Q stringlengths 4 3.96k | A stringlengths 1 3k | Result stringclasses 4
values |
|---|---|---|
Example 4.27 (M/M/s Queue). Imagine a bank with \( s \geq 1 \) tellers that serve customers who queue in a single line if all servers are busy. We imagine that customers arrive at the times of a Poisson process with rate \( \lambda \), and each requires an independent amount of service that has an exponential distribut... | The conditions that result from using the detailed balance condition are \[ {\lambda \pi }\left( {j - 1}\right) = {\mu j\pi }\left( j\right) \;\text{ for }j \leq s, \] \[ {\lambda \pi }\left( {j - 1}\right) = {\mu j\pi }\left( j\right) \;\text{ for }j \geq s. \] From this we conclude that \[ \pi \left( k\right) = \left... | Yes |
Theorem 4.12. If \( \lambda < {s\mu } \), then the \( M/M/s \) queue is positive recurrent. | Proof. To prove the last two conclusions we note that an \( M/M/s \) queue with \( s \) rate \( \mu \) servers is less efficient than an \( M/M/1 \) queue with 1 rate \( {s\mu } \) server, since the single server queue always has departures at rate \( {s\mu } \), while the \( s \) server queue sometimes has departures ... | Yes |
Theorem 4.13. If \( \lambda < \mu \) s, then the output process of the M/M/s queue in equilibrium is a rate \( \lambda \) Poisson process. | Proof of Theorem 4.13. By repeating the proof of (1.13) one can show | No |
Lemma 4.14. Fix \( T \) and let \( {Y}_{s} = {X}_{T - s} \) for \( 0 \leq s \leq T \) . Then \( {Y}_{s} \) is a Markov chain with transition probability\n\n\[ \n{\widehat{p}}_{t}\left( {i, j}\right) = \frac{\pi \left( j\right) {p}_{t}\left( {j, i}\right) }{\pi \left( i\right) }\n\] | Proof. If \( s + t \leq T \), then\n\n\[ \nP\left( {{Y}_{s + t} = j \mid {Y}_{s} = i}\right) = \frac{P\left( {{Y}_{s + t} = j,{Y}_{s} = i}\right) }{P\left( {{Y}_{s} = i}\right) } = \frac{P\left( {{X}_{T - \left( {s + t}\right) } = j,{X}_{T - s} = i}\right) }{P\left( {{X}_{T - s} = i}\right) }\n\]\n\n\[ \n= \frac{P\left... | Yes |
Theorem 4.16. Let \( N\left( t\right) \) be the number of departures between time 0 and time \( t \) for the \( M/M/1 \) queue \( X\left( t\right) \) started from its equilibrium distribution. Then \( \{ N\left( s\right) : 0 \leq \) \( s \leq t\} \) and \( X\left( t\right) \) are independent. | Why is this true? At first it may sound deranged to claim that the output process up to time \( t \) is independent of the queue length. However, if we reverse time, then the departures before time \( t \) turn into arrivals after \( t \), and these are obviously independent of the queue length at time \( t, X\left( t\... | No |
In this system customers at times of a Poisson process with rate \( \lambda \) arrive at service facility 1 where they each require an independent exponential amount of service with rate \( {\mu }_{1} \). When they complete service at the first site, they join a second queue to wait for an exponential amount of service... | The first queue is not affected by the second, so if \( \lambda < {\mu }_{1} \), then (4.35) tells us that the equilibrium probability of the number of customers in the first queue, \( {X}_{t}^{1} \), is given by the shifted geometric distribution\n\n\[ P\left( {{X}_{t}^{1} = m}\right) = {\left( \frac{\lambda }{{\mu }_... | Yes |
Lemma 4.17. If \( \pi \left( {m, n}\right) = c{\lambda }^{m + n}/\left( {{\mu }_{1}^{m}{\mu }_{2}^{n}}\right) \), where \( c = \left( {1 - \lambda /{\mu }_{1}}\right) \left( {1 - \lambda /{\mu }_{2}}\right) \) is a constant chosen to make the probabilities sum to 1, then \( \pi \) is a stationary distribution. | Proof. The first step in checking \( {\pi Q} = 0 \) is to compute the rate matrix \( Q \) . To do this it is useful to draw a picture which assumes \( m, n > 0 \n\n\n\n\nThe rate arrows plus the ordinary lines on the... | Yes |
When is the system stable? That is, when is there a stationary distribution? | (i) if \( {\lambda }_{1} + {p}_{2}{\mu }_{2} < {\mu }_{1} \) and \( {\lambda }_{2} + {p}_{1}{\mu }_{1} < {\mu }_{2} \), then each server can handle their maximum arrival rate and the system will have a stationary distribution.\n\n(ii) if \( {\lambda }_{1} + {p}_{2}{\mu }_{2} > {\mu }_{1} \) and \( {\lambda }_{2} + {p}_... | Yes |
Example 4.30 (Network of M/M/1 Queues). Assume now that there are stations \( 1 \leq i \leq K \) . Arrivals from outside the system occur to station \( i \) at rate \( {\lambda }_{i} \) and service occurs there at rate \( {\mu }_{i} \) . Departures go to station \( j \) with probability \( p\left( {i, j}\right) \) and ... | Generalizing (4.40), we investigate stability by solving the system of equations for the \( {r}_{j} \) that represent the arrival rate at station \( j \) . As remarked earlier, the departure rate from station \( j \) must equal the arrival rate, or a linearly growing queue would develop. Thinking about the arrival rate... | Yes |
Example 4.31. At a government agency entering customers always go to server 1. After completing service there, \( {30}\% \) leave the system while \( {70}\% \) go to server 2 . At server \( 2,{50}\% \) go to server \( 3,{20}\% \) of the customers have to return to server 1, and \( {30}\% \) leave the system. From serve... | The first step is to solve the equations\n\n\[ {r}_{j} = {\lambda }_{j} + \mathop{\sum }\limits_{{i = 1}}^{3}{r}_{i}p\left( {i, j}\right) \]\n\nBy (4.42) the solution is \( r = \lambda {\left( I - p\right) }^{-1} \), where\n\n\[ {\left( I - p\right) }^{-1} = \left( \begin{matrix} {45}/{3835}/{3835}/{76} & \\ 5/{1925}/{... | Yes |
A simple but important special case arises when the random variable \( Y \) and the set \( A \) are independent, i.e., for any set \( B \) we have\n\n\[ P\left( {Y \in B, A}\right) = P\left( {Y \in B}\right) P\left( A\right) \] | Noticing that this implies that \( P\left( {Y \in B,{A}^{c}}\right) = P\left( {Y \in B}\right) P\left( {A}^{c}\right) \) and comparing with the definition of independence of random variables in (A.13), we see that this holds if and only if \( Y \) and \( {1}_{A} \) are independent, so Theorem A. 1 implies\n\n\[ E\left(... | Yes |
Lemma 5.1. If \( X \) is a constant \( c \) on \( A \), then \( E\left( {{XY} \mid A}\right) = {cE}\left( {Y \mid A}\right) \) . | Proof. Since \( X = c \) on \( A,{XY}{1}_{A} = {cY}{1}_{A} \) . Taking expected values and pulling the constant out front, \( E\left( {{XY}{1}_{A}}\right) = E\left( {{cY}{1}_{A}}\right) = {cE}\left( {Y{1}_{A}}\right) \) . Dividing by \( P\left( A\right) \) now gives the result. | Yes |
Lemma 5.3. If \( B \) is the disjoint union of \( {A}_{1},\ldots ,{A}_{k} \), then\n\n\[ E\left( {Y;B}\right) = \mathop{\sum }\limits_{{j = 1}}^{k}E\left( {Y;{A}_{j}}\right) \] | Proof. Our assumption implies \( Y{1}_{B} = \mathop{\sum }\limits_{{j = 1}}^{k}Y{1}_{{A}_{j}} \), so taking expected values, we have\n\n\[ E\left( {Y;B}\right) = E\left( {Y{1}_{B}}\right) = E\left( {\mathop{\sum }\limits_{{j = 1}}^{k}Y{1}_{{A}_{j}}}\right) = \mathop{\sum }\limits_{{j = 1}}^{k}E\left( {Y{1}_{{A}_{j}}}\r... | Yes |
