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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... | 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{mat... | Yes |
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 ... | 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 lo... | Yes |
Consider a family with two children. Given that one of the children is a boy, what is the probability that both children are boys? | 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 \ | No |
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? | 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 a... | No |
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 en... | 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 t... | Yes |
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. | 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, ... | Yes |
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 probabili... | 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 tw... | Yes |
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... | 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\[ = {q}^{k}\text{.} \] | Yes |
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. | \[ u\left( {x,2,1/2}\right) = \left( \begin{matrix} x - 1 \\ 1 \end{matrix}\right) \frac{1}{{2}^{x}}, \] for \( x = 2,3,\ldots \) | Yes |
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} \... | \[ \frac{{e}^{-{.1}}{\left( {.1}\right) }^{j}}{j!} \] | Yes |
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? | 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... | Yes |
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 s... | 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 \ | No |
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 b... | 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\[ \n{F}_{X}\left( t\right) = 1 - {e}^{-{\lambda t}}. \n\]... | No |
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. T... | 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 ... | Yes |
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 the... | Proof. This result follows from Theorem 5.1 by using the Chain Rule. | No |
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 . | 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... | Yes |
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 distri... | 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 densi... | No |
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. ... | 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 ... | Yes |
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 i... | 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 ... | Yes |
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 | \[ 0\left( \frac{1}{8}\right) + 1\left( \frac{3}{8}\right) + 2\left( \frac{3}{8}\right) + 3\left( \frac{1}{8}\right) = \frac{3}{2}. \] | Yes |
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 dis... | \[ = \mathop{\sum }\limits_{{i = 1}}^{\infty }\frac{1}{{2}^{i}} + \mathop{\sum }\limits_{{i = 2}}^{\infty }\frac{1}{{2}^{i}} + \cdots \]\n\n\[ = 1 + \frac{1}{2} + \frac{1}{{2}^{2}} + \cdots \]\n\n\[ = 2\text{.} \] | Yes |
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? | 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 ... | Yes |
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\[ m\left( j\right) = P\left( {T = j}\right) = {q}^{j - 1}p. \] | 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 ... | Yes |
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. | \[ \frac{{69} + {69} + {66} + {68} + {71} + {65} + {67} + {66} + {66} + {67} + {70} + {72}}{12} = {67.9}. \] | Yes |
Now suppose an experiment consists of tossing a fair coin three times. Find the expected number of runs. | 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... | Yes |
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... | 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. | No |
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? | 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 re... | No |
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) . | 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 + ... | Yes |
Let \( Y \) be the number of fixed points in a random permutation of the set \( \{ a, b, c\} \) . To find the expected value of \( Y \) | 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... | Yes |
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\[ E\left( {S}_{n}\right) = {np}. \] | 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 th... | Yes |
Theorem 6.4 If \( X \) and \( Y \) are independent random variables, then\n\n\[ E\left( {X \cdot Y}\right) = E\left( X\right) E\left( Y\right) . \] | 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{\su... | Yes |
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) ... | Hence, \( E\left( {X \cdot Y}\right) = 0 \neq \) \( E\left( X\right) E\left( Y\right) \). | Yes |
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... | 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}\r... | Yes |
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 \( r \) com... | 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 ... | Yes |
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 other... | 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\... | No |
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}}\... | 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... | Yes |
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 sec... | \[ E\left( T\right) = \mathop{\sum }\limits_{{j = 2}}^{{12}}E\left( {T \mid X = j}\right) P\left( {X = j}\right) . \] 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 ... | Yes |
