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Example 1.25 (Symmetric reflecting random walk on the line). The state space is \( \{ 0,1,2\ldots, L\} \) . The chain goes to the right or left at each step with probability \( 1/2 \) , subject to the rules that if it tries to go to the left from 0 or to the right from \( L \) it stays put. For example, when \( L = 4 \... | It is clear in the example \( L = 4 \) that each column adds up to 1 . With a little thought one sees that this is true for any \( L \), so the stationary distribution is uniform, \( \pi \left( i\right) = \) \( 1/\left( {L + 1}\right) \) . | Yes |
Example 1.26 (Tiny Board Game). Consider a circular board game with only six spaces \( \{ 0,1,2,3,4,5\} \) . On each turn we roll a die with 1 on three sides,2 on two sides, and 3 on one side to decide how far to move. Here we consider 5 to be adjacent to 0, so if we are there and we roll a 2 then the result is \( 5 + ... | \[ \begin{matrix} & \mathbf{0} & \mathbf{1} & \mathbf{2} & \mathbf{3} & \mathbf{4} & \mathbf{5} & \\ \mathbf{0} & 0 & 1/3 & 1/3 & 1/6 & 0 & 0 & \\ \mathbf{1} & 0 & 0 & 1/2 & 1/3 & 1/6 & 0 & \\ \mathbf{2} & 0 & 0 & 0 & 1/2 & 1/3 & 1/6 & \\ \mathbf{3} & 1/6 & 0 & 0 & 0 & 1/2 & 1/3 & \\ \mathbf{4} & 1/3 & 1/6 & 0 & 0 & 0 ... | Yes |
Example 1.27 (Mathematician's Monopoly). The game Monopoly is played on a game board that has 40 spaces arranged around the outside of a square. The squares have names like Reading Railroad and Park Place but we will number the squares 0 (Go), 1 (Baltic Avenue), ... 39 (Boardwalk). In Monopoly you roll two dice and mov... | This example is larger but has the same structure as the previous example. Each row has the same entries but shift 1 unit to the right each time with the number that goes off the right edge emerging in the 0 column. This structure implies that each entry in the row appears once in each column and hence the sum of the e... | Yes |
Consider\n\n\[ \n\\begin{matrix} & 1 & 2 & 3 \\ 1 & {0.5} & {0.5} & 0 \\ 2 & {0.3} & {0.1} & {0.6} \\ 3 & {0.2} & {0.4} & {0.4} \\end{matrix} \n\]\n\nThere is no stationary distribution with detailed balance since \( \\pi \\left( 1\\right) p\\left( {1,3}\\right) = 0 \) but \( p\\left( {1,3}\\right) > 0 \) so we would h... | This chain is doubly stochastic so \( \\left( {1/3,1/3,1/3}\\right) \) is a stationary distribution. | Yes |
Birth and death chains are defined by the property that the state space is some sequence of integers \( \ell ,\ell + 1,\ldots r - 1, r \) and it is impossible to jump by more than one:\n\n\[ p\left( {x, y}\right) = 0\;\text{ when }\left| {x - y}\right| > 1 \]\n\nSuppose that the transition probability has\n\n\[ p\left(... | \[ \pi \left( {x + 1}\right) = \frac{{p}_{x}}{{q}_{x + 1}} \cdot \pi \left( x\right) \] | Yes |
Example 1.31 (Ehrenfest chain). For concreteness, suppose there are three balls. In this case the transition probability is\n\n\[ \n\\begin{matrix} & 0 & 1 & 2 & 3 \\ 0 & 0 & 3/3 & 0 & 0 \\ 1 & 1/3 & 0 & 2/3 & 0 \\ 2 & 0 & 2/3 & 0 & 1/3 \\ 3 & 0 & 0 & 3/3 & 0 \\end{matrix} \n\] | Setting \( \pi \left( 0\right) = c \) and using (1.12) we have\n\n\[ \n\pi \left( 1\right) = {3c},\;\pi \left( 2\right) = \pi \left( 1\right) = {3c}\;\pi \left( 3\right) = \pi \left( 2\right) /3 = c.\n\]\n\nThe sum of the \( \pi \) ’s is \( {8c} \), so we pick \( c = 1/8 \) to get\n\n\[ \n\pi \left( 0\right) = 1/8,\;\p... | Yes |
Example 1.32 (Three machines, one repairman). Suppose that an office has three machines that each break with probability 0.1 each day, but when there is at least one broken, then with probability 0.5 the repairman can fix one of them for use the next day. If we ignore the possibility of two machines breaking on the sam... | Rows 0 and 3 are easy to see. To explain row 1, we note that the state will only decrease by 1 if one machine breaks and the repairman fails to repair the one he is working on, an event of probability (0.1)(0.5), while the state can only increase by 1 if he succeeds and there is no new failure, an event of probability ... | Yes |
Example 1.33 (Random walks on graphs). A graph is described by giving two things: (i) a set of vertices \( V \) (which we suppose is a finite set) and (ii) an adjacency matrix \( A\left( {u, v}\right) \), which is 1 if there is an edge connecting \( u \) and \( v \) and 0 otherwise. By convention we set \( A\left( {v, ... | The degree of a vertex \( u \) is equal to the number of neighbors it has. In symbols, \[ d\left( u\right) = \mathop{\sum }\limits_{v}A\left( {u, v}\right) \] since each neighbor of \( u \) contributes 1 to the sum. To help explain the concept, we have indicated the degrees on our example. We write the degree this way ... | Yes |
Example 1.34 (Random walk of a knight on a chess board). A chess board is an 8 by 8 grid of squares. A knight moves by walking two steps in one direction and then one step in a perpendicular direction. | By patiently examining all of the possibilities, one sees that the degrees of the vertices are given by the following table. Lines have been drawn to make the symmetries more apparent.\n\n\n\nThe sum of the degrees is ... | No |
Theorem 1.25. Fix \( n \) and let \( {Y}_{m} = {X}_{n - m} \) for \( 0 \leq m \leq n \) . Then \( {Y}_{m} \) is a Markov chain with transition probability\n\n\[ \n\widehat{p}\left( {i, j}\right) = P\left( {{Y}_{m + 1} = j \mid {Y}_{m} = i}\right) = \frac{\pi \left( j\right) p\left( {j, i}\right) }{\pi \left( i\right) }... | Proof. We need to calculate the conditional probability.\n\n\[ \nP\left( {{Y}_{m + 1} = {i}_{m + 1} \mid {Y}_{m} = {i}_{m},{Y}_{m - 1} = {i}_{m - 1}\ldots {Y}_{0} = {i}_{0}}\right) \n\]\n\n\[ \n= \frac{P\left( {{X}_{n - \left( {m + 1}\right) } = {i}_{m + 1},{X}_{n - m} = {i}_{m},{X}_{n - m + 1} = {i}_{m - 1}\ldots {X}_... | Yes |
Example 1.35 (Geometric distribution). Suppose \( \pi \left( x\right) = {\theta }^{x}\left( {1 - \theta }\right) \) for \( x = \) \( 0,1,2,\ldots \) . To generate the jumps we will use a symmetric random walk \( q(x, x + \) 1) \( = q\left( {x, x - 1}\right) = 1/2 \) . Since \( q \) is symmetric \( r\left( {x, y}\right)... | \[
p\left( {x, x - 1}\right) = 1/2\;p\left( {x, x + 1}\right) = \theta /2\;p\left( {x, x}\right) = \left( {1 - \theta }\right) /2.
\]
When \( x = 0,\pi \left( {-1}\right) = 0 \) so
\[
p\left( {0, - 1}\right) = 0\;p\left( {0,1}\right) = \theta /2\;p\left( {0,0}\right) = 1 - \left( {\theta /2}\right) .
