases or decreases in value each day by 1 dollar, each with probability \( 1/2 \) . Then we can identify this simplified model with our familiar game of heads or tails. We assume that a buyer, Mr. Ace, adopts the following strategy. He buys the stock on the first day at its price \( V \) . He then waits until the price of the stock increases by one to \( V + 1 \) and sells. He then continues to watch the stock until its price falls back to \( V \) . He buys again and waits until it goes up to \( V + 1 \) and sells. Thus he holds the stock in intervals during which it increases by 1 dollar. In each such interval, he makes a profit of 1 dollar. However, we assume that he can do this only for a finite number of trading days. Thus he can lose if, in the last interval that he holds the stock, it does not get back up to \( V + 1 \) ; and this is the only way he can lose. In Figure 6.4 we illustrate a typical history if Mr. Ace must stop in twenty days. Mr. Ace holds the stock under his system during the days indicated by broken lines. We note that for the history shown in Figure 6.4, his system nets him a gain of 4 dollars.
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We have written a program StockSystem to simulate the fortune of Mr. Ace if he uses his sytem over an \( n \) -day period. If one runs this program a large number of times, for \( n = {20} \), say, one finds that his expected winnings are very close to 0, but the probability that he is ahead after 20 days is significantly greater than \( 1/2 \) . For small values of \( n \), the exact distribution of winnings can be calculated. The distribution for the case \( n = {20} \) is shown in Figure 6.5. Using this distribution, it is easy to calculate that the expected value of his winnings is exactly 0 . This is another instance of the fact that a fair game (a martingale) remains fair under quite general systems of play.
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Yes
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Consider one roll of a die. Let \( X \) be the number that turns up. To find \( V\left( X\right) \), we must first find the expected value of \( X \) .
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This is\n\n\[ \mu = E\left( X\right) = 1\left( \frac{1}{6}\right) + 2\left( \frac{1}{6}\right) + 3\left( \frac{1}{6}\right) + 4\left( \frac{1}{6}\right) + 5\left( \frac{1}{6}\right) + 6\left( \frac{1}{6}\right) \]\n\n\[ = \frac{7}{2}\text{.} \]\n\nTo find the variance of \( X \), we form the new random variable \( {\left( X - \mu \right) }^{2} \) and compute its expectation. We can easily do this using the following table.\n\n<table><thead><tr><th>\( x \)</th><th>\( m\left( x\right) \)</th><th>\( {\left( x - 7/2\right) }^{2} \)</th></tr></thead><tr><td>1</td><td>\( 1/6 \)</td><td>\( {25}/4 \)</td></tr><tr><td>2</td><td>\( 1/6 \)</td><td>\( 9/4 \)</td></tr><tr><td>3</td><td>\( 1/6 \)</td><td>\( 1/4 \)</td></tr><tr><td>4</td><td>\( 1/6 \)</td><td>\( 1/4 \)</td></tr><tr><td>5</td><td>\( 1/6 \)</td><td>\( 9/4 \)</td></tr><tr><td>6</td><td>\( 1/6 \)</td><td>\( {25}/4 \)</td></tr></table>\n\nTable 6.6: Variance calculation.\n\nFrom this table we find \( E\left( {\left( X - \mu \right) }^{2}\right) \) is\n\n\[ V\left( X\right) = \frac{1}{6}\left( {\frac{25}{4} + \frac{9}{4} + \frac{1}{4} + \frac{1}{4} + \frac{9}{4} + \frac{25}{4}}\right) \]\n\n\[ = \frac{35}{12} \]\n\nand the standard deviation \( D\left( X\right) = \sqrt{{35}/{12}} \approx {1.707} \) .
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Yes
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Theorem 6.6 If \( X \) is any random variable with \( E\left( X\right) = \mu \), then\n\n\[ V\left( X\right) = E\left( {X}^{2}\right) - {\mu }^{2}. \]
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Proof. We have\n\n\[ V\left( X\right) = E\left( {\left( X - \mu \right) }^{2}\right) = E\left( {{X}^{2} - {2\mu X} + {\mu }^{2}}\right) \]\n\n\[ = E\left( {X}^{2}\right) - {2\mu E}\left( X\right) + {\mu }^{2} = E\left( {X}^{2}\right) - {\mu }^{2}. \]
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Yes
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Theorem 6.7 If \( X \) is any random variable and \( c \) is any constant, then\n\n\[ V\left( {cX}\right) = {c}^{2}V\left( X\right) \]\n\nand\n\n\[ V\left( {X + c}\right) = V\left( X\right) . \]
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Proof. Let \( \mu = E\left( X\right) \) . Then \( E\left( {cX}\right) = {c\mu } \), and\n\n\[ V\left( {cX}\right) = E\left( {\left( cX - c\mu \right) }^{2}\right) = E\left( {{c}^{2}{\left( X - \mu \right) }^{2}}\right) \]\n\n\[ = {c}^{2}E\left( {\left( X - \mu \right) }^{2}\right) = {c}^{2}V\left( X\right) . \]\n\nTo prove the second assertion, we note that, to compute \( V\left( {X + c}\right) \), we would replace \( x \) by \( x + c \) and \( \mu \) by \( \mu + c \) in Equation 6.1. Then the \( c \) ’s would cancel, leaving \( V\left( X\right) \) .
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Yes
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Theorem 6.8 Let \( X \) and \( Y \) be two independent random variables. Then\n\n\[ V\left( {X + Y}\right) = V\left( X\right) + V\left( Y\right) . \]
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Proof. Let \( E\left( X\right) = a \) and \( E\left( Y\right) = b \) . Then\n\n\[ V\left( {X + Y}\right) = E\left( {\left( X + Y\right) }^{2}\right) - {\left( a + b\right) }^{2} \]\n\n\[ = E\left( {X}^{2}\right) + {2E}\left( {XY}\right) + E\left( {Y}^{2}\right) - {a}^{2} - {2ab} - {b}^{2}. \]\n\nSince \( X \) and \( Y \) are independent, \( E\left( {XY}\right) = E\left( X\right) E\left( Y\right) = {ab} \) . Thus,\n\n\[ V\left( {X + Y}\right) = E\left( {X}^{2}\right) - {a}^{2} + E\left( {Y}^{2}\right) - {b}^{2} = V\left( X\right) + V\left( Y\right) . \]
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Yes
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Theorem 6.9 Let \( {X}_{1},{X}_{2},\ldots ,{X}_{n} \) be an independent trials process with \( E\left( {X}_{j}\right) = \) \( \mu \) and \( V\left( {X}_{j}\right) = {\sigma }^{2} \) . Let\n\n\[ \n{S}_{n} = {X}_{1} + {X}_{2} + \cdots + {X}_{n} \n\]\n\nbe the sum, and\n\n\[ \n{A}_{n} = \frac{{S}_{n}}{n} \n\]\n\nbe the average. Then\n\n\[ \nE\left( {S}_{n}\right) = {n\mu } \n\]\n\n\[ \nV\left( {S}_{n}\right) = n{\sigma }^{2} \n\]\n\n\[ \n\sigma \left( {S}_{n}\right) = \sigma \sqrt{n} \n\]\n\n\[ \nE\left( {A}_{n}\right) = \mu , \n\]\n\n\[ \nV\left( {A}_{n}\right) = \frac{{\sigma }^{2}}{n} \n\]\n\n\[ \n\sigma \left( {A}_{n}\right) = \frac{\sigma }{\sqrt{n}}. \n\]
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Proof. Since all the random variables \( {X}_{j} \) have the same expected value, we have\n\n\[ \nE\left( {S}_{n}\right) = E\left( {X}_{1}\right) + \cdots + E\left( {X}_{n}\right) = {n\mu }, \n\]\n\n\[ \nV\left( {S}_{n}\right) = V\left( {X}_{1}\right) + \cdots + V\left( {X}_{n}\right) = n{\sigma }^{2}, \n\]\n\nand\n\n\[ \n\sigma \left( {S}_{n}\right) = \sigma \sqrt{n} \n\]\n\nWe have seen that, if we multiply a random variable \( X \) with mean \( \mu \) and variance \( {\sigma }^{2} \) by a constant \( c \), the new random variable has expected value \( {c\mu } \) and variance \( {c}^{2}{\sigma }^{2} \) . Thus,\n\n\[ \nE\left( {A}_{n}\right) = E\left( \frac{{S}_{n}}{n}\right) = \frac{n\mu }{n} = \mu , \n\]\n\nand\n\n\[ \nV\left( {A}_{n}\right) = V\left( \frac{{S}_{n}}{n}\right) = \frac{V\left( {S}_{n}\right) }{{n}^{2}} = \frac{n{\sigma }^{2}}{{n}^{2}} = \frac{{\sigma }^{2}}{n}. \n\]\n\nFinally, the standard deviation of \( {A}_{n} \) is given by\n\n\[ \n\sigma \left( {A}_{n}\right) = \frac{\sigma }{\sqrt{n}}. \n\]
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Yes
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Let \( T \) denote the number of trials until the first success in a Bernoulli trials process. Then \( T \) is geometrically distributed. What is the variance of \( T \) ?
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In Example 4.15, we saw that\n\n\[ \n{m}_{T} = \left( \begin{matrix} 1 & 2 & 3 & \cdots \\ p & {qp} & {q}^{2}p & \cdots \end{matrix}\right) .\n\]\n\nIn Example 6.4, we showed that\n\n\[ \nE\left( T\right) = 1/p\n\]\n\nThus,\n\n\[ \nV\left( T\right) = E\left( {T}^{2}\right) - 1/{p}^{2},\n\]\n\nso we need only find\n\n\[ \nE\left( {T}^{2}\right) = {1p} + {4qp} + 9{q}^{2}p + \cdots\n\]\n\n\[ \n= p\left( {1 + {4q} + 9{q}^{2} + \cdots }\right) .\n\]\n\nTo evaluate this sum, we start again with\n\n\[ \n1 + x + {x}^{2} + \cdots = \frac{1}{1 - x}.\n\]\n\nDifferentiating, we obtain\n\n\[ \n1 + {2x} + 3{x}^{2} + \cdots = \frac{1}{{\left( 1 - x\right) }^{2}}.\n\]\n\nMultiplying by \( x \) ,\n\n\[ \nx + 2{x}^{2} + 3{x}^{3} + \cdots = \frac{x}{{\left( 1 - x\right) }^{2}}.\n\]\n\nDifferentiating again gives\n\n\[ \n1 + {4x} + 9{x}^{2} + \cdots = \frac{1 + x}{{\left( 1 - x\right) }^{3}}.\n\]\n\nThus,\n\n\[ \nE\left( {T}^{2}\right) = p\frac{1 + q}{{\left( 1 - q\right) }^{3}} = \frac{1 + q}{{p}^{2}}\n\]\n\nand\n\n\[ \nV\left( T\right) = E\left( {T}^{2}\right) - {\left( E\left( T\right) \right) }^{2}\n\]\n\n\[ \n= \frac{1 + q}{{p}^{2}} - \frac{1}{{p}^{2}} = \frac{q}{{p}^{2}}.\n\]\n\nFor example, the variance for the number of tosses of a coin until the first head turns up is \( \left( {1/2}\right) /{\left( 1/2\right) }^{2} = 2 \) . The variance for the number of rolls of a die until the first six turns up is \( \left( {5/6}\right) /{\left( 1/6\right) }^{2} = {30} \) . Note that, as \( p \) decreases, the variance increases rapidly. This corresponds to the increased spread of the geometric distribution as \( p \) decreases (noted in Figure 5.1).
