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Example 1.9(a) Consider a particle that moves along a set of \( m + \) 1 nodes, labelled \( 0,1,\ldots, m \), that are arranged around a circle (see Figure 192) At each step the particle is equally likely to move one position in either the clockwise or counterclockwise direction. That is, if \( {X}_{n} \) is the positi... | Solution Surprisingly enough, the probability that node \( i \) is the last node visited can be determined without any computations. To do so, consider the first time that the particle is at one of the two\n\n\n\nFigur... | Yes |
Suppose in Example 19(A) that the particle is not equally likely to move in either direction but rather moves at each step in the clockwise direction with probability \( p \) and in the counterclockwise direction with probability \( q = 1 - p \) If \( 5 < \) \( p < 1 \) then we will show that the probability that state... | To determine the probability that state \( i \) is the last state visited, condition on whether \( i - 1 \) or \( i + 1 \) is visited first Now, if \( i - 1 \) is visited first then the probability that \( i \) will be the last state visited is the same as the probability that a gambler who wins each 1 unit bet with pr... | Yes |
Example 1.9(c) A graph consisting of a central vertex, labeled 0, and rays emanating from that vertex is called a star graph (see Figure 193) Let \( r \) denote the number of rays of a star graph and let ray \( i \) consist of \( {n}_{t} \) vertices, for \( i = 1,\;, r \) Suppose that a particle moves along the vertice... | Solution Let \( L \) denote the first leaf visited. Conditioning on \( R \) , the first ray visited, yields\n\n(1 9 1)\n\n\[ P\{ L = i\} = \mathop{\sum }\limits_{{j = 1}}^{r}\frac{1}{r}P\{ L = i \mid \text{ first ray visited is }j\} \]\n\nNow, if \( j \) is the first ray visited (that is, the first move of the particle... | Yes |
Example 2.3(a) Suppose that travelers arrive at a train depot in accordance with a Poisson process with rate \( \lambda \) If the train departs at time \( t \), let us compute the expected sum of the waiting times of travelers arriving in \( \left( {0, t}\right) \) . That is, we want \( E\left\lbrack {\mathop{\sum }\li... | \[ E\left\lbrack {\mathop{\sum }\limits_{{t = 1}}^{{N\left( t\right) }}\left( {t - {S}_{t}}\right) \mid N\left( t\right) = n}\right\rbrack = E\left\lbrack {\mathop{\sum }\limits_{{t = 1}}^{n}\left( {t - {S}_{t}}\right) \mid N\left( t\right) = n}\right\rbrack \] \[ = {nt} - E\left\lbrack {\mathop{\sum }\limits_{{t = 1}}... | Yes |
Example 2.3(b) The Infinite Server Poisson Queue. Suppose that customers arrive at a service station in accordance with a Poisson process with rate \( \lambda \) . Upon arrival the customer is immediately served by one of an infinite number of possible servers, and the service times are assumed to be independent with a... | To compute the joint distribution of the number of customers that have completed their service and the number that are in service at \( t \), call an entering customer a type-I customer if it completes its service by time \( t \) and a type-II customer if it does not complete its service by time \( t \) . Now, if the c... | Yes |
Suppose that a device is subject to shocks that occur in accordance with a Poisson process having rate \( \lambda \) . The \( i \) th shock gives rise to a damage \( {D}_{t} \) . The \( {D}_{t}, i \geq 1 \), are assumed to be independent and identically distributed and also to be independent of \( \{ N\left( t\right), ... | We can determine \( E\left\lbrack {D\left( t\right) }\right\rbrack \) as follows:\n\n\[ E\left\lbrack {D\left( t\right) \mid N\left( t\right) = n}\right\rbrack = E\left\lbrack {\mathop{\sum }\limits_{{t = 1}}^{{N\left( t\right) }}{D}_{t}{e}^{-\alpha \left( {t - {S}_{t}}\right) } \mid N\left( t\right) = n}\right\rbrack ... | Yes |
Example 2.4(a) Record Values. Let \( {X}_{1},{X}_{2} \), denote a sequence of independent and identically distributed nonnegative continuous random variables whose hazard rate function is given by \( \lambda \left( t\right) \). (That is, \( \lambda \left( t\right) = f\left( t\right) /\bar{F}\left( t\right) \), where \(... | We claim that \( \{ N\left( t\right), t \geq 0\} \) will be a nonhomogeneous Poisson process with intensity function \( \lambda \left( t\right) \) To verify this claim note that there will be a record value between \( t \) and \( t + h \) if, and only if, the first \( {X}_{t} \) whose value is greater than \( t \) lies... | Yes |
The Output Process of an Infinite Server Poisson Queue \( \left( {M/G/\infty }\right) \) . It turns out that the output process of the \( M/ \) \( \bar{G}/\infty \) queue-that is, of the infinite server queue having Poisson arrivals and general service distribution \( G \) -is a nonhomogeneous Poisson process having in... | To prove this we shall first argue that\n\n(1) the number of departures in \( \left( {s, s + t}\right) \) is Poisson distributed with mean \( \lambda {\int }_{s}^{s + t}G\left( y\right) {dy} \), and\n\n(2) the numbers of departures in disjoint time intervals are independent.\n\nTo prove statement (1), call an arrival t... | Yes |
\[ {P}_{0} = {e}^{-\lambda } \] \[ {P}_{n} = \frac{\lambda }{n}\mathop{\sum }\limits_{{j - 1}}^{n}j{\alpha }_{j}{P}_{n - j},\;n \geq 1 \] | Proof That \( {P}_{0} = {e}^{-\lambda } \) is immediate, so take \( n > 0 \) Let \[ h\left( x\right) = \left\{ \begin{array}{ll} 0 & \text{ if }x \neq n \\ 1/n & \text{ if }x = n \end{array}\right. \] Since \( {Wh}\left( W\right) = I\{ W = n\} \), which is defined to equal 1 if \( W = n \) and 0 otherwise, we obtain up... | Yes |
Example 2.5(b) Let \( W \) be a compound Poisson random variable with Poisson parameter \( \lambda = 4 \) and with\n\n\[ P\left\{ {{X}_{i} = i}\right\} = 1/4,\;i = 1,2,3,4. \]\n\nTo determine \( P\{ W = 5\} \), | we use the recursion of Corollary 2.5.4 as follows:\n\n\[ {P}_{0} = {e}^{-\lambda } = {e}^{-4} \]\n\n\[ {P}_{1} = \lambda {\alpha }_{1}{P}_{0} = {e}^{-4} \]\n\n\[ {P}_{2} = \frac{\lambda }{2}\left\{ {{\alpha }_{1}{P}_{1} + 2{\alpha }_{2}{P}_{0}}\right\} = \frac{3}{2}{e}^{-4} \]\n\n\[ {P}_{3} = \frac{\lambda }{3}\left\{... | Yes |
Example 3.3(a) A container contains an infinite collection of coins. Each coin has its own probability of landing heads, and these probabilities are the values of independent random variables that are uniformly distributed over \( \\left( {0,1}\\right) \). Suppose we are to flip coins sequentially, at any time either f... | Solution. We will exhibit a strategy that results in the long-run proportion of heads being equal to 1 . To begin, let \( N\\left( n\\right) \) denote the number of tails in the first \( n \) flips, and so the long-run proportion of heads, call it \( {P}_{h} \), is given by\n\n\[ \n{P}_{h} = \\mathop{\\lim }\\limits_{{... | Yes |
Example 3.3(c) Let \( {X}_{n}, n = 1,2,\ldots \), be independent and such that\n\n\[ P\left\{ {{X}_{n} = - 1}\right\} = P\left\{ {{X}_{n} = 1}\right\} = \frac{1}{2}. \]\n\nThen\n\n\[ N = \min \left\{ {n : \;{X}_{1} + \cdots + {X}_{n} = 1}\right\} \]\nis a stopping time It can be regarded as the stopping time for a gamb... | ## THEOREM 3.3.2 (Wald's Equation).\n\nIf \( {X}_{1},{X}_{2} \), are independent and identically distributed random variables having finite expectations, and if \( N \) is a stopping time for \( {X}_{1},{X}_{2} \), such that \( E\left\lbrack N\right\rbrack < \infty \), then\n\n\[ E\left\lbrack {\mathop{\sum }\limits_{1... | No |