\]
To check rev... | Yes |
Example 1.36 (Binomial distribution). Suppose \( \pi \left( x\right) \) is Binomial \( \left( {N,\theta }\right) \) . In this case we can let \( q\left( {x, y}\right) = 1/\left( {N + 1}\right) \) for all \( 0 \leq x, y \leq N \) . Since \( q \) is symmetric \( r\left( {x, y}\right) = \min \{ 1,\pi \left( y\right) /\pi ... | This is closely related to the method of rejection sampling, in which one generates independent random variables \( {U}_{i} \) uniform on \( \{ 0,1,\ldots, N\} \) and keep \( {U}_{i} \) with probability \( \pi \left( {U}_{i}\right) /{\pi }^{ * } \) where \( {\pi }^{ * } = \mathop{\max }\limits_{{0 \leq x \leq n}}\pi \l... | No |
Lemma 1.26. If there is a stationary distribution, then all states \( y \) that have \( \pi \left( y\right) > 0 \) are recurrent. | Proof. Lemma 1.12 tells us that \( {E}_{x}N\left( y\right) = \mathop{\sum }\limits_{{n = 1}}^{\infty }{p}^{n}\left( {x, y}\right) \), so\n\n\[ \mathop{\sum }\limits_{x}\pi \left( x\right) {E}_{x}N\left( y\right) = \mathop{\sum }\limits_{x}\pi \left( x\right) \mathop{\sum }\limits_{{n = 1}}^{\infty }{p}^{n}\left( {x, y}... | Yes |
Theorem 1.20. Suppose \( p \) is irreducible and recurrent. Let \( x \in S \) and let \( {T}_{x} = \) \( \inf \left\{ {n \geq 1 : {X}_{n} = x}\right\} \) . \[ {\mu }_{x}\left( y\right) = \mathop{\sum }\limits_{{n = 0}}^{\infty }{P}_{x}\left( {{X}_{n} = y,{T}_{x} > n}\right) \] defines a stationary measure with \( 0 < {... | Proof. To formalize this intuition, let \( {\bar{p}}_{n}\left( {x, y}\right) = {P}_{x}\left( {{X}_{n} = y,{T}_{x} > n}\right) \) and interchange sums to get \[ \mathop{\sum }\limits_{y}{\mu }_{x}\left( y\right) p\left( {y, z}\right) = \mathop{\sum }\limits_{{n = 0}}^{\infty }\mathop{\sum }\limits_{y}{\bar{p}}_{n}\left(... | Yes |
Theorem 1.21. Suppose \( p \) is irreducible and recurrent. Let \( {N}_{n}\left( y\right) \) be the number of visits to \( y \) at times \( \leq n \) . As \( n \rightarrow \infty \)\n\n\[ \frac{{N}_{n}\left( y\right) }{n} \rightarrow \frac{1}{{E}_{y}{T}_{y}} \] | Proof. We have already shown (1.15). To turn this into the desired result, we note that from the definition of \( R\left( k\right) \) it follows that \( R\left( {{N}_{n}\left( y\right) }\right) \leq n < R\left( {{N}_{n}\left( y\right) + 1}\right) \) . Dividing everything by \( {N}_{n}\left( y\right) \) and then multipl... | Yes |
Theorem 1.22. If \( p \) is an irreducible and has stationary distribution \( \pi \), then\n\n\[ \pi \left( y\right) = 1/{E}_{y}{T}_{y} \] | Proof. Suppose \( {X}_{0} \) has distribution \( \pi \) . From Theorem 1.21 it follows that\n\n\[ \frac{{N}_{n}\left( y\right) }{n} \rightarrow \frac{1}{{E}_{y}{T}_{y}} \]\n\nTaking expected value and using the fact that \( {N}_{n}\left( y\right) \leq n \), it can be shown that this implies\n\n\[ \frac{{E}_{\pi }{N}_{n... | Yes |
Theorem 1.23. Suppose \( p \) is irreducible, has stationary distribution \( \pi \), and \( \mathop{\sum }\limits_{x}\left| {f\left( x\right) }\right| \pi \left( x\right) < \infty \) then\n\n\[ \frac{1}{n}\mathop{\sum }\limits_{{m = 1}}^{n}f\left( {X}_{m}\right) \rightarrow \mathop{\sum }\limits_{x}f\left( x\right) \pi... | Sketch of proof. Suppose that the chain starts at \( x \) . Let \( {T}_{0} = 0 \) and \( {T}_{k} = \min \{ n > \) \( \left. {{T}_{k - 1} : {X}_{n} = x}\right\} \) be the time of the \( k \) th return to \( x \) . By the strong Markov property, the random variables\n\n\[ {Y}_{k} = \mathop{\sum }\limits_{{m = {T}_{k - 1}... | Yes |
At a local 2 year college, \( {60}\% \) of freshmen become sophomores, \( {25}\% \) remain freshmen, and \( {15}\% \) drop out. seventy percent of sophomores graduate and transfer to a 4 year college, twenty percent remain sophomores and ten percent drop out. What fraction of new students eventually graduate? | We use a Markov chain with state space \( 1 = \) freshman, \( 2 = \) sophomore, \( \mathrm{G} = \) graduate, \( \mathrm{D} = \) dropout. The transition probability is\n\n\[ \begin{matrix} & \mathbf{1} & \mathbf{2} & \mathbf{G} & \mathbf{D} \\ \mathbf{1} & {0.25} & {0.6} & 0 & {0.15} \\ \mathbf{2} & 0 & {0.2} & {0.7} & ... | Yes |
In tennis the winner of a game is the first player to win four points, unless the score is 4-3, in which case the game must continue until one player is ahead by two points and wins the game. Suppose that the server win the point with probability 0.6 and successive points are independent. What is the probability the se... | We formulate the game as a Markov chain in which the state is the difference of the scores. The state space is \( 2,1,0, - 1, - 2 \) with 2 (win for server) and -2 (win for opponent). The transition probability is\n\n<table><tr><td></td><td>2</td><td>1</td><td>0</td><td>\( - 1 \)</td><td>\( - 2 \)</td></tr><tr><td>2</t... | Yes |
Theorem 1.27. Consider a Markov chain with finite state space \( S \) . Let \( a \) and \( b \) be two points in \( S \), and let \( C = S - \{ a, b\} \) . Suppose \( h\left( a\right) = 1, h\left( b\right) = 0 \), and that for \( x \in C \) we have\n\n\[ h\left( x\right) = \mathop{\sum }\limits_{y}p\left( {x, y}\right)... | Proof. Let \( T = {V}_{a} \land {V}_{b} \) . It follows from Lemma 1.3 that \( {P}_{x}\left( {T < \infty }\right) = 1 \) for all \( x \in C \) . (1.16) implies that \( h\left( x\right) = {E}_{x}h\left( {X}_{1}\right) \) when \( x \neq a, b \) . The Markov property implies\n\n\[ h\left( x\right) = {E}_{x}h\left( {X}_{T ... | Yes |
Example 1.41 (Matching pennies). Bob, who has 15 pennies, and Charlie, who has 10 pennies, decide to play a game. They each flip a coin. If the two coins match, Bob gets the 2 pennies (for a profit of 1). If the two coins are different, then Charlie gets the 2 pennies. They quit when someone has all of the pennies. Wha... | The answer will turn out to be \( {15}/{25} \), Bob’s fraction of the total supply of pennies. To explain this, let \( {X}_{n} \) be the number of pennies Bob has after \( n \) plays. \( {X}_{n} \) is a fair game, i.e., \( x = {E}_{x}{X}_{1} \), or in words the expected number of pennies Bob has is constant in time. Le... | Yes |
Example 1.42 (Wright-Fisher model with no mutation). The state space is \( S = \) \( \{ 0,1,\ldots N\} \) and the transition probability is\n\n\[ p\left( {x, y}\right) = \frac{N}{y}{\left( \frac{x}{N}\right) }^{y}{\left( \frac{N - x}{N}\right) }^{N - y} \]\n\nThe right-hand side is the binomial \( \left( {N, x/N}\right... | Taking \( a = N \) and \( b = 0 \), we have \( h\left( a\right) = 1 \) and \( h\left( b\right) = 0 \) . Since \( {P}_{x}\left( {{V}_{a} \land {V}_{b} < }\right. \) \( \infty ) > 0 \) for all \( 0 < x < N \), it follows from Lemma 1.27 that\n\n\[ {P}_{x}\left( {{V}_{N} < {V}_{0}}\right) = x/N \]\n\n(1.18)\n\ni.e., the p... | Yes |
