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Lemma 8.10.4. If each subsequence of \( {\mu }_{n} \) has a further subsequence that converges to \( \nu \) then \( {\mu }_{n} \Rightarrow \nu \) . | Proof. Note that if \( f \) is a bounded continuous function, the sequence of real numbers \( \int f\left( \omega \right) {\mu }_{n}\left( {d\omega }\right) \) have the property that every subsequence has a further subsequence that converges to \( \int f\left( \omega \right) \nu \left( {d\omega }\right) \) . Exercise 8... | No |
Theorem 8.10.5. The sequence \( {\mu }_{n} \) is tight if and only if for each \( \epsilon > 0 \) there are \( {n}_{0}, M \) and \( \delta \) so that\n\n\[ \text{(i)}{\mu }_{n}\left( {\left| {\omega \left( 0\right) }\right| > M}\right) \leq \epsilon \text{for all}n \geq {n}_{0} \]\n\n\[ \text{(ii)}{\mu }_{n}\left( {{\o... | Proof. We begin by recalling (see e.g., Royden (1988), page 169) | No |
Theorem 8.10.6. Arzela-Ascoli Theorem. A subset A of \( C \) has compact closure if and only if \( \mathop{\sup }\limits_{{\omega \in A}}\left| {\omega \left( 0\right) }\right| < \infty \) and \( \mathop{\lim }\limits_{{\delta \rightarrow 0}}\mathop{\sup }\limits_{{\omega \in A}}{\operatorname{osc}}_{\delta }\left( \om... | To prove the necessity of (i) and (ii), we note that if \( {\mu }_{n} \) is tight and \( \epsilon > 0 \) we can choose a compact set \( K \) so that \( {\mu }_{n}\left( K\right) \geq 1 - \epsilon \) for all \( n \) . By Theorem 8.10.6, \( K \subset \{ X\left( 0\right) \leq M\} \) for large \( M \) and if \( \epsilon > ... | Yes |
For \( 1 \leq n \leq \infty \) let\n\n\[ \n{f}_{n}\left( t\right) = \left\{ \begin{array}{ll} 0 & t \in \lbrack 0,\left( {n + 1}\right) /{2n}) \\ 1 & t \in \left\lbrack {\left( {n + 1}\right) /{2n},1}\right\rbrack \end{array}\right.\n\]\n\nwhere \( \left( {n + 1}\right) /{2n} = 1/2 \) for \( n = \infty \) . We certainl... | Let \( \Lambda \) be the class of strictly increasing continuous mappings of \( \left\lbrack {0,1}\right\rbrack \) onto itself. Such functions necessarily have \( \lambda \left( 0\right) = 0 \) and \( \lambda \left( 1\right) = 1 \) . For \( f, g \in D \) define \( d\left( {f, g}\right) \) to be the infimum of those pos... | Yes |
For \( 1 \leq n < \infty \) let\n\n\[ \n{g}_{n}\left( t\right) = \left\{ \begin{array}{ll} 0 & t \in \lbrack 0,1/2) \\ 1 & t \in \lbrack 1/2,\left( {n + 1}\right) /{2n}) \\ 0 & t \in \left\lbrack {\left( {n + 1}\right) /{2n},1}\right\rbrack \end{array}\right. \n\]\n\nThe pointwise limit of \( {g}_{n} \) is \( {g}_{\inf... | In order to have \( \epsilon < 1 \) in the definition of \( d\left( {{g}_{n},{g}_{m}}\right) \) we must have \( \lambda \left( {1/2}\right) = 1/2 \) and \( \lambda \left( {\left( {n + 1}\right) /{2n}}\right) = \left( {m + 1}\right) /{2m} \) so\n\n\[ \nd\left( {{g}_{n},{g}_{m}}\right) = \left| {\frac{1}{2n} - \frac{1}{2... | No |
Theorem 8.11.2. Kolmogorov’s test. If \( h\left( t\right) \uparrow \) and \( {t}^{-1/2}h\left( t\right) \downarrow \) then \( h \) is upper or lower class according as\n\n\[{\int }_{0}^{1}{t}^{-3/2}h\left( t\right) \exp \left( {-{h}^{2}\left( t\right) /{2t}}\right) {dt}\;\text{ converges or diverges }\] | Approximating \( h \) from above by piecewise constant functions, it is easy to show that if the integral in Theorem 8.11.2 converges, \( h\left( t\right) \) is an upper class function. The proof of the other direction is much more difficult; see Motoo (1959) or Section 4.12 of Itô and McKean (1965). | No |
Theorem 8.11.3. If \( {X}_{1},{X}_{2},\ldots \) are i.i.d. with \( E{X}_{i} = 0 \) and \( E{X}_{i}^{2} = 1 \) then\n\n\[ \mathop{\limsup }\limits_{{n \rightarrow \infty }}{S}_{n}/{\left( 2n\log \log n\right) }^{1/2} = 1 \] | Proof. By Theorem 8.7.2, we can write \( {S}_{n} = B\left( {T}_{n}\right) \) with \( {T}_{n}/n \rightarrow 1 \) a.s. As in the proof of Donsker's theorem, this is all we will use in the argument below. Theorem 8.11.3 will follow from Theorem 8.11.1 once we show\n\n\[ \left( {{S}_{\left\lbrack t\right\rbrack } - {B}_{t}... | Yes |
Theorem 8.11.4. Strassen’s (1964) invariance principle. Let \( {X}_{1},{X}_{2},\ldots {be} \) i.i.d. with \( E{X}_{i} = 0 \) and \( E{X}_{i}^{2} = 1 \), let \( {S}_{n} = {X}_{1} + \cdots + {X}_{n} \), and let \( {S}_{\left( n \cdot \right) } \) be the usual linear interpolation. The limit set (i.e., the collection of l... | Jensen’s inequality implies \( f{\left( 1\right) }^{2} \leq {\int }_{0}^{1}g{\left( y\right) }^{2}{dy} \leq 1 \) with equality if and only if \( f\left( t\right) = t \), so Theorem 8.11.4 contains Theorem 8.11.3 as a special case and provides some information about how the large value of \( {S}_{n} \) came about. | Yes |
A more sophisticated way of measuring a set is the area of a plane set as in Examples (f) and \( \left( {\mathrm{f}}^{\prime }\right) \) of \( §{1.1} \), or the volume of a solid. It is said that the measurement of land areas was the origin of geometry and trigonometry in ancient times. While the nomads were still coun... | \[ P\left( A\right) = \frac{\left| A\right| }{\left| \Omega \right| }.\] | Yes |
When a die is thrown there are six possible outcomes. If we compare the process of throwing a particular number [face] with that of picking a particular apple in Example 1, we are led to take \( \Omega = \{ 1,2,3,4,5,6\} \) and define\n\n\[ P\left( {\{ k\} }\right) = \frac{1}{6},\;k = 1,2,3,4,5,6. \] | Here we are treating the six outcomes as \ | No |
A classical enunciation of probability runs as follows. The probability of an event is the ratio of the number of cases favorable to that event to the total number of cases, provided that all these are equally likely | To translate this into our language: the sample space is a finite set of possible cases: \( \left\{ {{\omega }_{1},{\omega }_{2},\ldots ,{\omega }_{m}}\right\} \), each \( {\omega }_{i} \) being a \ | No |
What does the set \( {AB} \) represent? | It is the set of integers that are divisible by 3 and by 4 . If you have not entirely forgotten your school arithmetic, you know this is just the set of multiples of \( 3 \cdot 4 = {12} \) . Hence \( P\left( {AB}\right) = 1/{12} \) . Now we can use (viii) to get \( P\left( {A \cup B}\right) \) :\n\n\[ P\left( {A \cup B... | Yes |
What is the probability of the set of numbers divisible by 3, not divisible by 5 , and divisible by 4 or 6 ? | Using the preceding notation, the set in question is \( A{D}^{c}\left( {B \cup C}\right) \), where \( D = {A}_{5} \) . Using distributive law, we can write this as \( A{D}^{c}B \cup A{D}^{c}C \) . We also have\n\n\[ \n\left( {A{D}^{c}B}\right) \left( {A{D}^{c}C}\right) = A{D}^{c}{BC} = {ABC} - {ABCD}.\n\]\n\nHence by (... | Yes |