Lemma 3.4.3\n\n\[ \nP\\left\\{ {{S}_{N\\left( t\\right) } \\leq s}\\right\\} = \\bar{F}\\left( t\\right) + {\\int }_{0}^{s}\\bar{F}\\left( {t - y}\\right) {dm}\\left( y\\right) ,\\;t \\geq s \\geq 0.\n\] | Proof\n\n\[ \nP\\left\\{ {{S}_{N\\left( t\\right) } \\leq s}\\right\\} = \\mathop{\\sum }\\limits_{{n = 0}}^{\\infty }P\\left\\{ {{S}_{n} \\leq s,{S}_{n + 1} > t}\\right\\} \n\]\n\n\[ \n= \\bar{F}\\left( t\\right) + \\mathop{\\sum }\\limits_{{n = 1}}^{\\infty }P\\left\\{ {{S}_{n} \\leq s,{S}_{n + 1} > t}\\right\\} \n\]... | Yes |
Suppose we want to determine the rate at which the pattern occurs. By the strong law for delayed renewal processes (part (i) of Theorem 3.5.1) this will equal the reciprocal of the mean time between patterns. But by Blackwell's theorem (part (iv) of Theorem 3.5.1) this is just the limiting probability of a renewal at t... | \( {\left( E\left\lbrack \text{time between patterns}\right\rbrack \right) }^{-1} = \mathop{\lim }\limits_{{n \rightarrow \infty }}P\{ \) pattern at time \( n\} \)\n\n\[ = \mathop{\prod }\limits_{{i = 1}}^{k}P\left\{ {X = {x}_{1}}\right\} \]\n\nHence the rate at which the pattern occurs is \( \mathop{\prod }\limits_{{i... | Yes |
Suppose we want to compute the mean time between system breakdowns. To do so let us first look at the probability of a breakdown in \( \left( {t, t + h}\right) \) for large \( t \) and small \( h \) . Now one way for a breakdown to occur in \( \left( {t, t + h}\right) \) is to have exactly 1 component up at time \( t \... | \[ \mathop{\lim }\limits_{{t \rightarrow \infty }}P\{ \text{ breakdown in }\left( {t, t + h}\right) \} = \mathop{\sum }\limits_{{t = 1}}^{n}\left\{ {\frac{{\lambda }_{t}}{{\lambda }_{t} + {\mu }_{t}}\mathop{\prod }\limits_{{j \neq t}}\frac{{\mu }_{j}}{{\lambda }_{j} + {\mu }_{j}}}\right\} \frac{1}{{\lambda }_{t}}h + o\... | Yes |
For an alternating renewal process (see Section 34.1) suppose that we earn at a rate of one per unit time when the system is on (and thus the reward for a cycle equals the on time of that cycle). Then the total reward earned by \( t \) is just the total on time in \( \left\lbrack {0, t}\right\rbrack \), and thus by The... | \[ \text{ amount of on time in }\frac{\left\lbrack 0, t\right\rbrack }{t} \rightarrow \frac{E\left\lbrack X\right\rbrack }{E\left\lbrack X\right\rbrack + E\left\lbrack Y\right\rbrack }, \] where \( X \) is an on time and \( Y \) an off time in a cycle. Thus by Theorem 3.4.4 when the cycle distribution is nonlattice the... | Yes |
Let \( A\left( t\right) \) denote the age at \( t \) of a renewal process, and suppose we are interested in computing\n\n\[ \mathop{\lim }\limits_{{t \rightarrow \infty }}{\int }_{0}^{t}A\left( s\right) {ds}/t \] | To do so assume that we are being paid money at any time at a rate equal to the age of the renewal process at that time. That is, at time \( s \) we are being paid at a rate \( A\left( s\right) \), and so \( {\int }_{0}^{t}A\left( s\right) {ds} \) represents our total earnings by time \( t \) As everything starts over ... | Yes |
For a regenerative process with \( E\left\lbrack {S}_{1}\right\rbrack < \infty \), with probability 1, | Proof Suppose that a reward is earned at rate 1 whenever the process is in state \( j \) This generates a renewal reward process and the proposition follows directly from Theorem 361 | No |
For the equilibrium renewal process, what is the relationship between \( \lambda \) and \( c \)? | In the case of the equilibrium renewal process, it follows from Example \( {38}\left( \mathrm{\;A}\right) \) and Theorem 3.5.2 that \( \lambda = \) \( c = 1/\mu \) | Yes |
Example 4.1(a) The M/G/1 Queue. Suppose that customers arrive at a service center in accordance with a Poisson process with rate \( \lambda \) . There is a single server and those arrivals finding the server free go immediately into service, all others wait in line until their service turn The service times of successi... | When \( {X}_{n} > 0 \), the \( n \) th departure leaves behind \( {X}_{n} \) customers-of which one enters service and the other \( {X}_{n} - 1 \) wait in line. Hence, at the next departure the system will contain the \( {X}_{n} - 1 \) customers that were in line in addition to any arrivals during the service time of t... | Yes |
Example 4.1(b) The G/M/1 Queue. Suppose that customers arrive at a single-server service center in accordance with an arbitrary renewal process having interarrival distribution \( G \) . Suppose further that the service distribution is exponential with rate \( \mu \) .\n\nIf we let \( {X}_{n} \) denote the number of cu... | \[ {P}_{t{t}^{t + 1 - j}} = {\int }_{0}^{\infty }{e}^{-{\mu t}}\frac{{\left( \mu t\right) }^{j}}{j!}{dG}\left( t\right) ,\;j = 0,1,\ldots, i, \]\n\nwhich follows since if an arrival finds \( i \) in the system, then the next arrival will find \( i + 1 \) minus the number served, and the probability that \( j \) will be... | Yes |
Example 4.1(b) The Absolute Value of the Simple Random Walk. The random walk \( \\left\\{ {{S}_{n}, n \\geq 1}\\right\\} \), where \( {S}_{n} = \\mathop{\\sum }\\limits_{1}^{n}{X}_{i} \), is said to be a simple random walk if for some \( p,0 < p < 1 \) ,\n\n\[ \nP\\left\\{ {{X}_{t} = 1}\\right\\\} = p \n\]\n\n\[ \nP\\l... | ## PROPOSITION 4.1.1\n\nIf \( \\left\\{ {{S}_{n}, n \\geq 1}\\right\\} \) is a simple random walk, then\n\n\[ \nP\\left\\{ {{S}_{n} = i\\left| \\right| {S}_{n}\\left| { = i,}\\right| {S}_{n - 1}\\left| { = {i}_{n - 1},\\ldots ,}\\right| {S}_{1} \\mid = {i}_{1}}\\right\\\} = \\frac{{p}^{\\prime }}{{p}^{\\prime } + {q}^{... | Yes |
The Simple Random Walk. The Markov chain whose state space is the set of all integers and has transition probabilities\n\n\[ \n{P}_{{tt} - 1} = p = 1 - {P}_{{tt} - 1},\;i = 0, \pm 1,\ldots ,\n\]\n\nwhere \( 0 < p < 1 \), is called the simple random walk One interpretation of this process is that it represents the wande... | Since it is impossible to be even (using the gambling model interpretation) after an odd number of plays, we must, of course, have that\n\n\[ \n{P}_{00}^{{2n} + 1} = 0,\;n = 1,2,\ldots\n\]\n\nOn the other hand, the gambler would be even after \( {2n} \) trials if, and only if, he won \( n \) of these and lost \( n \) o... | Yes |
Example 4.3(a) Limiting Probabilities for the Embedded M/G/1 Queue. Consider the embedded Markov chain of the \( M/G/1 \) system as in Example 4.1(A) and let\n\n\[ \n{a}_{1} = {\int }_{0}^{\infty }{e}^{-{\lambda x}}\frac{{\left( \lambda x\right) }^{1}}{j!}{dG}\left( x\right) \n\]\n\nThat is, \( {a}_{i} \) is the probab... | To solve, we introduce the generating functions\n\n\[ \n\pi \left( s\right) = \mathop{\sum }\limits_{{j = 0}}^{\infty }{\pi }_{j}{s}^{j},\;A\left( s\right) = \mathop{\sum }\limits_{{j = 0}}^{\infty }{a}_{j}{s}^{j} \n\]\n\nMultiplying both sides of (4.3.5) by \( {s}^{j} \) and summing over \( j \) yields\n\n\[ \n\pi \le... | Yes |
Consider the embedded Markov chain for the G/M/1 queueing system as presented in Example 4.1(B). The limiting probabilities \( {\pi }_{k}, k = 0,1,\ldots \) can be obtained as the unique solution of\n\n\[ \n{\pi }_{k} = \mathop{\sum }\limits_{i}{\pi }_{i}{P}_{ik},\;k \geq 0, \]\n\n\[ \n\mathop{\sum }\limits_{k}{\pi }_{... | To solve the above let us try a solution of the form \( {\pi }_{k} = c{\mathbf{\beta }}^{k} \) . Substitution into (4.3.6) leads to\n\n\[ \nc{\mathbf{\beta }}^{k} = c\mathop{\sum }\limits_{{t = k - 1}}^{\infty }{\mathbf{\beta }}^{t}{\int }_{0}^{\infty }{e}^{-{\mu t}}\frac{{\left( \mu t\right) }^{t + 1 - k}}{\left( {i +... | Yes |