Lemma 5.4. If \( B \) is the disjoint union of \( {A}_{1},\ldots ,{A}_{k} \), then\n\n\[ E\left( {Y \mid B}\right) = \mathop{\sum }\limits_{{j = 1}}^{k}E\left( {Y \mid {A}_{j}}\right) \cdot \frac{P\left( {A}_{j}\right) }{P\left( B\right) } \]\n\nIn particular when \( B = \Omega \) we have \( {EY} = \mathop{\sum }\limit... | Proof. Using the definition of conditional expectation, Lemma 5.3, then doing some arithmetic and using the definition again, we have\n\n\[ E\left( {Y \mid B}\right) = E\left( {Y;B}\right) /P\left( B\right) = \mathop{\sum }\limits_{{j = 1}}^{k}E\left( {Y;{A}_{j}}\right) /P\left( B\right) \]\n\n\[ = \mathop{\sum }\limit... | Yes |
Let \( {X}_{1},{X}_{2},\ldots \) be i.i.d. with \( E{X}_{i} = \mu \) . Let \( {S}_{n} = {S}_{0} + {X}_{1} + \cdots + {X}_{n} \) be a random walk. \( {M}_{n} = {S}_{n} - {n\mu } \) is a martingale with respect to \( {X}_{n} \) . | To check this, note that \( {M}_{n + 1} - {M}_{n} = {X}_{n + 1} - \mu \) is independent of \( {X}_{n},\ldots ,{X}_{0},{M}_{0} \), so by (5.1) the conditional mean of the difference is just the mean:\n\n\[ E\left( {{M}_{n + 1} - {M}_{n} \mid {A}_{v}}\right) = E{X}_{n + 1} - \mu = 0 \] | Yes |
Let \( {X}_{1},{X}_{2},\ldots \) be independent and identically distributed with \( E{X}_{i} = 0 \) and \( \operatorname{var}\left( {X}_{i}\right) = E\left( {X}_{i}^{2}\right) = {\sigma }^{2} \) . Then \( {M}_{n} = {S}_{n}^{2} - n{\sigma}^{2} \) is a martingale with respect to \( {X}_{n} \) . | Proof. We begin with a little algebra:\n\n\[ \n{M}_{n + 1} - {M}_{n} = {\left( {S}_{n} + {X}_{n + 1}\right) }^{2} - {S}_{n}^{2} - {\sigma }^{2} = 2{X}_{n + 1}{S}_{n} + {X}_{n + 1}^{2} - {\sigma }^{2} \n\]\n\nTaking the conditional probability now and using (5.3) and (5.1)\n\n\[ \nE\left( {2{X}_{n + 1}{S}_{n} + {X}_{n +... | Yes |
To build a discrete time model of the stock market we let \( {X}_{1},{X}_{2},\ldots \) be independent \( \geq 0 \) with \( E{X}_{i} = 1 \) . Then \( {M}_{n} = {M}_{0}{X}_{1}\cdots {X}_{n} \) is a martingale with respect to \( {X}_{n} \) . | To prove this we note that (5.3)\n\n\[ E\left( {{M}_{n + 1} - {M}_{n} \mid {A}_{v}}\right) = {M}_{n}E\left( {{X}_{n + 1} - 1 \mid {A}_{v}}\right) = 0 \] | Yes |
Example 5.5 (Exponential Martingale). Let \( {Y}_{1},{Y}_{2},\ldots \) be independent and identically distributed with \( \phi \left( \theta \right) = E\exp \left( {\theta {Y}_{1}}\right) < \infty \) . Let \( {S}_{n} = {S}_{0} + {Y}_{1} + \cdots + {Y}_{n} \) . Then \( {M}_{n} = \exp \left( {\theta {S}_{n}}\right) /\phi... | Proof. If we let \( {X}_{i} = \exp \left( {\theta {Y}_{i}}\right) /\phi \left( \theta \right) \), then \( {M}_{n} = {M}_{0}{X}_{1}\cdots {X}_{n} \) with \( E{X}_{i} = 1 \) and this reduces to the previous example. | Yes |
Let \( {X}_{1},{X}_{2},\ldots \) be independent with\n\n\[ P\left( {{X}_{i} = 1}\right) = p\;\text{ and }\;P\left( {{X}_{i} = - 1}\right) = 1 - p \]\n\nwhere \( p \in \left( {0,1}\right) \) and \( p \neq 1/2 \) . Let \( {S}_{n} = {S}_{0} + {X}_{1} + \cdots + {X}_{n}.{M}_{n} = {\left( \frac{1 - p}{p}\right) }^{{S}_{n}} ... | Proof. Pick \( \theta \) so that \( {e}^{\theta } = \left( {1 - p}\right) /p \) and hence \( {e}^{\theta } = p/\left( {1 - p}\right) \) . In this case\n\n\[ \phi \left( \theta \right) = E{e}^{\theta {X}_{i}} = \frac{1 - p}{p} \cdot p + \frac{p}{1 - p} \cdot \left( {1 - p}\right) = 1 \]\n\nso it follows from Example 5.5... | Yes |
Theorem 5.1. Let \( {X}_{n} \) be a Markov chain with transition probability \( p \) and let \( f\left( {x, n}\right) \) be a function of the state \( x \) and the time \( n \) so that\n\n\[ f\left( {x, n}\right) \geq \mathop{\sum }\limits_{y}p\left( {x, y}\right) f\left( {y, n + 1}\right) \]\n\nThen \( {M}_{n} = f\lef... | Proof. By the Markov property and our assumption on \( f \)\n\n\[ E\left( {f\left( {{X}_{n + 1}, n + 1}\right) \mid {A}_{v}}\right) \geq \mathop{\sum }\limits_{y}p\left( {{x}_{n}, y}\right) f\left( {y, n + 1}\right) = f\left( {{x}_{n}, n}\right) \]\n\nwhich proves the desired result. | Yes |
Consider a Markov chain \( {X}_{n} \) with finite state space \( S \) . Let \( A \subset S \), and let \( {V}_{A} = \min \left\{ {n \geq 0 : {X}_{n} \in A}\right\} \) . If for \( x \in C = S - A \) we have\n\n\[ h\left( x\right) \geq \mathop{\sum }\limits_{y}p\left( {x, y}\right) h\left( y\right) \]\n\nthen \( h\left( ... | Proof. Let \( {\bar{X}}_{n} \) be a modification of \( {X}_{n} \) in which all sites in \( A \) are absorbing and apply Theorem 5.1. | No |
Continuing to use the assumptions of the last example, we assume that for \( x \in C \) we have\n\n\[ g\left( x\right) \geq \alpha + \mathop{\sum }\limits_{y}p\left( {x, y}\right) g\left( y\right) \] | Adding \( n \) to each side of the equation, and applying Theorem 5.1 to \( f\left( {x, n}\right) = g \) \( \left( x\right) + {\alpha n} \) to \( {\bar{X}}_{n} \) defined in the previous example, we see that \( g\left( {X}_{{V}_{A} \land n}\right) + \alpha \left( {{V}_{A} \land n}\right) \) is a supermartingale. | Yes |
Theorem 5.6. If \( {M}_{m} \) is a supermartingale and \( m \leq n \), then \( E{M}_{m} \geq E{M}_{n} \) . | Proof. It is enough to show that the expected value decreases with each time step, i.e., \( E{M}_{k} \geq E{M}_{k + 1} \) . To do this, we will again use the notation from (5.5)\n\n\[ \n{A}_{v} = \left\{ {{X}_{n} = {x}_{n},{X}_{n - 1} = {x}_{n - 1},\ldots ,{X}_{0} = {x}_{0},{M}_{0} = m}\right\} \n\]\n\nand note that li... | Yes |
Example 5.9 (Doubling Strategy). Suppose you are playing a game in which you will win or lose \( \$ 1 \) on each play. If you win you bet \( \$ 1 \) on the next play but if you lose, then you bet twice the previous amount. The idea behind the system can be seen by looking at what happens if we lose four times in a row ... | In this example our net profit when we win is \( \$ 1 \) . Since \( 1 + 2 + \cdots + {2}^{k} = {2}^{k + 1} - 1 \) , this is true if we lose \( k \) times in a row before we win. Thus every time we win our net profit is up by \( \$ 1 \) from the previous time we won. | Yes |
Theorem 5.9. Suppose that \( {M}_{n} \) is a supermartingale with respect to \( {X}_{n},{H}_{n} \) is predictable, and \( 0 \leq {H}_{n} \leq {c}_{n} \) where \( {c}_{n} \) is a constant that may depend on \( n \) . Then\n\n\[ \n{W}_{n} = {W}_{0} + \mathop{\sum }\limits_{{m = 1}}^{n}{H}_{m}\left( {{M}_{m} - {M}_{m - 1}... | Proof. The change in our wealth from time \( n \) to time \( n + 1 \) is\n\n\[ \n{W}_{n + 1} - {W}_{n} = {H}_{n + 1}\left( {{Y}_{n + 1} - {Y}_{n}}\right) \n\]\n\nContinuing to use the notation introduced in (5.5)\n\n\[ \n{A}_{v} = \left\{ {{X}_{n} = {x}_{n},{X}_{n - 1} = {x}_{n - 1},\ldots ,{X}_{0} = {x}_{0},{M}_{0} = ... | Yes |
One possible gambling strategy is to bet \( \$ 1 \) each time until you stop playing at time \( T \) . In symbols, we let\n\n\[ \n{H}_{m} = \left\{ \begin{array}{ll} 1 & \text{ if }T \geq m \\ 0 & \text{ otherwise } \end{array}\right. \n\] | To check that this is an admissible gambling strategy we note that\n\n\[ \n\left\{ {{H}_{m} = 0}\right\} = \{ T \geq m{\} }^{c} = \{ T \leq m - 1\} = { \cup }_{k = 1}^{m - 1}\{ T = k\} \n\]\n\nBy the definition of a stopping time, the event \( \{ T = k\} \) can be determined from the values of \( {M}_{0},{X}_{0},\ldots... | Yes |