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Yes
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Theorem 6.10 If \( X \) and \( Y \) are real-valued random variables and \( c \) is any constant, then\n\n\[ E\left( {X + Y}\right) = E\left( X\right) + E\left( Y\right) ,\]\n\n\[ E\left( {cX}\right) = {cE}\left( X\right) .
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The proof is very similar to the proof of Theorem 6.2, and we omit it.
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No
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Example 6.20 Let \( X \) be uniformly distributed on the interval \( \left\lbrack {0,1}\right\rbrack \) . Then
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\[ E\left( X\right) = {\int }_{0}^{1}{xdx} = 1/2 \]
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Yes
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Example 6.21 Let \( Z = \left( {x, y}\right) \) denote a point chosen uniformly and randomly from the unit disk, as in the dart game in Example 2.8 and let \( X = {\left( {x}^{2} + {y}^{2}\right) }^{1/2} \) be the distance from \( Z \) to the center of the disk. The density function of \( X \) can easily be shown to equal \( f\left( x\right) = {2x} \), so by the definition of expected value,
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\[ E\left( X\right) = {\int }_{0}^{1}{xf}\left( x\right) {dx} \] \[ = {\int }_{0}^{1}x\left( {2x}\right) {dx} \] \[ = \frac{2}{3}\text{.} \]
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Yes
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In the example of the couple meeting at the Inn (Example 2.16), each person arrives at a time which is uniformly distributed between 5:00 and 6:00 PM. The random variable \( Z \) under consideration is the length of time the first person has to wait until the second one arrives. It was shown that\n\n\[ \n{f}_{Z}\left( z\right) = 2\left( {1 - z}\right) \n\]\n\nfor \( 0 \leq z \leq 1 \) . Hence,
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\[ \nE\left( Z\right) = {\int }_{0}^{1}z{f}_{Z}\left( z\right) {dz} \n\]\n\n\[ \n= {\int }_{0}^{1}{2z}\left( {1 - z}\right) {dz} \n\]\n\n\[ \n= {\left\lbrack {z}^{2} - \frac{2}{3}{z}^{3}\right\rbrack }_{0}^{1} \n\]\n\n\[ \n= \frac{1}{3}\text{.} \n\]
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Yes
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Theorem 6.11 If \( X \) is a real-valued random variable and if \( \phi : \mathbf{R} \rightarrow \mathbf{R} \) is a continuous real-valued function with domain \( \left\lbrack {a, b}\right\rbrack \), then\n\n\[ E\left( {\phi \left( X\right) }\right) = {\int }_{-\infty }^{+\infty }\phi \left( x\right) {f}_{X}\left( x\right) {dx} \]\n\nprovided the integral exists.
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For a proof of this theorem, see Ross. \( {}^{21} \)
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No
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Theorem 6.12 Let \( X \) and \( Y \) be independent real-valued continuous random variables with finite expected values. Then we have\n\n\[ E\left( {XY}\right) = E\left( X\right) E\left( Y\right) . \]
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Proof. We will prove this only in the case that the ranges of \( X \) and \( Y \) are contained in the intervals \( \left\lbrack {a, b}\right\rbrack \) and \( \left\lbrack {c, d}\right\rbrack \), respectively. Let the density functions of \( X \) and \( Y \) be denoted by \( {f}_{X}\left( x\right) \) and \( {f}_{Y}\left( y\right) \), respectively. Since \( X \) and \( Y \) are independent, the joint density function of \( X \) and \( Y \) is the product of the individual density functions. Hence\n\n\[ E\left( {XY}\right) = {\int }_{a}^{b}{\int }_{c}^{d}{xy}{f}_{X}\left( x\right) {f}_{Y}\left( y\right) {dydx} \]\n\n\[ = {\int }_{a}^{b}x{f}_{X}\left( x\right) {dx}{\int }_{c}^{d}y{f}_{Y}\left( y\right) {dy} \]\n\n\[ = E\left( X\right) E\left( Y\right) \text{.} \]
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Yes
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Let \( Z = \left( {X, Y}\right) \) be a point chosen at random in the unit square. Let \( A = {X}^{2} \) and \( B = {Y}^{2} \) . Then Theorem 4.3 implies that \( A \) and \( B \) are independent.
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Using Theorem 6.11, the expectations of \( A \) and \( B \) are easy to calculate:\n\n\[ E\left( A\right) = E\left( B\right) = {\int }_{0}^{1}{x}^{2}{dx} \]\n\n\[ = \frac{1}{3}\text{.} \]\n\nUsing Theorem 6.12, the expectation of \( {AB} \) is just the product of \( E\left( A\right) \) and \( E\left( B\right) \) , or \( 1/9 \) . The usefulness of this theorem is demonstrated by noting that it is quite a bit more difficult to calculate \( E\left( {AB}\right) \) from the definition of expectation. One finds that the density function of \( {AB} \) is\n\n\[ {f}_{AB}\left( t\right) = \frac{-\log \left( t\right) }{4\sqrt{t}} \]\n\nso\n\n\[ E\left( {AB}\right) = {\int }_{0}^{1}t{f}_{AB}\left( t\right) {dt} \]\n\n\[ = \frac{1}{9}\text{.} \]
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Yes
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Again let \( Z = \left( {X, Y}\right) \) be a point chosen at random in the unit square, and let \( W = X + Y \) . Then \( Y \) and \( W \) are not independent, and we have
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\[ E\left( Y\right) = \frac{1}{2} \] \[ E\left( W\right) = 1, \] \[ E\left( {YW}\right) = E\left( {{XY} + {Y}^{2}}\right) = E\left( X\right) E\left( Y\right) + \frac{1}{3} = \frac{7}{12} \neq E\left( Y\right) E\left( W\right) . \]
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Yes
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Theorem 6.15 If \( X \) is a real-valued random variable with \( E\left( X\right) = \mu \), then (cf. Theorem 6.6)
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\[ V\left( X\right) = E\left( {X}^{2}\right) - {\mu }^{2}. \]
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No
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If \( X \) is uniformly distributed on \( \left\lbrack {0,1}\right\rbrack \), then, using Theorem 6.15, we have
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\[ V\left( X\right) = {\int }_{0}^{1}{\left( x - \frac{1}{2}\right) }^{2}{dx} = \frac{1}{12}. \]
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Yes
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Example 6.26 Let \( X \) be an exponentially distributed random variable with parameter \( \lambda \) . Then the density function of \( X \) is \[ {f}_{X}\left( x\right) = \lambda {e}^{-{\lambda x}}. \]
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From the definition of expectation and integration by parts, we have \[ E\left( X\right) = {\int }_{0}^{\infty }x{f}_{X}\left( x\right) {dx} \] \[ = \lambda {\int }_{0}^{\infty }x{e}^{-{\lambda x}}{dx} \] \[ = - {\left. x{e}^{-{\lambda x}}\right| }_{0}^{\infty } + {\int }_{0}^{\infty }{e}^{-{\lambda x}}{dx} \] \[ = 0 + {\left. \frac{{e}^{-{\lambda x}}}{-\lambda }\right| }_{0}^{\infty } = \frac{1}{\lambda }. \]
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Yes
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Example 6.27 Let \( Z \) be a standard normal random variable with density function\n\n\[ \n{f}_{Z}\left( x\right) = \frac{1}{\sqrt{2\pi }}{e}^{-{x}^{2}/2}.\n\]\n\nSince this density function is symmetric with respect to the \( y \) -axis, then it is easy\n\nto show that\n\[ \n{\int }_{-\infty }^{\infty }x{f}_{Z}\left( x\right) {dx}\n\]\n\nhas value 0.
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The reader should recall however, that the expectation is defined to be the above integral only if the integral\n\n\[ \n{\int }_{-\infty }^{\infty }\left| x\right| {f}_{Z}\left( x\right) {dx}\n\]\n\n is finite. This integral equals\n\n\[ \n2{\int }_{0}^{\infty }x{f}_{Z}\left( x\right) {dx}\n\]\nwhich one can easily show is finite. Thus, the expected value of \( Z \) is 0.
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No
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Example 6.28 Let \( X \) be a continuous random variable with the Cauchy density function\n\n\[ \n{f}_{X}\left( x\right) = \frac{a}{\pi }\frac{1}{{a}^{2} + {x}^{2}}. \n\]\n\nThen the expectation of \( X \) does not exist, because the integral\n\n\[ \n\frac{a}{\pi }{\int }_{-\infty }^{+\infty }\frac{\left| x\right| {dx}}{{a}^{2} + {x}^{2}} \n\]\n\ndiverges.
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Thus the variance of \( X \) also fails to exist. Densities whose variance is not defined, like the Cauchy density, behave quite differently in a number of important respects from those whose variance is finite. We shall see one instance of this difference in Section 8.2.