Example 4.3(c) The Age of a Renewal Process. Initially an item is put into use, and when it fails it is replaced at the beginning of the next time period by a new item Suppose that the lives of the items are independent and each will fail in its \( i \) th period of use with probability \( {P}_{i}, i \geq 1 \), where t... | Hence the limiting probabilities are such that\n\n(4.3.9)\n\n\[ {\pi }_{1} = \mathop{\sum }\limits_{t}{\pi }_{t}\lambda \left( i\right) \]\n\n(4.3 10)\n\n\[ {\pi }_{t + 1} = {\pi }_{t}\left( {1 - \lambda \left( i\right) }\right) ,\;i \geq 1. \]\n\nIterating (4.3.10) yields\n\n\[ {\pi }_{t + 1} = {\pi }_{t}\left( {1 - \... | Yes |
Example 4.3(b) Suppose that during each time period, every member of a population independently dies with probability \( p \), and also that the number of new members that join the population in each time period is a Poisson random variable with mean \( \lambda \) . If we let \( {X}_{n} \) denote the number of members ... | To find the stationary probabilities of this chain, suppose that \( {X}_{0} \) is distributed as a Poisson random variable with parameter \( \alpha \) . Since each of these \( {X}_{0} \) individuals will independently be alive at the beginning of the next period with probability \( 1 - p \), it follows that the number ... | Yes |
Consider a gambler who at each play of the game has probability \( p \) of winning 1 unit and probability \( q = 1 - p \) of losing 1 unit. Assuming successive plays of the game are independent, what is the probability that, starting with \( i \) units, the gambler’s fortune will reach \( N \) before reaching 0 | Let \( {f}_{t} \equiv {f}_{tN} \) denote the probability that, starting with \( i,0 \leq \) \( i \leq N \), the gambler’s fortune will eventually reach \( N \) By conditioning on the outcome of the initial play of the game (or, equivalently, by using Proposition 4.4.2), we obtain\n\n\[ {f}_{t} = p{f}_{t + 1} + q{f}_{t ... | Yes |
Example 4.4(b) Consider the gambler’s ruin problem with \( p = 4 \)\n\nand \( n = 6 \) . Starting in state 3, determine\n\n(a) the expected amount of time spent in state 3 .\n\n(b) the expected number of visits to state 2\n\n(c) the probability of ever visiting state 4 . | Solution The matrix \( \mathbf{Q} \), which specifies \( {P}_{ij}, i,{j\varepsilon }\{ 1,2,3,4,5\} \) is as follows.\n\n\[ \mathbf{Q} = \begin{array}{l} 1 \\ 2 \\ 3 \\ 4 \\ 5 \end{array}\left\lbrack \begin{array}{lllll} 0 & {.4} & 0 & 0 & 0 \\ 0 & {.4} & 0 & 0 & 0 \\ {.6} & 0 & {.4} & 0 & 0 \\ 0 & {.6} & 0 & {.4} & 0 \... | No |
Example 4.7(A) An Ergodic Random Walk. We can argue, without any need for computations, that an ergodic chain with \( {P}_{t, t + 1} + \) \( {P}_{t, t - 1} = 1 \) is time reversible. This follows by noting that the number of transitions from \( i \) to \( i + 1 \) must at all times be within 1 of the number from \( i +... | This follows by noting that the number of transitions from \( i \) to \( i + 1 \) must at all times be within 1 of the number from \( i + 1 \) to \( i \) . This is so since between any two transitions from \( i \) to \( i + 1 \) there must be one from \( i + 1 \) to \( i \) (and conversely) since the only way to re-ent... | Yes |
Example 4.7(b) The Metropolis Algorithm. Let \( {a}_{j}, j = 1,\ldots, m \) be positive numbers, and let \( A = \mathop{\sum }\limits_{{j = 1}}^{m}{a}_{j} \) Suppose that \( m \) is large and that \( A \) is difficult to compute, and suppose we ideally want to simulate the values of a sequence of independent random var... | Let \( \mathbf{Q} \) be any irreducible transition probability matrix on the integers \( 1,\ldots, n \) such that \( {q}_{ij} = {q}_{ij} \) for all \( i \) and \( j \) . Now define a Markov chain \( \left\{ {{X}_{n}, n \geq 0}\right\} \) as follows. If \( {X}_{n} = i \), then generate a random variable that is equal to... | Yes |
Example 4.7(b) A List Problem. Suppose we are given a set of \( n \) elements-numbered 1 through \( n \) -that are to be arranged in some ordered list. At each unit of time a request is made to retrieve one of these elements-element \( i \) being requested (independently of the past) with probability \( {P}_{t} \) . Af... | For any given probability vector \( \underline{P} = \left( {{P}_{1},\ldots ,{P}_{n}}\right) \), the above can be modeled as a Markov chain with \( n \) ! states with the state at any time being the list order at that time. By using Theorem 47.1 it is easy to show that this chain is time reversible. For instance, suppos... | Yes |
Let us reconsider Example 4.3(C), which deals with the age of a discrete time renewal process. That is, let \( {X}_{n} \) denote the age at time \( n \) of a renewal process whose interarrival times are all integers. Since the state of this Markov chain always increases by one until it hits a value chosen by the intera... | Thus letting \( {P}_{t} \) denote the probability that an interarrival is \( i \) , \( i \geq 1 \), it seems likely that\n\n\[ \n{P}_{1t}^{ * } = {P}_{i},\;{P}_{i, t - 1}^{ * } = 1,\;i > 1 \]\n\nSince\n\n\[ \n{P}_{t1} = \frac{{P}_{t}}{\mathop{\sum }\limits_{{j = t}}^{\infty }{P}_{j}} = 1 - {P}_{{t1} + 1},\;i \geq 1, \]... | No |
Example 4.8(a) Consider a machine that can be in one of three states: good condition, fair condition, or broken down. Suppose that a machine in good condition will remain this way for a mean time \( {\mu }_{1} \) and will then go to either the fair condition or the broken condition with respective probabilities \( \fra... | Solution. Letting the states be \( 1,2,3 \), we have that the \( \pi \), satisfy\n\n\[ \n{\pi }_{1} + {\pi }_{2} + {\pi }_{3} = 1 \n\]\n\n\[ \n{\pi }_{1} = \frac{2}{3}{\pi }_{3} \n\]\n\n\[ \n{\pi }_{2} = \frac{3}{4}{\pi }_{1} + \frac{1}{3}{\pi }_{3} \n\]\n\n\[ \n{\pi }_{3} = \frac{1}{4}{\pi }_{1} + {\pi }_{2} \n\]\n\nT... | Yes |
Consider a Yule process with \( X\left( 0\right) = 1 \) . Let us compute the expected sum of the ages of the members of the population at time \( t \) The sum of the ages at time \( t \), call it \( A\left( t\right) \), can be expressed as\n\n\[ A\left( t\right) = {a}_{0} + t + \mathop{\sum }\limits_{{t = 1}}^{{X\left(... | \[ E\left\lbrack {A\left( t\right) \mid X\left( t\right) = n + 1}\right\rbrack = {a}_{0} + t + E\left\lbrack {\mathop{\sum }\limits_{{t = 1}}^{n}\left( {t - {S}_{t}}\right) \mid X\left( t\right) = n + 1}\right\rbrack \]\n\n\[ = {a}_{0} + t + n{\int }_{0}^{t}\left( {t - x}\right) \frac{\lambda {e}^{-\lambda \left( {t - ... | Yes |
Consider a two-state continuous-time Markov chain that spends an exponential time with rate \( \lambda \) in state 0 before going to state 1, where it spends an exponential time with rate \( \mu \) before returning to state 0. The forward equations yield\n\n\[ \n{P}_{00}^{\prime }\left( t\right) = \mu {P}_{01}\left( t\... | \[ \n= - \left( {\lambda + \mu }\right) {P}_{00}\left( t\right) + \mu \n\]\n\nwhere the last equation follows from \( {P}_{01}\left( t\right) = 1 - {P}_{00}\left( t\right) \) . Hence,\n\n\[ \n{e}^{\left( {\lambda + \mu }\right) t}\left\lbrack {{P}_{00}^{\prime }\left( t\right) + \left( {\lambda + \mu }\right) {P}_{00}\... | Yes |
For a pure birth process, the forward equations reduce to\n\n\[ {P}_{tt}^{\prime }\left( t\right) = - {\lambda }_{t}{P}_{tt}\left( t\right) \]\n\n\[ {P}_{{t}_{j}}^{\prime }\left( t\right) = {\lambda }_{j - 1}{P}_{{t}_{j} - 1}\left( t\right) - {\lambda }_{j}{P}_{{t}_{j}}\left( t\right) ,\;j > i. \] | Integrating the top equation of (543) and then using \( {P}_{u}\left( 0\right) = 1 \) yields\n\n\[ {P}_{t\prime }\left( t\right) = {e}^{-{\lambda }_{t}t} \]\n\nThe above, of course, is true as \( {P}_{u}\left( t\right) \) is the probability that the time until a transition from state \( i \) is greater than \( t \) The... | Yes |
