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Example 4.29. 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 server $ 3... | 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 (4.30) the solution is $ r = \lambda {\left( I - p\right) }^{-1} $, where\n\n\[ {\left( I - p\right) }^{-1} = \left( \begin{array}{rrr} {45}/{38} & {35}/{38} & {35}/{76} \\ ... | 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 \( Y \) and \( {1}_{A} \) are independent, so Theorem A. 1 implies\n\n\[ E\left( {Y... | 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\[ E\left( {Y \mid B}\right) = \mathop{\sum }\limits_{{j = 1}}^{k}E\left( {Y \mid {A}_{j}}\right) \cdot \frac{P\left( {A}_{j}\right) }{P\left( B\right) } \] | 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 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 |
Theorem 5.5. 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\[ f\left( {x, n}\right) = \mathop{\sum }\limits_{y}p\left( {x, y}\right) f\left( {y, n + 1}\right) \]\n\nThen \( {M}_{n} = f\left( ... | Proof. By the Markov property and our assumption on \( f \)\n\n\[ E\left( {f\left( {{X}_{n + 1}, n + 1}\right) \mid {A}_{v}}\right) = \mathop{\sum }\limits_{y}p\left( {{x}_{n}, y}\right) f\left( {y, n + 1}\right) = f\left( {{x}_{n}, n}\right) \]\n\nwhich proves the desired result. | Yes |
Example 5.3 (Gambler’s ruin). 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... | Proof. Using Theorem 5.5 with \( h\left( x\right) = {\left( \left( 1 - p\right) /p\right) }^{x} \), we need only check that \( h\left( x\right) = \mathop{\sum }\limits_{y}p\left( {x, y}\right) h\left( y\right) \) . To do this we note that\n\n\[ \mathop{\sum }\limits_{y}p\left( {x, y}\right) h\left( y\right) = p \cdot {... | Yes |
Let \( {Y}_{1},{Y}_{2},\ldots \) be independent with\n\n\[ P\left( {{Y}_{i} = 1}\right) = P\left( {{Y}_{i} = - 1}\right) = 1/2 \]\n\nand let \( {X}_{n} = {X}_{0} + {Y}_{1} + \cdots + {Y}_{n} \). Then \( {M}_{n} = {X}_{n}^{2} - n \) is a martingale with respect to \( {X}_{n} \). By Theorem 5.5 with \( f\left( {x, n}\rig... | To do this we work out the squares to conclude the left-hand side is\n\n\[ \frac{1}{2}\left\lbrack {{x}^{2} + {2x} + 1 + {x}^{2} - {2x} + 1}\right\rbrack - 1 = {x}^{2} \] | 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\n\n\[\nE\left( {{M}_{n + 1} - {M}_{n} \mid {A}_{v}}\right) = {M}_{n}E\left( {{X}_{n + 1} - 1 \mid {A}_{v}}\right) = 0\n\] | Yes |
Example 5.6 (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 |
Lemma 5.6. If \( {M}_{n} \) is a martingale and \( \phi \) is a convex function then \( \phi \left( {M}_{n}\right) \) is a submartingale. If \( {M}_{n} \) is a submartingale and \( \phi \) is a nondecreasing convex function then \( \phi \left( {M}_{n}\right) \) is a submartingale. | Proof. Using Lemma 5.2 and the definition of a martingale, we have\n\n\[ E\left( {\phi \left( {M}_{n + 1}\right) \mid {A}_{v}}\right) \geq \phi \left( {E\left( {{M}_{n + 1} \mid {A}_{v}}\right) }\right) = \phi \left( {M}_{n}\right) \]\n\nIn the proof of the second statement, the submartingale property of \( {M}_{n} \) ... | Yes |