Consider a gambling game in which on any turn you win \\( \\$ 1 \\) with probability \\( p \\neq 1/2 \\) or lose \\( \\$ 1 \\) with probability \\( 1 - p \\) . Suppose further that you will quit playing if your fortune reaches \\( \\$ N \\) . Of course, if your fortune reaches \\( \\$ 0 \\), then the casino makes you s... | To solve this we rearrange to get \\( p\\left( {h\\left( {x + 1}\\right) - h\\left( x\\right) }\\right) = q\\left( {h\\left( x\\right) - h\\left( {x - 1}\\right) }\\right) \\) and conclude\n\n\\[ \nh\\left( {x + 1}\\right) - h\\left( x\\right) = \\frac{q}{p} \\cdot \\left( {h\\left( x\\right) - h\\left( {x - 1}\\right)... | Yes |
If we bet \$ 1 on red on a roulette wheel with \( {18}\mathrm{{red}},{18} \) black, and 2 green (0 and 00) holes, we win \$1 with probability \( {18}/{38} = {0.4737} \) and lose \( \$ 1 \) with probability \( {20}/{38} \). Suppose we bring \( \$ {50} \) to the casino with the hope of reaching \( \$ {100} \) before goin... | Here \( \theta = q/p = {20}/{18} \), so (1.22) implies\n\n\[ \n{P}_{50}\left( {{V}_{100} < {V}_{0}}\right) = \frac{{\left( \frac{20}{18}\right) }^{50} - 1}{{\left( \frac{20}{18}\right) }^{100} - 1} \]\n\nUsing \( {\left( {20}/{18}\right) }^{50} = {194} \), we have\n\n\[ \n{P}_{50}\left( {{V}_{100} < {V}_{0}}\right) = \... | Yes |
On the average how many years does a student take to graduate or drop out? | Let \( g\left( x\right) \) be the expected time for a student starting in state \( x.g\left( G\right) = \) \( g\left( D\right) = 0 \) . By considering what happens on one step\n\n\[ g\left( 1\right) = 1 + {0.25g}\left( 1\right) + {0.6g}\left( 2\right) \]\n\n\[ g\left( 2\right) = 1 + {0.2g}\left( 2\right) \]\n\nwhere th... | Yes |
Let \( g\left( x\right) \) be the expected time to complete the game when the current state is \( x \) . By considering what happens on one step | \[ g\left( x\right) = 1 + \mathop{\sum }\limits_{y}p\left( {x, y}\right) g\left( y\right) \] Since \( g\left( 2\right) = g\left( {-2}\right) = 0 \), if we let \( r\left( {x, y}\right) \) be the restriction of the transition probability to \( 1,0, - 1 \) we have \[ g\left( x\right) - \mathop{\sum }\limits_{y}r\left( {x,... | Yes |
Theorem 1.28. Consider a Markov chain with finite state space \( S \) . Let \( A \subset S \) and \( {V}_{A} = \inf \left\{ {n \geq 0 : {X}_{n} \in A}\right\} \) . We suppose \( C = S - A \) is finite, and that \( {P}_{x}\left( {{V}_{A} < \infty }\right) > 0 \) for any \( x \in C \) . Suppose \( g\left( a\right) = 0 \)... | Proof. It follows from Lemma 1.3 that \( {E}_{x}{V}_{A} < \infty \) for all \( x \in C \) . (1.25) implies that \( g\left( x\right) = 1 + {E}_{x}g\left( {X}_{1}\right) \) when \( x \notin A \) . The Markov property implies\n\n\[ g\left( x\right) = {E}_{x}\left( {T \land n}\right) + {E}_{x}g\left( {{X}_{T \land n}n}\rig... | Yes |
Let \( {T}_{TT} \) be the (random) number of times we need to flip a coin before we have gotten Tails on two consecutive tosses. To compute the expected value of \( {T}_{TT} \) we will introduce a Markov chain with states \( 0,1,2 = \) the number of Tails we have in a row. | Since getting a Tails increases the number of Tails we have in a row by 1 , but getting a Heads sets the number of Tails we have in a row to 0 , the transition matrix is\n\n\[ \begin{matrix} & 0 & 1 & 2 \\ 0 & 1/2 & 1/2 & 0 \\ 1 & 1/2 & 0 & 1/2 \\ 2 & 0 & 0 & 1 \end{matrix} \]\n\nSince we are not interested in what hap... | Yes |
Let \( {T}_{HT} \) be the (random) number of times we need to flip a coin before we have gotten a Heads followed by a Tails. Consider \( {X}_{n} \) is Markov chain with transition probability: | To compute the expected waiting time for our original problem, we note that after the first two tosses we have each of the four possibilities with probability \( 1/4 \) so\n\n\[ E{T}_{HT} = 2 + \frac{1}{4}\left( {0 + 2 + 2 + 4}\right) = 4 \] | No |
Consider the gambler's ruin chain in which \( p\left( {i, i + 1}\right) = p\left( {i, i - 1}\right) = 1/2 \) . Let \( \tau = \min \left\{ {n : {X}_{n} \notin \left( {0, N}\right) }\right\} \) . We claim that\n\n\[ \n{E}_{x}\tau = x\left( {N - x}\right) \n\] | Verify the guess. Let \( g\left( x\right) = x\left( {N - x}\right) \) . Clearly, \( g\left( 0\right) = g\left( N\right) = 0 \) . If \( 0 < x < N \) then by considering what happens on the first step we have\n\n\[ \ng\left( x\right) = 1 + \frac{1}{2}g\left( {x + 1}\right) + \frac{1}{2}g\left( {x - 1}\right) \n\]\n\nIf \... | Yes |
Consider the gambler's ruin chain in which \( p\left( {i, i + 1}\right) p \) and \( p\left( {i, i - 1}\right) = q \), where \( p \neq q \) . Let \( \tau = \min \left\{ {n : {X}_{n} \notin }\right. \) \( \left( {0, N}\right) \} \) . We claim that\n\n\[ \n{E}_{x}\tau = \frac{x}{q - p} - \frac{N}{q - p} \cdot \frac{1 - {\... | This time the derivation is somewhat tedious, so we will just verify the guess. We want to show that \( g\left( x\right) = 1 + {pg}\left( {x + 1}\right) + {qg}\left( {x - 1}\right) \) . Plugging the formula into the right-hand side:\n\n\[ \n= 1 + p\frac{x + 1}{q - p} + q\frac{x - 1}{q - p} - \frac{N}{q - p}\left\lbrack... | No |
Imagine a particle that moves on \( \{ 0,1,2,\ldots \} \) according to the following rules. It takes a step to the right with probability \( p \) . It attempts to take a step to the left with probability \( 1 - p \), but if it is at 0 and tries to jump to the left, it stays at 0 , since there is no -1 to jump to. In sy... | The reflecting random walk is clearly irreducible. To check that it is aperiodic note that \( p\left( {0,0}\right) > 0 \) implies 0 has period 1, and then Lemma 1.18 implies that all states have period 1. Using the convergence theorem, Theorem 1.19, now we see that\n\n1. When \( p < 1/2, P\left( {{X}_{n} = j}\right) \r... | Yes |
Theorem 1.29. For an irreducible chain the following are equivalent:\n\n(i) Some state is positive recurrent.\n\n(ii) There is a stationary distribution \( \pi \) .\n\n(iii) All states are positive recurrent. | Proof. The stationary measure constructed in Theorem 1.20 has total mass\n\n\[ \mathop{\sum }\limits_{y}\mu \left( y\right) = \mathop{\sum }\limits_{{n = 0}}^{\infty }\mathop{\sum }\limits_{y}{P}_{x}\left( {{X}_{n} = y,{T}_{x} > n}\right) \]\n\n\[ = \mathop{\sum }\limits_{{n = 0}}^{\infty }{P}_{x}\left( {{T}_{x} > n}\r... | Yes |
What is the probability the species avoids extinction? | Here \ | No |