Example 5.4(b) A Two Sex Population Growth Model. Consider a population of males and females and suppose that each female in the population independently gives birth at an exponential rate \( \\lambda \), and that each birth is (independently of all else) a female with probability \( p \) and a male with probability \(... | Solution (a) To find \( E\\left\\lbrack {X\\left( t\\right) }\\right\\rbrack \), note that \( \\{ X\\left( t\\right), t \\geq 0\\} \) is itself a continuous-time Markov chain. Letting \( {M}_{X}\\left( t\\right) = E\\left\\lbrack {X\\left( t\\right) \\mid X\\left( 0\\right) = i}\\right\\rbrack \) we will derive a diffe... | Yes |
Example 5.8(a) Let us reconsider the two-state chain of Example \( {54}\left( \mathrm{\;A}\right) \), which has\n\n\[ \n{P}_{01} = {P}_{10} = 1 \]\n\n\[ \n{\nu }_{0} = \lambda ,\;{\nu }_{1} = \mu .\n\]\n\nLetting \( \nu = \lambda + \mu \), the uniformized version of the above is to consider it a continuous-time Markov ... | Hence,\n\n\[ \n{P}_{00}\left( t\right) = \mathop{\sum }\limits_{{n = 0}}^{\infty }{P}_{00}^{n}{e}^{-\left( {\lambda + \mu }\right) t}\frac{{\left\lbrack \left( \lambda + \mu \right) t\right\rbrack }^{n}}{n!} \]\n\n\[ \n= {e}^{-\left( {\lambda + \mu }\right) t} + \left\lbrack {1 - {e}^{-\left( {\lambda + \mu }\right) t}... | Yes |
Suppose that a sequence of independent and identically distributed discrete random variables is observed sequentially, one at each day. What is the expected number that must be observed until some given sequence appears? More specifically, suppose that each outcome is either 0,1, or 2 with respective probabilities \( \... | To compute \( E\left\lbrack N\right\rbrack \) imagine a sequence of gamblers, each initially having 1 unit, playing at a fair gambling casino Gambler \( i \) begins betting at the beginning of day \( i \) and bets her 1 unit that the value on that day will equal 0 . If she wins (and thus has 2 units), she then bets the... | Yes |
Consider an individual who starts at position 0 and at each step either moves 1 position to the right with probability \( p \) or one to the left with probability \( 1 - p \) Assume that the successive movements are independent. If \( p > 1/2 \) find the expected number of steps it takes until the individual reaches po... | Let \( {X}_{1} \) equal 1 or -1 depending on whether step \( j \) is to the right or the left. If \( N \) is the number of steps it takes to reach \( i \), then\n\n\[ \mathop{\sum }\limits_{{j = 1}}^{N}{X}_{j} = i \]\n\nHence, since \( E\left\lbrack {X}_{l}\right\rbrack = {2p} - 1 \), we obtain from Wald’s equation tha... | Yes |
Example 6.2(c) Players \( X, Y \), and \( Z \) contest the following game At each stage two of them are randomly chosen in sequence, with the first one chosen being required to give 1 coin to the other All of the possible choices are equally likely and successive choices are independent of the past This continues until... | Solution. Suppose that the game does not end when one of the players has all \( s \) coins but rather that the final two contestants continue to play, keeping track of their winnings by allowing for negative fortunes Let \( {X}_{n},{Y}_{n} \), and \( {Z}_{n} \) denote, respectively, the amount of money that \( X, Y \),... | Yes |
In Example 15(C) we showed, by mathematical induction, that the expected number of rounds that it takes for all \( n \) people in the matching problem to obtain their own hats is equal to \( n \) We will now present a martingale argument for this result. | Let \( R \) denote the number of rounds until all people have a match. Let \( {X}_{i} \) denote the number of matches on the \( i \) th round, for \( i = 1 \) , , \( R \), and define \( {X}_{i} \) to equal 1 for \( i > R \). We will use the zero-mean martingale \( \left\{ {{Z}_{k}, k \geq 1}\right\} \), where \[ {Z}_{k... | Yes |
Let \( {X}_{1},\ldots ,{X}_{n} \) be random variables such that \( E\left\lbrack {X}_{1}\right\rbrack = 0 \) and \( E\left\lbrack {{X}_{i} \mid {X}_{1},\ldots ,{X}_{i - 1}}\right\rbrack = 0, i \geq 1 \) Then, from Example 5 of Section 61, we see that \( \mathop{\sum }\limits_{{i = 1}}^{j}{X}_{i}, j = 1,., n \) is a zer... | \[ P\left\{ {\mathop{\sum }\limits_{{i = 1}}^{n}{X}_{t} \geq a}\right\} \leq \exp \left\{ {-2{a}^{2}/\mathop{\sum }\limits_{{t = 1}}^{n}{\left( {\alpha }_{t} + {\beta }_{t}\right) }^{2}}\right\} \] and \[ P\left\{ {\mathop{\sum }\limits_{{i = 1}}^{n}{X}_{i} \leq - a}\right\} \leq \exp \left\{ {-2{a}^{2}/\mathop{\sum }\... | Yes |
Example 6.3(b) Suppose that \( n \) balls are put in \( m \) urns in such a manner that each ball, independently, is equally likely to go into any of the urns. We will use Azuma's inequality to obtain bounds on the tail probabilities of \( X \), the number of empty urns To begin, letting \( I\{ A\} \) be the indicator ... | To determine a bound on \( \\left| {{Z}_{t} - {Z}_{t - 1}}\\right| \) first note that \( \\left| {{Z}_{1} - {Z}_{0}}\\right| = 0 \) . Now, for \( i \\geq 2 \), let \( D \) denote the number of distinct values in the set \( {X}_{1},;,{X}_{t - 1} \) That is, \( D \) is the number of urns having at least one ball after th... | Yes |
Example 6.3(c) Suppose that \( n \) balls are to be placed in \( m \) urns, with each ball independently going into urn \( j \) with probability \( {p}_{j} \) , \( j = 1,\ldots, m \) . Let \( {Y}_{k} \) denote the number of urns with exactly \( k \) balls, \( 0 \leq k < n \), and use the preceding corollary to obtain a... | Solution To begin, note that\n\n\[ E\left\lbrack {Y}_{k}\right\rbrack = E\left\lbrack {\mathop{\sum }\limits_{{i = 1}}^{m}I\{ \text{ urn }i\text{ has exactly }k\text{ balls }\} }\right\rbrack \]\n\n\[ = \mathop{\sum }\limits_{{i = 1}}^{m}\left( \begin{array}{l} n \\ k \end{array}\right) {p}_{i}^{k}{\left( 1 - {p}_{i}\r... | Yes |
Example 6.3(b) Consider a set of \( n \) components that are to be used in performing certain experiments Let \( {X}_{i} \) equal 1 if component \( i \) is in functioning condition and let it equal 0 otherwise, and suppose that the \( {X}_{t} \) are independent with \( E\left\lbrack {X}_{t}\right\rbrack = {p}_{t} \) Su... | Solution Since\n\n\[ X = \mathop{\sum }\limits_{{j = 1}}^{m}I\{ \text{ experiment }j\text{ can be performed }\} \]\n\nwe see that\n\n\[ E\left\lbrack X\right\rbrack = \mathop{\sum }\limits_{{j = 1}}^{m}\mathop{\prod }\limits_{{\iota \in {A}_{\iota }}}{p}_{\iota } \]\n\nIf we let \( h\left( \mathbf{X}\right) \) equal th... | Yes |
Example 6.4(b) A Gambling Result. Consider a gambler playing a fair game, that is, if \( {Z}_{n} \) is the gambler’s fortune after the \( n \) th play, then \( \left\{ {{Z}_{n}, n \geq 1}\right\} \) is a martingale Now suppose that no credit is given, and so the gambler's fortune is not allowed to become negative, and ... | Since \( \left\{ {Z}_{n}\right\} \) is a nonnegative martingale we see by the convergence theorem that\n\n(6.4.4) \( \mathop{\lim }\limits_{{n \rightarrow \infty }}{Z}_{n}\; \) exists and is finite, with probability 1\n\nBut \( \left| {{Z}_{n + 1} - {Z}_{n}}\right| \geq 1 \) for \( n < N \), and so (64.4) implies that\... | Yes |