Theorem 5.11. Suppose \( {M}_{n} \) is a martingale and \( T \) a stopping time with \( P\left( {T < \infty }\right) = 1 \) and \( \left| {M}_{T \land n}\right| \leq K \) for some constant \( K \) . Then \( E{M}_{T} = E{M}_{0} \) . | Proof. Theorem 5.10 implies\n\n\[ E{M}_{0} = E{M}_{T \land n} = E\left( {{M}_{T};T \leq n}\right) + E\left( {{M}_{n};T > n}\right) .\n\]\n\nThe second term \( \leq {KP}\left( {T > n}\right) \) and\n\n\[ \left| {E\left( {{M}_{T};T \leq n}\right) - E\left( {M}_{T}\right) }\right| \leq {KP}\left( {T > n}\right) \]\n\nSinc... | Yes |
Example 5.12 (Exit Distribution for Simple Random Walk). Let \( {M}_{n} = {S}_{n} \) and \( T = \min \left\{ {n : {S}_{n} \notin \left( {a, b}\right) }\right\} .{M}_{T \land n} \in \left\lbrack {a, b}\right\rbrack \) so it is bounded. Thus using Theorem 5.11 we can conclude | \[ x = {E}_{x}{S}_{T} = a{P}_{x}\left( {{S}_{T} = a}\right) + b\left( {1 - P\left( {{S}_{T} = a}\right) }\right) ,\] and then solve to conclude \[ {P}_{x}\left( {{S}_{T} = a}\right) = \frac{b - x}{b - a}\;{P}_{x}\left( {{S}_{T} = b}\right) = \frac{x - a}{b - a} \] | Yes |
Example 5.13 (Gambler’s Ruin). Let \( {S}_{n} = {S}_{0} + {X}_{1} + \cdots {X}_{n} \) where \( {X}_{1},{X}_{2},\ldots {X}_{n} \) are independent with\n\n\[ P\left( {{X}_{i} = 1}\right) = p\;\text{ and }\;P\left( {{X}_{i} = - 1}\right) = q = 1 - p. \]\n\nSuppose \( p \in \left( {0,1}\right), p \neq 1/2 \) and let \( h\l... | \[ {M}_{T \land n} \leq \max \left\{ {{\left( \left( 1 - p\right) /p\right) }^{a},{\left( \left( 1 - p\right) /p\right) }^{a}}\right\} \]\n\nso using Theorem 5.11.\n\n\[ {\left( q/p\right) }^{x} = {E}_{x}{\left( q/x\right) }^{S\left( \tau \right) } = {\left( q/p\right) }^{a}P\left( {{S}_{\tau } = a}\right) + {\left( q/... | Yes |
Example 5.14 (General Birth and Death Chains). The state space is \( \{ 0,1,2,\ldots \} \) and the transition probability has\n\n\[ p\left( {x, x + 1}\right) = {p}_{x} \]\n\n\[ p\left( {x, x - 1}\right) = {q}_{x}\;\text{ for }x > 0 \]\n\n\[ p\left( {x, x}\right) = 1 - {p}_{x} - {q}_{x}\;\text{ for }x \geq 0 \]\n\nwhile... | \[\n\phi \left( x\right) = {E}_{x}\phi \left( {X}_{\tau }\right) = \phi \left( a\right) {P}_{x}\left( {{X}_{\tau } = a}\right) + \phi \left( b\right) \left\lbrack {1 - {P}_{x}\left( {{X}_{\tau } = a}\right) }\right\rbrack\n\]\n\nand a little algebra gives\n\n\[ {P}_{x}\left( {{V}_{a} < {V}_{b}}\right) = \frac{\phi \lef... | Yes |
Let \( {S}_{n} \) be symmetric simple random walk. Let \( T = \min \left\{ {n : {S}_{n} \notin \left( {a, b}\right) }\right\} \) where \( a < 0 < b \) . Our goal here is to prove a generalization of (1.29):\n\n\[ {E}_{x}T = \left( {x - a}\right) \left( {b - x}\right) \] | It suffices to show \( {E}_{0}T = - {ab} \) . Example 5.3 implies that \( {S}_{n}^{2} - n \) is a martingale.\n\nIf we argue casually \( 0 = {E}_{0}\left( {{S}_{T}^{2} - T}\right) \) so using (5.9)\n\n\[ {E}_{0}\left( T\right) = {E}_{0}\left( {S}_{T}^{2}\right) = {a}^{2}{P}_{0}\left( {{S}_{T} = a}\right) + {b}^{2}{P}_{... | Yes |
Theorem 5.12 (Wald’s Equation). If \( T \) is a stopping time with \( {ET} < \infty \), then\n\n\[ E\left( {{S}_{T} - {S}_{0}}\right) = {\mu ET} \] | Proof. To make the last calculation rigorous, we need to stop at time \( T \land n \) and let \( n \rightarrow \infty \) . Theorem 5.10 gives\n\n\[ E{S}_{0} = E\left( {S}_{T \land n}\right) - {\mu E}\left( {T \land n}\right) \]\n\nAs \( n \uparrow \infty ,{E}_{0}\left( {T \land n}\right) \uparrow {E}_{0}T \) by (5.12).... | Yes |
Consider asymmetric simple random walk \( {S}_{n} = {S}_{0} + {X}_{1} + \cdots + {X}_{n} \) where \( {X}_{1},{X}_{2},\ldots {X}_{n} \) be independent with\n\n\[ P\left( {{X}_{i} = 1}\right) = p\;\text{ and }\;P\left( {{X}_{i} = - 1}\right) = q = 1 - p \]\n\nSuppose \( 0 < p < 1/2 \), and let \( {V}_{0} = \min \left\{ {... | Noting that \( E{X}_{i} = p - q \) and using Theorem 5.10 we have\n\n\[ x = {E}_{x}\left\lbrack {S\left( {{V}_{0} \land n}\right) - \left( {p - q}\right) \left( {{V}_{0} \land n}\right) }\right\rbrack \geq \left( {q - p}\right) {E}_{x}\left( {{V}_{0} \land n}\right) \]\n\nLetting \( n \rightarrow \infty \) we see that ... | Yes |
Example 5.17 (Left-Continuous Random Walk). Suppose that \( {X}_{1},{X}_{2},\ldots \) are independent integer valued random variables with \( E{X}_{i} > 0, P\left( {{X}_{i} \geq - 1}\right) = 1 \), and \( P\left( {{X}_{i} = - 1}\right) > 0 \) . These walks are called left-continuous since they cannot jump over any inte... | To see that such an \( \alpha \) exists, note that (i) \( \phi \left( 0\right) = 1 \) and \[ {\phi }^{\prime }\left( \theta \right) = \frac{d}{d\theta }E{e}^{\theta {x}_{i}} = E\left( {{x}_{i}{e}^{\theta {x}_{i}}}\right) \;\text{ so }\;{\phi }^{\prime }\left( 0\right) = E{x}_{i} > 0 \] and it follows that \( \phi \left... | Yes |
Theorem 5.13. Consider a left-continuous random walk with positive mean. Let \( x > 0 \) and \( {V}_{0} = \min \left\{ {n : {S}_{n} = 0}\right\} \) . \[ {P}_{x}\left( {{V}_{0} < \infty }\right) = {e}^{\alpha x} \] | Proof. Again if one argues casually \[ {e}^{\alpha x} = {E}_{x}\left( {\exp \left( {\alpha {V}_{0}}\right) }\right) = {P}_{x}\left( {{V}_{0} < \infty }\right) \] but we have to prove that there is no contribution from \( \left\{ {{V}_{0} = \infty }\right\} \) . To do this note that Theorem 5.10 gives \[ {e}^{\alpha x} ... | No |
Example 5.18 (Extinction Probability for Branching Process). Consider for simplicity a continuous time branching process in which individuals live for an exponentially distributed amount of time and then die leaving \( k \) offspring with probability \( {p}_{k} \) . Let \( {Y}_{1},{Y}_{2},\ldots \) be independent with ... | To have a random walk, we will use this definition for all \( n \) . Since \( {Y}_{n + 1} \geq 0 \) this is a left-continuous random walk with jumps \( {X}_{k} = {Y}_{k} - 1 \) . If \( E{Y}_{k} > 1 \), then \( E{X}_{k} > 0 \) so we can use Theorem 5.13 to conclude that if \( \alpha < 0 \) has \( E{e}^{\alpha {X}_{k}} =... | Yes |
Let \( B \) for bankrupt be the event that the wealth of the insurance company is negative at some time \( n \). We will show\n\n\[ P\left( B\right) \leq \exp \left( {-{2\mu }{S}_{0}/{\sigma }^{2}}\right) \] | Proof. We begin by computing \( \phi \left( \theta \right) = E\exp \left( {\theta {X}_{i}}\right) \) . To do this we need a little algebra\n\n\[ - \frac{{\left( x - \mu \right) }^{2}}{2{\sigma }^{2}} + \theta \left( {x - \mu }\right) + {\theta \mu } = - \frac{{\left( x - \mu - {\sigma }^{2}\theta \right) }^{2}}{2{\sigm... | Yes |
Lemma 5.15. Let \( \epsilon = \left( {E{s}_{0} - E{t}_{1}}\right) /2 \) . There is a \( K \) so that \( {E}_{x}\left( {{X}_{1} - x}\right) \leq - \epsilon \) for \( x > K \) . | Proof. To do this we note that\n\n\[ E{\left( x + {s}_{0} - {t}_{1}\right) }^{ + } - x = E\left( {x + {s}_{0} - {t}_{1};{s}_{0} - {t}_{1} \geq - x}\right) - x \]\n\n\[ = E\left( {{s}_{0} - {t}_{1};{s}_{0} - {t}_{1} \geq - x}\right) - {xP}\left( {{s}_{0} - {t}_{1} < - x}\right) \]\n\n\[ \leq E\left( {{s}_{0} - {t}_{1};{... | Yes |