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\[ E\left( {{S}_{n} \mid {S}_{n - 1} = a,\ldots ,{S}_{1} = r}\right) = \frac{1}{2}\left( {a + 1}\right) + \frac{1}{2}\left( {a - 1}\right) = a. \] | 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... | Yes |
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 \( V \) ... | 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 significan... | Yes |
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 \( X \) . | 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 \( {\le... | Yes |
Theorem 6.6 If \( X \) is any random variable with \( E\left( X\right) = \mu \), then\n\n\[ V\left( X\right) = E\left( {X}^{2}\right) - {\mu }^{2}. \] | 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\[ = E\left( {X}^{2}\right) - {2\mu E}\left( X\right) + {\mu }^{2} = E\left( {X}^{2}\right) - {\mu }^{2}. \] | Yes |
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\[ V\left( {X + c}\right) = V\left( X\right) . \] | 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 p... | Yes |
Theorem 6.8 Let \( X \) and \( Y \) be two independent random variables. Then\n\n\[ V\left( {X + Y}\right) = V\left( X\right) + V\left( Y\right) . \] | 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 ... | Yes |
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 av... | 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\... | Yes |
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 \( T \) ? | 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\[... | Yes |
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) . | The proof is very similar to the proof of Theorem 6.2, and we omit it. | No |
Example 6.20 Let \( X \) be uniformly distributed on the interval \( \left\lbrack {0,1}\right\rbrack \) . Then | \[ E\left( X\right) = {\int }_{0}^{1}{xdx} = 1/2 \] | Yes |
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 eq... | \[ E\left( X\right) = {\int }_{0}^{1}{xf}\left( x\right) {dx} \] \[ = {\int }_{0}^{1}x\left( {2x}\right) {dx} \] \[ = \frac{2}{3}\text{.} \] | Yes |
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( ... | \[ \nE\left( Z\right) = {\int }_{0}^{1}z{f}_{Z}\left( z\right) {dz} \n\]\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\[ \n= \frac{1}{3}\text{.} \n\] | Yes |
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\ri... | For a proof of this theorem, see Ross. \( {}^{21} \) | No |
Theorem 6.12 Let \( X \) and \( Y \) be independent real-valued continuous random variables with finite expected values. Then we have\n\n\[ E\left( {XY}\right) = E\left( X\right) E\left( Y\right) . \] | 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}\lef... | Yes |
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. | 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 \... | Yes |
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 | \[ E\left( Y\right) = \frac{1}{2} \] \[ E\left( W\right) = 1, \] \[ E\left( {YW}\right) = E\left( {{XY} + {Y}^{2}}\right) = E\left( X\right) E\left( Y\right) + \frac{1}{3} = \frac{7}{12} \neq E\left( Y\right) E\left( W\right) . \] | Yes |
Theorem 6.15 If \( X \) is a real-valued random variable with \( E\left( X\right) = \mu \), then (cf. Theorem 6.6) | \[ V\left( X\right) = E\left( {X}^{2}\right) - {\mu }^{2}. \] | No |
If \( X \) is uniformly distributed on \( \left\lbrack {0,1}\right\rbrack \), then, using Theorem 6.15, we have | \[ V\left( X\right) = {\int }_{0}^{1}{\left( x - \frac{1}{2}\right) }^{2}{dx} = \frac{1}{12}. \] | Yes |
Example 6.26 Let \( X \) be an exponentially distributed random variable with parameter \( \lambda \) . Then the density function of \( X \) is \[ {f}_{X}\left( x\right) = \lambda {e}^{-{\lambda x}}. \] | 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 +... | Yes |
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(... | 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 sho... | No |
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}... | 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. | Yes |
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\[ \n{A}_{n} = \frac{{S}_{n}}{n} \n\]\... | 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). \( ▱... | No |
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: | \[ m = \left( \begin{matrix} 1 & 2 & 3 & 4 & 5 & 6 \\ 1/6 & 1/6 & 1/6 & 1/6 & 1/6 & 1/6 \end{matrix}\right) . \] 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{... | Yes |
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 dis... | 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. | Yes |
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? | 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 ... | Yes |
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? | 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... | Yes |
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... | 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\[... | Yes |
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\... | This integral requires some effort, and we give here only the result (see Section 10.3, or Dwass3):\n\n\[ \n{f}_{Z}\\left( z\\right) = \\frac{2}{\\pi \\left( {4 + {z}^{2}}\\right) }.\n\] | 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? | 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}{l... | Yes |
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 \( {o}_{i} \)... | 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 ... | Yes |
Suppose the \( {X}_{i} \) are uniformly distributed on the interval \( \left\lbrack {0,1}\right\rbrack \) . Then | \[ {f}_{{X}_{i}}\left( x\right) = \left\{ \begin{array}{ll} 1, & \text{ if }0 \leq x \leq 1 \\ 0, & \text{ otherwise } \end{array}\right. \] 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{\su... | Yes |
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\[ P\left( {\left| {X - \mu }\right| \geq \epsilon }\right) \leq \frac{V\left( X\right) }{{\epsilon }^{2}}. \] | 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\righ... | Yes |