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Yes
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Corollary 6.1 If \( {X}_{1},{X}_{2},\ldots ,{X}_{n} \) is an independent trials process of real-valued random variables, with \( E\left( {X}_{i}\right) = \mu \) and \( V\left( {X}_{i}\right) = {\sigma }^{2} \), and if\n\n\[ \n{S}_{n} = {X}_{1} + {X}_{2} + \cdots + {X}_{n}, \n\]\n\n\[ \n{A}_{n} = \frac{{S}_{n}}{n} \n\]\n\nthen\n\n\[ \nE\left( {S}_{n}\right) = {n\mu } \n\]\n\n\[ \nE\left( {A}_{n}\right) = \mu , \n\]\n\n\[ \nV\left( {S}_{n}\right) = n{\sigma }^{2}, \n\]\n\n\[ \nV\left( {A}_{n}\right) = \frac{{\sigma }^{2}}{n}. \n\]
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It follows that if we set\n\n\[ \n{S}_{n}^{ * } = \frac{{S}_{n} - {n\mu }}{\sqrt{n{\sigma }^{2}}} \n\]\n\nthen\n\n\[ \nE\left( {S}_{n}^{ * }\right) = 0, \n\]\n\n\[ \nV\left( {S}_{n}^{ * }\right) = 1 \n\]\n\nWe say that \( {S}_{n}^{ * } \) is a standardized version of \( {S}_{n} \) (see Exercise 12 in Section 6.2). \( ▱ \)
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No
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A die is rolled twice. Let \( {X}_{1} \) and \( {X}_{2} \) be the outcomes, and let \( {S}_{2} = {X}_{1} + {X}_{2} \) be the sum of these outcomes. Then \( {X}_{1} \) and \( {X}_{2} \) have the common distribution function:
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\[ m = \left( \begin{matrix} 1 & 2 & 3 & 4 & 5 & 6 \\ 1/6 & 1/6 & 1/6 & 1/6 & 1/6 & 1/6 \end{matrix}\right) . \] The distribution function of \( {S}_{2} \) is then the convolution of this distribution with itself. Thus, \[ P\left( {{S}_{2} = 2}\right) = m\left( 1\right) m\left( 1\right) \] \[ = \frac{1}{6} \cdot \frac{1}{6} = \frac{1}{36} \] \[ P\left( {{S}_{2} = 3}\right) = m\left( 1\right) m\left( 2\right) + m\left( 2\right) m\left( 1\right) \] \[ = \frac{1}{6} \cdot \frac{1}{6} + \frac{1}{6} \cdot \frac{1}{6} = \frac{2}{36} \] \[ P\left( {{S}_{2} = 4}\right) = m\left( 1\right) m\left( 3\right) + m\left( 2\right) m\left( 2\right) + m\left( 3\right) m\left( 1\right) \] \[ = \frac{1}{6} \cdot \frac{1}{6} + \frac{1}{6} \cdot \frac{1}{6} + \frac{1}{6} \cdot \frac{1}{6} = \frac{3}{36}. \] Continuing in this way we would find \( P\left( {{S}_{2} = 5}\right) = 4/{36}, P\left( {{S}_{2} = 6}\right) = 5/{36} \) , \( P\left( {{S}_{2} = 7}\right) = 6/{36}, P\left( {{S}_{2} = 8}\right) = 5/{36}, P\left( {{S}_{2} = 9}\right) = 4/{36}, P\left( {{S}_{2} = {10}}\right) = 3/{36} \) , \( P\left( {{S}_{2} = {11}}\right) = 2/{36} \), and \( P\left( {{S}_{2} = {12}}\right) = 1/{36} \) .
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Yes
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If a card is dealt at random to a player, then the point count for this card has distribution\n\n\[ \n{p}_{X} = \\left( \\begin{matrix} 0 & 1 & 2 & 3 & 4 \\ {36}/{52} & 4/{52} & 4/{52} & 4/{52} & 4/{52} \\end{matrix}\\right) .\n\]\n\nLet us regard the total hand of 13 cards as 13 independent trials with this common distribution. (Again this is not quite correct because we assume here that we are always choosing a card from a full deck.) Then the distribution for the point count \( C \) for the hand can be found from the program NFoldConvolution by using the distribution for a single card and choosing \( n = {13} \) . A player with a point count of 13 or more is said to have an opening bid. The probability of having an opening bid is then\n\n\[ \nP\\left( {C \\geq {13}}\\right) \\text{.} \n\]
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Since we have the distribution of \( C \), it is easy to compute this probability. Doing this we find that\n\n\[ \nP\\left( {C \\geq {13}}\\right) = {.2845} \n\]\n\nso that about one in four hands should be an opening bid according to this simplified model.
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Yes
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Example 7.3 Suppose we choose independently two numbers at random from the interval \( \\left\\lbrack {0,1}\\right\\rbrack \) with uniform probability density. What is the density of their sum?
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Let \( X \) and \( Y \) be random variables describing our choices and \( Z = X + Y \) their sum. Then we have\n\n\[ \n{f}_{X}\left( x\\right) = {f}_{Y}\left( x\\right) = \\left\\{ \\begin{array}{ll} 1 & \\text{ if }0 \\leq x \\leq 1 \\\\ 0 & \\text{ otherwise } \\end{array}\\right.\n\]\n\nand the density function for the sum is given by\n\n\[ \n{f}_{Z}\left( z\\right) = {\\int }_{-\\infty }^{+\\infty }{f}_{X}\left( {z - y}\\right) {f}_{Y}\left( y\\right) {dy}.\n\]\n\nSince \( {f}_{Y}\left( y\\right) = 1 \) if \( 0 \\leq y \\leq 1 \) and 0 otherwise, this becomes\n\n\[ \n{f}_{Z}\left( z\\right) = {\\int }_{0}^{1}{f}_{X}\left( {z - y}\\right) {dy}\n\]\n\nNow the integrand is 0 unless \( 0 \\leq z - y \\leq 1 \) (i.e., unless \( z - 1 \\leq y \\leq z \) ) and then it is 1 . So if \( 0 \\leq z \\leq 1 \), we have\n\n\[ \n{f}_{Z}\left( z\\right) = {\\int }_{0}^{z}{dy} = z\n\]\n\nwhile if \( 1 < z \\leq 2 \), we have\n\n\[ \n{f}_{Z}\left( z\\right) = {\\int }_{z - 1}^{1}{dy} = 2 - z\n\]\n\nand if \( z < 0 \) or \( z > 2 \) we have \( {f}_{Z}\left( z\\right) = 0 \) (see Figure 7.2). Hence,\n\n\[ \n{f}_{Z}\left( z\\right) = \\left\\{ \\begin{array}{ll} z, & \\text{ if }0 \\leq z \\leq 1 \\\\ 2 - z, & \\text{ if }1 < z \\leq 2 \\\\ 0, & \\text{ otherwise. } \\end{array}\\right.\n\]\n\nNote that this result agrees with that of Example 2.4.
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Yes
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Suppose we choose two numbers at random from the interval \( \lbrack 0,\infty ) \) with an exponential density with parameter \( \lambda \) . What is the density of their sum?
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Let \( X, Y \), and \( Z = X + Y \) denote the relevant random variables, and \( {f}_{X},{f}_{Y} \) , and \( {f}_{Z} \) their densities. Then\n\n\[ \n{f}_{X}\left( x\right) = {f}_{Y}\left( x\right) = \left\{ \begin{array}{ll} \lambda {e}^{-{\lambda x}}, & \text{ if }x \geq 0 \\ 0, & \text{ otherwise } \end{array}\right.\n\]\n\nand so, if \( z > 0 \) ,\n\n\[ \n{f}_{Z}\left( z\right) = {\int }_{-\infty }^{+\infty }{f}_{X}\left( {z - y}\right) {f}_{Y}\left( y\right) {dy}\n\]\n\n\[ \n= {\int }_{0}^{z}\lambda {e}^{-\lambda \left( {z - y}\right) }\lambda {e}^{-{\lambda y}}{dy}\n\]\n\n\[ \n= {\int }_{0}^{z}{\lambda }^{2}{e}^{-{\lambda z}}{dy}\n\]\n\n\[ \n= {\lambda }^{2}z{e}^{-{\lambda z}}\n\]\n\nwhile if \( z < 0,{f}_{Z}\left( z\right) = 0 \) (see Figure 7.3). Hence,\n\n\[ \n{f}_{Z}\left( z\right) = \left\{ \begin{array}{ll} {\lambda }^{2}z{e}^{-{\lambda z}}, & \text{ if }z \geq 0 \\ 0, & \text{ otherwise. } \end{array}\right.\n\]
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Yes
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It is an interesting and important fact that the convolution of two normal densities with means \( {\mu }_{1} \) and \( {\mu }_{2} \) and variances \( {\sigma }_{1} \) and \( {\sigma }_{2} \) is again a normal density, with mean \( {\mu }_{1} + {\mu }_{2} \) and variance \( {\sigma }_{1}^{2} + {\sigma }_{2}^{2} \) . We will show this in the special case that both random variables are standard normal.
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Suppose \( X \) and \( Y \) are two independent random variables, each with the standard normal density (see Example 5.8). We have\n\n\[ \n{f}_{X}\left( x\right) = {f}_{Y}\left( y\right) = \frac{1}{\sqrt{2\pi }}{e}^{-{x}^{2}/2}, \n\]\n\nand so\n\n\[ \n{f}_{Z}\left( z\right) = {f}_{X} * {f}_{Y}\left( z\right) \n\]\n\n\[ \n= \frac{1}{2\pi }{\int }_{-\infty }^{+\infty }{e}^{-{\left( z - y\right) }^{2}/2}{e}^{-{y}^{2}/2}{dy} \n\]\n\n\[ \n= \frac{1}{2\pi }{e}^{-{z}^{2}/4}{\int }_{-\infty }^{+\infty }{e}^{-{\left( y - z/2\right) }^{2}}{dy} \n\]\n\n\[ \n= \frac{1}{2\pi }{e}^{-{z}^{2}/4}\sqrt{\pi }\left\lbrack {\frac{1}{\sqrt{\pi }}{\int }_{-\infty }^{\infty }{e}^{-{\left( y - z/2\right) }^{2}}{dy}}\right\rbrack . \n\]\n\nThe expression in the brackets equals 1 , since it is the integral of the normal density function with \( \mu = 0 \) and \( \sigma = \sqrt{2} \) . So, we have\n\n\[ \n{f}_{Z}\left( z\right) = \frac{1}{\sqrt{4\pi }}{e}^{-{z}^{2}/4}. \n\]
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Yes
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Choose two numbers at random from the interval \( \\left( {-\\infty , + \\infty }\\right) \) with the Cauchy density with parameter \( a = 1 \) (see Example 5.10). Then\n\n\[ \n{f}_{X}\\left( x\\right) = {f}_{Y}\\left( x\\right) = \\frac{1}{\\pi \\left( {1 + {x}^{2}}\\right) },\n\]\n\nand \( Z = X + Y \) has density\n\n\[ \n{f}_{Z}\\left( z\\right) = \\frac{1}{{\\pi }^{2}}{\\int }_{-\\infty }^{+\\infty }\\frac{1}{1 + {\\left( z - y\\right) }^{2}}\\frac{1}{1 + {y}^{2}}{dy}.\n\]
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This integral requires some effort, and we give here only the result (see Section 10.3, or Dwass3):\n\n\[ \n{f}_{Z}\\left( z\\right) = \\frac{2}{\\pi \\left( {4 + {z}^{2}}\\right) }.\n\]
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No
|
Suppose \( X \) and \( Y \) are two independent standard normal random variables. Now suppose we locate a point \( P \) in the \( {xy} \) -plane with coordinates \( \left( {X, Y}\right) \) and ask: What is the density of the square of the distance of \( P \) from the origin?