Lemma 1.30. The extinction probability \( \rho \) is the smallest solution of the equation \( \phi \left( x\right) = x \) with \( 0 \leq x \leq 1 \) (Fig. 1.4). | Proof. Extending the reasoning for (1.33) we see that in order for the process to hit 0 by time \( n \), all of the processes started by first-generation individuals must hit 0 by time \( n - 1 \), so\n\n\[ P\left( {{X}_{n} = 0}\right) = \mathop{\sum }\limits_{{k = 0}}^{\infty }{p}_{k}P{\left( {X}_{n - 1} = 0\right) }^... | Yes |
2. If \( \mu > 1 \), then there is positive probability of avoiding extinction. | Proof. In view of Lemma 1.30, we only have to show there is a root \( < 1 \) . We begin by discarding a trivial case. If \( {p}_{0} = 0 \), then \( \phi \left( 0\right) = 0,0 \) is the smallest root, and there is no probability of dying out. If \( {p}_{0} > 0 \), then \( \phi \left( 0\right) = {p}_{0} > 0 \) . Differen... | Yes |
Theorem 1.7. If \( C \) is a finite closed and irreducible set, then all states in \( C \) are recurrent. | The keys to the proof of Theorem 1.7 are: (i) If \( x \) is recurrent and \( {\rho }_{xy} > 0 \) then \( y \) is recurrent, and (ii) In a finite closed set there has to be at least one recurrent state. To prove these results, it was useful to know that if \( N\left( y\right) \) is the number of visits to \( y \) at tim... | No |
Theorem 1.20. Suppose \( p \) is irreducible and recurrent. Let \( x \in S \) and let \( {T}_{x} = \) \( \inf \left\{ {n \geq 1 : {X}_{n} = x}\right\} \) . \[ {\mu }_{x}\left( y\right) = \mathop{\sum }\limits_{{n = 0}}^{\infty }{P}_{x}\left( {{X}_{n} = y,{T}_{x} > n}\right) \] defines a stationary measure with \( 0 < {... | If the state space \( S \) is finite and irreducible there is a unique stationary distribution. More generally if \( {E}_{x}{T}_{x} < \infty \), i.e., \( x \) is positive recurrent then \( {\mu }_{x}\left( y\right) /{E}_{x}{T}_{x} \) is a stationary distribution. Since \( {\mu }_{x}\left( x\right) = 1 \) we see that \[... | Yes |
Theorem 1.27. Consider a Markov chain with finite state space \( S \) . Let \( a \) and \( b \) be two points in \( S \), and let \( C = S - \{ a, b\} \) . Suppose \( h\left( a\right) = 1, h\left( b\right) = 0 \), and that for \( x \in C \) we have\n\n\[ h\left( x\right) = \mathop{\sum }\limits_{y}p\left( {x, y}\right)... | Let \( r\left( {x, y}\right) \) be the part of the matrix \( p\left( {x, y}\right) \) with \( x, y \in C \) . Since \( h\left( a\right) = 1 \) and \( h\left( b\right) = 0 \), the equation for \( h \) can be written for \( x \in C \) as\n\n\[ h\left( x\right) = r\left( {x, a}\right) + \mathop{\sum }\limits_{y}r\left( {x... | Yes |
Theorem 1.28. Consider a Markov chain with finite state space \( S \) . Let \( A \subset S \) and \( {V}_{A} = \inf \left\{ {n \geq 0 : {X}_{n} \in A}\right\} \) . Suppose \( g\left( a\right) = 0 \) for all \( a \in A \), and that for \( x \in C = S - A \) we have\n\n\[ g\left( x\right) = 1 + \mathop{\sum }\limits_{y}p... | Since \( g\left( x\right) = 0 \) for \( x \in A \) the equation for \( g \) can be written for \( x \in C \) as\n\n\[ g\left( x\right) = 1 + \mathop{\sum }\limits_{y}r\left( {x, y}\right) g\left( y\right) \]\n\nso if we let 1 be a column vector consisting of all 1 's then the last equation says \( \left( {I - r}\right)... | Yes |
Theorem 2.1. Let \( {\tau }_{1},{\tau }_{2},\ldots \) be independent exponential \( \left( \lambda \right) \) . The sum \( {T}_{n} = {\tau }_{1} + \) \( \cdots + {\tau }_{n} \) has a gamma \( \left( {n,\lambda }\right) \) distribution. That is, the density function of \( {T}_{n} \) is given by\n\n\[ \n{f}_{{T}_{n}}\lef... | Proof. The proof is by induction on \( n \) . When \( n = 1,{T}_{1} \) has an exponential \( \left( \lambda \right) \) distribution. Recalling that the 0th power of any positive number is 1 , and by convention we set \( 0! = 1 \), the formula reduces to\n\n\[ \n{f}_{{T}_{1}}\left( t\right) = \lambda {e}^{-{\lambda t}} ... | Yes |
Lemma 2.2. \( N\left( s\right) \) has a Poisson distribution with mean \( {\lambda s} \) . | Proof. Now \( N\left( s\right) = n \) if and only if \( {T}_{n} \leq s < {T}_{n + 1} \) ; i.e., the \( n \) th customer arrives before time \( s \) but the \( \left( {n + 1}\right) \) th after \( s \) . Breaking things down according to the value of \( {T}_{n} = t \) and noting that for \( {T}_{n + 1} > s \), we must h... | Yes |
Theorem 2.3. For any \( k \geq 1 \)\n\n\[ \n{EX}\left( {X - 1}\right) \cdots \left( {X - k + 1}\right) = {\lambda }^{k} \n\]\n\n(2.13)\n\nand hence \( \operatorname{var}\left( X\right) = \lambda \) | Proof. \( X\left( {X - 1}\right) \cdots \left( {X - k + 1}\right) = 0 \) if \( X \leq k - 1 \) so\n\n\[ \n{EX}\left( {X - 1}\right) \cdots \left( {X - k + 1}\right) = \mathop{\sum }\limits_{{j = k}}^{\infty }{e}^{-\lambda }\frac{{\lambda }^{j}}{j!}j\left( {j - 1}\right) \cdots \left( {j - k + 1}\right) \n\]\n\n\[ \n= {... | Yes |
Theorem 2.4. If \( {X}_{i} \) are independent Poissson \( \left( {\lambda }_{i}\right) \) then\n\n\[ {X}_{1} + \cdots + {X}_{k} = \operatorname{Poisson}\left( {{\lambda }_{1} + \cdots + {\lambda }_{n}}\right) . \] | Proof. It suffices to prove the result for \( k = 2 \), for then the general result follows by induction.\n\n\[ P\left( {{X}_{1} + {X}_{2} = n}\right) = \mathop{\sum }\limits_{{m = 0}}^{n}P\left( {{X}_{1} = m}\right) P\left( {{X}_{2} = n - m}\right) \]\n\n\[ = \mathop{\sum }\limits_{{m = 0}}^{n}{e}^{-{\lambda }_{1}}\fr... | Yes |
Lemma 2.5. \( N\left( {t + s}\right) - N\left( s\right), t \geq 0 \) is a rate \( \lambda \) Poisson process and independent of \( N\left( r\right) ,0 \leq r \leq s \) . | Why is this true? Suppose for concreteness (and so that we can use Fig. 2.1 at the beginning of this section again) that by time \( s \) there have been four arrivals \( {T}_{1},{T}_{2},{T}_{3},{T}_{4} \) that occurred at times \( {t}_{1},{t}_{2},{t}_{3},{t}_{4} \) . We know that the waiting time for the fifth arrival ... | Yes |
Lemma 2.6. \( N\left( t\right) \) has independent increments: if \( {t}_{0} < {t}_{1} < \ldots < {t}_{n} \), then\n\n\[ N\left( {t}_{1}\right) - N\left( {t}_{0}\right), N\left( {t}_{2}\right) - N\left( {t}_{1}\right) ,\ldots N\left( {t}_{n}\right) - N\left( {t}_{n - 1}\right) \;\text{ are independent } \] | Why is this true? Lemma 2.5 implies that \( N\left( {t}_{n}\right) - N\left( {t}_{n - 1}\right) \) is independent of \( N\left( r\right), r \leq {t}_{n - 1} \) and hence of \( N\left( {t}_{n - 1}\right) - N\left( {t}_{n - 2}\right) ,\ldots N\left( {t}_{1}\right) - N\left( {t}_{0}\right) \) . The desired result now foll... | Yes |