Lemma 5.16. Consider a Markov chain on \( \lbrack 0,\infty ) \) . Suppose that there is a \( \phi \left( x\right) \geq 0 \) with \( \phi \left( x\right) \rightarrow \infty \) as \( x \rightarrow \infty \), and a \( K \) so that if \( x > K \), then \( {E}_{x}\phi \left( {X}_{1}\right) \leq \phi \left( x\right) \) . Let... | Proof. It follows from Example 5.8 that \( \phi \left( {X}_{{U}_{K} \land n}\right) + \epsilon \left( {{U}_{K} \land n}\right) \) is a supermartingale. Stopping at time \( {U}_{K} \land n \) we have\n\n\[ \phi \left( x\right) \geq {E}_{x}\phi \left( {X\left( {{U}_{K} \land n}\right) }\right) + {\epsilon E}\left( {{U}_{... | Yes |
Lemma 5.17. Suppose in addition to the assumptions in Lemma 5.16 that \( M = \sup \left\{ {{E}_{x}\phi \left( {X}_{1}\right) : x \leq K}\right\} < \infty \) . There are constants \( {C}_{1} \) and \( {C}_{2} \) so that\n\n\[ \n{E}_{x}{V}_{0} \leq \frac{\phi \left( x\right) + {C}_{1}}{\epsilon } + {C}_{2} \n\] | Proof. Let \( K\left( x\right) = \min \left\{ {k : {p}^{k}\left( {x,0}\right) > 0}\right\}, L = \mathop{\max }\limits_{{x \leq K}}K\left( x\right) \), and \( \rho = \min \left\{ {{P}_{x}\left( {{V}_{0} \leq }\right. }\right. \) \( L) : x \leq K\} \) . Let \( {R}_{0} = \min \left\{ {n : {X}_{n} \leq K}\right\} \) . Let\... | Yes |
Theorem 6.2. There is no arbitrage if and only if there is a strictly positive probability vector so that all the stock prices are martingale. | Proof. One direction is easy. If (i) is true, then for any strictly positive probability vector \( \mathop{\sum }\limits_{{i = 1}}^{m}\mathop{\sum }\limits_{{j = 1}}^{n}{x}_{i}{a}_{i, j}{p}_{j} > 0 \), so (ii) is false.\n\nSuppose now that (i) is false. The linear combinations \( \mathop{\sum }\limits_{{i = 1}}^{m}{x}_... | Yes |
A stock is selling at \( \$ {60} \) today. A month from now it will either be at \( \$ {80} \) or \( \$ {50} \), i.e., \( u = 4/3 \) and \( d = 5/6 \) . We assume an interest rate of \( r = 1/{18} \) so the risk neutral probability is\n\n\[ \n\frac{{19}/{18} - 5/6}{4/3 - 5/6} = \frac{4}{9} \n\]\n\nConsider now a call o... | By (6.6) the value is\n\n\[ \n{V}_{0} = \frac{18}{19} \cdot \frac{4}{9} \cdot {15} = \frac{120}{19} = {6.3158} \n\]\n\nBeing a savvy businessman you offer to sell this for \$6.50. You are delighted when a customer purchases 10,000 calls for \( \$ {65},{000} \), but then become worried about the fact that if the stock g... | Yes |
Theorem 6.3. If \( {W}_{0} = {V}_{0} \) and we use the investment strategy in (6.13), then we have \( {W}_{n} = {V}_{n} \) . | Proof. We proceed by induction. By assumption the result is true when \( n = 0 \) . Let \( a \) be a string of \( H \) and \( T \) of length \( n \) . (6.14) implies\n\n\[ \n{W}_{n + 1}\left( {aH}\right) = {\Delta }_{n}\left( a\right) {S}_{n + 1}\left( {aH}\right) + \left( {1 + r}\right) \left( {{W}_{n}\left( a\right) ... | Yes |
Theorem 6.4. In the binomial model, under the risk neutral probability measure \( {M}_{n} = {S}_{n}/{\left( 1 + r\right) }^{n} \) is a martingale with respect to \( {S}_{n} \) . | Proof. Let \( {p}^{ * } \) and \( 1 - {p}^{ * } \) be defined by (6.5). Given a string \( a \) of heads and tails of length \( n \) \n\n\[ \n{P}^{ * }\left( a\right) = {\left( {p}^{ * }\right) }^{H\left( a\right) }{\left( 1 - {p}^{ * }\right) }^{T\left( a\right) } \n\] \n\nwhere \( H\left( a\right) \) and \( T\left( a\... | Yes |
Theorem 6.5. Assume that the holdings \( {\Delta }_{n}\left( a\right) \) are given by (6.13) and let \( {W}_{n} \) be the wealth process defined by (6.14). Under \( {P}^{ * },{W}_{n}/{\left( 1 + r\right) }^{n} \) is a martingale, and hence the value has \( {V}_{0} = {E}^{ * }\left( {{V}_{n}/{\left( 1 + r\right) }^{n}}\... | Proof. The second conclusion follows from the first and Theorem 6.3. A little arithmetic with (6.14) shows that\n\n\[ \frac{{W}_{n + 1}}{{\left( 1 + r\right) }^{n + 1}} = \frac{{W}_{n}}{{\left( 1 + r\right) }^{n}} + {\Delta }_{n}\left( {\frac{{S}_{n + 1}}{{\left( 1 + r\right) }^{n + 1}} - \frac{{S}_{n}}{{\left( 1 + r\r... | Yes |
In this option you can buy the stock at time 3 at its current price and then sell it at the highest price seen in the past for a profit of\n\n\[ \n{V}_{3} = \mathop{\max }\limits_{{0 \leq m \leq 3}}{S}_{m} - {S}_{3} \n\]\n\nOur goal is to compute the value \( {V}_{n}\left( a\right) \) and the replicating strategy \( {\... | Starting at the right edge, \( {S}_{3}\left( {HTT}\right) = 2 \) but the maximum in the past is \( 8 = {S}_{1}\left( H\right) \) so \( {V}_{3}\left( {HTT}\right) = 8 - 2 = 6 \).\n\nOn the tree, stock prices are above the nodes and option prices below. To explain the computation of the option price note that by (6.17).\... | Yes |
Example 6.3 (Put Option). We will use the binomial model in (6.16) but now suppose \( {S}_{0} = 8 \) and consider the put option with value \( {V}_{3} = {\left( {10} - {S}_{3}\right) }^{ + } \) . The value of this option depends only on the price so we can reduce the tree considered above to: | \n\nOn the tree itself stock prices are above the nodes and option prices below. To explain the computation of the option price note that by (6.17).\n\n\[ \n{V}_{2}\left( 2\right) = {0.4}\left\lbrack {{V}_{3}\left( 4... | Yes |
Example 6.4 (Put-Call Parity). Consider the binomial model with \( {S}_{0} = {32} \) , \( u = 3/2, d = 2/3 \), and \( r = 1/6 \) . By (6.5) the risk neutral probability | \[ {p}^{ * } = \frac{1 + r - d}{u - d} = \frac{7/6 - 2/3}{3/2 - 2/3} = \frac{3/6}{5/6} = {0.6} \] so by (6.12) the value satisfies \[ {V}_{n}\left( a\right) = \frac{1}{7}\left( {{3.6}{V}_{n + 1}\left( {aH}\right) + {2.4}{V}_{n + 1}\left( {aT}\right) }\right) \] We will now compute the values for the call and put with s... | Yes |
Theorem 6.6. The values \( {V}_{P} \) and \( {V}_{C} \) of the put and call options with the same strike \( K \) and expiration \( N \) are related by\n\n\[ \n{V}_{P} - {V}_{C} = \frac{K}{{\left( 1 + r\right) }^{N}} - {S}_{0} \n\]\n\nIn particular if \( K = {\left( 1 + r\right) }^{N}{S}_{0} \), then \( {V}_{P} = {V}_{C... | Proof. The key observation is that\n\n\[ \n{S}_{N} + {\left( K - {S}_{N}\right) }^{ + } - {\left( {S}_{N} - K\right) }^{ + } = K \n\]\n\nTo prove this consider the two cases \( {S}_{N} \geq K \) and \( {S}_{N} \leq K \) .\n\n\[ \n{S}_{N} + 0 - \left( {{S}_{N} - K}\right) = K \n\]\n\n\[ \n\left. {{S}_{N} + \left( {K - {... | Yes |
In these options when the price falls below a certain level the option is worthless no matter what the value of the stock is at the end. To illustrate consider the binomial model from Example 6.4: \( u = 3/2 \) , \( d = 2/3 \), and \( r = 1/6 \) . This time we suppose \( {S}_{0} - {24} \) and consider a call \( {\left(... | As we have computed the risk neutral probability is \( {p}^{ * } = {0.6} \) and the value recursion is\n\n\[ \n{V}_{n}\left( a\right) = \frac{6}{7}\left\lbrack {{.6}{V}_{n}\left( {aH}\right) + {.4}{V}_{n}\left( {aT}\right) }\right\rbrack \n\]\n\nwith the extra boundary condition that if the price is \( \leq {20} \) the... | Yes |