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Here, with the preceding notation, we have\n\n\[ \n{f}_{X}\left( x\right) = {f}_{Y}\left( x\right) = \frac{1}{\sqrt{2\pi }}{e}^{-{x}^{2}/2}.\n\]\n\nMoreover, if \( {X}^{2} \) denotes the square of \( X \), then (see Theorem 5.1 and the discussion following)\n\n\[ \n{f}_{{X}^{2}}\left( r\right) = \left\{ \begin{array}{ll} \frac{1}{2\sqrt{r}}\left( {{f}_{X}\left( \sqrt{r}\right) + {f}_{X}\left( {-\sqrt{r}}\right) }\right) & \text{ if }r > 0, \\ 0 & \text{ otherwise. } \end{array}\right.\n\]\n\n\[ \n= \left\{ \begin{array}{ll} \frac{1}{\sqrt{2\pi r}}\left( {e}^{-r/2}\right) & \text{ if }r > 0, \\ 0 & \text{ otherwise. } \end{array}\right.\n\]\n\nThis is a gamma density with \( \lambda = 1/2,\beta = 1/2 \) (see Example 7.4). Now let \( {R}^{2} = {X}^{2} + {Y}^{2} \) . Then\n\n\[ \n{f}_{{R}^{2}}\left( r\right) = {\int }_{-\infty }^{+\infty }{f}_{{X}^{2}}\left( {r - s}\right) {f}_{{Y}^{2}}\left( s\right) {ds}\n\]\n\n\[ \n= \frac{1}{2\pi }{\int }_{0}^{r}{e}^{-\left( {r - s}\right) /2}{\left( r - s\right) }^{-1/2}{e}^{-s}{s}^{-1/2}{ds}\n\]\n\n\[ \n= \left\{ \begin{array}{ll} \frac{1}{2}{e}^{-r/2}, & \text{ if }r \geq 0 \\ 0, & \text{ otherwise. } \end{array}\right.\n\]\n\nHence, \( {R}^{2} \) has a gamma density with \( \lambda = 1/2,\beta = 1 \) . We can interpret this result as giving the density for the square of the distance of \( P \) from the center of a target if its coordinates are normally distributed.\n\nThe density of the random variable \( R \) is obtained from that of \( {R}^{2} \) in the usual way (see Theorem 5.1), and we find\n\n\[ \n{f}_{R}\left( r\right) = \left\{ \begin{array}{ll} \frac{1}{2}{e}^{-{r}^{2}/2} \cdot {2r} = r{e}^{-{r}^{2}/2}, & \text{ if }r \geq 0, \\ 0, & \text{ otherwise. } \end{array}\right.\n\]\n\nPhysicists will recognize this as a Rayleigh density. Our result here agrees with our simulation in Example 5.9.
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Yes
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Example 7.8 Suppose we are given a single die. We wish to test the hypothesis that the die is fair. Thus, our theoretical distribution is the uniform distribution on the integers between 1 and 6 . So, if we roll the die \( n \) times, the expected number of data points of each type is \( n/6 \) . Thus, if \( {o}_{i} \) denotes the actual number of data points of type \( i \), for \( 1 \leq i \leq 6 \), then the expression\n\n\[ \nV = \mathop{\sum }\limits_{{i = 1}}^{6}\frac{{\left( {o}_{i} - n/6\right) }^{2}}{n/6} \n\]\n\nis approximately chi-squared distributed with 5 degrees of freedom.
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Now suppose that we actually roll the die 60 times and obtain the data in Table 7.1. If we calculate \( V \) for this data, we obtain the value 13.6. The graph of the chi-squared density with 5 degrees of freedom is shown in Figure 7.4. One sees that values as large as 13.6 are rarely taken on by \( V \) if the die is fair, so we would reject the hypothesis that the die is fair. (When using this test, a statistician will reject the hypothesis if the data gives a value of \( V \) which is larger than \( {95}\% \) of the values one would expect to obtain if the hypothesis is true.)
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Yes
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Suppose the \( {X}_{i} \) are uniformly distributed on the interval \( \left\lbrack {0,1}\right\rbrack \) . Then
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\[ {f}_{{X}_{i}}\left( x\right) = \left\{ \begin{array}{ll} 1, & \text{ if }0 \leq x \leq 1 \\ 0, & \text{ otherwise } \end{array}\right. \] and \( {f}_{{S}_{n}}\left( x\right) \) is given by the formula \( {}^{4} \)\[ {f}_{{S}_{n}}\left( x\right) = \left\{ \begin{array}{ll} \frac{1}{\left( {n - 1}\right) !}\mathop{\sum }\limits_{{0 \leq j \leq x}}{\left( -1\right) }^{j}\left( \begin{array}{l} n \\ j \end{array}\right) {\left( x - j\right) }^{n - 1}, & \text{ if }0 < x < n, \\ 0, & \text{ otherwise. } \end{array}\right. \]
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Yes
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Theorem 8.1 (Chebyshev Inequality) Let \( X \) be a discrete random variable with expected value \( \mu = E\left( X\right) \), and let \( \epsilon > 0 \) be any positive real number. Then\n\n\[ P\left( {\left| {X - \mu }\right| \geq \epsilon }\right) \leq \frac{V\left( X\right) }{{\epsilon }^{2}}. \]
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Proof. Let \( m\left( x\right) \) denote the distribution function of \( X \) . Then the probability that \( X \) differs from \( \mu \) by at least \( \epsilon \) is given by\n\n\[ P\left( {\left| {X - \mu }\right| \geq \epsilon }\right) = \mathop{\sum }\limits_{{\left| {x - \mu }\right| \geq \epsilon }}m\left( x\right) . \]\n\nWe know that\n\n\[ V\left( X\right) = \mathop{\sum }\limits_{x}{\left( x - \mu \right) }^{2}m\left( x\right) ,\]\n\nand this is clearly at least as large as\n\n\[ \mathop{\sum }\limits_{{\left| {x - \mu }\right| \geq \epsilon }}{\left( x - \mu \right) }^{2}m\left( x\right) \]\n\nsince all the summands are positive and we have restricted the range of summation in the second sum. But this last sum is at least\n\n\[ \mathop{\sum }\limits_{{\left| {x - \mu }\right| \geq \epsilon }}{\epsilon }^{2}m\left( x\right) = {\epsilon }^{2}\mathop{\sum }\limits_{{\left| {x - \mu }\right| \geq \epsilon }}m\left( x\right) \]\n\n\[ = {\epsilon }^{2}P\left( {\left| {X - \mu }\right| \geq \epsilon }\right) . \]\n\nSo,\n\n\[ P\left( {\left| {X - \mu }\right| \geq \epsilon }\right) \leq \frac{V\left( X\right) }{{\epsilon }^{2}}. \]
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Yes
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Example 8.1 Let \( X \) by any random variable with \( E\left( X\right) = \mu \) and \( V\left( X\right) = {\sigma }^{2} \) . Then, if \( \epsilon = {k\sigma } \), Chebyshev’s Inequality states that\n\n\[ P\left( {\left| {X - \mu }\right| \geq {k\sigma }}\right) \leq \frac{{\sigma }^{2}}{{k}^{2}{\sigma }^{2}} = \frac{1}{{k}^{2}}. \]
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Thus, for any random variable, the probability of a deviation from the mean of more than \( k \) standard deviations is \( \leq 1/{k}^{2} \) . If, for example, \( k = 5,1/{k}^{2} = {.04} \).
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Yes
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Theorem 8.2 (Law of Large Numbers) Let \( {X}_{1},{X}_{2},\ldots ,{X}_{n} \) be an independent trials process, with finite expected value \( \mu = E\left( {X}_{j}\right) \) and finite variance \( {\sigma }^{2} = V\left( {X}_{j}\right) \) . Let \( {S}_{n} = {X}_{1} + {X}_{2} + \cdots + {X}_{n} \) . Then for any \( \epsilon > 0 \) ,\n\n\[ P\left( {\left| {\frac{{S}_{n}}{n} - \mu }\right| \geq \epsilon }\right) \rightarrow 0 \]\n\nas \( n \rightarrow \infty \) . Equivalently,\n\n\[ P\left( {\left| {\frac{{S}_{n}}{n} - \mu }\right| < \epsilon }\right) \rightarrow 1 \]\n\nas \( n \rightarrow \infty \) .
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Proof. Since \( {X}_{1},{X}_{2},\ldots ,{X}_{n} \) are independent and have the same distributions, we can apply Theorem 6.9. We obtain\n\n\[ V\left( {S}_{n}\right) = n{\sigma }^{2} \]\n\nand\n\n\[ V\left( \frac{{S}_{n}}{n}\right) = \frac{{\sigma }^{2}}{n}. \]\n\nAlso we know that\n\n\[ E\left( \frac{{S}_{n}}{n}\right) = \mu \]\n\nBy Chebyshev’s Inequality, for any \( \epsilon > 0 \) ,\n\n\[ P\left( {\left| {\frac{{S}_{n}}{n} - \mu }\right| \geq \epsilon }\right) \leq \frac{{\sigma }^{2}}{n{\epsilon }^{2}}. \]\n\nThus, for fixed \( \epsilon \) ,\n\n\[ P\left( {\left| {\frac{{S}_{n}}{n} - \mu }\right| \geq \epsilon }\right) \rightarrow 0 \]\n\nas \( n \rightarrow \infty \), or equivalently,\n\n\[ P\left( {\left| {\frac{{S}_{n}}{n} - \mu }\right| < \epsilon }\right) \rightarrow 1 \]\n\nas \( n \rightarrow \infty \) .