in how many ways can six dice appear when they are rolled? And in how many ways can they show all different faces? | Each die here represents a multiple choice of six possibilities. For the first problem these 6 choices can be freely combined so the rule applies directly to give the answer \( {6}^{6} = {46656} \) . For the second problem the choices cannot be freely combined since they are required to all be different. Offhand the ru... | No |
Six mountain climbers decide to divide into three groups for the final assault on the peak. The groups will be of size \( 1,2,3 \), respectively, and all manners of deployment are considered. What is the total number of possible grouping and deploying? | The number of ways of splitting in \( {G}_{1},{G}_{2},{G}_{3} \), where the subscript denotes the size of group, is given by (3.3.2):\n\n\[ \frac{6!}{1!2!3!} = {60} \]\n\nHaving formed these three groups, there remains the decision of which group leads, which is in the middle, and which backs up. This is solved by Case... | Yes |
If a deck of poker cards is thoroughly shuffled, what is the probability that the four aces are found in a row? | There are 52 cards among which are 4 aces. A thorough shuffling signifies that all permutations of the cards are equally likely. For the whole deck, there are (52)! outcomes by Case IIa. In how many of these do the four aces stick together? Here we use tip (b) to break up the problem according to where the aces are fou... | Yes |
Problem 3. Fifteen new students are to be evenly distributed among three classes. Suppose that there are 3 whiz-kids among the 15. What is the probability that each class gets one? One class gets them all? | It should be clear that this is the partition problem discussed under Case IIIb, with \( m = {15},{m}_{1} = {m}_{2} = {m}_{3} = 5 \) . Hence the total number of outcomes is given by\n\n\[ \n\frac{{15}!}{5!5!5!}\text{. }\n\]\n\nTo count the number of these assignments in which each class gets one whiz-kid, we will first... | Yes |
What is the probability that among \( n \) people there are at least two who have the same birthday? | \[ {p}_{n} = 1 - \frac{{\left( {365}\right) }_{n}}{{\left( {365}\right) }^{n}} \] | No |
Problem 6. (Matching). Four cards numbered 1 to 4 are laid face down on a table and a person claiming clairvoyance will name them by his extrasensory power. If he is a faker and just guesses at random, what is the probability that he gets at least one right? | There is a neat solution to this famous problem by a formula to be established later in \( §{6.2} \) . But for a small number like 4, brute force will do and in the process we shall learn something new. Now the faker simply picks any one of the 4 ! permutations, and these are considered equally likely. Using tip (b), w... | No |
In how many ways can \( n \) balls be put into \( n \) numbered boxes so that exactly one box is empty? | Hypothesis 1. The balls are indistinguishable. Then it is clearly just a matter of picking the empty box and the one that must have two balls. This is a sampling problem under Case II, and the answer is \( {\left( n\right) }_{2} = n\left( {n - 1}\right) \).\n\nHypothesis 2. The balls are distinguishable. Then after the... | Yes |
Let \( \Omega \) be gaseous molecules in a given container. We can still represent \( \Omega \) as in (4.1.1) even though \( n \) is now a very large number such as \( {10}^{25} \) . Let \( m = \) mass, \( v = \) velocity, \( M = \) momentum, \( E = \) kinetic energy. Then we have the corresponding functions: | \[ \omega \rightarrow m\left( \omega \right) \] \[ \omega \rightarrow v\left( \omega \right) \] \[ \omega \rightarrow M\left( \omega \right) = m\left( \omega \right) v\left( \omega \right) \] \[ \omega \rightarrow E\left( \omega \right) = \frac{1}{2}m\left( \omega \right) v{\left( \omega \right) }^{2}. \] In experiment... | Yes |
Let \( \Omega \) be the outcome space of throwing a die twice. Then it consists of \( {6}^{2} = {36} \) points listed below: | <table><tr><td>\( \left( {1,1}\right) \)</td><td>\( \left( {1,2}\right) \)</td><td>\( \left( {1,3}\right) \)</td><td>\( \left( {1,4}\right) \)</td><td>\( \left( {1,5}\right) \)</td><td>\( \left( {1,6}\right) \)</td></tr><tr><td>\( \left( {2,1}\right) \)</td><td>\( \left( {2,2}\right) \)</td><td>\( \left( {2,3}\right) \... | Yes |
Proposition 1. If \( X \) and \( Y \) are random variables, then so are\n\n\[ X + Y,\;X - Y,\;{XY},\;X/Y\left( {Y \neq 0}\right) ,\]\n\nand \( {aX} + {bY} \) where \( a \) and \( b \) are two numbers. | This is immediate from the general definition, since, e.g.,\n\n\[ \omega \rightarrow X\left( \omega \right) + Y\left( \omega \right) \]\n\nis a function on \( \Omega \) as well as \( X \) and \( Y \) . The situation is exactly the same as in calculus: if \( f \) and \( g \) are functions, then so are\n\n\[ f + g,\;f - ... | Yes |
What is the probability that the book is a financial loss? | It is that of the event represented by the set\n\n\[ \{ {5X} - {3000} < 0\} = \{ X < {600}\} . \] | Yes |
What is the total number of claims received in the year? It is given by \( N \), where\n\n\[ N = \max \left\{ {n \mid {S}_{n} \leq {365}}\right\} \] | Obviously \( N \) is also random but it is determined by the sequence of \( {S}_{n} \) ’s; in theory we need to know the entire sequence because \( N \) may be arbitrarily large. Knowing \( N \) and the sequence of \( {C}_{n} \) ’s, we can determine the total amount of claims in that year:\n\n\[ {C}_{1} + \cdots + {C}_... | No |
Define \( X\left( \omega \right) \) to be the price per acre of the lot \( \omega \) . Then \( E\left( X\right) \) is the average price per acre of the whole parcel and is given by | \[ \left( {800}\right) \frac{5}{100} + \left( {900}\right) \frac{10}{100} + \left( {1000}\right) \frac{10}{100} + \left( {1200}\right) \frac{10}{100} + \left( {800}\right) \frac{15}{100} + \left( {900}\right) \frac{20}{100} + \left( {800}\right) \frac{30}{100} = {890} \] namely \$890 per acre. This can also be computed... | Yes |
Suppose \( L \) is a positive integer, and\n\n\[ \n{p}_{n} = \frac{1}{L},\;1 \leq n \leq L.\n\]\n\nThen automatically all other \( {p}_{n} \) ’s must be zero because \( \mathop{\sum }\limits_{{n = 1}}^{L}{p}_{n} = L \cdot \frac{1}{L} = 1 \) and the conditions in (4.3.10) must be satisfied. | Next, we have\n\n\[ \nE\left( X\right) = \frac{1}{L}\mathop{\sum }\limits_{{n = 1}}^{L}n = \frac{1}{L} \cdot \frac{L\left( {L + 1}\right) }{2} = \frac{L + 1}{2}.\n\]\nThe preceding sum is done by a formula for arithmetical progression which you have probably learned in school. | No |