Theorem 2.7. If \( \{ N\left( s\right), s \geq 0\} \) is a Poisson process, then\n\n(i) \( N\left( 0\right) = 0 \) ,\n\n(ii) \( N\left( {t + s}\right) - N\left( s\right) = \operatorname{Poisson}\left( {\lambda t}\right) \), and\n\n(iii) \( N\left( t\right) \) has independent increments.\n\nConversely, if \( \left( i\ri... | Why is this true? Lemmas 2.2 and 2.6 prove (ii) and (iii). To start to prove the converse, let \( {T}_{n} \) be the time of the \( n \) th arrival. The first arrival occurs after time \( t \) if and only if there were no arrivals in \( \left\lbrack {0, t}\right\rbrack \) . So using the formula for the Poisson distribut... | Yes |
Theorem 2.8. If \( n \) is large the binomial \( \left( {n,\lambda /n}\right) \) distribution is approximately Poisson( \( \lambda \) ). | Proof. Exchanging the numerators of the first two fractions and breaking the last term into two, (2.14) becomes\n\n\[ \frac{{\lambda }^{k}}{k!} \cdot \frac{n\left( {n - 1}\right) \cdots \left( {n - k + 1}\right) }{{n}^{k}} \cdot {\left( 1 - \frac{\lambda }{n}\right) }^{n}{\left( 1 - \frac{\lambda }{n}\right) }^{-k} \]\... | Yes |
Theorem 2.9. Let \( {X}_{n, m},1 \leq m \leq n \) be independent random variables with \( P\left( {{X}_{m} = 1}\right) = {p}_{m} \) and \( P\left( {{X}_{m} = 0}\right) = 1 - {p}_{m} \) . Let\n\n\[ \n{S}_{n} = {X}_{1} + \cdots + {X}_{n},\;{\lambda }_{n} = E{S}_{n} = {p}_{1} + \cdots + {p}_{n}, \n\]\n\nand \( {Z}_{n} = \... | Why is this true? If \( X \) and \( Y \) are integer valued random variables then for any set \( A \)\n\n\[ \n\left| {P\left( {X \in A}\right) - P\left( {Y \in A}\right) }\right| \leq \frac{1}{2}\mathop{\sum }\limits_{n}\left| {P\left( {X = n}\right) - P\left( {Y = n}\right) }\right| \n\]\n\nThe right-hand side is call... | Yes |
Theorem 2.10. Let \( {Y}_{1},{Y}_{2},\ldots \) be independent and identically distributed, let \( N \) be an independent nonnegative integer valued random variable, and let \( S = {Y}_{1} + \cdots + \) \( {Y}_{N} \) with \( S = 0 \) when \( N = 0 \). (i) If \( E\left| {Y}_{i}\right| ,{EN} < \infty \), then \( {ES} = {E... | Proof. When \( N = n, S = {X}_{1} + \cdots + {X}_{n} \) has \( {ES} = {nE}{Y}_{i} \) . Breaking things down according to the value of \( N \) ,\n\n\[ \n{ES} = \mathop{\sum }\limits_{{n = 0}}^{\infty }E\left( {S \mid N = n}\right) \cdot P\left( {N = n}\right) \n\]\n\n\[ \n= \mathop{\sum }\limits_{{n = 0}}^{\infty }{nE}{... | Yes |
Suppose that the number of customers at a liquor store in a day has a Poisson distribution with mean 81 and that each customer spends an average of $8 with a standard deviation of $6. It follows from (i) in Theorem 2.10 that the mean revenue for the day is \( {81} \cdot \$ 8 = \$ {648} \) . | Using (iii), we see that the variance of the total revenue is\n\n\[ \n{81} \cdot \left\{ {{\left( \$ 6\right) }^{2} + {\left( \$ 8\right) }^{2}}\right\} = 8,{100} \n\]\n\nTaking square roots we see that the standard deviation of the revenue is \( \$ {90} \) compared with a mean of \$648. | Yes |
Theorem 2.11. \( {N}_{j}\left( t\right) \) are independent Poisson processes with rate \( {\lambda P}\left( {{Y}_{i} = j}\right) \) . | Proof. To begin we suppose that \( P\left( {{Y}_{i} = 1}\right) = p \) and \( P\left( {{Y}_{i} = 2}\right) = 1 - p \), so there are only two Poisson processes to consider: \( {N}_{1}\left( t\right) \) and \( {N}_{2}\left( t\right) \) . We will check the second definition given in Theorem 2.7. It should be clear that th... | Yes |
Example 2.4 (M/G/ \( \infty \) queue). In modeling telephone traffic, we can, as a first approximation, suppose that the number of phone lines is infinite, i.e., everyone who tries to make a call finds a free line. This certainly is not always true but analyzing a model in which we pretend this is true can help us to d... | The argument for arrivals at the Great Hall implies that the beginnings of calls follow a Poisson process. As for the calls themselves, while many people on the telephone show a lack of memory, there is no reason to suppose that the duration of a call has an exponential distribution. So we use a general distribution fu... | Yes |
Theorem 2.13. Suppose \( {N}_{1}\left( t\right) ,\ldots {N}_{k}\left( t\right) \) are independent Poisson processes with rates \( {\lambda }_{1},\ldots ,{\lambda }_{k} \), then \( {N}_{1}\left( t\right) + \cdots + {N}_{k}\left( t\right) \) is a Poisson process with rate \( {\lambda }_{1} + \cdots + {\lambda }_{k} \) . | Proof. Again we consider only the case \( k = 2 \) and check the second definition given in Theorem 2.7. It is clear that the sum has independent increments and \( {N}_{1}\left( 0\right) + \) \( {N}_{2}\left( 0\right) = 0 \) . The fact that the increments have the right Poisson distribution follows from Theorem 2.4. | No |
Given a Poisson process of red arrivals with rate \( \lambda \) and an independent Poisson process of green arrivals with rate \( \mu \), what is the probability that we will get six red arrivals before a total of four green ones? | The first step is to note that the event in question is equivalent to having at least six red arrivals in the first 9 . If this happens, then we have at most three green arrivals before the sixth red one. On the other hand if there are five or fewer red arrivals in the first 9 , then we have had at least four red arriv... | Yes |
Theorem 2.14. If we condition on \( N\left( t\right) = n \), then the vector \( \left( {{T}_{1},{T}_{2},\ldots {T}_{n}}\right) \) has the same distribution as \( \left( {{V}_{1},{V}_{2},\ldots {V}_{n}}\right) \) and hence the set of arrival times \( \left\{ {{T}_{1},{T}_{2},\ldots ,{T}_{n}}\right\} \) has the same dist... | Why is this true? We begin by finding the joint density function of \( \left( {{T}_{1},{T}_{2},{T}_{3}}\right) \) given that there were three arrivals before time \( t \) . The probability is 0 unless \( 0 < \) \( {v}_{1} < {v}_{2} < {v}_{3} < t \) . To compute the answer in this case, we note that \( P\left( {N\left( ... | Yes |
Theorem 2.15. Ifs \( < t \) and \( 0 \leq m \leq n \), then\n\n\[ P\left( {N\left( s\right) = m \mid N\left( t\right) = n}\right) = \left( \begin{matrix} n \\ m \end{matrix}\right) {\left( \frac{s}{t}\right) }^{m}{\left( 1 - \frac{s}{t}\right) }^{n - m} \]\n\nThat is, the conditional distribution of \( N\left( s\right)... | Proof. The number of arrivals by time \( s \) is the same as the number of \( {U}_{i} < s \) . The events \( \left\{ {{U}_{i} < s}\right\} \) these events are independent and have probability \( s/t \), so the number of \( {U}_{i} < s \) will be binomial \( \left( {n, s/t}\right) \) . | Yes |
Theorem 3.1. Let \( \mu = E{t}_{i} \) be mean interarrival time. If \( P\left( {{t}_{i} > 0}\right) > 0 \) then with probability one,\n\n\[ N\left( t\right) /t \rightarrow 1/\mu \;\text{ as }t \rightarrow \infty \] | ## Proof of Theorem 3.1. We use the | No |