For a concrete example, suppose as we did in Example 6.3 that the stock price follows the binomial model with \( {S}_{0} = 8, u = 2, d = 1/2, r = 1/4 \) and consider a put option with strike 10, that is \( {g}_{n} = {\left( {10} - {s}_{n}\right) }^{ + } \) . The risk neutral probability \( {p}^{ * } = {0.5} \) and the ... | On the drawing above, the two numbers above each line are the price of the stock and the value of the option. Below the line are the value of the option if exercised, and the value computed by the recursion if we continue for one more period. A star indicates the large of the two, which is the value of the option at th... | Yes |
Continue now the set-up of the previous example but consider the call option \( {\left( {S}_{n} - {10}\right) }^{ + } \). The computations are the same but the result is boring: the optimal strategy is to always continue, so there is no difference between the American and the European option. | To spare the reader the chore of doing the arithmetic we give the recursion:\n\n\[ \n{V}_{2}\left( 2\right) = \max \{ 0,0\} = 0 \]\n\n\[ \n{V}_{2}\left( 8\right) = \max \{ 0,{0.4}\left( {0 + 6}\right) = {2.4}\} = {2.4} \]\n\n\[ \n{V}_{2}\left( {32}\right) = \max \{ {22},{0.4}({54} + {60} = {24}\} = {24} \]\n\n\[ \n{V}_... | No |
Theorem 6.7. If \( g \) is a nonnegative convex function with \( g\left( 0\right) = 0 \), then for the American option with payoff \( g\left( {S}_{n}\right) \) it is optimal to wait until the end of exercise. | Proof. Since \( {S}_{n}/{\left( 1 + r\right) }^{n} \) is a martingale under \( {P}^{ * } \n\n\[ g\left( {S}_{n}\right) = g\left( {{E}_{n}^{ * }\left( \frac{{S}_{n + 1}}{1 + r}\right) }\right) \]\n\nUnder the risk neutral probability\n\n\[ {S}_{n}\left( a\right) = {p}_{n}^{ * }\left( a\right) \frac{{S}_{n + 1}\left( {aH... | Yes |
Theorem 6.8. Central Limit Theorem. Let \( {X}_{1},{X}_{2},\ldots \) be i.i.d. with \( E{X}_{i} = 0 \) and \( \operatorname{var}\left( {X}_{i}\right) = 1 \) Then for all \( x \) we have\n\n\[ P\left( {\frac{1}{\sqrt{n}}\mathop{\sum }\limits_{{m = 1}}^{n}{X}_{m} \leq x}\right) \rightarrow P\left( {\chi \leq x}\right) \]... | The conclusion in (6.23) is often written as\n\n\[ \frac{1}{\sqrt{n}}\mathop{\sum }\limits_{{m = 1}}^{n}{X}_{m} \Rightarrow \chi \]\n\nwhere \( \Rightarrow \) is read \ | No |
Theorem 6.9. Write \( {E}^{ * } \) for expected values when \( \mu = r - {\sigma }^{2}/2 \) in (6.24). The value of a European option \( g\left( {S}_{T}\right) \) is given by \( {E}^{ * }{e}^{-{rT}}g\left( {S}_{T}\right) \) . | Proof. We prove this by taking limits of the discrete approximation. The risk neutral probabilities, \( {p}_{h}^{ * } \), are given by\n\n\[ \n{p}_{h}^{ * } = \frac{1 + {rh} - d}{u - d} \n\]\n\n(6.26)\n\nUsing the formulas for \( u \) and \( d \) in (6.21 and recalling that \( {e}^{x} = 1 + x + {x}^{2}/2 + \cdots \) ,\... | Yes |
Theorem 6.10. The price of the European call option \( {\left( {S}_{t} - K\right) }^{ + } \) is given by\n\n\[ \n{S}_{0}\Phi \left( {d}_{1}\right) - {e}^{-{rt}}{K\Phi }\left( {d}_{2}\right) \n\]\n\nwhere the constants\n\n\[ \n{d}_{1} = \frac{\ln \left( {{S}_{0}/K}\right) + \left( {r + {\sigma }^{2}/2}\right) t}{\sigma ... | Proof. Using the fact that \( \log \left( {{S}_{t}/{S}_{0}}\right) \) has a normal \( \left( {{\mu t},{\sigma }^{2}t}\right) \) distribution with \( \mu = r - {\sigma }^{2}/2 \), we see that\n\n\[ \n{E}^{ * }\left( {{e}^{-{rt}}{\left( {S}_{t} - K\right) }^{ + }}\right) = {e}^{-{rt}}{\int }_{\log \left( {K/{S}_{0}}\righ... | Yes |
Example 6.8 (A Google Call Options). On the morning of December 5, 2011, Google stock was selling for \$620 a share and a March 12 call option with strike \( K = {635} \) was selling for \( \$ {33.10} \) . To compare this with the prediction of the Black-Scholes formula we assume an interest rate of \( r = {0.01} \) pe... | With the help of a little spreadsheet we find that the formula predicts a price of \$32.93. | Yes |
Put-call parity allows us to compute the value of the put option, \( {V}_{P} \) from the value of the call option \( {V}_{C} \) by the formula:\n\n\[ {V}_{P} - {V}_{C} = {e}^{-{rT}}K - {S}_{0} \] | In the example for March 12 Google options, \( \exp \left( {-{rt}}\right) = {0.9966} \) so we might as well ignore that factor. As the next table shows the formula works well in practice\n\n<table><thead><tr><th>strike</th><th>\( {V}_{P} \)</th><th>\( {V}_{C} \)</th><th>\( {S}_{0} + {V}_{P} - {V}_{C} \)</th></tr></thea... | Yes |
Proposition 1.5. Every transition matrix on a finite state space has a random mapping representation. | Proof. Let \( P \) be the transition matrix of a Markov chain with state space \( \mathcal{X} = \left\{ {{x}_{1},\ldots ,{x}_{n}}\right\} \) . Take \( \Lambda = \left\lbrack {0,1}\right\rbrack \) ; our auxiliary random variables \( Z,{Z}_{1},{Z}_{2},\ldots \) will be uniformly chosen in this interval. Set \( {F}_{j, k}... | Yes |
Proposition 1.7. If \( P \) is aperiodic and irreducible, then there is an integer \( {r}_{0} \) such that \( {P}^{r}\left( {x, y}\right) > 0 \) for all \( x, y \in \mathcal{X} \) and \( r \geq {r}_{0} \) . | Proof. We use the following number-theoretic fact: any set of non-negative integers which is closed under addition and which has greatest common divisor 1 must contain all but finitely many of the non-negative integers. (See Lemma 1.30 in the Notes of this chapter for a proof.) For \( x \in \mathcal{X} \), recall that ... | Yes |
Consider the graph \( G \) shown in Figure 1.4. The transition matrix of simple random walk on \( G \) is | \[ P = \left( \begin{matrix} 0 & \frac{1}{2} & \frac{1}{2} & 0 & 0 \\ \frac{1}{3} & 0 & \frac{1}{3} & \frac{1}{3} & 0 \\ \frac{1}{4} & \frac{1}{4} & 0 & \frac{1}{4} & \frac{1}{4} \\ 0 & \frac{1}{2} & \frac{1}{2} & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \end{matrix}\right) \] | Yes |
Proposition 1.19. If \( P \) is an irreducible transition matrix and \( \pi \) is the unique probability distribution solving \( \pi = {\pi P} \), then for all states \( z \) , \[ \pi \left( z\right) = \frac{1}{{\mathbf{E}}_{z}{\tau }_{z}^{ + }}. \] | Proof. Let \( {\widetilde{\pi }}_{z}\left( y\right) \) equal \( \widetilde{\pi }\left( y\right) \) as defined in (1.19), and write \( {\pi }_{z}\left( y\right) = {\widetilde{\pi }}_{z}\left( y\right) /{\mathbf{E}}_{z}{\tau }_{z}^{ + } \) . Proposition 1.14 implies that \( {\pi }_{z} \) is a stationary distribution, so ... | Yes |
Proposition 1.20. Let \( P \) be the transition matrix of a Markov chain with state space \( \mathcal{X} \) . Any distribution \( \pi \) satisfying the detailed balance equations (1.29) is stationary for \( P \) . | Proof. Sum both sides of (1.29) over all \( y \) :\n\n\[ \mathop{\sum }\limits_{{y \in \mathcal{X}}}\pi \left( y\right) P\left( {y, x}\right) = \mathop{\sum }\limits_{{y \in \mathcal{X}}}\pi \left( x\right) P\left( {x, y}\right) = \pi \left( x\right) ,\]\n\nsince \( P \) is stochastic. | Yes |
Proposition 1.23. Let \( \left( {X}_{t}\right) \) be an irreducible Markov chain with transition matrix \( P \) and stationary distribution \( \pi \) . Write \( \left( {\widehat{X}}_{t}\right) \) for the time-reversed chain with transition matrix \( \widehat{P} \) . Then \( \pi \) is stationary for \( \widehat{P} \), a... | Proof. To check that \( \pi \) is stationary for \( \widehat{P} \), we simply compute\n\n\[ \n\mathop{\sum }\limits_{{y \in \mathcal{X}}}\pi \left( y\right) \widehat{P}\left( {y, x}\right) = \mathop{\sum }\limits_{{y \in \mathcal{X}}}\pi \left( y\right) \frac{\pi \left( x\right) P\left( {x, y}\right) }{\pi \left( y\rig... | Yes |