Lemma 5.7. If \( {M}_{n} \) is a martingale then\n\n\[ E\left( {{M}_{n + 1}^{2} \mid {A}_{v}}\right) - {M}_{n}^{2} = E\left( {{\left( {M}_{n + 1} - {M}_{n}\right) }^{2} \mid {A}_{v}}\right) \] | Proof. Expanding out the square on the right-hand side, then using (5.3) and Lemma 5.3 gives\n\n\[ E\left( {{M}_{n + 1}^{2} - 2{M}_{n + 1}{M}_{n} + {M}_{n}^{2} \mid {A}_{v}}\right) = E\left( {{M}_{n + 1}^{2} \mid {A}_{v}}\right) - 2{M}_{n}E\left( {{M}_{n + 1} \mid {A}_{v}}\right) + {M}_{n}^{2} \]\n\n\[ = E\left( {{M}_{... | Yes |
Lemma 5.8 (Orthogonality of martingale increments). If \( {M}_{n} \) is a martingale and \( 0 \leq i \leq j \leq k < n \) then\n\n\[ E\left\lbrack {\left( {{M}_{n} - {M}_{k}}\right) {M}_{j}}\right\rbrack = 0 \]\n\nand \( E\left\lbrack {\left( {{M}_{n} - {M}_{k}}\right) \left( {{M}_{j} - {M}_{i}}\right) }\right\rbrack =... | Proof. The second result follows by subtracting the result for \( j = i \) from the one for \( j \) . Let \( {A}_{v} = \left\{ {{X}_{k} = {x}_{k},\ldots ,{X}_{0} = {x}_{0},{M}_{0} = m}\right\} \) . Using Lemma 5.4 then Lemma 5.1 and the martingale property\n\n\[ E\left\lbrack {\left( {{M}_{n} - {M}_{k}}\right) {M}_{j}}... | Yes |
Theorem 5.9. 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 5.7 (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 a... | \[ \text{outcome L L L L L W} \] \[ \text{bet 1 2 4 8 16} \] \[ \text{net profit}\; - 1\; - 3\; - 7\; - {15}\;1 \] 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 ... | Yes |
Theorem 5.12. 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\nAs in the proof of Theorem 5.9 let\n\n\[ \n{A}_{v} = \left\{ {{X}_{n} = {x}_{n},{X}_{n - 1} = {x}_{n - 1},\ldots ,{X}_{0} = {x}_{0},{M}_{0} = {m}_{0}}\right\}... | Yes |
One possible gambling strategy is to bet \( \$ 1 \) each time until you stop playing at time \( T \) . In symbols, we let \( {H}_{m} = 1 \) if \( T \geq m \) and 0 otherwise. To check that this is an admissible gambling strategy we note that the set on which \( {H}_{m} \) is 0 is | \[ \{ T \geq m{\} }^{c} = \{ T \leq m - 1\} = { \cup }_{k = 1}^{m - 1}\{ T = k\} \] By the definition of a stopping time, the event \( \{ T = k\} \) can be determined from the values of \( {M}_{0},{X}_{0},\ldots ,{X}_{k} \) . Since the union is over \( k \leq m - 1,{H}_{m} \) can be determined from the values of \( {M}... | Yes |
Example 5.10 (Gambler’s ruin). Let \( {X}_{1},{X}_{2},\ldots {X}_{n} \) be independent with \[ P\left( {{X}_{i} = 1}\right) = p\;\text{ and }\;P\left( {{X}_{i} = - 1}\right) = q = 1 - p \] Suppose \( 1/2 < p < 1 \) and let \( h\left( x\right) = {\left( q/p\right) }^{x} \) . Example 5.3 implies that \( {M}_{n} = \) \( h... | To provide a proof for (5.11), we use Theorems 5.13 and 5.11, to conclude that \[ {\left( q/p\right) }^{x} = {E}_{x}{M}_{\tau \land n} = {\left( q/p\right) }^{a}P\left( {\tau \leq n,{S}_{\tau } = a}\right) + {\left( q/p\right) }^{b}P\left( {\tau \leq n,{S}_{\tau } = b}\right) \] \[ + E\left( {{\left( q/p\right) }^{{S}_... | Yes |
Theorem 5.14. Suppose \( {M}_{n} \) is a martingale and \( T \) a stopping time with \( P(T < \) \( \infty ) = 1 \) and \( \left| {M}_{T \land n}\right| \leq K \) for some constant \( K \) . Then \( E{M}_{T} = E{M}_{0} \) . | Proof. Theorem 5.13 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 |