Theorem 3.3. With probability one,\n\n\[ \frac{R\\left( t\\right) }{t} \\rightarrow \\frac{E{r}_{i}}{E{t}_{i}} \] | Proof. Multiplying and dividing by \( N\\left( t\\right) \), we have\n\n\[ \\frac{R\\left( t\\right) }{t} = \\left( {\\frac{1}{N\\left( t\\right) }\\mathop{\\sum }\\limits_{{i = 1}}^{{N\\left( t\\right) }}{r}_{i}}\\right) \\frac{N\\left( t\\right) }{t} \\rightarrow E{r}_{i} \\cdot \\frac{1}{E{t}_{i}} \]\n\nwhere in the... | No |
Suppose that the lifetime of a car is a random variable with density function \( h \) . Our methodical Mr. Brown buys a new car as soon as the old one breaks down or reaches \( T \) years. Suppose that a new car costs \( A \) dollars and that an additional cost of \( B \) dollars to repair the vehicle is incurred if it... | The duration of the \( i \) th cycle, \( {t}_{i} \), has\n\n\[ E{t}_{i} = {\int }_{0}^{T}{th}\left( t\right) {dt} + T{\int }_{T}^{\infty }h\left( t\right) {dt} \]\n\nsince the length of the cycle will be \( {t}_{i} \) if the car’s life is \( {t}_{i} < T \), but \( T \) if the car’s life \( {t}_{i} \geq T \) . The rewar... | Yes |
Theorem 3.4. In an alternating renewal process, the limiting fraction of time in state 1 is\n\n\\[ \n\\frac{{\\mu }_{F}}{{\\mu }_{F} + {\\mu }_{G}}\n\\] | Proof. In order to compute the limiting fraction of time the machine is working we let \\({t}_{i} = {s}_{i} + {u}_{i}\\) be the duration of the \\(i\\) th cycle, and let the reward \\({r}_{i} = {s}_{i}\\) , the amount of time the machine was working during the \\(i\\) th cycle. In this case, Theorem 3.3 implies that\n\... | Yes |
Theorem 3.5. Suppose \( \lambda < \mu \) . If the queue starts with some finite number \( k \geq 1 \) customers who need service, then it will empty out with probability one. Furthermore, the limiting fraction of time the server is busy is \( \geq \lambda /\mu \) . | Proof. Let \( {T}_{n} = {t}_{1} + \cdots + {t}_{n} \) be the time of the \( n \) th arrival. The strong law of large numbers, Theorem 3.2 implies that\n\n\[ \frac{{T}_{n}}{n} \rightarrow \frac{1}{\lambda } \]\n\nLet \( {Z}_{0} \) be the sum of the service times of the customers in the system at time 0 and let \( {s}_{i... | No |
Theorem 3.6 (Little’s formula). \( L = {\lambda }_{a}W \) . | Why is this true? Suppose each customer pays \( \$ 1 \) for each minute of time she is in the system. When \( \ell \) customers are in the system, we are earning \( \$ \ell \) per minute, so in the long run we earn an average of \( \$ L \) per minute. On the other hand, if we imagine that customers pay for their entire... | No |
Consider the \( {GI}/G/1 \) queue and suppose that we are only interested in the customer’s average waiting time in the queue, \( {W}_{Q} \). If we know the average waiting time \( W \) in the system, this can be computed by simply subtracting out the amount of time the customer spends in service | \[ {W}_{Q} = W - E{s}_{i} \] | No |
We define the workload in the system at time \( t,{Z}_{t} \), to be the sum of the remaining service times of all customers in the system, and define the long run average workload to be\n\n\[ Z = \mathop{\lim }\limits_{{t \rightarrow \infty }}\frac{1}{t}{\int }_{0}^{t}{Z}_{s}{ds} \] | As in the proof of Little's formula we will derive our result by computing the rate at which revenue is earned in two ways. This time we suppose that each customer in the queue or in service pays at a rate of \( \$ y \) when his remaining service time is \( y \) ; i.e., we do not count the remaining waiting time in the... | Yes |
(a) In the long run what is the fraction of time, \( \pi \left( 0\right) \), that the server is idle? \( \lambda = \) \( 1/6, E{s}_{i} = 5 = 1/\mu \), so by (3.5) \( \pi \left( 0\right) = 1 - \left( {1/6}\right) /\left( {1/5}\right) = 1/6 \) . | (a) In the long run what is the fraction of time, \( \pi \left( 0\right) \), that the server is idle? \( \lambda = \) \( 1/6, E{s}_{i} = 5 = 1/\mu \), so by (3.5) \( \pi \left( 0\right) = 1 - \left( {1/6}\right) /\left( {1/5}\right) = 1/6 \) . | Yes |
In Monopoly one rolls two dice and then moves that number of squares. As in Example 1.27 we will ignore Go to Jail, Chance, and other squares that make the chain complicated. The average number of spaces moved in one roll is \( E{t}_{1} = 7 \) so in the long run we land exactly on Go in \( 1/7 \) of the trips around th... | <table><tr><td>0</td><td>1</td><td></td><td>3</td><td>4</td><td>5</td><td>6</td><td>7</td><td>8</td><td>9</td><td>10</td><td>11</td></tr><tr><td>\( \frac{1}{7} \)</td><td>\( \frac{1}{7} \)</td><td>\( \frac{35}{252} \)</td><td>\( \frac{33}{252} \)</td><td>\( \frac{30}{252} \)</td><td>\( \frac{26}{252} \)</td><td>\( \fra... | Yes |
Theorem 3.9. As \( t \rightarrow \infty \)\n\n\[ \frac{1}{t}{\int }_{0}^{t}{1}_{\left\{ {A}_{s} > x,{Z}_{s} > y\right\} }{ds} \rightarrow \frac{1}{E{t}_{1}}{\int }_{x + y}^{\infty }P\left( {{t}_{i} > z}\right) {dz} \]\n | Proof. Let \( {I}_{x, y}\left( s\right) = 1 \) if \( {A}_{s} > x \) and \( {Z}_{s} > y \) . It is easy to see that\n\n\[ {\int }_{{T}_{i - 1}}^{{T}_{i}}{I}_{c}\left( s\right) {ds} = {\left( {t}_{i} - \left( x + y\right) \right) }^{ + } \]\n\nTo check this we consider two cases. Ignoring the contribution from the last i... | Yes |
Example 3.10 (Exponential). In this case the limiting density given in (3.11) is | \[ \frac{\lambda {e}^{-\lambda \left( {x + y}\right) }}{1/\lambda } = \lambda {e}^{-{\lambda a}} \cdot \lambda {e}^{-{\lambda z}} \] | No |
Example 3.11 (Uniform on \( \\left( {0, b}\\right) \) ). Plugging into (3.11) gives for \( a, z > 0, a + \) \( z < b \) : | \[ \frac{1/b}{b/2} = \frac{2}{{b}^{2}} \] | No |
Let \( N\left( t\right), t \geq 0 \) be a Poisson process with rate \( \lambda \) and let \( {Y}_{n} \) be a discrete time Markov chain with transition probability \( u\left( {i, j}\right) \) . Then \( {X}_{t} = {Y}_{N\left( t\right) } \) is a continuous-time Markov chain. In words, \( {X}_{t} \) takes one jump accordi... | Why is this true? Intuitively, this follows from the lack of memory property of the exponential distribution. If \( {X}_{s} = i \), then independent of what has happened in the past, the time to the next jump will be exponentially distributed with rate \( \lambda \) and will go to state \( j \) with probability \( u\le... | No |
Theorem 4.1 (Chapman-Kolmogorov equation).\n\n\[ \mathop{\sum }\limits_{k}{p}_{s}\left( {i, k}\right) {p}_{t}\left( {k, j}\right) = {p}_{s + t}\left( {i, j}\right) \] | Proof. Breaking things down according to the state at time \( s \), we have\n\n\[ P\left( {{X}_{s + t} = j \mid {X}_{0} = i}\right) = \mathop{\sum }\limits_{k}P\left( {{X}_{s + t} = j,{X}_{s} = k \mid {X}_{0} = i}\right) \]\n\nUsing the definition of conditional probability and the Markov property, the above is\n\n\[ =... | Yes |