What is the expectation of \( X \) ? According to (4.4.1), it is given by the formula \[ \mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{n}{{2}^{n}} = ? \] | Let us learn how to sum this series, though properly speaking this does not belong to this course. We begin with the fountainhead of many of such series: \[ \frac{1}{1 - x} = 1 + x + {x}^{2} + \cdots + {x}^{n} + \cdots = \mathop{\sum }\limits_{{n = 0}}^{\infty }{x}^{n}\;\text{ for }\;\left| x\right| < 1. \] This is a g... | Yes |
A perfect coin is tossed \( n \) times. Let \( {S}_{n} \) denote the number of heads obtained. In the notation of \( §{2.4} \), we have \( {S}_{n} = {X}_{1} + \cdots + {X}_{n} \) . We know from \( §{3.2} \) that\n\n\[ \n{p}_{k} = P\left( {{S}_{n} = k}\right) = \frac{1}{{2}^{n}}\left( \begin{array}{l} n \\ k \end{array}... | If we believe in probability, then we know \( \mathop{\sum }\limits_{{k = 0}}^{\infty }{p}_{k} = 1 \) from (4.3.10). Hence\n\n\[ \n\mathop{\sum }\limits_{{k = 0}}^{n}\frac{1}{{2}^{n}}\left( \begin{array}{l} n \\ k \end{array}\right) = 1\;\text{ or }\;\mathop{\sum }\limits_{{k = 0}}^{n}\left( \begin{array}{l} n \\ k \en... | Yes |
Spin a needle on a circular dial. When it stops it points at a random angle \( \theta \) (measured from the horizontal, say). Under normal conditions it is reasonable to suppose that \( \theta \) is uniformly distributed between \( {0}^{ \circ } \) and \( {360}^{ \circ } \) (cf. Example 7 of \( §{4.4} \) ). This means ... | \[ f\left( u\right) = \left\{ \begin{array}{ll} \frac{1}{360} & \text{ for }0 \leq u \leq {360} \\ 0 & \text{ otherwise. } \end{array}\right. \] Thus for any \( {\theta }_{1} < {\theta }_{2} \) we have \[ P\left( {{\theta }_{1} \leq \theta \leq {\theta }_{2}}\right) = {\int }_{{\theta }_{1}}^{{\theta }_{2}}\frac{1}{360... | Yes |
Suppose you station yourself at a spot on a relatively serene country road and watch the cars that pass by that spot. With your stopwatch you can clock the time before the first car passes. This is a random variable \( T \) called the waiting time. Under certain circumstances it is a reasonable hypothesis that \( T \) ... | \[ f\left( u\right) = \lambda {e}^{-{\lambda u}},\;u \geq 0. \] It goes without saying that \( f\left( u\right) = 0 \) for \( u < 0 \) . The corresponding distribution function is called the exponential distribution with parameter \( \lambda \) , obtained by integrating \( f \) as in (4.5.4): \[ F\left( x\right) = {\in... | Yes |
Suppose in a problem involving the random variable \( T \) above, what we really want to measure is its logarithm (to the base \( e \) ):\n\n\[ S = \log T \]\n\nThis is also a random variable (cf. Proposition 2 in \( §{4.2} \) ); it is negative if \( T > 1 \), zero if \( T = 1 \), and positive if \( T > 1 \). What are ... | Now the function\n\n\[ x \rightarrow \log x \]\n\nis monotone and its inverse is\n\n\[ x \rightarrow {e}^{x} \]\n\nso that\n\n\[ S \leq x \Leftrightarrow \log T \leq x \Leftrightarrow T \leq {e}^{x}. \]\n\nHence by (4.5.11)\n\n\[ {F}_{S}\left( x\right) = P\{ S \leq x\} = P\left\{ {T \leq {e}^{x}}\right\} = 1 - {e}^{-\l... | Yes |
A certain river floods every year. Suppose the low-water mark is set at 1, and the high-water mark \( Y \) has the distribution function\n\n\[ F\left( y\right) = P\left( {Y \leq y}\right) = 1 - \frac{1}{{y}^{2}},\;1 \leq y < \infty . \] | Observe that \( F\left( 1\right) = 0 \), that \( F\left( y\right) \) increases with \( y \), and that \( F\left( y\right) \rightarrow 1 \) as \( y \rightarrow \infty \) . This is as it should be from the meaning of \( P\left( {Y \leq y}\right) \) . To get the density function we differentiate:\n\n\[ f\left( y\right) = ... | Yes |
All students on a certain college campus are polled as to their reaction to a certain presidential candidate. Let \( D \) denote those who favor him. Now the student population \( \Omega \) may be cross-classified in various ways, for instance according to sex, age, race, etc. Let\n\n\[ A = \text{female,}B = \text{blac... | \[ P\left( {D \mid {A}^{c}{BC}}\right) = \frac{P\left( {{A}^{c}{BCD}}\right) }{P\left( {{A}^{c}{BC}}\right) } \] \ndenotes the proportion of male black students of voting age who favor the candidate;\n\n\[ P\left( {{D}^{c} \mid {A}^{c}C}\right) = \frac{P\left( {{A}^{c}C{D}^{c}}\right) }{P\left( {{A}^{c}C}\right) } \] \... | Yes |
A perfect die is thrown twice. Given [knowing] that the total obtained is 7, what is the probability that the first point obtained is \( k \) , \( 1 \leq k \leq 6 \) ? | Look at the list in Example 3 of §4.1. The outcomes with total equal to 7 are those on the \ | No |
Consider the waiting time \( X \) in Example 8 of \( §{4.4} \), for a biased coin. Knowing that it has fallen tails three times, what is the probability that it will fall heads within the next two trials? | This is the conditional probability\n\n\[ P\left( {X \leq 5 \mid X \geq 4}\right) = \frac{P\left( {4 \leq X \leq 5}\right) }{P\left( {X \geq 4}\right) }.\]\n\nWe know that\n\n\[ P\left( {X = n}\right) = {q}^{n - 1}p,\;n = 1,2,\ldots ; \]\n\nfrom which we can calculate\n\n\[ P\left( {X \geq 4}\right) = \mathop{\sum }\li... | Yes |
If \( X \) has the geometrical distribution, then for any nonnegative integer \( n \) we have \( P\left( {X > n}\right) = {q}^{n} \). | This can be shown by summing a geometrical series as in (5.1.8), but it is obvious if we remember that \ | No |
If a family is chosen at random from \(\Omega\) and found to have a boy in it, what is the probability that it has another boy, namely that it is of the type \(\left( {b, b}\right) \) ? | Let us put\n\n\[ A = \{ \omega \mid \text{ there is a boy in }\omega \} \]\n\n\[ B = \{ \omega \mid \text{ there are two boys in }\omega \} . \]\n\nThen \( B \subset A \) and so \( {AB} = B \), thus\n\n\[ P\left( {B \mid A}\right) = \frac{P\left( B\right) }{P\left( A\right) } = \frac{\frac{1}{4}}{\frac{3}{4}} = \frac{1... | Yes |
Proposition 1. For arbitrary events \( {A}_{1},{A}_{2},\ldots ,{A}_{n} \), we have\n\n\[ P\left( {{A}_{1}{A}_{2}\ldots {A}_{n}}\right) = P\left( {A}_{1}\right) P\left( {{A}_{2} \mid {A}_{1}}\right) P\left( {{A}_{3} \mid {A}_{1}{A}_{2}}\right) \ldots P\left( {{A}_{n} \mid {A}_{1}{A}_{2}\ldots {A}_{n - 1}}\right) \]\n\np... | Proof: Under the proviso, all conditional probabilities in (5.2.1) are well defined since\n\n\[ P\left( {A}_{1}\right) \geq P\left( {{A}_{1}{A}_{2}}\right) \geq \cdots \geq P\left( {{A}_{1}{A}_{2}\ldots {A}_{n - 1}}\right) > 0. \]\n\nNow the right side of (5.2.1) is explicitly\n\n\[ \frac{P\left( {A}_{1}\right) }{P\lef... | Yes |