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Yes
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Consider \( n \) rolls of a die. Let \( {X}_{j} \) be the outcome of the \( j \) th roll. Then \( {S}_{n} = {X}_{1} + {X}_{2} + \cdots + {X}_{n} \) is the sum of the first \( n \) rolls. This is an independent trials process with \( E\left( {X}_{j}\right) = 7/2 \) . Thus, by the Law of Large Numbers, for any \( \epsilon > 0 \)
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\[ P\left( {\left| {\frac{{S}_{n}}{n} - \frac{7}{2}}\right| \geq \epsilon }\right) \rightarrow 0 \] as \( n \rightarrow \infty \) . An equivalent way to state this is that, for any \( \epsilon > 0 \
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Yes
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Let \( {X}_{1},{X}_{2},\ldots ,{X}_{n} \) be a Bernoulli trials process with probability .3 for success and .7 for failure. Let \( {X}_{j} = 1 \) if the \( j \) th outcome is a success and 0 otherwise. Then, \( E\left( {X}_{j}\right) = {.3} \) and \( V\left( {X}_{j}\right) = \left( {.3}\right) \left( {.7}\right) = {.21} \) . If\n\n\[ \n{A}_{n} = \frac{{S}_{n}}{n} = \frac{{X}_{1} + {X}_{2} + \cdots + {X}_{n}}{n} \n\]\n\nis the average of the \( {X}_{i} \), then \( E\left( {A}_{n}\right) = {.3} \) and \( V\left( {A}_{n}\right) = V\left( {S}_{n}\right) /{n}^{2} = {.21}/n \) . Chebyshev’s Inequality states that if, for example, \( \epsilon = {.1} \),
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\[ \nP\left( {\left| {{A}_{n} - {.3}}\right| \geq {.1}}\right) \leq \frac{.21}{n{\left( {.1}\right) }^{2}} = \frac{21}{n}. \n\]\n\nThus, if \( n = {100} \), \n\n\[ \nP\left( {\left| {{A}_{100} - {.3}}\right| \geq {.1}}\right) \leq {.21} \n\]\n\nor if \( n = {1000} \), \n\n\[ \nP\left( {\left| {{A}_{1000} - {.3}}\right| \geq {.1}}\right) \leq {.021}. \n\]
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Yes
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Theorem 8.3 (Chebyshev Inequality) Let \( X \) be a continuous random variable with density function \( f\left( x\right) \) . Suppose \( X \) has a finite expected value \( \mu = E\left( X\right) \) and finite variance \( {\sigma }^{2} = V\left( X\right) \) . Then for any positive number \( \epsilon > 0 \) we have\n\n\[ P\left( {\left| {X - \mu }\right| \geq \epsilon }\right) \leq \frac{{\sigma }^{2}}{{\epsilon }^{2}}. \]
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The proof is completely analogous to the proof in the discrete case, and we omit it.
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No
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Example 8.4 Let \( X \) be any continuous random variable with \( E\left( X\right) = \mu \) and \( V\left( X\right) = {\sigma }^{2} \) . Then, if \( \epsilon = {k\sigma } = k \) standard deviations for some integer \( k \), then
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\[ P\left( {\left| {X - \mu }\right| \geq {k\sigma }}\right) \leq \frac{{\sigma }^{2}}{{k}^{2}{\sigma }^{2}} = \frac{1}{{k}^{2}}, \]
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Yes
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Suppose we choose at random \( n \) numbers from the interval \( \left\lbrack {0,1}\right\rbrack \) with uniform distribution. Then if \( {X}_{i} \) describes the \( i \) th choice, we have\n\n\[ \mu = E\left( {X}_{i}\right) = {\int }_{0}^{1}{xdx} = \frac{1}{2}, \]\n\n\[ {\sigma }^{2} = V\left( {X}_{i}\right) = {\int }_{0}^{1}{x}^{2}{dx} - {\mu }^{2} \]\n\n\[ = \frac{1}{3} - \frac{1}{4} = \frac{1}{12}\text{.} \]
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Hence,\n\n\[ E\left( \frac{{S}_{n}}{n}\right) = \frac{1}{2} \]\n\n\[ V\left( \frac{{S}_{n}}{n}\right) = \frac{1}{12n} \]\n\nand for any \( \epsilon > 0 \) ,\n\n\[ P\left( {\left| {\frac{{S}_{n}}{n} - \frac{1}{2}}\right| \geq \epsilon }\right) \leq \frac{1}{{12n}{\epsilon }^{2}}. \]
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Yes
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Suppose we choose \( n \) real numbers at random, using a normal distribution with mean 0 and variance 1 . Then
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\[ \mu = E\left( {X}_{i}\right) = 0, \] \[ {\sigma }^{2} = V\left( {X}_{i}\right) = 1 \] Hence, \[ E\left( \frac{{S}_{n}}{n}\right) = 0, \] \[ V\left( \frac{{S}_{n}}{n}\right) = \frac{1}{n} \] and, for any \( \epsilon > 0 \) , \[ P\left( {\left| {\frac{{S}_{n}}{n} - 0}\right| \geq \epsilon }\right) \leq \frac{1}{n{\epsilon }^{2}}. \]
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Yes
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Let \( g\left( x\right) \) be a continuous function defined for \( x \in \left\lbrack {0,1}\right\rbrack \) with values in \( \left\lbrack {0,1}\right\rbrack \). Here is a better way to estimate the same area. Let us choose a large number of independent values \( {X}_{n} \) at random from \( \left\lbrack {0,1}\right\rbrack \) with uniform density, set \( {Y}_{n} = g\left( {X}_{n}\right) \), and find the average value of the \( {Y}_{n} \). Then this average is our estimate for the area.
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To see this, note that if the density function for \( {X}_{n} \) is uniform,\n\n\[ \mu = E\left( {Y}_{n}\right) = {\int }_{0}^{1}g\left( x\right) f\left( x\right) {dx} \]\n\n\[ = {\int }_{0}^{1}g\left( x\right) {dx} \]\n\n\[ = \text{average value of}g\left( x\right) \text{,} \]\n\nwhile the variance is\n\n\[ {\sigma }^{2} = E\left( {\left( {Y}_{n} - \mu \right) }^{2}\right) = {\int }_{0}^{1}{\left( g\left( x\right) - \mu \right) }^{2}{dx} < 1, \]\n\nsince for all \( x \) in \( \left\lbrack {0,1}\right\rbrack, g\left( x\right) \) is in \( \left\lbrack {0,1}\right\rbrack \), hence \( \mu \) is in \( \left\lbrack {0,1}\right\rbrack \), and so \( \left| {g\left( x\right) - \mu }\right| \leq 1 \). Now let \( {A}_{n} = \left( {1/n}\right) \left( {{Y}_{1} + {Y}_{2} + \cdots + {Y}_{n}}\right) \). Then by Chebyshev’s Inequality, we have\n\n\[ P\left( {\left| {{A}_{n} - \mu }\right| \geq \epsilon }\right) \leq \frac{{\sigma }^{2}}{n{\epsilon }^{2}} < \frac{1}{n{\epsilon }^{2}}. \]\n\nThis says that to get within \( \epsilon \) of the true value for \( \mu = {\int }_{0}^{1}g\left( x\right) {dx} \) with probability at least \( p \), we should choose \( n \) so that \( 1/n{\epsilon }^{2} \leq 1 - p \) (i.e., so that \( n \geq 1/{\epsilon }^{2}\left( {1 - p}\right) \) ). Note that this method tells us how large to take \( n \) to get a desired accuracy.
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Yes
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Theorem 9.1 (Central Limit Theorem for Binomial Distributions) For the binomial distribution \( b\left( {n, p, j}\right) \) we have\n\n\[ \mathop{\lim }\limits_{{n \rightarrow \infty }}\sqrt{npq}b\left( {n, p,\langle {np} + x\sqrt{npq}\rangle }\right) = \phi \left( x\right) ,\]\n\nwhere \( \phi \left( x\right) \) is the standard normal density.
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The proof of this theorem can be carried out using Stirling's approximation from Section 3.1. We indicate this method of proof by considering the case \( x = 0 \) . In this case, the theorem states that\n\n\[ \mathop{\lim }\limits_{{n \rightarrow \infty }}\sqrt{npq}b\left( {n, p,\langle {np}\rangle }\right) = \frac{1}{\sqrt{2\pi }} = {.3989}\ldots . \]\n\nIn order to simplify the calculation, we assume that \( {np} \) is an integer, so that \( \langle {np}\rangle = \) \( {np} \) . Then\n\n\[ \sqrt{npq}b\left( {n, p,{np}}\right) = \sqrt{npq}{p}^{np}{q}^{nq}\frac{n!}{\left( {np}\right) !\left( {nq}\right) !}. \]\n\nRecall that Stirling's formula (see Theorem 3.3) states that\n\n\[ n! \sim \sqrt{2\pi n}{n}^{n}{e}^{-n}\;\text{ as }\;n \rightarrow \infty . \]\n\nUsing this, we have\n\n\[ \sqrt{npq}b\left( {n, p,{np}}\right) \sim \frac{\sqrt{npq}{p}^{np}{q}^{nq}\sqrt{2\pi n}{n}^{n}{e}^{-n}}{\sqrt{2\pi np}\sqrt{2\pi nq}{\left( np\right) }^{np}{\left( nq\right) }^{nq}{e}^{-{np}}{e}^{-{nq}}}, \]\n\nwhich simplifies to \( 1/\sqrt{2\pi } \) .
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Yes
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Let us estimate the probability of exactly 55 heads in 100 tosses of a coin.
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For this case \( {np} = {100} \cdot 1/2 = {50} \) and \( \sqrt{npq} = \sqrt{{100} \cdot 1/2 \cdot 1/2} = 5 \) . Thus \( {x}_{55} = \left( {{55} - {50}}\right) /5 = 1 \) and\n\n\[ P\left( {{S}_{100} = {55}}\right) \sim \frac{\phi \left( 1\right) }{5} = \frac{1}{5}\left( {\frac{1}{\sqrt{2\pi }}{e}^{-1/2}}\right) \]\n\n\[ = {.0484}\text{.} \]
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Yes
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Theorem 9.2 (Central Limit Theorem for Bernoulli Trials) Let \( {S}_{n} \) be the number of successes in \( n \) Bernoulli trials with probability \( p \) for success, and let \( a \) and \( b \) be two fixed real numbers. Then\n\n\[ \mathop{\lim }\limits_{{n \rightarrow \infty }}P\left( {a \leq \frac{{S}_{n} - {np}}{\sqrt{npq}} \leq b}\right) = {\int }_{a}^{b}\phi \left( x\right) {dx}. \]\n
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This theorem can be proved by adding together the approximations to \( b\left( {n, p, k}\right) \) given in Theorem 9.1. It is also a special case of the more general Central Limit Theorem (see Section 10.3).