Proposition 1.28. If \( \pi \) is stationary for the finite transition matrix \( P \), then \( \pi \left( {y}_{0}\right) = 0 \) for all inessential states \( {y}_{0} \) . | Proof. Let \( \mathcal{C} \) be an essential communicating class. Then\n\n\[ \n{\pi P}\left( \mathcal{C}\right) = \mathop{\sum }\limits_{{z \in \mathcal{C}}}\left( {\pi P}\right) \left( z\right) = \mathop{\sum }\limits_{{z \in \mathcal{C}}}\left\lbrack {\mathop{\sum }\limits_{{y \in \mathcal{C}}}\pi \left( y\right) P\l... | Yes |
Proposition 1.29. The transition matrix \( P \) has a unique stationary distribution if and only if there is a unique essential communicating class. | Proof. Suppose that there is a unique essential communicating class \( \mathcal{C} \) . Recall that \( {P}_{\mathcal{C}} \) is the restriction of the matrix \( P \) to the states in \( \mathcal{C} \), and that \( {P}_{\mid \mathcal{C}} \) is a transition matrix, irreducible on \( \mathcal{C} \) with a unique stationary... | Yes |
Assume that a gambler making fair unit bets on coin flips will abandon the game when her fortune falls to 0 or rises to \( n \) . Let \( {X}_{t} \) be gambler’s fortune at time \( t \) and let \( \tau \) be the time required to be absorbed at one of 0 or \( n \) . Assume that \( {X}_{0} = k \), where \( 0 \leq k \leq n... | Proof. Let \( {p}_{k} \) be the probability that the gambler reaches a fortune of \( n \) before ruin, given that she starts with \( k \) dollars. We solve simultaneously for \( {p}_{0},{p}_{1},\ldots ,{p}_{n} \) . Clearly \( {p}_{0} = 0 \) and \( {p}_{n} = 1 \), while\n\n\[ \n{p}_{k} = \frac{1}{2}{p}_{k - 1} + \frac{1... | No |
Consider a collector attempting to collect a complete set of coupons. Assume that each new coupon is chosen uniformly and independently from the set of \( n \) possible types, and let \( \tau \) be the (random) number of coupons collected when the set first contains every type. Then\n\n\[ \mathbf{E}\left( \tau \right) ... | The expectation \( \mathbf{E}\left( \tau \right) \) can be computed by writing \( \tau \) as a sum of geometric random variables. Let \( {\tau }_{k} \) be the total number of coupons accumulated when the collection first contains \( k \) distinct coupons. Then\n\n\[ \tau = {\tau }_{n} = {\tau }_{1} + \left( {{\tau }_{2... | Yes |
Proposition 2.4. Let \( \tau \) be a coupon collector random variable, as in Proposition 2.3. For any \( c > 0 \) , \[ \mathbf{P}\{ \tau > \lceil n\log n + {cn}\rceil \} \leq {e}^{-c}. \] | Proof. Let \( {A}_{i} \) be the event that the \( i \) -th type does not appear among the first \( \lceil n\log n + {cn}\rceil \) coupons drawn. Observe first that \[ \mathbf{P}\{ \tau > \lceil n\log n + {cn}\rceil \} = \mathbf{P}\left( {\mathop{\bigcup }\limits_{{i = 1}}^{n}{A}_{i}}\right) \leq \mathop{\sum }\limits_{... | Yes |
Proposition 2.8. Every birth-and-death chain is reversible. | Proof. A function \( w \) on \( \mathcal{X} \) satisfies the detailed balance equations (1.29) if and only if\n\n\[ \n{p}_{k - 1}{w}_{k - 1} = {q}_{k}{w}_{k} \n\]\n\nfor \( 1 \leq k \leq n \) . For our birth-and-death chain, a solution is given by \( {w}_{0} = 1 \) and\n\n\[ \n{w}_{k} = \mathop{\prod }\limits_{{i = 1}}... | Yes |
Proposition 2.12. Let \( P \) be the transition matrix of a random walk on a finite group \( G \) and let \( U \) be the uniform probability distribution on \( G \) . Then \( U \) is a stationary distribution for \( P \) . | Proof. Let \( \mu \) be the increment distribution of the random walk. For any \( g \in G, \)\n\n\[\n\mathop{\sum }\limits_{{h \in G}}U\left( h\right) P\left( {h, g}\right) = \frac{1}{\left| G\right| }\mathop{\sum }\limits_{{k \in G}}P\left( {{k}^{-1}g, g}\right) = \frac{1}{\left| G\right| }\mathop{\sum }\limits_{{k \i... | Yes |
Proposition 2.13. Let \( \mu \) be a probability distribution on a finite group \( G \) . The random walk on \( G \) with increment distribution \( \mu \) is irreducible if and only if \( S = \{ g \in G : \mu \left( g\right) > 0\} \) generates \( G \) . | Proof. Let \( a \) be an arbitrary element of \( G \) . If the random walk is irreducible, then there exists an \( r > 0 \) such that \( {P}^{r}\left( {\mathrm{{id}}, a}\right) > 0 \) . In order for this to occur, there must be a sequence \( {s}_{1},\ldots ,{s}_{r} \in G \) such that \( a = {s}_{r}{s}_{r - 1}\ldots {s}... | Yes |
Proposition 2.14. The random walk on a finite group \( G \) with increment distribution \( \mu \) is reversible if \( \mu \) is symmetric. | Proof. Let \( U \) be the uniform probability distribution on \( G \) . For any \( g, h \in G \) , we have that\n\n\[ \nU\left( g\right) P\left( {g, h}\right) = \frac{\mu \left( {h{g}^{-1}}\right) }{\left| G\right| }\;\text{ and }\;U\left( h\right) P\left( {h, g}\right) = \frac{\mu \left( {g{h}^{-1}}\right) }{\left| G\... | Yes |
Proposition 2.16. Let \( P \) be the transition matrix of a transitive Markov chain on a finite state space \( \mathcal{X} \) . Then the uniform probability distribution on \( \mathcal{X} \) is stationary for \( P \) . | Proof. Fix \( x, y \in \mathcal{X} \) and let \( \varphi : \mathcal{X} \rightarrow \mathcal{X} \) be a transition-probability-preserving bijection for which \( \varphi \left( x\right) = y \) . Let \( U \) be the uniform probability on \( \mathcal{X} \) . Then\n\n\[ \mathop{\sum }\limits_{{z \in \mathcal{X}}}U\left( z\r... | Yes |
Proposition 4.2. Let \( \mu \) and \( \nu \) be two probability distributions on \( \mathcal{X} \) . Then\n\n\[ \parallel \mu - \nu {\parallel }_{\mathrm{{TV}}} = \frac{1}{2}\mathop{\sum }\limits_{{x \in \mathcal{X}}}\left| {\mu \left( x\right) - \nu \left( x\right) }\right| . \] | Proof. Let \( B = \{ x : \mu \left( x\right) \geq \nu \left( x\right) \} \) and let \( A \subset \mathcal{X} \) be any event. Then\n\n\[ \mu \left( A\right) - \nu \left( A\right) \leq \mu \left( {A \cap B}\right) - \nu \left( {A \cap B}\right) \leq \mu \left( B\right) - \nu \left( B\right) . \]\n\nThe first inequality ... | Yes |
Proposition 4.5. Let \( \mu \) and \( \nu \) be two probability distributions on \( \mathcal{X} \) . Then the total variation distance between them satisfies\n\n\[ \parallel \mu - \nu {\parallel }_{\mathrm{{TV}}} = \frac{1}{2}\sup \left\{ {\mathop{\sum }\limits_{{x \in \mathcal{X}}}f\left( x\right) \mu \left( x\right) ... | Proof. If \( \mathop{\max }\limits_{{x \in \mathcal{X}}}\left| {f\left( x\right) }\right| \leq 1 \), then\n\n\[ \frac{1}{2}\left| {\mathop{\sum }\limits_{{x \in \mathcal{X}}}f\left( x\right) \mu \left( x\right) - \mathop{\sum }\limits_{{x \in \mathcal{X}}}f\left( x\right) \nu \left( x\right) }\right| \leq \frac{1}{2}\m... | Yes |