Proposition 2. Suppose that\n\n\\[ \n\\Omega = \\mathop{\\sum }\\limits_{n}{A}_{n} \n\\]\n\nis a partition of the sample space into disjoint sets. Then for any set \\( B \\) we have\n\n\\[ \nP\\left( B\\right) = \\mathop{\\sum }\\limits_{n}P\\left( {A}_{n}\\right) P\\left( {B \\mid {A}_{n}}\\right) .\n\\]\n\n\\( \\left... | Proof: First we write\n\n\\[ \nB = {\\Omega B} = \\left( {\\mathop{\\sum }\\limits_{n}{A}_{n}}\\right) B = \\mathop{\\sum }\\limits_{n}{A}_{n}B \n\\]\n\nby simple set theory, in particular (1.3.6); then we deduce\n\n\\[ \nP\\left( B\\right) = P\\left( {\\mathop{\\sum }\\limits_{n}{A}_{n}B}\\right) = \\mathop{\\sum }\\l... | Yes |
Proposition 3. Under the assumption and notation of Proposition 2, we have also\n\n\[ P\left( {{A}_{n} \mid B}\right) = \frac{P\left( {A}_{n}\right) P\left( {B \mid {A}_{n}}\right) }{\mathop{\sum }\limits_{n}P\left( {A}_{n}\right) P\left( {B \mid {A}_{n}}\right) } \]\n\n\( \left( {5.2.5}\right) \)\n\nprovided \( P\left... | Proof: The denominator above is equal to \( P\left( B\right) \) by Proposition 2, so the equation may be multiplied out to read\n\n\[ P\left( B\right) P\left( {{A}_{n} \mid B}\right) = P\left( {A}_{n}\right) P\left( {B \mid {A}_{n}}\right) . \]\n\nThis is true since both sides are equal to \( P\left( {{A}_{n}B}\right) ... | Yes |
What is the probability of throwing six perfect die and getting six different faces? | Number the dice from 1 to 6, and put\n\n\[ \n{A}_{1} = \text{any face for Die 1,}\n\]\n\n\( {A}_{2} = \) Die 2 shows a different face from Die 1,\n\n\( {A}_{3} = \) Die 3 shows a different face from Die 1 and Die 2,\n\netc. Then we have, assuming that the dice act independently,\n\n\[ \nP\left( {A}_{1}\right) = 1,\;P\l... | Yes |
What is the probability that the dog will be found in the park? | Let \( A, B, C \) be the hypotheses above, and let \( D = \) \ | No |
Urn one contains 2 black and 3 red balls; urn two contains 3 black and 2 red balls. We toss an unbiased coin to decide on the urn to draw from but we do not know which is which. Suppose the first ball drawn is black and it is put back; what is the probability that the second ball drawn from the same urn is also black? | Call the two urns \( {U}_{1} \) and \( {U}_{2} \) ; the a priori probability that either one is chosen by the coin-tossing is \( 1/2 \) :\n\n\[ P\left( {U}_{1}\right) = \frac{1}{2},\;P\left( {U}_{2}\right) = \frac{1}{2}. \]\n\nDenote the event that the first ball is black by \( {B}_{1} \), that the second ball is black... | No |
Suppose that the sun has risen \( n \) times in succession; what is the probability that it will rise once more? | It is assumed that the a priori probability for a sunrise on any day is a constant whose value is unknown to us. Due to our total ignorance it will be assumed to take all possible values in \( \left\lbrack {0,1}\right\rbrack \) with equal likelihood. That is to say, this probability will be treated as a random variable... | No |
Theorem 2 (Poisson’s Theorem). Suppose in an urn containing b black and \( r \) red balls, \( n \) balls have been drawn first and discarded without their colors being noted. If \( m \) balls are drawn next, the probability that there are \( k \) black balls among them is the same as if we had drawn these \( m \) balls... | Here is Poisson’s quick argument: if \( n + m \) balls are drawn out, the probability of a combination made up of \( n \) black and red balls in given proportions followed by \( m \) balls of which \( k \) are black and \( m - k \) are red must be the same as that of a similar combination in which the \( m \) balls pre... | No |
Theorem 3. The probability of drawing (from the beginning) any specified sequence of \( k \) black balls and \( n - k \) red balls is equal to\n\n\[ \n\frac{b\left( {b + c}\right) \cdots \left( {b + \left( {k - 1}\right) c}\right) r\left( {r + c}\right) \cdots \left( {r + \left( {n - k - 1}\right) c}\right) }{\left( {b... | Proof: This is really an easy application of Proposition 1 in \( §{5.2} \), but in a scrambled way. We have shown it above in the case \( k = 2 \) and \( n = 3 \) . If you will try a few more cases with say \( n = 4, k = 2 \) or \( n = 5, k = 3 \), you will probably see how it goes in the general case more quickly than... | No |
Theorem 4. The probability of drawing (from the beginning) \( k \) black balls in \( n \) drawings is equal to the number in (5.4.2) multiplied by \( \left( \begin{array}{l} n \\ k \end{array}\right) \) . In terms of generalized binomial coefficients [see (5.4.4) below], it is equal to\n\n\[ \frac{\left( \begin{matrix}... | Proof: There are \( \left( \begin{array}{l} n \\ k \end{array}\right) \) ways of permuting \( k \) black and \( n - k \) red balls; see §3.2. According to (5.4.2), every specified sequence of drawing \( k \) black and \( n - k \) red balls has the same probability. These various permutations correspond to disjoint even... | Yes |
Proposition 4. We have for arbitrary countable sets \( {S}_{1},\ldots ,{S}_{n} \) :\n\n\[ P\left( {{X}_{1} \in {S}_{1},\ldots ,{X}_{n} \in {S}_{n}}\right) = P\left( {{X}_{1} \in {S}_{1}}\right) \ldots P\left( {{X}_{n} \in {S}_{n}}\right) . \]\n\n\( \left( {5.5.2}\right) \) | Proof: The left member of (5.5.2) is equal to\n\n\[ \mathop{\sum }\limits_{{{x}_{1} \in {S}_{1}}}\cdots \mathop{\sum }\limits_{{{x}_{n} \in {S}_{n}}}P\left( {{X}_{1} = {x}_{1},\ldots ,{X}_{n} = {x}_{n}}\right) \]\n\n\[ = \mathop{\sum }\limits_{{{x}_{1} \in {S}_{1}}}\cdots \mathop{\sum }\limits_{{{x}_{n} \in {S}_{n}}}P\... | Yes |
Proposition 5. The events\n\n\[ \n\\left\\{ {{X}_{1} \\in {S}_{1}}\\right\\} ,\\ldots ,\\left\\{ {{X}_{n} \\in {S}_{n}}\\right\\} \n\]\n\nare independent. | Proof: It is important to recall that the definition of independent events requires not only the relation (5.5.2), but also similar relations for all subsets of \( \\left( {{X}_{1},\\ldots ,{X}_{n}}\\right) \) . However, these also hold because the subsets are also sets of independent random variables, as just shown. | No |
Proposition 6. Let \( {\varphi }_{1},\ldots ,{\varphi }_{n} \) be an arbitrary real-valued function on \( \left( {-\infty ,\infty }\right) \) ; then the random variables\n\n\[ \n{\varphi }_{1}\left( {X}_{1}\right) ,\ldots ,{\varphi }_{n}\left( {X}_{n}\right) \n\]\n\nare independent. | Proof: Let us omit the subscripts on \( X \) and \( \varphi \) and ask the question: for a given real number \( y \), what are the values of \( x \) such that\n\n\[ \n\varphi \left( x\right) = y\;\text{ and }\;X = x?\n\]\n\nThe set of such values must be countable since \( X \) is countably valued; call it \( S \) ; of... | Yes |