Example 4.5 (Pure birth processes with power law rates). Suppose \( q\left( {i, i + 1}\right) = \) \( \lambda {i}^{p} \) and all the other \( q\left( {i, j}\right) = 0 \) . In this case the jump to \( n + 1 \) is made at time \( {T}_{n} = {t}_{1} + \cdots + {t}_{n} \), where \( {t}_{n} \) is exponential with rate \( {n... | Proof. \( \operatorname{var}\left( {T}_{n}\right) = \mathop{\sum }\limits_{{m = 1}}^{n}1/{m}^{2}{\beta }^{2} \leq C = \mathop{\sum }\limits_{{m = 1}}^{\infty }1/{m}^{2}{\beta }^{2} \) . Chebyshev’s inequality implies\n\n\[ P\left( {{T}_{n} \leq E{T}_{n}/2}\right) \leq {4C}/{\left( E{T}_{n}\right) }^{2} \rightarrow 0 \]... | Yes |
Let \( X\left( t\right) \) be the number of arrivals up to time \( t \) in a Poisson process with rate \( \lambda \) . In order to go from \( i \) arrivals at time \( s \) to \( j \) arrivals at time \( t + s \) we must have \( j \geq i \) and have exactly \( j - i \) arrivals in \( t \) units of time, so\n\n\[ \n{p}_{... | To check the differential equation, we have to first figure out what it is. Using the more explicit form of the backwards equation, (4.6), and plugging in our rates, we have\n\n\[ \n{p}_{t}^{\prime }\left( {i, j}\right) = \lambda {p}_{t}\left( {i + 1, j}\right) - \lambda {p}_{t}\left( {i, j}\right) \n\]\n\nTo check thi... | Yes |
Example 4.8 (Two-state chains). For concreteness, we can suppose that the state space is \( \{ 1,2\} \) . In this case, there are only two flip rates \( q\left( {1,2}\right) = \lambda \) and \( q\left( {2,1}\right) = \mu \) , so when we fill in the diagonal with minus the sum of the flip rates on that row we get\n\n\[ ... | Writing out the backward equation in matrix form, (4.7), now we have\n\n\[ \left( \begin{array}{ll} {p}_{t}^{\prime }\left( {1,1}\right) & {p}_{t}^{\prime }\left( {1,2}\right) \\ {p}_{t}^{\prime }\left( {2,1}\right) & {p}_{t}^{\prime }\left( {2,2}\right) \end{array}\right) = \left( \begin{matrix} - \lambda & \lambda \\... | Yes |
In this model each particle splits into two at rate \( \beta \), so \( q\left( {i, i + 1}\right) = {\beta i} \) . To find the transition probability of the Yule process we will guess and verify that\n\n\[ \n{p}_{t}\left( {1, j}\right) = {e}^{-{\beta t}}{\left( 1 - {e}^{-{\beta t}}\right) }^{j - 1}\;\text{ for }j \geq 1... | To check (4.13), we will use the forward equation (4.9) to conclude that if \( j \geq 1 \) then\n\n\[ \n{p}_{t}^{\prime }\left( {1, j}\right) = - {\beta j}{p}_{t}\left( {1, j}\right) + \beta \left( {j - 1}\right) {p}_{t}\left( {1, j - 1}\right) \n\]\n\nwhere \( {p}_{t}\left( {1,0}\right) = 0 \) . The use of the forward... | Yes |
Lemma 4.2. If \( {X}_{t} \) is irreducible and \( t > 0 \) then \( {p}_{t}\left( {i, j}\right) > 0 \) . | Proof. Since \( {p}_{s}\left( {i, j}\right) \geq \exp \left( {-{\lambda }_{j}s}\right) > 0 \) and \( {p}_{t + s}\left( {i, j}\right) \geq {p}_{t}\left( {i, j}\right) {p}_{s}\left( {j, j}\right) \) it suffices to show that this holds for small \( t \) . Since\n\n\[ \mathop{\lim }\limits_{{h \rightarrow 0}}{p}_{h}\left( ... | Yes |
Lemma 4.3. \( \pi \) is a stationary distribution if and only if \( {\pi Q} = 0 \) . | Why is this true? Filling in the definition of \( Q \) and rearranging, the condition \( {\pi Q} = 0 \) becomes\n\n\[ \mathop{\sum }\limits_{{k \neq j}}\pi \left( k\right) q\left( {k, j}\right) = \pi \left( j\right) {\lambda }_{j} \]\n\nIf we think of \( \pi \left( k\right) \) as the amount of sand at \( k \), the righ... | Yes |
Example 4.10 (L.A. weather chain). There are three states: \( 1 = \) sunny, \( 2 = \) smoggy, \( 3 = \) rainy. The weather stays sunny for an exponentially distributed number of days with mean 3 , then becomes smoggy. It stays smoggy for an exponentially distributed number of days with mean 4 , then rain comes. The rai... | Adding the three equations gives \( 0 = 0 \) so we delete the third equation and add \( {\pi }_{1} + {\pi }_{2} + {\pi }_{3} = 1 \) to get an equation that can be written in matrix form as\n\n\[ \left( {{\pi }_{1}{\pi }_{2}{\pi }_{3}}\right) A = \left( \begin{array}{lll} 0 & 0 & 1 \end{array}\right) \;\text{ where }\;A... | Yes |
Theorem 4.5. If (4.16) holds, then \( \pi \) is a stationary distribution. | Proof. Summing 4.16 over all \( k \neq j \) and recalling the definition of \( {\lambda }_{j} \) gives\n\n\[ \mathop{\sum }\limits_{{k \neq j}}\pi \left( k\right) q\left( {k, j}\right) = \pi \left( j\right) \mathop{\sum }\limits_{{k \neq j}}q\left( {j, k}\right) = \pi \left( j\right) {\lambda }_{j} \]\n\nRearranging we... | Yes |
Example 4.12 (Birth and death chains). Suppose that \( S = \{ 0,1,\ldots, N\} \) with \( N \leq \infty \) and\n\n\[ q\left( {n, n + 1}\right) = {\lambda }_{n}\;\text{ for }n < N \]\n\n\[ q\left( {n, n - 1}\right) = {\mu }_{n}\;\text{ for }n > 0 \]\n\nHere \( {\lambda }_{n} \) represents the birth rate when there are \(... | Using this again we have \( \pi \left( {n - 1}\right) = \left( {{\lambda }_{n - 2}/{\mu }_{n - 1}}\right) \pi \left( {n - 2}\right) \) and it follows that\n\n\[ \pi \left( n\right) = \frac{{\lambda }_{n - 1}}{{\mu }_{n}} \cdot \frac{{\lambda }_{n - 2}}{{\mu }_{n - 1}} \cdot \pi \left( {n - 2}\right) \]\n\nRepeating the... | Yes |
Example 4.13 (Two state chains). Suppose that the state space is \( \{ 1,2\}, q\left( {1,2}\right) = \lambda \) , and \( q\left( {2,1}\right) = \mu \), where both rates are positive. The equations \( {\pi Q} = 0 \) can be written as\n\n\[ \left( \begin{array}{ll} {\pi }_{1} & {\pi }_{2} \end{array}\right) \left( \begin... | The first equation says \( - \lambda {\pi }_{1} + \mu {\pi }_{2} = 0 \) . Taking into account that we must have \( {\pi }_{1} + {\pi }_{2} = 1 \), it follows that\n\n\[ {\pi }_{1} = \frac{\mu }{\lambda + \mu }\;\text{ and }\;{\pi }_{2} = \frac{\lambda }{\lambda + \mu } \] | Yes |
Example 4.14 (Barbershop). A barber can cut hair at rate 3, where the units are people per hour, i.e., each haircut requires an exponentially distributed amount of time with mean \( {20}\mathrm{\;{min}} \) . Suppose customers arrive at times of a rate 2 Poisson process, but will leave if both chairs in the waiting room... | Solution. We define our state to be the number of customers in the system, so \( S = \) \( \{ 0,1,2,3\} \) . From the problem description it is clear that\n\n\[ q\left( {i, i - 1}\right) = 3\;\text{ for }i = 1,2,3 \]\n\n\[ q\left( {i, i + 1}\right) = 2\;\text{ for }i = 0,1,2 \]\n\nThe detailed balance conditions say\n\... | Yes |