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No
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Dartmouth College would like to have 1050 freshmen. This college cannot accommodate more than 1060. Assume that each applicant accepts with probability .6 and that the acceptances can be modeled by Bernoulli trials. If the college accepts 1700, what is the probability that it will have too many acceptances?
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If it accepts 1700 students, the expected number of students who matriculate is \( {.6} \cdot {1700} = {1020} \). The standard deviation for the number that accept is \( \sqrt{{1700} \cdot {.6} \cdot {.4}} \approx {20} \). Thus we want to estimate the probability\n\n\[ P\left( {{S}_{1700} > {1060}}\right) = P\left( {{S}_{1700} \geq {1061}}\right) \]\n\n\[ = P\left( {{S}_{1700}^{ * } \geq \frac{{1060.5} - {1020}}{20}}\right) \]\n\n\[ = P\left( {{S}_{1700}^{ * } \geq {2.025}}\right) \text{.} \]\n\nFrom Table 9.4, if we interpolate, we would estimate this probability to be \( {.5} - {.4784} = {.0216} \). Thus, the college is fairly safe using this admission policy.
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Yes
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One frequently reads that a poll has been taken to estimate the proportion of people in a certain population who favor one candidate over another in a race with two candidates. (This model also applies to races with more than two candidates \( A \) and \( B \), and two ballot propositions.) Clearly, it is not possible for pollsters to ask everyone for their preference. What is done instead is to pick a subset of the population, called a sample, and ask everyone in the sample for their preference. Let \( p \) be the actual proportion of people in the population who are in favor of candidate \( A \) and let \( q = 1 - p \). If we choose a sample of size \( n \) from the population, the preferences of the people in the sample can be represented by random variables \( {X}_{1},{X}_{2},\ldots ,{X}_{n} \), where \( {X}_{i} = 1 \) if person \( i \) is in favor of candidate \( A \), and \( {X}_{i} = 0 \) if person \( i \) is in favor of candidate \( B \). Let \( {S}_{n} = {X}_{1} + {X}_{2} + \cdots + {X}_{n} \). If each subset of size \( n \) is chosen with the same probability, then \( {S}_{n} \) is hypergeometrically distributed. If \( n \) is small relative to the size of the population (which is typically true in practice), then \( {S}_{n} \) is approximately binomially distributed, with parameters \( n \) and \( p \).
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The pollster wants to estimate the value \( p \). An estimate for \( p \) is provided by the value \( \bar{p} = {S}_{n}/n \), which is the proportion of people in the sample who favor candidate \( B \). The Central Limit Theorem says that the random variable \( \bar{p} \) is approximately normally distributed. (In fact, our version of the Central Limit Theorem says that the distribution function of the random variable\n\n\[ \n{S}_{n}^{ * } = \frac{{S}_{n} - {np}}{\sqrt{npq}}\n\]\n\nis approximated by the standard normal density.) But we have\n\n\[ \n\bar{p} = \frac{{S}_{n} - {np}}{\sqrt{npq}}\sqrt{\frac{pq}{n}} + p,\n\]\n\ni.e., \( \bar{p} \) is just a linear function of \( {S}_{n}^{ * } \). Since the distribution of \( {S}_{n}^{ * } \) is approximated by the standard normal density, the distribution of the random variable \( \bar{p} \) must also be bell-shaped. We also know how to write the mean and standard deviation of \( \bar{p} \) in terms of \( p \) and \( n \). The mean of \( \bar{p} \) is just \( p \), and the standard deviation is\n\n\[ \n\sqrt{\frac{pq}{n}}.\n\]\n\nThus, it is easy to write down the standardized version of \( \bar{p} \) ; it is\n\n\[ \n{\bar{p}}^{ * } = \frac{\bar{p} - p}{\sqrt{{pq}/n}}.\n\]\n\nSince the distribution of the standardized version of \( \bar{p} \) is approximated by the standard normal density, we know, for example, that \( {95}\% \) of its values will lie within two standard deviations of its mean, and the same is true of \( \bar{p} \). So we have\n\n\[ \nP\left( {p - 2\sqrt{\frac{pq}{n}} < \bar{p} < p + 2\sqrt{\frac{pq}{n}}}\right) \approx {.954}.\n\]\n\nNow the pollster does not know \( p \) or \( q \), but he can use \( \bar{p} \) and \( \bar{q} = 1 - \bar{p} \) in their place without too much danger. With this idea in mind, the above statement is equivalent to the statement\n\n\[ \nP\left( {\bar{p} - 2\sqrt{\frac{\bar{p}\bar{q}}{n}} < p < \bar{p} + 2\sqrt{\frac{\bar{p}\bar{q}}{n}}}\right) \approx {.954}.\n\]\n\nThe resulting interval\n\n\[ \n\left( {\bar{p} - \frac{2\sqrt{\bar{p}\bar{q}}}{\sqrt{n}},\bar{p} + \frac{2\sqrt{\bar{p}\bar{q}}}{\sqrt{n}}}\right)\n\]\n\nis called the 95 percent confidence interval for the unknown value of \( p \). The name is suggested by the fact that if we use this method to estimate \( p \) in a large number of samples we should expect that in about 95 percent of the samples the true value of \( p \) is contained in the confidence interval obtained from the sample.
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Yes
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Theorem 9.3 Let \( {X}_{1},{X}_{2},\ldots ,{X}_{n} \) be an independent trials process and let \( {S}_{n} = \) \( {X}_{1} + {X}_{2} + \cdots + {X}_{n} \) . Assume that the greatest common divisor of the differences of all the values that the \( {X}_{j} \) can take on is 1 . Let \( E\left( {X}_{j}\right) = \mu \) and \( V\left( {X}_{j}\right) = {\sigma }^{2} \) . Then for \( n \) large,\n\n\[ P\left( {{S}_{n} = j}\right) \sim \frac{\phi \left( {x}_{j}\right) }{\sqrt{n{\sigma }^{2}}} \]\n\nwhere \( {x}_{j} = \left( {j - {n\mu }}\right) /\sqrt{n{\sigma }^{2}} \), and \( \phi \left( x\right) \) is the standard normal density.
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The program CLTIndTrialsLocal implements this approximation. When we run this program for 6 rolls of a die, and ask for the probability that the sum of the rolls equals 21, we obtain an actual value of .09285 , and a normal approximation value of .09537. If we run this program for 24 rolls of a die, and ask for the probability that the sum of the rolls is 72 , we obtain an actual value of .01724 and a normal approximation value of .01705 . These results show that the normal approximations are quite good.
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Yes
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A die is rolled 420 times. What is the probability that the sum of the rolls lies between 1400 and 1550?
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The sum is a random variable\n\n\[ \n{S}_{420} = {X}_{1} + {X}_{2} + \cdots + {X}_{420}, \]\n\nwhere each \( {X}_{j} \) has distribution\n\n\[ \n{m}_{X} = \left( \begin{matrix} 1 & 2 & 3 & 4 & 5 & 6 \\ 1/6 & 1/6 & 1/6 & 1/6 & 1/6 & 1/6 \end{matrix}\right) \]\n\nWe have seen that \( \mu = E\left( X\right) = 7/2 \) and \( {\sigma }^{2} = V\left( X\right) = {35}/{12} \) . Thus, \( E\left( {S}_{420}\right) = \) \( {420} \cdot 7/2 = {1470},{\sigma }^{2}\left( {S}_{420}\right) = {420} \cdot {35}/{12} = {1225} \), and \( \sigma \left( {S}_{420}\right) = {35} \) . Therefore,\n\n\[ \nP\left( {{1400} \leq {S}_{420} \leq {1550}}\right) \approx P\left( {\frac{{1399.5} - {1470}}{35} \leq {S}_{420}^{ * } \leq \frac{{1550.5} - {1470}}{35}}\right) \]\n\n\[ \n= \;P\left( {-{2.01} \leq {S}_{420}^{ * } \leq {2.30}}\right) \]\n\n\[ \n\approx \mathrm{{NA}}\left( {-{2.01},{2.30}}\right) = {.9670}\text{.} \]\n\nWe note that the program CLTIndTrialsGlobal could be used to calculate these probabilities.
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No
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We wish to estimate the probability that these two average grades differ by less than .05 for a given student. We now assume that \( p = 1/{20} \) . We also assume that the total error is the sum \( {S}_{30} \) of 30 independent random variables each with distribution\n\n\[ \n{m}_{X} : \\left\\{ \\begin{matrix} - 5 & - 4 & - 3 & - 2 & - 1 & 0 & 1 & 2 & 3 & 4 & 5 \\\\ \\frac{1}{100} & \\frac{1}{80} & \\frac{1}{60} & \\frac{1}{40} & \\frac{1}{20} & \\frac{463}{600} & \\frac{1}{20} & \\frac{1}{40} & \\frac{1}{60} & \\frac{1}{80} & \\frac{1}{100} \\end{matrix}\\right\\} .\n\]
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One can easily calculate that \( E\\left( X\\right) = 0 \) and \( {\\sigma }^{2}\\left( X\\right) = {1.5} \) . Then we have\n\n\[ \n\\begin{matrix} P\\left( {-{.05} \\leq \\frac{{S}_{30}}{30} \\leq {.05}}\\right) & = P\\left( {-{1.5} \\leq {S}_{30} \\leq {1.5}}\\right) \n\\end{matrix} \n\]\n\n\[ \n= P\\left( {\\frac{-{1.5}}{\\sqrt{{30} \\cdot {1.5}}} \\leq {S}_{30}^{ * } \\leq \\frac{1.5}{\\sqrt{{30} \\cdot {1.5}}}}\\right) \n\]\n\n\[ \n= P\\left( {-{.224} \\leq {S}_{30}^{ * } \\leq {.224}}\\right) \n\]\n\n\[ \n\\approx \\mathrm{{NA}}\\left( {-{.224},{.224}}\\right) = {.1772}. \n\]\n\nThis means that there is only a \( {17.7}\\% \) chance that a given student’s grade point average is accurate to within .05.