Proposition 4.7. Let \( \mu \) and \( \nu \) be two probability distributions on \( \mathcal{X} \) . Then\n\n\[ \parallel \mu - \nu {\parallel }_{\mathrm{{TV}}} = \inf \{ \mathbf{P}\{ X \neq Y\} : \left( {X, Y}\right) \text{ is a coupling of }\mu \text{ and }\nu \} .\n\] | Proof. First, we note that for any coupling \( \left( {X, Y}\right) \) of \( \mu \) and \( \nu \) and any event \( A \subset \mathcal{X} \)\n\n\[ \mu \left( A\right) - \nu \left( A\right) = \mathbf{P}\{ X \in A\} - \mathbf{P}\{ Y \in A\} \]\n\n\[ \leq \mathbf{P}\{ X \in A, Y \notin A\} \]\n\n\[ \leq \mathbf{P}\{ X \neq... | Yes |
Proposition 4.15. For a reversible Markov chain,\n\n\[ \n{d}^{\left( \infty \right) }\left( {2t}\right) = {\left\lbrack {d}^{\left( 2\right) }\left( t\right) \right\rbrack }^{2} = \mathop{\max }\limits_{{x \in \mathcal{X}}}{q}_{2t}\left( {x, x}\right) - 1.\n\] | Proof. First observe that\n\n\[ \n{P}^{2t}\left( {x, y}\right) = \mathop{\sum }\limits_{{z \in \mathcal{X}}}{P}^{t}\left( {x, z}\right) {P}^{t}\left( {z, y}\right) .\n\]\n\nDividing both sides by \( \pi \left( y\right) \) and using reversibility yields\n\n\[ \n{q}_{2t}\left( {x, y}\right) = \mathop{\sum }\limits_{{z \i... | Yes |
Proposition 5.7. Let \( Q \) be an irreducible transition matrix and consider the lazy chain with transition matrix \( P = \left( {Q + I}\right) /2 \) . The distributions at time \( t \) and \( t + 1 \) satisfy\n\n\[{\begin{Vmatrix}{P}^{t}\left( x, \cdot \right) - {P}^{t + 1}\left( x, \cdot \right) \end{Vmatrix}}_{\mat... | Proof. Let \( \left( {{N}_{t},{M}_{t}}\right) \) be a coupling of the Binomial \( \left( {t,\frac{1}{2}}\right) \) distribution with the Binomial \( \left( {t + 1,\frac{1}{2}}\right) \) distribution, and let \( \left( {Z}_{t}\right) \) be a Markov chain with transition matrix \( Q \) started from \( x \) and independen... | Yes |
Proposition 6.1. Let \( \left( {X}_{t}\right) \) be the random walk on \( {\mathcal{S}}_{n} \) corresponding to the top-to-random shuffle on \( n \) cards. Given at time \( t \) that there are \( k \) cards under the original bottom card, each of the \( k \) ! possible orderings of these cards are equally likely. There... | Proof. When \( t = 0 \), there are no cards under the original bottom card, and the claim is trivially valid. Now suppose that the claim holds at time \( t \) . There are two possibilities at time \( t + 1 \) : either a card is placed under the original bottom card, or not. In the second case, the cards under the origi... | Yes |
Proposition 6.11. If \( \tau \) is a strong stationary time for starting state \( x \), then\n\n\[ \n{\begin{Vmatrix}{P}^{t}\left( x, \cdot \right) - \pi \end{Vmatrix}}_{\mathrm{{TV}}} \leq {\mathbf{P}}_{x}\{ \tau > t\} .\n\] | We break the proof into two lemmas. It will be convenient to introduce a parameter \( {s}_{x}\left( t\right) \), called separation distance and defined by\n\n\[ \n{s}_{x}\left( t\right) \mathrel{\text{:=}} \mathop{\max }\limits_{{y \in \mathcal{X}}}\left\lbrack {1 - \frac{{P}^{t}\left( {x, y}\right) }{\pi \left( y\righ... | No |
Proposition 6.14. If there exists a halting state for starting state \( x \), then \( \tau \) is an optimal strong stationary time for \( x \), i.e.\n\n\[ \n{s}_{x}\left( t\right) = {\mathbf{P}}_{x}\{ \tau > t\} \n\]\n\nand it is stochastically dominated under \( {\mathbf{P}}_{x} \) by every other strong stationary tim... | Proof. If \( y \) is a halting state for starting state \( x \) and the stopping time \( \tau \) , then inequality (6.9) is an equality for every \( t \) . Therefore, if there exists a halting state for starting state \( x \), then (6.8) is also an equality. | No |
Proposition 6.21. For every starting state \( x \), there exists a strong stationary time \( \tau \) such that, for all \( t \geq 0 \) , \[ {s}_{x}\left( t\right) = {\mathbf{P}}_{x}\{ \tau > t\} \] | Proof. Fix \( x \in \mathcal{X} \), and let \( {a}_{t} \mathrel{\text{:=}} \mathop{\min }\limits_{y}\frac{{P}^{t}\left( {x, y}\right) }{\pi \left( y\right) } = 1 - {s}_{x}\left( t\right) \) . Note that \( {a}_{t} \) is nondecreasing. (See Exercise 6.4.) If there exists a strong stationary time \( \tau \) satisfying (6.... | No |
Proposition 7.9. For \( f : \mathcal{X} \rightarrow \mathbb{R} \), define \( {\sigma }_{ \star }^{2} \mathrel{\text{:=}} \max \left\{ {{\operatorname{Var}}_{\mu }\left( f\right) ,{\operatorname{Var}}_{\nu }\left( f\right) }\right\} \) . If\n\n\[ \left| {{E}_{\nu }\left( f\right) - {E}_{\mu }\left( f\right) }\right| \ge... | Proof of Proposition 7.9. Suppose without loss of generality that \( {E}_{\mu }\left( f\right) \leq \) \( {E}_{\nu }\left( f\right) \) . If \( A = \left( {{E}_{\mu }\left( f\right) + r{\sigma }_{ \star }/2,\infty }\right) \), then Chebyshev’s inequality yields that\n\n\[ \mu {f}^{-1}\left( A\right) \leq \frac{4}{{r}^{2... | Yes |
Proposition 8.4. Let \( 0 < \varepsilon < 1 \) . For the random transposition chain on an \( n \) -card deck,\n\n\[ \n{t}_{\operatorname{mix}}\left( \varepsilon \right) \geq \frac{n - 1}{2}\log \left( {\frac{1 - \varepsilon }{6}n}\right) .\n\] | Proof. It is well known (and easily proved using indicators) that the expected number of fixed points in a uniform random permutation in \( {\mathcal{S}}_{n} \) is 1, regardless of the value of \( n \) .\n\nLet \( F\left( \sigma \right) \) denote the number of fixed points of the permutation \( \sigma \) . If \( \sigma... | Yes |
Proposition 8.5. Let \( \tau \) be the time required for the two decks to coincide. Then, no matter the initial configurations of the two decks, \( \mathbf{E}\left( \tau \right) < \frac{{\pi }^{2}}{6}{n}^{2} \) . | Proof. Decompose\n\n\[ \tau = {\tau }_{1} + \cdots + {\tau }_{n} \]\n\nwhere \( {\tau }_{i} \) is the number of transpositions between the first time that \( {a}_{t} \) is greater than or equal to \( i - 1 \) and the first time that \( {a}_{t} \) is greater than or equal to \( i \) . (Since \( {a}_{0} \) can be greater... | Yes |
Proposition 8.6. In the random transposition shuffle, let \( {R}_{t} \) and \( {L}_{t} \) be the cards chosen by the right and left hands, respectively, at time t. Assume that when \( t = 0 \), no cards have been marked. At time \( t \), mark card \( {R}_{t} \) if both of the following are true:\n\n- \( {R}_{t} \) is u... | Proof. It is clear that \( \tau \) is a stopping time. To show that it is a strong stationary time, we prove the following subclaim by induction on \( t \) . Let \( {V}_{t} \subseteq \left\lbrack n\right\rbrack \) be the set of cards marked at or before time \( t \), and let \( {U}_{t} \subseteq \left\lbrack n\right\rb... | Yes |
Proposition 8.11. Let \( \tau \) be the number of inverse riffle shuffles required for all cards to have different bitstring labels. Then \( \tau \) is a strong stationary time. | Proof. Condition on the event that \( \tau = t \) . Since the bitstrings are generated by independent fair coin flips, every possible assignment \( {}^{1} \) of strings of length \( t \) to cards is equally likely. Since the labeling bitstrings are distinct, the permutation is fully determined by the labels. Hence the ... | Yes |