Example 4.15 (Machine repair model). A factory has three machines in use and one repairman. Suppose each machine works for an exponential amount of time with mean 60 days between breakdowns, but each breakdown requires an exponential repair time with mean 4 days. What is the long-run fraction of time all three machines... | Solution. Let \( {X}_{t} \) be the number of working machines. Since there is one repairman we have \( q\left( {i, i + 1}\right) = 1/4 \) for \( i = 0,1,2 \) . On the other hand, the failure rate is proportional to the number of machines working, so \( q\left( {i, i - 1}\right) = i/{60} \) for \( i = \) \( 1,2,3 \) . S... | Yes |
Example 4.16 (M/M/ \( \propto \) queue). In this case \( q\left( {n, n + 1}\right) = \lambda \) and \( q\left( {n, n - 1}\right) = {n\mu } \) so | \[ \pi \left( n\right) = \pi \left( 0\right) \frac{{\left( \lambda /\mu \right) }^{n}}{n!} \] If we take \( \pi \left( 0\right) = {e}^{-\lambda /\mu } \) then this becomes the Poisson distribution with mean \( \lambda /\mu \) . | Yes |
Theorem 4.6. If \( {a}_{n} \rightarrow 0 \) as \( n \rightarrow \infty \), then there is a stationary distribution. | Proof. It follows from (4.17) that if \( n \geq s \), then\n\n\[ \pi \left( {n + 1}\right) = \frac{{\lambda }_{n}}{{\mu }_{n + 1}} \cdot \pi \left( n\right) = {a}_{n} \cdot \frac{\lambda }{s\mu } \cdot \pi \left( n\right) \]\n\nIf \( N \) is large enough and \( n \geq N \), then \( {a}_{n}\lambda /\left( {s\mu }\right)... | Yes |
Branching process has jump rates \( q\left( {i, i + 1}\right) = {\lambda i} \) and \( q\left( {i, i - 1}\right) = \) \( {\mu i} \) . 0 is an absorbing state but for \( i \geq 1 \) the \( i \) ’s cancel and we have\n\n\[ r\left( {i, i + 1}\right) = \frac{\lambda }{\lambda + \mu }\;r\left( {i, i - 1}\right) = \frac{\mu }... | For another derivation let \( \rho = {P}_{1}\left( {{T}_{0} < \infty }\right) \) . By considering what happens when the chain leaves 0 we have\n\n\[ \rho = \frac{\mu }{\lambda + \mu } \cdot 1 + \frac{\lambda }{\lambda + \mu } \cdot {\rho }^{2} \]\n\nsince starting from state 2 extinction occurs if and only if each indi... | Yes |
Example 4.20 (M/M/1 queue). This is particularly simple because the time in each state \( i > 0 \) is exponential with rate \( \lambda + \mu \) so the new result follows from the one in discrete time given in (1.28) | \[ {E}_{1}{T}_{0} = \frac{1}{\lambda + \mu } \cdot \frac{\lambda + \mu }{\mu - \lambda } = \frac{1}{\mu - \lambda } \] | Yes |
Example 4.21 (Barbershop chain). (continuation of Example 4.14) The transition rates are\n\n\[ \nq\left( {i, i - 1}\right) = 3\;\text{ for }i = 1,2,3 \]\n\n\[ \nq\left( {i, i + 1}\right) = 2\;\text{ for }i = 0,1,2 \]\n\nso the embedded chain is\n\n\[ \n{13}/5\;0\;2/5\;0 \]\n\n\[ \n\begin{array}{lllll} 2 & 0 & 3/5 & 0 &... | Inserting the last equation in the second one:\n\n\[ \ng\left( 2\right) = \frac{1}{5} + \frac{3}{5}g\left( 1\right) + \frac{2}{15} + \frac{2}{5}g\left( 2\right) \]\n\nor \( \left( {3/5}\right) g\left( 2\right) = \left( {1/3}\right) + \left( {3/5}\right) g\left( 1\right) \) . Multiplying by 2/3’s and inserting this in t... | Yes |
What is the expected time until all three students are gone? | If we describe the state of the Markov chain by the rates of the students that are left, with \( \varnothing \) to denote an empty office, then the \( Q \) -matrix is\n\n\[ \begin{array}{rrrrrrrrr} {123} & {12} & {13} & {23} & 1 & 2 & 3 & \varnothing & \\ {123} & - 6 & 3 & 2 & 1 & 0 & 0 & 0 & 0 \\ {12} & 0 & - 3 & 0 & ... | Yes |
In this system customers arrive to a single server facility at the times of a Poisson process with rate \( \lambda \), and each requires an independent amount of service that has an exponential distribution with rate \( \mu \). From the description it should be clear that the transition rates are\n\n\[ q\left( {n, n + ... | To find the value of \( \pi \left( 0\right) \), we recall that when \( \left| \theta \right| < 1,\mathop{\sum }\limits_{{n = 0}}^{\infty }{\theta }^{n} = 1/\left( {1 - \theta }\right) \). From this we see that if \( \lambda < \mu \), then\n\n\[ \mathop{\sum }\limits_{{n = 0}}^{\infty }\pi \left( n\right) = \mathop{\sum... | Yes |
Lemma 4.7. Let \( {X}_{t} \) be a Markov chain with a stationary distribution \( \pi \) that satisfies the detailed balance condition. Let \( {Y}_{t} \) be the chain constrained to stay in a subset \( A \) of the state space. That is, jumps which take the chain out of \( A \) are not allowed, but allowed jumps occur at... | Proof. If \( x, y \in A \) then detailed balance for \( {X}_{t} \) implies \( \pi \left( x\right) q\left( {x, y}\right) = \pi \left( y\right) q\left( {y, x}\right) \) . From this it follows that \( v\left( x\right) \bar{q}\left( {x, y}\right) = v\left( y\right) \bar{q}\left( {y, x}\right) \) so \( v \) satisfies the de... | Yes |
Example 4.25 (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... | From this we conclude that \[ \pi \left( k\right) = \left\{ \begin{array}{ll} \frac{c}{k!}{\left( \frac{\lambda }{\mu }\right) }^{k} & k \leq s \\ \frac{c}{s!{s}^{k - s}}{\left( \frac{\lambda }{\mu }\right) }^{k} & k \geq s \end{array}\right. \] (4.26) where \( c \) is a constant that makes the sum equal to 1 . From th... | Yes |
Theorem 4.8. If \( \lambda < \mu \) s, then the output process of the M/M/s queue in equilibrium is a rate \( \lambda \) Poisson process. | Proof. for \( s = 1 \) . Our first step in making the result in Theorem 4.8 seem reasonable is to check by hand that if there is one server and the queue is in equilibrium, then the time of the first departure, \( D \), has an exponential distribution with rate \( \lambda \) . There are two cases to consider.\n\nCase 1... | No |
Lemma 4.9. 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 4.11. 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 (4.23) 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 4.12. 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\nThe rate arrows plus the ordinary lines on the p... | Yes |
When is the system stable? That is, when is there a stationary distribution? To get started on this question suppose that both servers are busy. | In this case work arrives at station 1 at rate \( {\lambda }_{1} + {p}_{2}{\mu }_{2} \), and work arrives at station 2 at rate \( {\lambda }_{2} + {p}_{1}{\mu }_{1} \) . It should be intuitively clear that:\n\n(i) If \( {\lambda }_{1} + {p}_{2}{\mu }_{2} < {\mu }_{1} \) and \( {\lambda }_{2} + {p}_{1}{\mu }_{1} < {\mu ... | Yes |
Example 4.28 (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) \... | To have a chance of stability we must suppose\n\n(A) For each \( i \) it is possible for a customer entering at \( i \) to leave the system. That is, for each \( i \) there is a sequence of states \( i = {j}_{0},{j}_{1},\ldots {j}_{n} \) with \( p\left( {{j}_{m - 1},{j}_{m}}\right) > 0 \) for \( 1 \leq m \leq n \) and ... | Yes |
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