Example 6.5(a) Let \( {S}_{n} \) equal the number of heads in the first \( n \) independent flips of a coin that lands heads with probability \( p \), and let us consider the probability that after some specified number of flips the proportion of heads will ever differ from \( p \) by more than \( \varepsilon \) . That... | Now, if we let \( {X}_{t} \) equal 1 if the \( i \) th flip lands heads and 0 otherwise, then\n\n\[ {Z}_{n} \\equiv {S}_{n} - {np} = \\mathop{\\sum }\\limits_{{i = 1}}^{n}\\left( {{X}_{i} - p}\\right) \]\n\nis a martingale with 0 mean As,\n\n\[ - p \\leq {Z}_{n} - {Z}_{n - 1} \\leq 1 - p \]\n\nit follows that \( \\left... | Yes |
In the symmetric simple random walk the expected number of visits to state \( k \) before returning to the origin is equal to 1 for all \( k \neq 0 \) | For \( k > 0 \) let \( Y \) denote the number of visits to state \( k \) prior to the first return to the origin Then \( Y \) can be expressed as\n\n\[ Y = \mathop{\sum }\limits_{{n = 1}}^{\infty }{I}_{n} \]\n\nwhere\n\n\[ {I}_{n} = \left\{ \begin{array}{ll} 1 & \text{ if a visit to state }k\text{ occurs at time }n\tex... | Yes |
(De Finetti’s Theorem) To every infinite sequence of exchangeable random variables \( {X}_{1},{X}_{2} \), taking values either 0 or 1, there corresponds a probability distribution \( G \) on \( \left\lbrack {0,1}\right\rbrack \) such that, for all \( 0 \leq k \leq n \) ,\n\n\[ P\left\{ {{X}_{1} = {X}_{2} = \cdot = {X}_... | Proof Let \( m \geq n \) We start by computing the above probability by first conditioning on\n\n\[ {S}_{m} = \mathop{\sum }\limits_{{t = 1}}^{m}{X}_{t} \]\n\nThis yields\n\n\[ \text{(722)}\;P\left\{ {{X}_{1} = \; = {X}_{k} = 1,{X}_{k + 1} = \; = {X}_{n} = 0}\right\} \]\n\n\[ = \mathop{\sum }\limits_{{j = 0}}^{m}P\left... | Yes |
Example 8.3(a) The Value of a Stock Option. Suppose one has the option of purchasing, at a time \( T \) in the future, one unit of a stock at a fixed price \( K \) . Supposing that the present value of the stock is \( y \) and that its price varies according to geometric Brownian motion, let us compute the expected wor... | \[ E\left\lbrack {\max \left( {Y\left( T\right) - K,0}\right) }\right\rbrack = {\int }_{0}^{\infty }P\{ Y\left( T\right) - K > a\} {da} \] \[ = {\int }_{0}^{\infty }P\left\{ {y{e}^{x\left( T\right) } - K > a}\right\} {da} \] \[ = {\int }_{0}^{\infty }P\left\{ {X\left( T\right) > \log \frac{K + a}{y}}\right\} {da} \] \[... | Yes |
Suppose we have the option of buying, at some time in the future, one unit of a stock at a fixed price \( A \), independent of its current market price. The current market price of the stock is taken to be 0 , and we suppose that it changes in accordance with a Brownian motion process having a negative drift coefficien... | Let us consider the policy that exercises the option when the market price is \( x \). Our expected gain under such a policy is\n\n\[ P\left( x\right) \left( {x - A}\right) \]\n\nwhere \( P\left( x\right) \) is the probability that the process will ever reach \( x \). From (8.45) we see that\n\n\[ P\left( x\right) = {e... | Yes |
\[ {p}^{ * } \leq {p}^{* * } \] | Proof For any \( p > {p}^{* * } \), it follows that player II would quit if \( X\left( t\right) = p \) and player I doubles Hence at \( X\left( t\right) = p \) player I can guarantee himself an expected gain of the present stake by doubling, and since player II can always guarantee that I never receives more (by quitti... | Yes |
Lemma 2\n\n\[ {p}^{ * } = {p}^{* * } \] | Proof Suppose \( {p}^{ * } < {p}^{* * } \) We will obtain a contradiction by showing that player I has a better strategy than \( {p}^{ * } \) Specifically, rather than doubling at \( {p}^{ * } \), player I can do better by waiting for \( X\left( t\right) \) to either hit 0 or \( {p}^{* * } \) If it hits \( {p}^{* * } \... | Yes |
Example 8.4(c) Controlling a Production Process. In this example, we consider a production process that tends to deteriorate with time Specifically, we suppose that the production process changes its state in accordance with a Wiener process with drift coefficient \( \mu ,\mu > 0 \) When the state of the process is \( ... | Let \( f\left( x\right) \) denote the expected time that it takes the process to reach \( x \) We derive a differential equation for \( f\left( x\right) \) by conditioning on \( Y = X\left( h\right) - X\left( 0\right) \) the change in time \( h \) This yields\n\n\[ f\left( x\right) = h + E\left\lbrack {f\left( {x - Y}\... | Yes |
Example 8.8(a) An Auto Regressive Process. Let \( {Z}_{0},{Z}_{1},{Z}_{2},\ldots \) be uncorrelated random variables with \( E\left\lbrack {Z}_{n}\right\rbrack = 0, n \geq 0 \), and\n\n\[ \operatorname{Var}\left( {Z}_{n}\right) = \left\{ \begin{array}{ll} {\sigma }^{2}/\left( {1 - {\lambda }^{2}}\right) & n = 0 \\ {\si... | Iterating (8.8.2) yields\n\n\[ {X}_{n} = \lambda \left( {\lambda {X}_{n - 2} + {Z}_{n - 1}}\right) + {Z}_{n} \]\n\n\[ = {\lambda }^{2}{X}_{n - 2} + \lambda {Z}_{n - 1} + {Z}_{n} \]\n\n\[ \text{:} \]\n\n\[ = \mathop{\sum }\limits_{{i = 0}}^{n}{\lambda }^{n - i}{Z}_{t} \]\n\nand so\n\n\[ \operatorname{Cov}\left( {{X}_{n}... | Yes |
Example 9.2(a) Stochastic Ordering of Vectors. Let \( {X}_{1},\ldots ,{X}_{n} \) be independent and \( {Y}_{1},\ldots ,{Y}_{n} \) be independent If \( {X}_{t}{ \geq }_{\mathrm{{st}}}{Y}_{t} \), then for any increasing \( f \) | Proof Let \( {X}_{1},.,{X}_{n} \) be independent and use Proposition 9.22 to generate independent \( {Y}_{1}^{ * },\ldots ,{Y}_{n}^{ * } \), where \( {Y}_{t}^{ * } \) has the distribu- tion of \( {Y}_{t} \) and \( {Y}_{t}^{ * } \leq {X}_{t} \) Then \( f\left( {{X}_{1},\ldots ,{X}_{n}}\right) \geq f\left( {{Y}_{1}^{ * }... | Yes |
Example 9.2(c) Comparing Renewal Processes. Let \( {N}_{t} = \left\{ {{N}_{t}\left( t\right) }\right. \) , \( t \geq 0\}, i = 1,2 \), denote renewal processes having interarrival distributions \( F \) and \( G \), respectively. If \( \bar{F} \geq \bar{G} \), then \[ \left\{ {{N}_{1}\left( t\right), t \geq 0}\right\} \u... | To prove the above we use coupling as follows. Let \( {X}_{1},{X}_{2},\ldots \) be independent and distributed according to \( F \) . Then the renewal process generated by the \( {X}_{t} \) -call it \( {N}_{1}^{ * } \) -has the same probability distributions as \( {N}_{1} \) . Now generate independent random variables ... | Yes |
Example 9.2(b) Let \( {X}_{1},{X}_{2}, \) . be a sequence of independent Bernoulli random variables, and let \( {p}_{t} = P\left\{ {{X}_{t} = 1}\right\}, i \geq 1 \) If \( {p}_{t} \geq \) \( p \) for all \( i \), show that with probability 1\n\n\[ \mathop{\liminf }\limits_{n}\mathop{\sum }\limits_{{i = 1}}^{n}{X}_{i}/n... | Solution We start by coupling the sequence \( {X}_{i}, i \geq 1 \) with a sequence of independent and identically distributed Bernoulli random variables \( {Y}_{i}, i \geq 1 \), such that \( P\left\{ {{Y}_{i} = 1}\right\} = p \) and \( {X}_{i} \geq {Y}_{i} \) for all \( i \) . To accomplish this, let \( {U}_{i}, i \geq... | Yes |
A central problem in statistics is that of making inferences about the unknown distribution of a given random variable In the simplest case, we suppose that \( X \) is a continuous random variable having a density function known to be either \( f \) or \( g \) Based on the observed value of \( X \), we must decide on e... | ## Neyman-Pearson Lemma\n\nAmong all decision rules \( \phi \) satisfying (941), the one that maximizes \( \int g\left( x\right) \phi \left( x\right) {dx} \) is \( {\phi }^{ * } \) given by\n\n\[{\phi }^{ * }\left( x\right) = \left\{ \begin{array}{ll} 0 & \text{ if }f\left( x\right) /g\left( x\right) \geq c \\ 1 & \tex... | Yes |