For the riffle shuffle on an \( n \) -card deck, \( {t}_{\operatorname{mix}} \leq 2{\log }_{2}\left( {{4n}/3}\right) \) for sufficiently large \( n \) . | Consider inverse riffle shuffling an \( n \) -card deck and let \( \tau \) be the stopping time defined in Proposition 8.11. If \( \tau \leq t \), then different labels have been assigned to all \( n \) cards after \( t \) inverse riffle shuffles. Hence\n\n\[ \mathbf{P}\left( {\tau \leq t}\right) = \mathop{\prod }\limi... | Yes |
Proposition 8.13. Fix \( 0 < \varepsilon ,\delta < 1 \) . Consider riffle shuffling an \( n \) -card deck. For sufficiently large \( n \) , | Proof. There are at most \( {2}^{n} \) possible states accessible in one step of the time-reversed chain, since we can generate a move using \( n \) independent unbiased bits. Thus \( {\log }_{2}\Delta \leq n \), where \( \Delta \) is the maximum out-degree defined in (7.1). The state space has size \( n \) !, and Stir... | Yes |
Proposition 9.1. Let \( \left( {X}_{t}\right) \) be a Markov chain with irreducible transition matrix \( P \), let \( B \subset \mathcal{X} \), and let \( {h}_{B} : B \rightarrow \mathbb{R} \) be a function defined on \( B \) . The function \( h : \mathcal{X} \rightarrow \mathbb{R} \) defined by \( h\left( x\right) \ma... | Proof. We first show that \( h\left( x\right) = {\mathbf{E}}_{x}{h}_{B}\left( {X}_{{\tau }_{B}}\right) \) is a harmonic extension of \( {h}_{B} \) . Clearly \( h\left( x\right) = {h}_{B}\left( x\right) \) for all \( x \in B \) . Suppose that \( x \in \mathcal{X} \smallsetminus B \) . Then\n\n\[ h\left( x\right) = {\mat... | Yes |
Proposition 9.4 (Node law/cycle law/strength). If \( \theta \) is a flow from a to \( z \) satisfying the cycle law\n\n\[ \mathop{\sum }\limits_{{i = 1}}^{m}r\left( \overrightarrow{{e}_{i}}\right) \theta \left( \overrightarrow{{e}_{i}}\right) = 0 \]\n\n\( \left( {9.10}\right) \)\n\nfor any cycle \( \overrightarrow{{e}_... | Proof. The function \( f = \theta - I \) satisfies the node law at all nodes and the cycle law. Suppose \( f\left( \overrightarrow{{e}_{1}}\right) > 0 \) for some oriented edge \( \overrightarrow{{e}_{1}} \) . By the node law, \( {e}_{1} \) must lead to some oriented edge \( \overrightarrow{{e}_{2}} \) with \( f\left( ... | Yes |
Proposition 9.5. For any \( a, z \in \mathcal{X} \) with \( a \neq z \) , \[ {\mathbf{P}}_{a}\left\{ {{\tau }_{z} < {\tau }_{a}^{ + }}\right\} = \frac{1}{c\left( a\right) \mathcal{R}\left( {a \leftrightarrow z}\right) } = \frac{\mathcal{C}\left( {a \leftrightarrow z}\right) }{c\left( a\right) }.\] | Proof. Applying Proposition 9.1 to \( B = \{ a, z\} \) and \( {h}_{B} = {\mathbf{1}}_{\{ z\} } \) yields that \[ x \mapsto {\mathbf{E}}_{x}{h}_{B}\left( {X}_{{\tau }_{B}}\right) = {\mathbf{P}}_{x}\left\{ {{\tau }_{z} < {\tau }_{a}}\right\} \] is the unique harmonic function on \( \mathcal{X} \smallsetminus \{ a, z\} \)... | Yes |
Proposition 9.16. If \( \left\{ {\Pi }_{k}\right\} \) are disjoint edge-cutsets which separate nodes a and \( z \), then\n\n\[ \mathcal{R}\left( {a \leftrightarrow z}\right) \geq \mathop{\sum }\limits_{k}{\left( \mathop{\sum }\limits_{{e \in {\Pi }_{k}}}c\left( e\right) \right) }^{-1}. \]\n\n\( \left( {9.24}\right) \)\... | Proof. Let \( \theta \) be a unit flow from \( a \) to \( z \) . For any \( k \), by the Cauchy-Schwarz inequality\n\n\[ \mathop{\sum }\limits_{{e \in {\Pi }_{k}}}c\left( e\right) \cdot \mathop{\sum }\limits_{{e \in {\Pi }_{k}}}r\left( e\right) \theta {\left( e\right) }^{2} \geq {\left( \mathop{\sum }\limits_{{e \in {\... | Yes |
Proposition 9.17. Let \( a = \left( {1,1}\right) \) be the lower left-hand corner of \( {B}_{n} \), and let \( z = \left( {n, n}\right) \) be the upper right-hand corner of \( {B}_{n} \) . Suppose each edge of \( {B}_{n} \) has unit conductance. The effective resistance \( \mathcal{R}\left( {a \leftrightarrow z}\right)... | We separate the proof into the lower and upper bounds.\n\nProof of lower bound in (9.25). Let \( {\Pi }_{k} \) be the edge set\n\n\[ {\Pi }_{k} = \left\{ {\{ v, w\} \in E\left( {B}_{n}\right) : \parallel v{\parallel }_{\infty } = k,\parallel w{\parallel }_{\infty } = k + 1}\right\} ,\]\n\nwhere \( {\begin{Vmatrix}\left... | Yes |
Proposition 10.7 (Commute Time Identity). Let \( \\left( {G,\\{ c\\left( e\\right) \\} }\\right) \) be a network, and let \( \\left( {X}_{t}\\right) \) be the random walk on this network. For any nodes \( a \) and \( b \) in \( V \) , | Proof. By (10.13), \[ \\frac{{G}_{{\\tau }_{a, b}}\\left( {a, a}\\right) }{{\\mathbf{E}}_{a}\\left( {\\tau }_{a, b}\\right) } = \\pi \\left( a\\right) = \\frac{c\\left( a\\right) }{{c}_{G}}. \] By definition, after visiting \( b \), the chain does not visit \( a \) until time \( {\\tau }_{a, b} \), so \( {G}_{{\\tau }_... | No |
Proposition 10.10. For a random walk on a transitive connected network \( \langle G,\{ c\left( e\right) \} \rangle \), for any vertices \( a, b \in V \) ,\n\n\[ \n{\mathbf{E}}_{a}\left( {\tau }_{b}\right) = {\mathbf{E}}_{b}\left( {\tau }_{a}\right) \n\] | Proof. Suppose \( \xi \) and \( \eta \) are finite strings with letters in \( V \), that is, \( \xi \in {V}^{m} \) and \( \eta \in {V}^{n} \) . We say that \( \xi \preccurlyeq \eta \) if and only if \( \xi \) is a subsequence of \( \eta \) .\n\nLet \( {\tau }_{ab} \) be the time required to first visit \( a \) and then... | Yes |
Proposition 10.20. Let \( G = \left( {V, E}\right) \) be an Eulerian directed graph. Let \( m = \) \( \left| E\right| \), and assume that there exists a directed path of length \( \ell \) from vertex \( x \) to vertex y. Then\n\n\[{\mathbf{E}}_{x}\left( {\tau }_{y}\right) + {\mathbf{E}}_{y}\left( {\tau }_{x}\right) \le... | Proof. It is enough to prove this for the case where there is a directed edge \( \left( {x, y}\right) \), since otherwise \( {\mathbf{E}}_{x}\left( {\tau }_{y}\right) \) is bounded by the sum of the expected hitting times along the path from \( x \) to \( y \), and similarly for \( {\mathbf{E}}_{y}\left( {\tau }_{x}\ri... | Yes |
Proposition 10.21. Consider the simple random walk on the torus \( {\mathbb{Z}}_{n}^{d} \) . There exist constants \( 0 < {c}_{d} \leq {C}_{d} < \infty \) such that if \( x \) and \( y \) are at distance \( k \geq 1 \), then\n\n\[ \n{c}_{d}{n}^{d} \leq {\mathbf{E}}_{x}\left( {\tau }_{y}\right) \leq {C}_{d}{n}^{d}\;\tex... | Proof of Proposition 10.21. First, the lower bounds. For \( j \geq 0 \), let \( {\Pi }_{j} \) be the edge-boundary of the cube of side-length \( {2j} \) centered at \( x \), i.e., the set of edges connecting the cube to its complement. For \( 1 \leq j \leq k/d \), the edges in \( {\Pi }_{j} \) form an edge-cutset separ... | Yes |
Proposition 10.25. Let \( P \) be the transition matrix for a finite reversible chain on state space \( \mathcal{X} \) with stationary distribution \( \pi \) .\n\n(i) For all \( t \geq 0 \) and \( x \in \mathcal{X} \) we have \( {P}^{{2t} + 2}\left( {x, x}\right) \leq {P}^{2t}\left( {x, x}\right) \) . | Proof. (i) Since \( {P}^{{2t} + 2}\left( {x, x}\right) = \mathop{\sum }\limits_{{y, z \in \mathcal{X}}}{P}^{t}\left( {x, y}\right) {P}^{2}\left( {y, z}\right) {P}^{t}\left( {z, x}\right) \), we have\n\n\[ \pi \left( x\right) {P}^{{2t} + 2}\left( {x, x}\right) = \mathop{\sum }\limits_{{y, z \in \mathcal{X}}}{P}^{t}\left... | Yes |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.