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Yes
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Theorem 9.5 (Central Limit Theorem) Let \( {X}_{1},{X}_{2},\ldots ,{X}_{n} \) ,... be a sequence of independent discrete random variables, and let \( {S}_{n} = {X}_{1} + {X}_{2} + \cdots + {X}_{n} \) . For each \( n \), denote the mean and variance of \( {X}_{n} \) by \( {\mu }_{n} \) and \( {\sigma }_{n}^{2} \), respectively. Define the mean and variance of \( {S}_{n} \) to be \( {m}_{n} \) and \( {s}_{n}^{2} \), respectively, and assume that \( {s}_{n} \rightarrow \infty \) . If there exists a constant \( A \), such that \( \left| {X}_{n}\right| \leq A \) for all \( n \), then for \( a < b \) , \[ \mathop{\lim }\limits_{{n \rightarrow \infty }}P\left( {a < \frac{{S}_{n} - {m}_{n}}{{s}_{n}} < b}\right) = \frac{1}{\sqrt{2\pi }}{\int }_{a}^{b}{e}^{-{x}^{2}/2}{dx}. \]
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The condition that \( \left| {X}_{n}\right| \leq A \) for all \( n \) is sometimes described by saying that the sequence \( \left\{ {X}_{n}\right\} \) is uniformly bounded. The condition that \( {s}_{n} \rightarrow \infty \) is necessary (see Exercise 15). We illustrate this theorem by generating a sequence of \( n \) random distributions on the interval \( \left\lbrack {a, b}\right\rbrack \) . We then convolute these distributions to find the distribution of the sum of \( n \) independent experiments governed by these distributions. Finally, we standardize the distribution for the sum to have mean 0 and standard deviation 1 and compare it with the normal density. The program CLTGeneral carries out this procedure. In Figure 9.9 we show the result of running this program for \( \left\lbrack {a, b}\right\rbrack = \left\lbrack {-2,4}\right\rbrack \), and \( n = 1,4 \), and 10 . We see that our first random distribution is quite asymmetric. By the time we choose the sum of ten such experiments we have a very good fit to the normal curve.
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No
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Suppose we choose \( n \) random numbers from the interval \( \left\lbrack {0,1}\right\rbrack \) with uniform density. Let \( {X}_{1},{X}_{2},\ldots ,{X}_{n} \) denote these choices, and \( {S}_{n} = {X}_{1} + {X}_{2} + \cdots + {X}_{n} \) their sum.
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We saw in Example 7.9 that the density function for \( {S}_{n} \) tends to have a normal shape, but is centered at \( n/2 \) and is flattened out. In order to compare the shapes of these density functions for different values of \( n \), we proceed as in the previous section: we standardize \( {S}_{n} \) by defining \[ {S}_{n}^{ * } = \frac{{S}_{n} - {n\mu }}{\sqrt{n}\sigma }. \] Then we see that for all \( n \) we have \[ E\left( {S}_{n}^{ * }\right) = 0, \] \[ V\left( {S}_{n}^{ * }\right) = 1. \] The density function for \( {S}_{n}^{ * } \) is just a standardized version of the density function for \( {S}_{n} \) (see Figure 9.13).
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No
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Example 9.8 Let us do the same thing, but now choose numbers from the interval \( \lbrack 0, + \infty ) \) with an exponential density with parameter \( \lambda \) . Then (see Example 6.26)
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\[ \mu = E\left( {X}_{i}\right) = \frac{1}{\lambda } \] \[ {\sigma }^{2} = V\left( {X}_{j}\right) = \frac{1}{{\lambda }^{2}}. \] Here we know the density function for \( {S}_{n} \) explicitly (see Section 7.2). We can use Corollary 5.1 to calculate the density function for \( {S}_{n}^{ * } \) . We obtain \[ {f}_{{S}_{n}}\left( x\right) = \frac{\lambda {e}^{-{\lambda x}}{\left( \lambda x\right) }^{n - 1}}{\left( {n - 1}\right) !}, \] \[ {f}_{{S}_{n}^{ * }}\left( x\right) = \frac{\sqrt{n}}{\lambda }{f}_{{S}_{n}}\left( \frac{\sqrt{n}x + n}{\lambda }\right) . \] The graph of the density function for \( {S}_{n}^{ * } \) is shown in Figure 9.14.
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Yes
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Theorem 9.6 (Central Limit Theorem) Let \( {S}_{n} = {X}_{1} + {X}_{2} + \cdots + {X}_{n} \) be the sum of \( n \) independent continuous random variables with common density function \( p \) having expected value \( \mu \) and variance \( {\sigma }^{2} \) . Let \( {S}_{n}^{ * } = \left( {{S}_{n} - {n\mu }}\right) /\sqrt{n}\sigma \) . Then we have, for all \( a < b \) ,\n\n\[ \mathop{\lim }\limits_{{n \rightarrow \infty }}P\left( {a < {S}_{n}^{ * } < b}\right) = \frac{1}{\sqrt{2\pi }}{\int }_{a}^{b}{e}^{-{x}^{2}/2}{dx}. \]
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We shall give a proof of this theorem in Section 10.3.
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No
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Suppose a surveyor wants to measure a known distance, say of \( 1\mathrm{{mile}} \) , using a transit and some method of triangulation. He knows that because of possible motion of the transit, atmospheric distortions, and human error, any one measurement is apt to be slightly in error. He plans to make several measurements and take an average. He assumes that his measurements are independent random variables with a common distribution of mean \( \mu = 1 \) and standard deviation \( \sigma = {.0002} \) (so, if the errors are approximately normally distributed, then his measurements are within 1 foot of the correct distance about \( {65}\% \) of the time). What can he say about the average?
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He can say that if \( n \) is large, the average \( {S}_{n}/n \) has a density function that is approximately normal, with mean \( \mu = 1 \) mile, and standard deviation \( \sigma = {.0002}/\sqrt{n} \) miles.
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Yes
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Now suppose our surveyor is measuring an unknown distance with the same instruments under the same conditions. He takes 36 measurements and averages them. How sure can he be that his measurement lies within .0002 of the true value?
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Again using the normal approximation, we get\n\n\[ P\left( {\left| {\frac{{S}_{n}}{n} - \mu }\right| < {.0002}}\right) = P\left( {\left| {S}_{n}^{ * }\right| < {.5}\sqrt{n}}\right) \]\n\n\[ \approx \frac{2}{\sqrt{2\pi }}{\int }_{-3}^{3}{e}^{-{x}^{2}/2}{dx} \]\n\n\[ \approx {.997}\text{.} \]\n\nThis means that the surveyor can be 99.7 percent sure that his average is within .0002 of the true value. To improve his confidence, he can take more measurements, or require less accuracy, or improve the quality of his measurements (i.e., reduce the variance \( {\sigma }^{2} \) ). In each case, the Central Limit Theorem gives quantitative information about the confidence of a measurement process, assuming always that the normal approximation is valid.
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Yes
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Example 10.1 Suppose \( X \) has range \( \{ 1,2,3,\ldots, n\} \) and \( {p}_{X}\left( j\right) = 1/n \) for \( 1 \leq j \leq n \) (uniform distribution). Then
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\[ g\left( t\right) = \mathop{\sum }\limits_{{j = 1}}^{n}\frac{1}{n}{e}^{tj} \] \[ = \frac{1}{n}\left( {{e}^{t} + {e}^{2t} + \cdots + {e}^{nt}}\right) \] \[ = \frac{{e}^{t}\left( {{e}^{nt} - 1}\right) }{n\left( {{e}^{t} - 1}\right) }. \] If we use the expression on the right-hand side of the second line above, then it is easy to see that \[ {\mu }_{1} = {g}^{\prime }\left( 0\right) = \frac{1}{n}\left( {1 + 2 + 3 + \cdots + n}\right) = \frac{n + 1}{2}, \] \[ {\mu }_{2} = {g}^{\prime \prime }\left( 0\right) = \frac{1}{n}\left( {1 + 4 + 9 + \cdots + {n}^{2}}\right) = \frac{\left( {n + 1}\right) \left( {{2n} + 1}\right) }{6}, \] and that \( \mu = {\mu }_{1} = \left( {n + 1}\right) /2 \) and \( {\sigma }^{2} = {\mu }_{2} - {\mu }_{1}^{2} = \left( {{n}^{2} - 1}\right) /{12} \)
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Yes
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Suppose now that \( X \) has range \( \{ 0,1,2,3,\ldots, n\} \) and \( {p}_{X}\left( j\right) = \) \( \left( \begin{matrix} n \\ j \end{matrix}\right) {p}^{j}{q}^{n - j} \) for \( 0 \leq j \leq n \) (binomial distribution). Then
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\[ g\left( t\right) = \mathop{\sum }\limits_{{j = 0}}^{n}{e}^{tj}\left( \begin{array}{l} n \\ j \end{array}\right) {p}^{j}{q}^{n - j} \]\n\[ = \mathop{\sum }\limits_{{j = 0}}^{n}\left( \begin{array}{l} n \\ j \end{array}\right) {\left( p{e}^{t}\right) }^{j}{q}^{n - j} \]\n\[ = {\left( p{e}^{t} + q\right) }^{n}\text{.} \]\n\nNote that\n\n\[ {\mu }_{1} = {g}^{\prime }\left( 0\right) = {\left. n{\left( p{e}^{t} + q\right) }^{n - 1}p{e}^{t}\right| }_{t = 0} = {np}, \]\n\n\[ {\mu }_{2} = {g}^{\prime \prime }\left( 0\right) = n\left( {n - 1}\right) {p}^{2} + {np}, \]\n\nso that \( \mu = {\mu }_{1} = {np} \), and \( {\sigma }^{2} = {\mu }_{2} - {\mu }_{1}^{2} = {np}\left( {1 - p}\right) \), as expected.