Example 9.7(a) Consider a system composed of \( n \) components, each of which is either working or failed. Let \( {X}_{t} \) equal 1 if component \( i \) is working and 0 if it is failed, and suppose that the \( {X}_{t} \) are independent and that \( P\left\{ {{X}_{i} = 1}\right\} = {p}_{i}, i = 1,\ldots, n \) . In ad... | As \( {Y}_{1},\ldots ,{Y}_{r} \) are all increasing functions of the independent random variables \( {X}_{1},\ldots ,{X}_{n} \), they are associated Hence,\n\n\[ P\{ S = 1\} = E\left\lbrack S\right\rbrack = E\left\lbrack {\mathop{\prod }\limits_{{t = 1}}^{r}{Y}_{t}}\right\rbrack \geq E\left\lbrack {Y}_{1}\right\rbrack ... | Yes |
Example 9.7(b) Any stochastic process \( \\{ X\\left( t\\right), t \\geq 0\\} \) having independent increments and \( X\\left( 0\\right) = 0 \), such as a Poisson process or a Brownian motion process, is associated. | To verify this assertion, let \( {t}_{1} < {t}_{2} < \\cdots < {t}_{n} \) Then as \( X\\left( {t}_{1}\\right) ,\\ldots, X\\left( {t}_{n}\\right) \) are all increasing functions of the independent random variables \( X\\left( {t}_{i}\\right) - X\\left( {t}_{i - 1}\\right), i = \) \( 1,\\ldots, n \) (where \( {t}_{0} = 0... | Yes |
Example 9.7(c) Let \( \\Lambda \) be a random variable and suppose that, conditional on \( \\Lambda = \\lambda \), the random variables \( {X}_{1},\\ldots ,{X}_{n} \) are independent with a common distribution \( {F}_{\\lambda } \) If \( {F}_{\\lambda }\\left( x\\right) \) is decreasing in \( \\lambda \) for all \( x \... | This result can be verified by an argument similar to the one used in Proposition 9.7.2. Namely, let \( {U}_{1},\\ldots ,{U}_{n} \) be independent uniform \( \\left( {0,1}\\right) \) random variables that are independent of \( \\Lambda \) . Then define \( {X}_{1},.,{X}_{n} \) recursively by \( {X}_{i} = {F}_{\\Lambda }... | No |
Suppose that \( W \) is a binomial random variable with parameters \( n \) and \( p \), where \( n \) is large and \( p \) is small. Let \( \lambda = \) \( {np} \) and interpret \( W \) as the number of successes in \( n \) independent trials where each is a success with probability \( p \) ; and so \( \left( \begin{ma... | \[ \left( \begin{matrix} W \\ i \end{matrix}\right) = \mathop{\sum }\limits_{J}{X}_{J} \] and thus \[ E\left\lbrack \left( \begin{matrix} W \\ i \end{matrix}\right) \right\rbrack = \left( \begin{matrix} n \\ i \end{matrix}\right) {p}^{t} \] \[ = \frac{n\left( {n - 1}\right) \cdots \left( {n - i + 1}\right) }{{i}^{\prim... | Yes |
Let us reconsider the matching problem, which was considered in Example 13(A) In this problem, \( n \) individuals mix their hats and then each randomly chooses one Let \( W \) denote the number of individuals that choose his or her own hat and note that \( \left( \begin{matrix} W \\ i \end{matrix}\right) \) represents... | Thus, from Brun's sieve, the number of matches approximately has a Poisson distribution with mean 1. | Yes |
If the \( {X}_{t} \) are independent Bernoulli random variables then we can let \( {V}_{t} = \mathop{\sum }\limits_{{j \neq t}}{X}_{j} \) . Therefore, | \[ E\left\lbrack \left| {W - {V}_{t}}\right| \right\rbrack = E\left\lbrack {X}_{t}\right\rbrack = {\lambda }_{t} \] and so, \[ \left| {P\{ W \in A\} - \mathop{\sum }\limits_{{t \in A}}{e}^{-\lambda }{\lambda }^{t}/i!}\right| \leq \min \left( {1,1/\lambda }\right) \mathop{\sum }\limits_{{t = 1}}^{n}{\lambda }_{t}^{2}. \... | No |
Example 10.2(b) Suppose that \( m \) balls are placed among \( n \) urns, with each ball independently going into urn \( i \) with probability \( {p}_{t} \) , \( i = 1,.\; \) , \( n \) . Let \( {X}_{t} \) be the indicator for the event that urn \( i \) is empty and let \( W = \mathop{\sum }\limits_{{i = 1}}^{n}{X}_{i} ... | As the \( {X}_{t} \) have a negative dependence, since urn \( i \) being empty makes it less likely that other urns will also be empty, we might suppose that we can construct a coupling that results in \( W \geq {V}_{t} \) . To show that this is indeed possible, imagine that the \( m \) balls are distributed according ... | Yes |
Example 10.3(a) (Bernoulli Convolutions). Suppose that \( {X}_{t} \) , \( i = 1,\ldots ,{10} \), are independent Bernoulli random variables with \( E\left\lbrack {X}_{t}\right\rbrack = i/{1000} \) . The following table compares the new approximation given in this section with the usual Poisson approximation. | <table><thead><tr><th>\( k \)</th><th>\( P\{ W = k\} \)</th><th>New Approx.</th><th>Usual Approx</th></tr></thead><tr><td>0</td><td>946302</td><td>946299</td><td>946485</td></tr><tr><td>1</td><td>.052413</td><td>052422</td><td>052056</td></tr><tr><td>2</td><td>001266</td><td>001257</td><td>001431</td></tr><tr><td>3</td... | Yes |
Example 10.3(b) Example 102(B) was concerned with the number of multinomial outcomes that never occur when each of \( m \) independent trials results in any of \( n \) possible outcomes with respective probabilities \( {p}_{1},\ldots ,{p}_{n} \) Let \( {X}_{t} \) equal 1 if none of the trials result in outcome \( i, i ... | \[ {a}_{t} = \mathop{\sum }\limits_{{j \neq t}}E\left\lbrack {{X}_{j} \mid {X}_{t} = 1}\right\rbrack = \mathop{\sum }\limits_{{j \neq t}}{\left( 1 - \frac{{p}_{j}}{1 - {p}_{t}}\right) }^{m} \] are more precise than the straight Poisson approximations when \( \lambda < 1 \) | Yes |
\[ P\{ W > 0\} \geq \mathop{\sum }\limits_{{i = 1}}^{n}{\lambda }_{t}/\left( {1 + {a}_{t}}\right) \] | Proof Since \( f\left( x\right) = 1/x \) is a convex function for \( x \geq 0 \) the result follows from Proposition 1031(a) upon applying Jensen's inequality | Yes |
Proposition 2.2 If \( W \) is a Brownian motion started at 0, \( a > 0 \), and \( {Y}_{t} = a{W}_{t/{a}^{2}} \), then \( {Y}_{t} \) is a Brownian motion started at 0. | Proof We use \( {\mathcal{G}}_{t} = {\mathcal{F}}_{t/{a}^{2}} \) for the filtration for \( Y \) . Clearly \( {Y}_{t} \) has continuous paths, \( {Y}_{0} = 0 \) , a.s., and \( {Y}_{t} \) is \( {\mathcal{G}}_{t} \) measurable. If \( s < t \) ,\n\n\[ \n{Y}_{t} - {Y}_{s} = a\left( {{W}_{t/{a}^{2}} - {W}_{s/{a}^{2}}}\right)... | Yes |
Proposition 2.3 If \( W \) is a Brownian motion, then \( W \) is a Gaussian process. | Proof Suppose \( W \) is a Brownian motion and let \( 0 = {t}_{0} < {t}_{1} < \cdots < {t}_{n} \) . Define\n\n\[ \n{Z}_{i} = \frac{{W}_{{t}_{i}} - {W}_{{t}_{i - 1}}}{\sqrt{{t}_{i} - {t}_{i - 1}}},\;i = 1,2,\ldots, n.\n\]\n\nBy Definition 2.1(4), \( {Z}_{i} \) is independent of \( {\mathcal{F}}_{{t}_{i - 1}} \), and hen... | Yes |
Theorem 2.4 If \( W \) is a process such that all the finite-dimensional distributions are jointly normal, \( \mathbb{E}{W}_{s} = 0 \) for all \( s,\operatorname{Cov}\left( {{W}_{s},{W}_{t}}\right) = s \) when \( s \leq t \), and the paths of \( {W}_{t} \) are continuous, then \( W \) is a Brownian motion. | Proof For \( {\mathcal{F}}_{t} \) we take the filtration generated by \( W \) . If we take \( s = t \), then \( \operatorname{Var}{W}_{t} = \) \( \operatorname{Cov}\left( {{W}_{t},{W}_{t}}\right) = t \) . In particular, \( \operatorname{Var}{W}_{0} = 0 \), and since \( \mathbb{E}{W}_{0} = 0 \), then \( {W}_{0} = 0 \), ... | No |
Example 3.2 Let \( {M}_{t} = {W}_{t} \), where \( {W}_{t} \) is a Brownian motion. Then \( {M}_{t} \) is a martingale. | To verify Definition 3.1(3), we write\n\n\[\n\mathbb{E}\left\lbrack {{M}_{t} \mid {\mathcal{F}}_{s}}\right\rbrack = {M}_{s} + \mathbb{E}\left\lbrack {{W}_{t} - {W}_{s} \mid {\mathcal{F}}_{s}}\right\rbrack = {M}_{s} + \mathbb{E}\left\lbrack {{W}_{t} - {W}_{s}}\right\rbrack = {M}_{s},\n\]\n\nusing the independent increme... | Yes |