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... | 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, \( k = 5,1/{k}^{2} = {.04} \). | Yes |
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 \( \epsi... | 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)... | Yes |
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 \( \epsilo... | \[ P\left( {\left| {\frac{{S}_{n}}{n} - \frac{7}{2}}\right| \geq \epsilon }\right) \rightarrow 0 \] as \( n \rightarrow \infty \) . An equivalent way to state this is that, for any \( \epsilon > 0 \ | Yes |
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... | \[ \nP\left( {\left| {{A}_{n} - {.3}}\right| \geq {.1}}\right) \leq \frac{.21}{n{\left( {.1}\right) }^{2}} = \frac{21}{n}. \n\]\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|... | Yes |
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\... | The proof is completely analogous to the proof in the discrete case, and we omit it. | No |
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 | \[ P\left( {\left| {X - \mu }\right| \geq {k\sigma }}\right) \leq \frac{{\sigma }^{2}}{{k}^{2}{\sigma }^{2}} = \frac{1}{{k}^{2}}, \] | Yes |
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 }... | 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\[ P\left( {\left| {\frac{{S}_{n}}{n} - \frac{1}{2}}\right| \geq \epsilon }\right) \leq \frac{1}{{12n}{\epsilon }^{2}}. \] | Yes |
Suppose we choose \( n \) real numbers at random, using a normal distribution with mean 0 and variance 1 . Then | \[ \mu = E\left( {X}_{i}\right) = 0, \] \[ {\sigma }^{2} = V\left( {X}_{i}\right) = 1 \] 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{\eps... | Yes |
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\rb... | 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 }^... | Yes |
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 th... | 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}{... | Yes |
Let us estimate the probability of exactly 55 heads in 100 tosses of a coin. | 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\[... | Yes |
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}}{\... | 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). | No |
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? | 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... | Yes |
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 ... | 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... | Yes |
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 \(... | 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 pro... | Yes |
A die is rolled 420 times. What is the probability that the sum of the rolls lies between 1400 and 1550? | 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 \... | No |
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 & -... | 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{-{... | Yes |
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 \( {\sigma }_{n}^{2} \), respe... | 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 \( n \) ... | No |
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. | 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 \[... | No |
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) | \[ \mu = E\left( {X}_{i}\right) = \frac{1}{\lambda } \] \[ {\sigma }^{2} = V\left( {X}_{j}\right) = \frac{1}{{\lambda }^{2}}. \] 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... | Yes |
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) /\sq... | We shall give a proof of this theorem in Section 10.3. | No |
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 measurem... | 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. | Yes |
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? | 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 surve... | Yes |
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 | \[ g\left( t\right) = \mathop{\sum }\limits_{{j = 1}}^{n}\frac{1}{n}{e}^{tj} \] \[ = \frac{1}{n}\left( {{e}^{t} + {e}^{2t} + \cdots + {e}^{nt}}\right) \] \[ = \frac{{e}^{t}\left( {{e}^{nt} - 1}\right) }{n\left( {{e}^{t} - 1}\right) }. \] If we use the expression on the right-hand side of the second line above, then it ... | Yes |
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 | \[ g\left( t\right) = \mathop{\sum }\limits_{{j = 0}}^{n}{e}^{tj}\left( \begin{array}{l} n \\ j \end{array}\right) {p}^{j}{q}^{n - j} \]\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{.} ... | Yes |
Suppose \( X \) has range \( \{ 1,2,3,\ldots \} \) and \( {p}_{X}\left( j\right) = {q}^{j - 1}p \) for all \( j \) (geometric distribution). Then | \[ g\left( t\right) = \mathop{\sum }\limits_{{j = 1}}^{\infty }{e}^{tj}{q}^{j - 1}p = \frac{p{e}^{t}}{1 - q{e}^{t}}. \] 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) = ... | Yes |
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 | \[ g\left( t\right) = \mathop{\sum }\limits_{{j = 0}}^{\infty }{e}^{tj}\frac{{e}^{-\lambda }{\lambda }^{j}}{j!} \] \[ = {e}^{-\lambda }\mathop{\sum }\limits_{{j = 0}}^{\infty }\frac{{\left( \lambda {e}^{t}\right) }^{j}}{j!} \] \[ = {e}^{-\lambda }{e}^{\lambda {e}^{t}} = {e}^{\lambda \left( {{e}^{t} - 1}\right) }. \] Th... | Yes |
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. | 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 ... | Yes |
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) ... | 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\[ \... | Yes |
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}... | 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 \( ... | Yes |
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 ... | 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}_... | No |
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