If two points are picked at random from the interval \( \\left\\lbrack {0,1}\\right\\rbrack \) , what is the probability that the distance between them is less than \( 1/2 \) ? | By now you should be able to interpret this kind of cryptogram. It means: if \( X \) and \( Y \) are two independent random variables each of which is uniformly distributed in \( \\left\\lbrack {0,1}\\right\\rbrack \), find the probability \( P\\left( {\\left| {X - Y}\\right| < 1/2}\\right) \) . Under the hypotheses th... | Yes |
Find the distribution functions of \( M \) and \( m \) . | Using (5.5.7), we have for each \( x \)\n\n\[ \n{F}_{\max }\left( x\right) = P\left( {M \leq x}\right) = P\left( {{X}_{1} \leq x;{X}_{2} \leq x;\ldots ;{X}_{n} \leq x}\right) \n\]\n\n\[ \n= P\left( {{X}_{1} \leq x}\right) P\left( {{X}_{2} \leq x}\right) \ldots P\left( {{X}_{n} \leq x}\right) \n\]\n\n\[ \n= {F}_{1}\left... | No |
Consider families with two children as in Example 5 of §5.1: \( \Omega = \{ \left( {bb}\right) ,\left( {bg}\right) ,\left( {gb}\right) ,\left( {gg}\right) \} \) . Let such a family be chosen at random and consider the three events below:\n\n\[ A = \text{first child is a boy;}\]\n\n\[ B = \text{the two children are of d... | A trivial computation then shows that \( P\left( {AB}\right) = P\left( A\right) P\left( B\right), P\left( {BC}\right) = \) \( P\left( B\right) P\left( C\right) \), but \( P\left( {AC}\right) = 0 \neq P\left( A\right) P\left( C\right) \) . Thus the pairs \( \{ A, B\} \) and \( \{ B, C\} \) are independent but the pair \... | Yes |
The arithmetical puzzle is easily solved by the following explicit formulas: | \[ P\left( {A \mid B}\right) = \frac{P\left( {AB}\right) }{P\left( B\right) } = \frac{P\left( {ABC}\right) + P\left( {{AB}{C}^{c}}\right) }{P\left( B\right) } \] \[ = \frac{P\left( {ABC}\right) }{P\left( {BC}\right) }\frac{P\left( {BC}\right) }{P\left( B\right) } + \frac{P\left( {{AB}{C}^{c}}\right) }{P\left( {B{C}^{c}... | Yes |
Suppose the probability of a carrier among the general population is \( p \), irrespective of sex. Now if a person has an affected brother or sister who died in childhood, then he has a history in the family and cannot be treated genetically as a member of the general population. The probability of his being a carrier ... | \[ P\left( {{AA} \mid {AA} \cup {Aa}}\right) = \frac{1}{3},\;P\left( {{Aa} \mid {AA} \cup {Aa}}\right) = \frac{2}{3}. \]\n\nIf he marries a woman who is not known to have a history of that kind in the family, then she is of genotype \( {AA} \) or \( {Aa} \) with probability \( 1 - p \) or \( p \) as for the general pop... | No |
Theorem 1. If \( X \) and \( Y \) are summable, then so is \( X + Y \) and we have\n\n\[ E\left( {X + Y}\right) = E\left( X\right) + E\left( Y\right) \] | Proof: Applying the definition (6.1.1) to \( X + Y \), we have\n\n\[ E\left( {X + Y}\right) = \mathop{\sum }\limits_{\omega }\left( {X\left( \omega \right) + Y\left( \omega \right) }\right) P\left( \omega \right) \]\n\n\[ = \mathop{\sum }\limits_{\omega }X\left( \omega \right) P\left( \omega \right) + \mathop{\sum }\li... | Yes |
A raffle lottery contains 100 tickets, of which there is one ticket bearing the prize \$10000, the rest being all zero. If I buy two tickets, what is my expected gain? | If I have only one ticket, my gain is represented by the random variable \( X \), which takes the value 10000 on exactly one \( \omega \) and 0 on all the rest. The tickets are assumed to be equally likely to win the prize; hence\n\n\[ \nX = \left\{ \begin{array}{ll} {10000} & \text{ with probability }\frac{1}{100}, \\... | Yes |
There are \( N \) coupons marked 1 to \( N \) in a bag. We draw one coupon after another with replacement. Suppose we wish to collect \( r \) different coupons; what is the expected number of drawings to get them? | The problem may be regarded as one of waiting time, namely: we wait for the \( r \) th new arrival. Let \( {X}_{1},{X}_{2},\ldots \) denote the successive waiting times for a new coupon. Thus \( {X}_{1} = 1 \) since the first is always new. Now \( {X}_{2} \) is the waiting time for any coupon that is different from the... | Yes |
Poincaré’s Formula. For arbitrary events \( {A}_{1},\ldots ,{A}_{n} \) we have\n\n\[ P\left( {\mathop{\bigcup }\limits_{{j = 1}}^{n}{A}_{j}}\right) = \mathop{\sum }\limits_{j}P\left( {A}_{j}\right) - \mathop{\sum }\limits_{{j, k}}P\left( {{A}_{j}{A}_{k}}\right) + \mathop{\sum }\limits_{{j, k, l}}P\left( {{A}_{j}{A}_{k}... | Proof: Let \( {\alpha }_{j} = {I}_{{A}_{j}} \) be the indicator of \( {A}_{j} \) . Then the indicator of \( {A}_{1}^{c}\cdots {A}_{n}^{c} \) is \( \mathop{\prod }\limits_{{j = 1}}^{n}\left( {1 - {\alpha }_{j}}\right) \) ; hence that of its complement is given by\n\n\[ {I}_{{A}_{1}} \cup \cdots \cup {A}_{n} = 1 - \matho... | Yes |
Two sets of cards both numbered 1 to \( n \) are randomly matched. What is the probability of at least one match? | Let \( {A}_{j} \) be the event that the \( j \) th cards are matched, regardless of the others. There are \( n \) ! permutations of the second set against the first set, which may be considered as laid out in natural order. If the \( j \) th cards match, that leaves \( \left( {n - 1}\right) \) ! permutations for the re... | Yes |
Example 5. Iron bars in the shape of slim cylinders are test-measured. Suppose the average length is 10 inches and average area of ends is 1 square inch. The average error made in the measurement of the length is .005 inch, that in the measurement of the area is .01 square inch. What is the average error made in estima... | Since weight is a constant times volume, it is sufficient to consider the latter: \( V = {LA} \) where \( L = \) length, \( A = \) area of ends. Let the errors be \( {\Delta L} \) and \( {\Delta A} \), respectively; then the error in \( V \) is given by\n\n\[{\Delta V} = \left( {L + {\Delta L}}\right) \left( {A + {\Del... | Yes |
Theorem 3. If \( E\left( {X}^{2}\right) \) is finite, then so is \( E\left( \left| X\right| \right) \) . We then have\n\n\[ \n{\sigma }^{2}\left( X\right) = E\left( {X}^{2}\right) - E{\left( X\right) }^{2} \n\] \n\n(6.3.5) \n\nconsequently, \n\n\[ \nE{\left( \left| X\right| \right) }^{2} \leq E\left( {X}^{2}\right) \n\... | Proof: Since \n\n\[ \n{X}^{2} - 2\left| X\right| + 1 = {\left( \left| X\right| - 1\right) }^{2} \geq 0 \n\] \n\nwe must have (why?) \( E\left( {{X}^{2} - 2\left| X\right| + 1}\right) \geq 0 \), and therefore \( E\left( {X}^{2}\right) + 1 \geq \) \( {2E}\left( \left| X\right| \right) \) by Theorem 1 and (6.1.5). This pr... | Yes |
Theorem 4. If \( X \) and \( Y \) are independent and both have finite variances, then\n\n\[{\sigma }^{2}\left( {X + Y}\right) = {\sigma }^{2}\left( X\right) + {\sigma }^{2}\left( Y\right)\] | Proof: By the preceding remark, we may suppose that \( X \) and \( Y \) both have mean zero. Then \( X + Y \) also has mean zero and the variances in (6.3.7) are the same as second moments. Now\n\n\[E\left( {XY}\right) = E\left( X\right) E\left( Y\right) = 0\]\n\nby Theorem 2, and\n\n\[E\left\{ {\left( X + Y\right) }^{... | Yes |