Example 3.3 Let \( {M}_{t} = {W}_{t}^{2} - t \), where \( {W}_{t} \) is a Brownian motion. To show \( {M}_{t} \) is a martingale, | \[ \mathbb{E}\left\lbrack {{M}_{t} \mid {\mathcal{F}}_{s}}\right\rbrack = \mathbb{E}\left\lbrack {{\left( {W}_{t} - {W}_{s} + {W}_{s}\right) }^{2} \mid {\mathcal{F}}_{s}}\right\rbrack - t \] \[ = {W}_{s}^{2} + \mathbb{E}\left\lbrack {{\left( {W}_{t} - {W}_{s}\right) }^{2} \mid {\mathcal{F}}_{s}}\right\rbrack + 2\mathbb... | Yes |
Example 3.4 Again let \( {W}_{t} \) be a Brownian motion, let \( a \in \mathbb{R} \), and let \( {M}_{t} = {e}^{a{W}_{t} - {a}^{2}t/2} \) . Since \( {W}_{t} - {W}_{s} \) is normal with mean zero and variance \( t - s \), we know \( \mathbb{E}{e}^{a\left( {{W}_{t} - {W}_{s}}\right) } = {e}^{{a}^{2}\left( {t - s}\right) ... | Then\n\n\[ \mathbb{E}\left\lbrack {{M}_{t} \mid {\mathcal{F}}_{s}}\right\rbrack = {e}^{-{a}^{2}t/2}{e}^{a{W}_{s}}\mathbb{E}\left\lbrack {{e}^{a\left( {{W}_{t} - {W}_{s}}\right) } \mid {\mathcal{F}}_{s}}\right\rbrack \]\n\n\[ = {e}^{-{a}^{2}t/2}{e}^{a{W}_{s}}\mathbb{E}\left\lbrack {e}^{a\left( {{W}_{t} - {W}_{s}}\right)... | Yes |
Recall that given a filtration \( \left\{ {\mathcal{F}}_{t}\right\} \), each \( {\mathcal{F}}_{t} \) is contained in \( \mathcal{F} \), where \( \left( {\Omega ,\mathcal{F},\mathbb{P}}\right) \) is our probability space. Let \( X \) be an integrable \( \mathcal{F} \) measurable random variable, and let \( {M}_{t} = \ma... | \[ \mathbb{E}\left\lbrack {{M}_{t} \mid {\mathcal{F}}_{s}}\right\rbrack = \mathbb{E}\left\lbrack {\mathbb{E}\left\lbrack {X \mid {\mathcal{F}}_{t}}\right\rbrack \mid {\mathcal{F}}_{s}}\right\rbrack = \mathbb{E}\left\lbrack {X \mid {\mathcal{F}}_{s}}\right\rbrack = {M}_{s}, \] and \( M \) is a martingale. | Yes |
Theorem 3.6 Suppose \( {M}_{t} \) is a martingale or non-negative submartingale with paths that are right continuous with left limits. Then\n\n(1)\n\n\[ \mathbb{P}\left( {\mathop{\sup }\limits_{{s \leq t}}\left| {M}_{s}\right| \geq \lambda }\right) \leq \mathbb{E}\left| {M}_{t}\right| /\lambda \] | Proof We will do the case where \( {M}_{t} \) is a martingale, the submartingale case being nearly identical. Let \( {\mathcal{D}}_{n} = \left\{ {{kt}/{2}^{n} : 0 \leq k \leq {2}^{n}}\right\} \) . If we set \( {N}_{k}^{\left( n\right) } = {M}_{{kt}/{2}^{n}} \) and \( {\mathcal{G}}_{k}^{\left( n\right) } = {\mathcal{F}}... | Yes |
Proposition 3.8 Suppose \( {\mathcal{F}}_{t} \) satisfies the usual conditions. Then\n\n(1) \( T \) is a stopping time if and only if \( \left( {T \leq t}\right) \in {\mathcal{F}}_{t} \) for all \( t \) . | Proof We will just prove part of (1), leaving the rest as Exercise 3.4. Note \( \left( {T \leq t}\right) = \) \( { \cap }_{n \geq N}\left( {T < t + 1/n}\right) \in {\mathcal{F}}_{t + 1/N} \) for each \( N \) . Thus \( \left( {T \leq t}\right) \in { \cap }_{N}{\mathcal{F}}_{t + 1/N} \subset {\mathcal{F}}_{t + } = {\math... | No |
Proposition 3.9 Suppose \( {\mathcal{F}}_{t} \) satisfies the usual conditions and \( {X}_{t} \) has continuous paths.\n\n(1) If \( A \) is open, then \( {T}_{A} \) is a stopping time.\n\n(2) If \( A \) is closed, then \( {T}_{A} \) is a stopping time. | Proof (1) \( \left( {{T}_{A} < t}\right) = { \cap }_{q \in {\mathbb{Q}}_{ + }, q < t}\left( {{X}_{q} \in A}\right) \), where \( {\mathbb{Q}}_{ + } \) denotes the set of non-negative rationals. Since \( \left( {{X}_{q} \in A}\right) \in {\mathcal{F}}_{q} \subset {\mathcal{F}}_{t} \), then \( \left( {{T}_{A} < t}\right) ... | Yes |
Proposition 3.10 Suppose \( \\left\\{ {\\mathcal{F}}_{t}\\right\\} \) is a filtration satisfying the usual conditions.\n\n(1) \( {\\mathcal{F}}_{T} \) is a \( \\sigma \) -field.\n\n(2) If \( S \\leq T \), then \( {\\mathcal{F}}_{S} \\subset {\\mathcal{F}}_{T} \) .\n\n(3) If \( {\\mathcal{F}}_{T + } = { \\cap }_{\\varep... | Proof If \( A \\in {\\mathcal{F}}_{T} \), then \( {A}^{\\mathrm{c}} \\cap \\left( {T \\leq t}\\right) = \\left( {T \\leq t}\\right) \\smallsetminus \\left\\[ {A \\cap \\left( {T \\leq t}\\right) }\\right\\rbrack \\in {\\mathcal{F}}_{t} \), so \( {A}^{\\mathrm{c}} \\in {\\mathcal{F}}_{T} \) . The rest of the proof of (1... | Yes |
Theorem 3.11 Let \( \\left\\{ {\\mathcal{F}}_{t}\\right\\} \) be a filtration satisfying the usual conditions. If \( {M}_{t} \) is a martingale or non-negative submartingale whose paths are right continuous, \( \\mathop{\\sup }\\limits_{{t \\geq 0}}\\mathbb{E}{M}_{t}^{2} < \\infty \), and \( T \) is a finite stopping t... | Proof We do the submartingale case, the martingale case being very similar. By Doob's inequality (Theorem 3.6(1)),\n\n\[ \n\\mathbb{E}\\left\\lbrack {\\mathop{\\sup }\\limits_{{s \\leq t}}{M}_{s}^{2}}\\right\\rbrack \\leq 4\\mathbb{E}{M}_{t}^{2} \n\]\n\nLetting \( t \\rightarrow \\infty \), we have \( \\mathbb{E}\\left... | Yes |
Proposition 3.14 Suppose \( \left\{ {\mathcal{F}}_{t}\right\} \) is a filtration satisfying the usual conditions and suppose \( {A}_{t} \) is an adapted process with paths that are increasing, are right continuous with left limits, and \( {A}_{\infty } = \mathop{\lim }\limits_{{t \rightarrow \infty }}{A}_{t} \) exists,... | Proof First suppose \( X \) and \( A \) are bounded. Let \( n > 1 \) and write \( \mathbb{E}{\int }_{0}^{\infty }{Xd}{A}_{s} \) as\n\n\[ \mathop{\sum }\limits_{{k = 1}}^{\infty }\mathbb{E}\left\lbrack {X\left( {{A}_{k/{2}^{n}} - {A}_{\left( {k - 1}\right) /{2}^{n}}}\right) }\right\rbrack \]\n\nConditioning the \( k \) ... | Yes |
Proposition 3.15 If \( {W}_{t} \) is a Brownian motion, then\n\n\[ \mathbb{P}\left( {\mathop{\sup }\limits_{{s \leq t}}{W}_{s} \geq \lambda }\right) \leq {e}^{-{\lambda }^{2}/{2t}},\;\lambda > 0, \]\n\n(3.8)\n\nand\n\n\[ \mathbb{P}\left( {\mathop{\sup }\limits_{{s \leq t}}\left| {W}_{s}\right| \geq \lambda }\right) \le... | Proof For any \( a \) the process \( \left\{ {e}^{a{W}_{t}}\right\} \) is a submartingale. To see this, since \( x \rightarrow {e}^{ax} \) is convex, the conditional expectation form of Jensen's inequality (Proposition A.21) implies\n\n\[ \mathbb{E}\left\lbrack {{e}^{a{W}_{t}} \mid {\mathcal{F}}_{s}}\right\rbrack \geq ... | Yes |
Proposition 3.16 Let \( W \) be a Brownian motion, let \( T = \inf \left\{ {t > 0 : {W}_{t} \notin \left\lbrack {-a, b}\right\rbrack }\right\} \), and let \( a, b > 0 \) . Then\n\n\[ \mathbb{P}\left( {{W}_{T} = - a}\right) = \frac{b}{a + b},\;\mathbb{P}\left( {{W}_{T} = b}\right) = \frac{a}{a + b}, \]\n\n(3.11)\n\nand\... | Proof Since \( {W}_{t}^{2} - t \) is a martingale with \( {W}_{0} = 0 \), it is easy to check that for each \( u \) , \( {W}_{t \land u}^{2} - \left( {t \land u}\right) \) is also a martingale. Applying Theorem 3.11, we see that \( \mathbb{E}{W}_{u \land T}^{2} = \mathbb{E}\left\lbrack {u \land T}\right\rbrack \) . As ... | Yes |
Corollary 3.17 Suppose \( {M}_{t} \) is a martingale with continuous paths and with \( {M}_{0} = 0 \), a.s., \( T = \inf \left\{ {t \geq 0 : {M}_{t} \notin \left\lbrack {-a, b}\right\rbrack }\right\} \), and \( T < \infty \), a.s. Then\n\n\[ \mathbb{P}\left( {{M}_{T} = - a}\right) = \frac{b}{a + b},\;\mathbb{P}\left( {... | We can also use martingales to get more subtle results. Suppose \( r > 0 \) . Since \( {e}^{r{W}_{t} - {r}^{2}t/2} \) is a martingale, as above\n\n\[ \mathbb{E}{e}^{r{W}_{T \land t} - {r}^{2}\left( {T \land t}\right) /2} = 1 \]\n\nThe exponent is bounded by \( {rb} \) if \( r > 0 \), so we can let \( t \rightarrow \inf... | Yes |