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Yes
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Suppose \( X \) has range \( \{ 1,2,3,\ldots \} \) and \( {p}_{X}\left( j\right) = {q}^{j - 1}p \) for all \( j \) (geometric distribution). Then
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\[ g\left( t\right) = \mathop{\sum }\limits_{{j = 1}}^{\infty }{e}^{tj}{q}^{j - 1}p = \frac{p{e}^{t}}{1 - q{e}^{t}}. \] Here \[ {\mu }_{1} = {g}^{\prime }\left( 0\right) = {\left. \frac{p{e}^{t}}{{\left( 1 - q{e}^{t}\right) }^{2}}\right| }_{t = 0} = \frac{1}{p}, \] \[ {\mu }_{2} = {g}^{\prime \prime }\left( 0\right) = {\left. \frac{p{e}^{t} + {pq}{e}^{2t}}{{\left( 1 - q{e}^{t}\right) }^{3}}\right| }_{t = 0} = \frac{1 + q}{{p}^{2}}, \] \( \mu = {\mu }_{1} = 1/p \), and \( {\sigma }^{2} = {\mu }_{2} - {\mu }_{1}^{2} = q/{p}^{2} \), as computed in Example 6.26.
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Yes
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Let \( X \) have range \( \{ 0,1,2,3,\ldots \} \) and let \( {p}_{X}\left( j\right) = {e}^{-\lambda }{\lambda }^{j}/j! \) for all \( j \) (Poisson distribution with mean \( \lambda \) ). Then
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\[ g\left( t\right) = \mathop{\sum }\limits_{{j = 0}}^{\infty }{e}^{tj}\frac{{e}^{-\lambda }{\lambda }^{j}}{j!} \] \[ = {e}^{-\lambda }\mathop{\sum }\limits_{{j = 0}}^{\infty }\frac{{\left( \lambda {e}^{t}\right) }^{j}}{j!} \] \[ = {e}^{-\lambda }{e}^{\lambda {e}^{t}} = {e}^{\lambda \left( {{e}^{t} - 1}\right) }. \] Then \[ {\mu }_{1} = {g}^{\prime }\left( 0\right) = {\left. {e}^{\lambda \left( {{e}^{t} - 1}\right) }\lambda {e}^{t}\right| }_{t = 0} = \lambda , \] \[ {\mu }_{2} = {g}^{\prime \prime }\left( 0\right) = {\left. {e}^{\lambda \left( {{e}^{t} - 1}\right) }\left( {\lambda }^{2}{e}^{2t} + \lambda {e}^{t}\right) \right| }_{t = 0} = {\lambda }^{2} + \lambda , \] \( \mu = {\mu }_{1} = \lambda \), and \( {\sigma }^{2} = {\mu }_{2} - {\mu }_{1}^{2} = \lambda \) .
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Yes
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Theorem 10.1 Let \( X \) be a discrete random variable with finite range \( \\left\\{ {{x}_{1},{x}_{2},\\ldots }\\right. \), \( \\left. {x}_{n}\\right\\} \), distribution function \( p \), and moment generating function \( g \). Then \( g \) is uniquely determined by \( p \), and conversely.
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Proof. We know that \( p \) determines \( g \), since\n\n\[ g\\left( t\\right) = \\mathop{\\sum }\\limits_{{j = 1}}^{n}{e}^{t{x}_{j}}p\\left( {x}_{j}\\right) .\n\]\n\nConversely, assume that \( g\\left( t\\right) \) is known. We wish to determine the values of \( {x}_{j} \) and \( p\\left( {x}_{j}\\right) \), for \( 1 \\leq j \\leq n \). We assume, without loss of generality, that \( p\\left( {x}_{j}\\right) > 0 \) for \( 1 \\leq j \\leq n \), and that\n\n\[ {x}_{1} < {x}_{2} < \\ldots < {x}_{n}.\n\]\n\nWe note that \( g\\left( t\\right) \) is differentiable for all \( t \), since it is a finite linear combination of exponential functions. If we compute \( {g}^{\\prime }\\left( t\\right) /g\\left( t\\right) \), we obtain\n\n\[ \\frac{{x}_{1}p\\left( {x}_{1}\\right) {e}^{t{x}_{1}} + \\ldots + {x}_{n}p\\left( {x}_{n}\\right) {e}^{t{x}_{n}}}{p\\left( {x}_{1}\\right) {e}^{t{x}_{1}} + \\ldots + p\\left( {x}_{n}\\right) {e}^{t{x}_{n}}}.\n\]\n\nDividing both top and bottom by \( {e}^{t{x}_{n}} \), we obtain the expression\n\n\[ \\frac{{x}_{1}p\\left( {x}_{1}\\right) {e}^{t\\left( {{x}_{1} - {x}_{n}}\\right) } + \\ldots + {x}_{n}p\\left( {x}_{n}\\right) }{p\\left( {x}_{1}\\right) {e}^{t\\left( {{x}_{1} - {x}_{n}}\\right) } + \\ldots + p\\left( {x}_{n}\\right) }.\n\]\n\nSince \( {x}_{n} \) is the largest of the \( {x}_{j} \)'s, this expression approaches \( {x}_{n} \) as \( t \) goes to \( \\infty \). So we have shown that\n\n\[ {x}_{n} = \\mathop{\\lim }\\limits_{{t \\rightarrow \\infty }}\\frac{{g}^{\\prime }\\left( t\\right) }{g\\left( t\\right) }.\n\]\n\nTo find \( p\\left( {x}_{n}\\right) \), we simply divide \( g\\left( t\\right) \) by \( {e}^{t{x}_{n}} \) and let \( t \) go to \( \\infty \). Once \( {x}_{n} \) and \( p\\left( {x}_{n}\\right) \) have been determined, we can subtract \( p\\left( {x}_{n}\\right) {e}^{t{x}_{n}} \) from \( g\\left( t\\right) \), and repeat the above procedure with the resulting function, obtaining, in turn, \( {x}_{n - 1},\\ldots ,{x}_{1} \) and \( p\\left( {x}_{n - 1}\\right) ,\\ldots, p\\left( {x}_{1}\\right) \).
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Yes
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If \( X \) and \( Y \) are independent discrete random variables with range \( \{ 0,1,2,\ldots, n\} \) and binomial distribution\n\n\[ \n{p}_{X}\left( j\right) = {p}_{Y}\left( j\right) = \left( \begin{array}{l} n \\ j \end{array}\right) {p}^{j}{q}^{n - j}, \n\]\n\nand if \( Z = X + Y \), then we know (cf. Section 7.1) that the range of \( X \) is\n\n\[ \n\{ 0,1,2,\ldots ,{2n}\} \n\]\n\nand \( X \) has binomial distribution\n\n\[ \n{p}_{Z}\left( j\right) = \left( {{p}_{X} * {p}_{Y}}\right) \left( j\right) = \left( \begin{matrix} {2n} \\ j \end{matrix}\right) {p}^{j}{q}^{{2n} - j}. \n\]
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Here we can easily verify this result by using generating functions. We know that\n\n\[ \n{g}_{X}\left( t\right) = {g}_{Y}\left( t\right) = \mathop{\sum }\limits_{{j = 0}}^{n}{e}^{tj}\left( \begin{array}{l} n \\ j \end{array}\right) {p}^{j}{q}^{n - j} \n\]\n\n\[ \n= {\left( p{e}^{t} + q\right) }^{n} \n\]\n\nand\n\n\[ \n{h}_{X}\left( z\right) = {h}_{Y}\left( z\right) = {\left( pz + q\right) }^{n}. \n\]\n\n\nHence, we have\n\n\[ \n{g}_{Z}\left( t\right) = {g}_{X}\left( t\right) {g}_{Y}\left( t\right) = {\left( p{e}^{t} + q\right) }^{2n}, \n\]\n\n\nor, what is the same,\n\n\[ \n{h}_{Z}\left( z\right) = {h}_{X}\left( z\right) {h}_{Y}\left( z\right) = {\left( pz + q\right) }^{2n} \n\]\n\n\[ \n= \mathop{\sum }\limits_{{j = 0}}^{{2n}}\left( \begin{matrix} {2n} \\ j \end{matrix}\right) {\left( pz\right) }^{j}{q}^{{2n} - j} \n\]\n\nfrom which we can see that the coefficient of \( {z}^{j} \) is just \( {p}_{Z}\left( j\right) = \left( \begin{matrix} {2n} \\ j \end{matrix}\right) {p}^{j}{q}^{{2n} - j} \) .
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Yes
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If \( X \) and \( Y \) are independent discrete random variables with the non-negative integers \( \{ 0,1,2,3,\ldots \} \) as range, and with geometric distribution function\n\n\[ \n{p}_{X}\left( j\right) = {p}_{Y}\left( j\right) = {q}^{j}p \n\]\n\nthen\n\n\[ \n{g}_{X}\left( t\right) = {g}_{Y}\left( t\right) = \frac{p}{1 - q{e}^{t}}, \n\]\n\nand if \( Z = X + Y \), then\n\n\[ \n{g}_{Z}\left( t\right) = {g}_{X}\left( t\right) {g}_{Y}\left( t\right) \n\]\n\n\[ \n= \frac{{p}^{2}}{1 - {2q}{e}^{t} + {q}^{2}{e}^{2t}}. \n\]
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If we replace \( {e}^{t} \) by \( z \), we get\n\n\[ \n{h}_{Z}\left( z\right) = \frac{{p}^{2}}{{\left( 1 - qz\right) }^{2}} \n\]\n\n\[ \n= {p}^{2}\mathop{\sum }\limits_{{k = 0}}^{\infty }\left( {k + 1}\right) {q}^{k}{z}^{k} \n\]\n\nand we can read off the values of \( {p}_{Z}\left( j\right) \) as the coefficient of \( {z}^{j} \) in this expansion for \( h\left( z\right) \), even though \( h\left( z\right) \) is not a polynomial in this case. The distribution \( {p}_{Z} \) is a negative binomial distribution (see Section 5.1).
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Yes
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Theorem 10.2 Consider a branching process with generating function \( h\left( z\right) \) for the number of offspring of a given parent. Let \( d \) be the smallest root of the equation \( z = h\left( z\right) \). If the mean number \( m \) of offspring produced by a single parent is \( \leq 1 \), then \( d = 1 \) and the process dies out with probability 1. If \( m > 1 \) then \( d < 1 \) and the process dies out with probability \( d \).
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We shall often want to know the probability that a branching process dies out by a particular generation, as well as the limit of these probabilities. Let \( {d}_{n} \) be the probability of dying out by the \( n \) th generation. Then we know that \( {d}_{1} = {p}_{0} \). We know further that \( {d}_{n} = h\left( {d}_{n - 1}\right) \) where \( h\left( z\right) \) is the generating function for the number of offspring produced by a single parent. This makes it easy to compute these probabilities.
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No
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