Returning to the matching problem in \( §{6.2} \), let us now compute the standard deviation of the number of matches. The \( {I}_{{A}_{j}} \) ’s in (6.2.8) are not independent, but formula (6.3.8) is applicable and yields | \[ E\left( {N}^{2}\right) = \mathop{\sum }\limits_{{j = 1}}^{n}E\left( {I}_{{A}_{j}}^{2}\right) + 2\mathop{\sum }\limits_{{1 \leq j < k \leq n}}E\left( {{I}_{{A}_{j}}{I}_{{A}_{k}}}\right) . \] Clearly, \[ E\left( {I}_{{A}_{j}}^{2}\right) = P\left( {A}_{j}\right) = \frac{1}{n} \] \[ E\left( {{I}_{{A}_{j}}{I}_{{A}_{k}}}\... | Yes |
Three identical dice are thrown. What is the probability of obtaining a total of 9 ? The dice are not supposed to be symmetrical, and the probability of turning up face \( j \) is equal to \( {p}_{j},1 \leq j \leq 6 \) ; same for all three dice. | Let us list the possible cases in terms of the \( X \) ’s and the \( N \) ’s, respectively:\n\n<table><thead><tr><th>\( {X}_{1} \)</th><th>\( {X}_{2} \)</th><th>\( {X}_{3} \)</th><th>\( {N}_{1} \)</th><th>\( {N}_{2} \)</th><th>\( {N}_{3} \)</th><th>\( {N}_{4} \)</th><th>\( {N}_{5} \)</th><th>\( {N}_{6} \)</th><th>Permu... | Yes |
Theorem 5. The probability distribution of a nonnegative integer-valued random variable is uniquely determined by its generating function. | Let \( Y \) be a random variable having the probability distribution \( \left\{ {{b}_{k}, k \geq }\right. \) \( 0\} \) where \( {b}_{k} = P\left( {Y = k}\right) \), and let \( h \) be its generating function:\n\n\[ h\left( z\right) = \mathop{\sum }\limits_{{k = 0}}^{\infty }{b}_{k}{z}^{k} \]\n\nSuppose that \( g\left( ... | Yes |
For the Bernoullian random variables \( {X}_{1},\ldots ,{X}_{n} \) (Example 6 of \( §{6.3} \) ), the common generating function is\n\n\[ g\left( z\right) = q + {pz} \]\n\nsince \( {a}_{0} = q,{a}_{1} = p \) in (6.5.1). Hence the generating function of \( {S}_{n} = \) \( {X}_{1} + \cdots + {X}_{n} \), where the \( X \)’... | Its power series is therefore known from the binomial theorem, namely,\n\n\[ g{\left( z\right) }^{n} = \mathop{\sum }\limits_{{k = 0}}^{n}\left( \begin{array}{l} n \\ k \end{array}\right) {q}^{n - k}{p}^{k}{z}^{k} \]\n\nOn the other hand, by definition of a generating function, we have\n\n\[ g{\left( z\right) }^{n} = \... | Yes |
For the waiting-time distribution (§4.4), we have \( {p}_{j} = \) \( {q}^{j - 1}p, j \geq 1 \) ; hence | \[ g\left( z\right) = \mathop{\sum }\limits_{{j = 1}}^{\infty }{q}^{j - {1p}{z}^{j}} = \frac{p}{q}\mathop{\sum }\limits_{{j = 1}}^{\infty }{\left( qz\right) }^{i} = \frac{p}{q}\frac{qz}{1 - {qz}} = \frac{pz}{1 - {qz}}. \] | Yes |
For the dice problem at the end of \( §{6.4} \), we have \( {p}_{j} = 1/6 \) for \( 1 \leq j \leq 6 \) if the dice are symmetrical. Hence the associated generating function is given by\n\n\[ g\left( z\right) = \frac{1}{6}\left( {z + {z}^{2} + {z}^{3} + {z}^{4} + {z}^{5} + {z}^{6}}\right) = \frac{z\left( {1 - {z}^{6}}\r... | The generating function of the total points obtained by throwing three dice is just \( {g}^{3} \) . This can be expanded into a power series as follows:\n\n\[ g{\left( z\right) }^{3} = \frac{{z}^{3}}{{6}^{3}}\frac{{\left( 1 - {z}^{6}\right) }^{3}}{{\left( 1 - z}\right) }^{3}} = \frac{{z}^{3}}{{6}^{3}}\left( {1 - 3{z}^{... | Yes |
Theorem 7. Theorems 5 and 6 remain true when the generating function is replaced by the Laplace transform (for nonnegative random variables) or the Fourier transform (for arbitrary random variables). | In the case of Theorem 6, this is immediate from (6.5.13) if the variable \( z \) there is replaced by \( {e}^{-\lambda } \) or \( {e}^{i\theta } \) . For Theorem 5 the analogues lie deeper and require more advanced analysis (see [Chung 1, Chapter 6]). The reader is asked to accept their truth by analogy from the discu... | No |
Consider the card-matching problem in \( §{6.2} \). If a person who claims ESP (extrasensory perception) is a fake and is merely trying to match the cards at random, will his average score be better or worse when the number of cards is increased? Intuitively, two opposite effects are apparent. On one hand, there will b... | Here is an ideal setting for (7.1.6) with \( \alpha = 1 \). In fact, we can make it conform exactly to the previous scheme by allowing duplication in the guessing. That is, if we think of a deck of \( n \) cards laid face down on the table, we are allowed to guess them one by one with total forgetfulness. Then we can g... | No |
Theorem 1. The total number of arrivals in a time interval of length \( t \) has the Poisson distribution \( \pi \left( {\alpha t}\right) \), for each \( t > 0 \) . | The reader should observe that the theorem asserts more than has been proved. For in our formulation above we have implicitly chosen an initial instant from which time is measured, namely the zero time for the first arrival time \( {T}_{1} \) . Thus the result was proved only for the total number of arrivals in the int... | No |
Theorem 2. If the intervals \( \left( {{s}_{1},{s}_{1} + {t}_{1}}\right) ,\left( {{s}_{2},{s}_{2} + {t}_{2}}\right) ,\ldots \) are disjoint, then the random variables in (7.2.16) are independent and have the Poisson distributions \( \pi \left( {\alpha {t}_{1}}\right) ,\pi \left( {\alpha {t}_{2}}\right) \ldots \) . | The proof of Theorem 2 depends again on the lack-of-memory property of the \( {T}_{j} \) ’s. We will indicate the main idea here without going into formal details. Going back to the sequence in (7.2.14), where we put \( s = {s}_{2} \), we now make the further observation that all the random variables there are not only... | No |
Consider the number of arrivals in two disjoint time intervals: \( {X}_{1} = N\left( {{s}_{1},{s}_{1} + {t}_{1}}\right) \) and \( {X}_{2} = N\left( {{s}_{2},{s}_{2} + {t}_{2}}\right) \) as in (7.2.16). What is the probability that the total number \( {X}_{1} + {X}_{2} \) is equal to \( n \) ? | By Theorem \( 2,{X}_{1} \) and \( {X}_{2} \) are independent random variables with the distributions \( \pi \left( {\alpha {t}_{1}}\right) \) and \( \pi \left( {\alpha {t}_{2}}\right) \), respectively. Hence\n\n\[ P\left( {{X}_{1} + {X}_{2} = n}\right) = \mathop{\sum }\limits_{{j + k = n}}P\left( {{X}_{1} = j}\right) P... | Yes |