Theorem 4.1 Let \( \left\{ {\mathcal{F}}_{t}\right\} \) be a filtration, not necessarily satisfying the usual conditions, and let \( W \) be a Brownian motion with respect to \( \left\{ {\mathcal{F}}_{t}\right\} \) . If \( u \) is a fixed time, then \( {Y}_{t} = {W}_{t + u} - {W}_{u} \) is a Brownian motion independent... | Proof Let \( {\mathcal{G}}_{t} = {\mathcal{F}}_{t + u} \) . It is clear that \( Y \) has continuous paths, is zero at time 0, and is adapted to \( \left\{ {\mathcal{G}}_{t}\right\} \) . Since \( {Y}_{t} - {Y}_{s} = {W}_{t + u} - {W}_{s + u} \), then \( {Y}_{t} - {Y}_{s} \) is a mean zero normal random variable with var... | Yes |
Theorem 4.2 Let \( \left\{ {\mathcal{F}}_{t}\right\} \) be a filtration, not necessarily satisfying the usual conditions, and let \( W \) be a Brownian motion adapted to \( \left\{ {\mathcal{F}}_{t}\right\} \) . If \( T \) is a finite stopping time, then \( {Y}_{t} = {W}_{T + t} - {W}_{T} \) is a Brownian motion indepe... | Proof We will first show that whenever \( m \geq 1,{t}_{1} < \cdots < {t}_{m}, f \) is a bounded continuous function on \( {\mathbb{R}}^{m} \), and \( A \in {\mathcal{F}}_{T} \), then\n\n\[ \mathbb{E}\left\lbrack {f\left( {{Y}_{{t}_{1}},\ldots ,{Y}_{{t}_{m}}}\right) ;A}\right\rbrack = \mathbb{E}\left\lbrack {f\left( {{... | No |
Theorem 4.4 Let \( {W}_{t} \) be a Brownian motion, \( b > 0, T = \inf \left\{ {t : {W}_{t} \geq b}\right\} \), and \( x < b \) . Then\n\n\[ \mathbb{P}\left( {\mathop{\sup }\limits_{{s \leq t}}{W}_{s} \geq b,{W}_{t} < x}\right) = \mathbb{P}\left( {{W}_{t} > {2b} - x}\right) . \]\n\n(4.5) | Proof Let \( {T}_{n} \) be defined by (3.2). We first show that\n\n\[ \mathbb{P}\left( {{T}_{n} \leq t,{W}_{t} - {W}_{{T}_{n}} < x - b}\right) = \mathbb{P}\left( {{T}_{n} \leq t,{W}_{t} - {W}_{{T}_{n}} > b - x}\right) . \]\n\n(4.6)\n\n\n\nWriting \( \left\lbrack x\right\rbrack \) for the integer part of \( x \), the le... | Yes |
Proposition 4.5 Let \( {W}_{t} \) be a Brownian motion with respect to a filtration \( \left\{ {\mathcal{F}}_{t}\right\} \) satisfying the usual conditions. Let \( T \) be a finite stopping time and \( s > 0 \) . If \( a < b \), then\n\n\[ \mathbb{P}\left( {{W}_{T + s} \in \left\lbrack {a, b}\right\rbrack \mid {\mathca... | Proof If \( A \in {\mathcal{F}}_{T} \), let \( k > 0 \) and write\n\n\[ \mathbb{P}\left( {{W}_{T + s} \in \left\lbrack {a, b}\right\rbrack, A}\right) \]\n\n\[ = \mathop{\sum }\limits_{{j = - \infty }}^{\infty }\mathbb{P}\left( {{W}_{T + s} \in \left\lbrack {a, b}\right\rbrack, A, j/k \leq {W}_{T} < \left( {j + 1}\right... | Yes |
Proposition 5.3 The random variables \( {T}_{1},{T}_{2} - {T}_{1},\ldots ,{T}_{i + 1} - {T}_{i},\ldots \) are independent exponential random variables with parameter \( \lambda \) . | Proof In view of Corollary 4.3 it suffices to show that \( {T}_{1} \) is an exponential random variable with parameter \( \lambda \) . If \( {T}_{1} > t \), then the first jump has not occurred by time \( t \), so \( {X}_{t} \) is still zero. Hence\n\n\[ \n\mathbb{P}\left( {{T}_{1} > t}\right) = \mathbb{P}\left( {{X}_{... | Yes |
Proposition 5.4 Let \( \\left\\{ {\\mathcal{F}}_{t}\\right\\} \) be a filtration satisfying the usual conditions. Suppose \( {X}_{0} = 0 \) , a.s., \( X \) has paths that are right continuous with left limits, \( {X}_{t} - {X}_{s} \) is independent of \( {\\mathcal{F}}_{s} \) if \( s < t \), and \( {X}_{t} - {X}_{s} \)... | Proof Let \( {T}_{0} = 0 \) and \( {T}_{i + 1} = \\inf \\left\\{ {t > {T}_{i} : \\Delta {X}_{t} = 1}\\right\\}, i = 1,2,\\ldots \) We will show that if we set \( {U}_{i} = {T}_{i} - {T}_{i - 1} \), then the \( {U}_{i} \) are i.i.d. exponential random variables and then appeal to Exercise 5.4.\n\nBy Corollary 4.3, the \... | Yes |
Theorem 6.1 There exists a process \( \left\{ {{W}_{t};0 \leq t \leq 1}\right\} \) that is Brownian motion. | Proof If we fix \( t \in \left\lbrack {0,\pi }\right\rbrack \) and compute the Fourier series for the function \( f\left( s\right) = s \land t \), it is an exercise in calculus to get the Fourier coefficients. We end up with\n\n\[ s \land t = \frac{st}{\pi } + \frac{2}{\pi }\mathop{\sum }\limits_{{k = 1}}^{\infty }\fra... | No |
Proposition 8.5 If \( \alpha < 1/2 \), then the paths of a one-dimensional Brownian motion \( \left\{ {{W}_{t};0 \leq }\right. \) \( t \leq 1\} \) are Hölder continuous of order \( \alpha \) with probability one. | Proof By the stationary increments property and scaling,\n\n\[ \mathbb{E}{\left| {W}_{t} - {W}_{s}\right| }^{p} = \mathbb{E}{\left| {W}_{t - s}\right| }^{p} = {\left| t - s\right| }^{p/2}\mathbb{E}{\left| {W}_{1}\right| }^{p}. \]\n\nIf \( \alpha < 1/2 \), choose \( p \) large enough so that \( \left( {\left( {p/2}\righ... | No |
Proposition 9.2 Let \( \\left\\{ {\\mathcal{F}}_{t}\\right\\} \) be a filtration satisfying the usual conditions and \( M \) a right continuous process. The following are equivalent:\n\n(1) \( {M}_{t} \) is a square integrable martingale.\n\n(2) \( M \) is a martingale with \( \\mathop{\\sup }\\limits_{{t \\geq 0}}\\ma... | Proof To show (1) implies (2), suppose \( M \) is a square integrable martingale. Then by Jensen's inequality for conditional expectations (Proposition A.21),\n\n\[ \n\\mathbb{E}{M}_{t}^{2} = \\mathbb{E}\\left\\[ {\\left( \\mathbb{E}\\left\\[ {M}_{\\infty } \\mid {\\mathcal{F}}_{t}\\right\\rbrack \\right) }^{2}\\right\... | Yes |
Proposition 9.3 If \( M \) is a square integrable martingale and \( S \leq T \) are finite stopping times, then \( \mathbb{E}\left\lbrack {{M}_{T} \mid {\mathcal{F}}_{S}}\right\rbrack = {M}_{S} \) . | Proof Let \( A \in {\mathcal{F}}_{S} \) and define \( U\left( \omega \right) = S\left( \omega \right) {1}_{A}\left( \omega \right) + T\left( \omega \right) {1}_{{A}^{c}}\left( \omega \right) \) . Thus \( U \) is equal to \( S \) if \( \omega \in A \) and otherwise is equal to \( T \) . Since \( A \in {\mathcal{F}}_{S} ... | Yes |
Proposition 9.5 Suppose \( \left\{ {\mathcal{F}}_{t}\right\} \) is a filtration satisfying the usual conditions and \( M \) is a process that is adapted to \( \left\{ {\mathcal{F}}_{t}\right\} \) such that \( {M}_{t} \) is integrable for each \( t \) . If \( \mathbb{E}{M}_{T} = 0 \) for every bounded stopping time \( T... | Proof Suppose \( s < t \) and \( A \in {\mathcal{F}}_{s} \) . Define \( T \) to be equal to \( s \) if \( \omega \in A \) and equal to \( t \) if \( \omega \notin A \) . As in the proof of Proposition 9.3, but even more simply, \( T \) is a stopping time, so\n\n\[ 0 = \mathbb{E}{M}_{T} = \mathbb{E}\left\lbrack {{M}_{s}... | Yes |
Proposition 9.6 Suppose \( {M}_{t} \) is a square integrable martingale. Then\n\n\[ \n\mathbb{E}\left\lbrack {{\left( {M}_{T} - {M}_{S}\right) }^{2} \mid {\mathcal{F}}_{S}}\right\rbrack = \mathbb{E}\left\lbrack {{M}_{T}^{2} - {M}_{S}^{2} \mid {\mathcal{F}}_{S}}\right\rbrack \n\] | Proof By Proposition 9.3\n\n\[ \n\mathbb{E}\left\lbrack {{\left( {M}_{T} - {M}_{S}\right) }^{2} \mid {\mathcal{F}}_{S}}\right\rbrack = \mathbb{E}\left\lbrack {{M}_{T}^{2} \mid {\mathcal{F}}_{S}}\right\rbrack - 2{M}_{S}\mathbb{E}\left\lbrack {{M}_{T} \mid {\mathcal{F}}_{S}}\right\rbrack + {M}_{S}^{2} \n\]\n\n\[ \n= \mat... | Yes |
Theorem 9.7 Suppose \( {M}_{0} = 0,{M}_{t} \) is a continuous local martingale, and the paths of \( {M}_{t} \) are locally of bounded variation. Then \( M \) is identically 0, a.s., that is, \( \mathbb{P}\left( {{M}_{t} = 0\text{for all}t}\right) = 1 \) . | Proof Using the definition of local martingale, it suffices to suppose \( M \) is a continuous uniformly integrable martingale. Let \( {t}_{0} \) be fixed and let \( {A}_{t} \) denote the total variation of the paths of \( M \) up to time \( t \) . If \( {T}_{N} = \inf \left\{ {t : {A}_{t} \geq N}\right\} \), we look a... | Yes |
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