Theorem 3. Let \( {X}_{j} \) be independent random variables with Poisson distributions \( \pi \left( {\alpha }_{j}\right) ,1 \leq j \leq n \) . Then \( {X}_{1} + \cdots + {X}_{n} \) has Poisson distribution \( \pi \left( {{\alpha }_{1} + \cdots + {\alpha }_{n}}\right) \) . | This follows from an easy induction, but we can also make speedy use of generating functions. If we denote the generating function of \( {X}_{i} \) by \( {g}_{{x}_{i}} \) , then by Theorem 6 of §6.5:\n\n\[ \n{g}_{{X}_{1} + \cdots + {X}_{n}}\left( z\right) = {g}_{{X}_{1}}\left( z\right) {g}_{{X}_{2}}\left( z\right) \cdo... | Yes |
In a unit period of time what is the number of cars counted with these license plates? | We are assuming that if \( n \) cars are counted the distribution of various license plates follows a multinomial distribution \( M\left( {n;{50};{p}_{1},\ldots ,{p}_{50}}\right) \) where the first three \( p \) ’s are given. Now the number of cars passing in the period of time is a random variable \( N \) such that\n\... | Yes |
Theorem 5. Suppose \( 0 < p < 1 \) ; put \( q = 1 - p \), and\n\n\[ \n{x}_{nk} = \frac{k - {np}}{\sqrt{npq}},\;0 \leq k \leq n.\n\]\n\n(7.3.11)\n\nClearly \( {x}_{nk} \) depends on both \( n \) and \( k \), but it will be written as \( {x}_{k} \) below.\n\nLet \( A \) be an arbitrary but fixed positive constant. Then i... | Proof: We have from (7.3.11)\n\n\[ \nk = {np} + \sqrt{npq}{x}_{k},\;n - k = {nq} - \sqrt{npq}{x}_{k}.\n\]\n\n(7.3.14)\n\nHence in the range indicated in (7.3.12),\n\n\[ \nk \sim {np},\;n - k \sim {nq}.\n\]\n\n\( \left( {7.3.15}\right) \)\n\nUsing Stirling's formula (7.3.3), we may write the left member of (7.3.13)\n\na... | Yes |
Theorem 6 (De Moivre-Laplace Theorem). For any two constants a and \( b, - \infty < a < b < + \infty \), we have\n\n\[ \mathop{\lim }\limits_{{n \rightarrow \infty }}P\left( {a < \frac{{S}_{n} - {np}}{\sqrt{npq}} \leq b}\right) = \frac{1}{\sqrt{2\pi }}{\int }_{a}^{b}{e}^{-{x}^{2}/2}{dx}. \] | Proof: Let \( k \) denote a possible value of \( {S}_{n} \) so that \( {S}_{n} = k \) means \( \left( {{S}_{n} - }\right. \) \( {np})/\sqrt{npq} = {x}_{k} \) by the transformation (7.3.11). Hence the probability on the left side of (7.3.19) is just\n\n\[ \mathop{\sum }\limits_{{a < {x}_{k} \leq b}}P\left( {{S}_{n} = k}... | Yes |
Theorem 7. Let \( {X}_{j} \) be independent random variables with normal distributions \( N\left( {{m}_{j},{\sigma }_{j}^{2}}\right) ,1 \leq j \leq n \) . Then \( {X}_{1} + \cdots + {X}_{n} \) has the normal distribution \( N\left( {\mathop{\sum }\limits_{{j = 1}}^{n}{m}_{j},\mathop{\sum }\limits_{{j = 1}}^{n}{\sigma }... | Proof: It is sufficient to prove this for \( n = 2 \), since the general case follows by induction. This is easily done by means of the moment-generating function. We have by the product theorem as in Theorem 6 of \( §{6.5} \)\n\n\[ \n{M}_{{X}_{1} + {X}_{2}}\left( \theta \right) = {M}_{{X}_{1}}\left( \theta \right) {M}... | Yes |
Theorem 8. For the sums \( {S}_{n} \) under the generalized conditions spelled out above, we have for any \( a < b \)\n\n\[ \mathop{\lim }\limits_{{n \rightarrow \infty }}P\left( {a < \frac{{S}_{n} - {nm}}{\sqrt{n}\sigma } \leq b}\right) = \frac{1}{\sqrt{2\pi }}{\int }_{a}^{b}{e}^{-{x}^{2}/2}{dx}. \]\n\n\( \left( {7.5.... | Proof: The powerful tool alluded to earlier is that of the characteristic function discussed in \( §{6.5} \) . [We could not have used the moment-generating function since it may not exist for \( {S}_{n} \) .] For the unit normal distribution \( \Phi \), its characteristic function \( g \) can be obtained by substituti... | No |
A physical quantity is measured many times for accuracy. Each measurement is subject to a random error. It is judged reasonable to assume that it is uniformly distributed between -1 and +1 in a conveniently chosen unit. Now if we take the arithmetical mean [average] of \( n \) measurements, what is the probability that... | Let the true value be denoted by \( m \) and the actual measurements obtained by \( {X}_{j},1 \leq j \leq n \) . Then the hypothesis says that\n\n\[ \n{X}_{j} = m + {\xi }_{j} \n\] \n\nwhere \( {\xi }_{j} \) is a random variable that has the uniform distribution in \( \left\lbrack {-1, + 1}\right\rbrack \) . Thus \n\n\... | Yes |
Theorem 10. Under the same conditions as in Theorem 8, we have for a fixed but arbitrary constant \( c > 0 \) ,\n\n\[ \mathop{\lim }\limits_{{n \rightarrow \infty }}P\left( {\left| {\frac{{S}_{n}}{n} - m}\right| < c}\right) = 1 \] | Proof: Since \( c \) is fixed, for any positive constant \( l \), we have\n\n\[ {l\sigma }\sqrt{n} < {cn} \]\n\nfor all sufficiently large values of \( n \) . Hence the event\n\n\[ \left\{ {\left| \frac{{S}_{n} - {mn}}{\sigma \sqrt{n}}\right| < l}\right\} \;\text{ certainly implies }\;\left\{ {\left| \frac{{S}_{n} - {m... | Yes |
Theorem 11. Suppose the random variable \( X \) has a finite second moment. Then for any constant \( c > 0 \) we have\n\n\[ P\left( {\left| X\right| \geq c}\right) \leq \frac{E\left( {X}^{2}\right) }{{c}^{2}}. \] | Proof: We will carry out the proof for a countably valued \( X \) and leave the analogous proof for the density case as an exercise. The idea of the proof is the same for a general random variable.\n\nSuppose that \( X \) takes the values \( {v}_{i} \) with probabilities \( {p}_{i} \), as in \( §{4.3} \) . Then we have... | No |
Theorem 12. Let \( \\left\\{ {{X}_{j}, j \\geq 1}\\right\\} \) be a sequence of independent random variables such that for each \( j \) ,\n\n\[ \nE\\left( {X}_{j}\\right) = {m}_{j},\\;{\\sigma }^{2}\\left( {X}_{j}\\right) = {\\sigma }_{j}^{2};\n\]\n\n(7.6.7)\n\nand furthermore suppose there exists a constant \( M < \\i... | Proof: If we write \( {X}_{j}^{0} = {X}_{j} - {m}_{j},{S}_{n}^{0} = \\mathop{\\sum }\\limits_{{j = 1}}^{n}{X}_{j}^{0} \), then the expression between the bars above is just \( {S}_{n}^{0}/n \) . Of course, \( E\\left( {S}_{n}^{0}\\right) = 0 \), whereas\n\n\[ \nE\\left( {\\left( {S}_{n}^{0}\\right) }^{2}\\right) = {\\s... | Yes |
Suppose \( c = 2\% \) and \( \epsilon = 5\% \) . Then (7.6.15) becomes\n\n\[ \Phi \left( \eta \right) \geq 1 - \frac{5}{200} = {.975} \] | From the table we see that this is satisfied if \( \eta \geq {1.96} \) . Thus\n\n\[ n \geq \frac{{\left( {1.96}\right) }^{2}{pq}}{{c}^{2}} = \frac{{\left( {1.96}\right) }^{2} \times {10000}}{4}{pq}. \]\n\nThe last term depends on \( p \), but \( p\left( {1 - p}\right) \leq 1/4 \) for all \( p \), as already noted, and ... | Yes |
If the particle starts inside the interval \( \left\lbrack {0, c}\right\rbrack \), what is the probability that it will ever reach the boundary? | Since the boundary consists of the two endpoints 0 and \( c \), the answer is given by (8.1.11) and is equal to 1 . In terms of the gamblers, this means that one of them is bound to be ruined sooner or later if the game is continued without a time limit; in other words, it cannot go on forever. Now you can object that ... | Yes |
Theorem 1. For any random walk (with arbitrary \( p \) ), the particle will almost surely* not remain in any finite interval forever. | As a consequence, we can define a random variable that denotes the waiting time until the particle reaches the boundary. This is sometimes referred to as \ | No |
Problem 3. If the particle starts from \( a\left( { \geq 1}\right) \), what is the probability that it will ever hit 0 ? | The answer is 1 if \( p \leq q \) ; and \( {\left( q/p\right) }^{a} \) if \( p > q \) . Observe that when \( p \leq q \) the particle is at least as likely to go left as to go right, so the first conclusion is most plausible. Indeed, in case \( p < q \) we can say more by invoking the law of large numbers in its strong... | Yes |
Theorem 2. Starting from any point in a symmetric random walk, the particle will almost surely hit any point any number of times. | Proof: Let us write \( i \Rightarrow j \) to mean that starting from \( i \) the particle will almost surely hit \( j \), where \( i \in I, j \in I \) . We have already proved that if \( i \neq j \), then \( i \Rightarrow j \) . Hence also \( j \Rightarrow i \) . But this implies \( j \Rightarrow j \) by the obvious di... | Yes |
Example 1. \( I = \{ \ldots , - 2, - 1,0,1,2,\ldots \} \) is the set of all integers. | \[ \Pi = \left\lbrack \begin{matrix} . & . & . & . & . & . & . & & & & \\ . & . & . & q & 0 & p & 0 & 0 & . & . & . \\ . & . & . & 0 & q & 0 & p & 0 & . & . & . \\ . & . & . & 0 & 0 & q & 0 & p & . & . & . \\ . & . & . & . & . & . & . & & & & \end{matrix}\right\rbrack ,\] where \( p + q = 1, p \geq 0, q \geq 0 \) . Thi... | No |
Let \( \left\{ {{\xi }_{n}, n \geq 0}\right\} \) be a sequence of independent integer-valued random variables such that all except possibly \( {\xi }_{0} \) have the same distribution given by \( \left\{ {{a}_{k}, k \in I}\right\} \), where \( I \) is the set of all integers. Define \( {X}_{n} \) as in (8.1.2): \( {X}_... | \[ P\left\{ {{X}_{n + 1} = j \mid A;{X}_{n} = i}\right\} = P\left\{ {{\xi }_{n + 1} = j - i \mid A;{X}_{n} = i}\right\} \] \[ = P\left\{ {{\xi }_{n + 1} = j - i}\right\} = {a}_{j - i}. \] Hence \( \left\{ {{X}_{n}, n \geq 0}\right\} \) constitutes a homogeneous Markov chain with the transition matrix \( \left\lbrack {p... | Yes |
Example 8. (Ehrenfest model). This may be regarded as a particular case of Example 7 in which we have \( I = \{ 0,1,\ldots, c\} \) and\n\n\[ \n{p}_{i, i + 1} = \frac{c - i}{c},\;{p}_{i, i - 1} = \frac{i}{c}.\n\]\n\n(8.3.16) | It can be realized by an urn scheme as follows. An urn contains \( c \) balls, each of which may be red or black; a ball is drawn at random from it and replaced by one of the other color. The state of the urn is the number of black balls in it. It is easy to see that the transition probabilities are as given above and ... | No |
Theorem 3. For any \( i \) and \( j \), and \( 1 \leq n < \infty \), we have\n\n\[ \n{p}_{ij}^{\left( n\right) } = \mathop{\sum }\limits_{{v = 1}}^{n}{f}_{ij}^{\left( v\right) }{p}_{ij}^{\left( n - v\right) }.\n\] | Proof: This result is worthy of a formal treatment in order to bring out the basic structure of a homogeneous Markov chain. Everything can be set down in a string of symbols:\n\n\[ \n{p}_{ij}^{\left( n\right) } = {P}_{i}\left\{ {{X}_{n} = j}\right\} = {P}_{i}\left\{ {{T}_{j} \leq n;{X}_{n} = j}\right\} = \mathop{\sum }... | Yes |
Theorem 4. For \( i \neq j \), and \( n \geq 1 \), we have\n\n\[ \n{p}_{ij}^{\left( n\right) } = \mathop{\sum }\limits_{{v = 0}}^{{n - 1}}{p}_{ii}^{\left( v\right) }{g}_{ij}^{\left( n - v\right) }.\n\] | Proof: We shall imitate the steps in the proof of Theorem 3 as far as possible; thus\n\n\[ \n{p}_{ij}^{\left( n\right) } = {P}_{i}\left\{ {{X}_{n} = j}\right\} = {P}_{i}\left\{ {0 \leq {U}_{i}^{\left( n\right) } \leq n - 1,{X}_{n} = j}\right\} \n\]\n\n\[ \n= \mathop{\sum }\limits_{{v = 0}}^{{n - 1}}{P}_{i}\left\{ {{U}_... | No |
For any state \( i \) we have \( {f}_{ii}^{ * } = 1 \) if and only if\n\n\[ \mathop{\sum }\limits_{{n = 0}}^{\infty }{p}_{ii}^{\left( n\right) } = \infty \]\n\n(8.4.15)\n\nif \( {f}_{ii}^{ * } < 1 \), then we have\n\n\[ \mathop{\sum }\limits_{{n = 0}}^{\infty }{p}_{ii}^{\left( n\right) } = \frac{1}{1 - {f}_{ii}^{ * }} ... | Proof: From (8.4.13) with \( i = j \) and solving for \( {P}_{ii}\left( z\right) \) we obtain\n\n\[ {P}_{ii}\left( z\right) = \frac{1}{1 - {F}_{ii}\left( z\right) }.\]\n\n(8.4.17)\n\nIf we put \( z = 1 \) above and observe that\n\n\[ {P}_{ii}\left( 1\right) = \mathop{\sum }\limits_{{n = 0}}^{\infty }{p}_{ii}^{\left( n\... | Yes |
Theorem 6. If \( i \) is recurrent and \( i \rightsquigarrow j \), then \( j \) is also recurrent. | Proof: There is nothing to provide if \( i = j \), hence we may suppose \( i \neq j \) . We have by (8.4.13) and (8.4.14)\n\n\[ \n{P}_{ij}\left( z\right) = {F}_{ij}\left( z\right) {P}_{jj}\left( z\right) ,\;{P}_{ij}\left( z\right) = {P}_{ii}\left( z\right) {G}_{ij}\left( z\right) ,\n\]\n\nfrom which we infer\n\n\[ \n{F... | Yes |
Theorem 7. For any state \( i \), we have\n\n\[ \n{q}_{ii} = \left\{ \begin{array}{ll} 1 & \text{ if }i\text{ is recurrent,} \\ 0 & \text{ if }i\text{ is nonrecurrent. } \end{array}\right.\n\] | Proof: Put \( {X}_{0} = i \), and \( \alpha = {f}_{ii}^{ * } \) . Then \( \alpha \) is the probability of at least one return to \( i \) . At the moment of the first return, the particle is in \( i \) and its prior history is irrelevant; hence from that moment on it will move as if making a fresh start from \( i \) (\ | No |
Theorem 8. If \( i \) is recurrent and \( i \rightsquigarrow j \), then\n\n\[ \n{q}_{ij} = {q}_{ji} = 1 \n\] | Proof: The conclusion implies that \( i \leftrightsquigarrow j \) and that \( j \) is recurrent by the corollary above. Thus the following proof contains a new proof of Theorem 6.\n\nLet us note that for any two events \( A \) and \( B \), we have \( A \subset {AB} \cup {B}^{c} \) , and consequently\n\n\[ \nP\left( A\r... | No |
Theorem 9. For any \( i \) and \( j \) we have\n\n\[ \mathop{\lim }\limits_{{n \rightarrow \infty }}\frac{1}{n + 1}\mathop{\sum }\limits_{{v = 0}}^{n}{p}_{ij}^{\left( v\right) } = \frac{1}{{m}_{jj}}. \]\n\n(8.6.4) | The argument indicated above can be made rigorous by invoking a general form of the strong law of large numbers (see §7.5), applied to the successive return times that form a sequence of independent and identically distributed random variables. Unfortunately the technical details are above the level of this book. There... | No |
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