Let \( {S}_{n} = {S}_{0} + {X}_{1} + \cdots + {X}_{n} \) where \( {X}_{1},{X}_{2},\ldots \) are independent with \( P\left( {{X}_{i} = 1}\right) = P\left( {{X}_{i} = - 1}\right) = 1/2 \) . Let \( \tau = \) \( \min \left\{ {n : {S}_{n} \notin \left( {a, b}\right) }\right\} \) where \( a < 0 < b \) . Our goal here is to ... | Example 5.4 implies that \( {S}_{n}^{2} - n \) is a martingale. Let \( \tau = \min \left\{ {n : {S}_{n} \notin \left( {a, b}\right) }\right\} \) . From the previous example we have that \( \tau \) is a stopping time with \( P\left( {\tau < \infty }\right) = 1 \) . Again if we argue casually \( 0 = {E}_{0}\left( {{S}_{\... | Yes |
Theorem 5.15 (Wald’s equation). If \( T \) is a stopping time with \( {ET} < \infty \), then\n\n\[ E\left( {{S}_{T} - {S}_{0}}\right) = {\mu ET} \] | Why is this true? Theorems 5.13 and 5.11 give\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 (5.13). To pass to the limit in the other term, we note that\n\n\[ E\left| {{S}_{T} - {S}_{T \land n}}\... | No |
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 integers when they are decreasing, which is going... | \[ {\phi }^{\prime }\left( \theta \right) = \frac{d}{d\theta }E{e}^{\theta {x}_{i}} = E\left( {{x}_{i}{e}^{\theta {x}_{i}}}\right) \;\text{ so }\;{\phi }^{\prime }\left( 0\right) = E{x}_{i} > 0 \] and it follows that \( \phi \left( \theta \right) < 1 \) for small negative \( \theta \). (ii) If \( \theta < 0 \), then \(... | Yes |
Theorem 5.16. Consider a left continuous random walk with positive mean. Let \( a < x \) and \( {V}_{a} = \min \left\{ {n : {S}_{n} = a}\right\} \) . \[ {P}_{x}\left( {{V}_{a} < \infty }\right) = {e}^{\alpha \left( {x - a}\right) } \] | Proof. Again if one argues casually \[ {e}^{\alpha x} = {E}_{x}\left( {\exp \left( {\alpha {V}_{a}}\right) }\right) = {e}^{\alpha a}{P}_{x}\left( {{V}_{a} < \infty }\right) \] but we have to prove that there is no contribution from \( \left\{ {{V}_{a} = \infty }\right\} \) . To do this note that Theorems 5.13 and 5.11 ... | 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 |
Theorem 5.17. If \( {X}_{n} \geq 0 \) is a supermartingale then \( {X}_{\infty } = \mathop{\lim }\limits_{{n \rightarrow \infty }}{X}_{n} \) exists and \( E{X}_{\infty } \leq E{X}_{0} \) | The bad martingale in Example 5.9 shows that we can have \( {X}_{0} = 1 \) and \( {X}_{\infty } = 0 \) . The key to the proof of this is the following maximal inequality. | No |
Lemma 5.18. Let \( {X}_{n} \geq 0 \) be a supermartingale and \( \lambda > 0 \) . \[ P\left( {\mathop{\max }\limits_{{n \geq 0}}{X}_{n} > \lambda }\right) \leq E{X}_{0}/\lambda \] | Proof. Let \( T = \min \left\{ {n \geq 0 : {X}_{n} > \lambda }\right\} \) . Theorem 5.13 implies that \[ E{X}_{0} \geq E\left( {X}_{T \land n}\right) \geq {\lambda P}\left( {T \leq n}\right) \] i.e., \( P\left( {T \leq n}\right) \leq E{X}_{0}/\lambda \) . Since this holds for all \( n \) the desired result follows. | Yes |
Example 5.14 (Polya's urn). Consider an urn that contains red and green balls. At time 0 there are \( \mathrm{k} \) balls with at least one ball of each color. At time \( n \) we draw out a ball chosen at random. We return it to the urn and add one more of the color chosen. Let \( {X}_{n} \) be the fraction of red ball... | \[ P\left( {{R}_{n + 1} = {R}_{n} + 1}\right) = {X}_{n}\;P\left( {{R}_{n + 1} = {R}_{n}}\right) = 1 - {X}_{n} \] Letting \( {A}_{v} = \left\{ {{X}_{n} = {x}_{n},\ldots {X}_{0} = {x}_{0}}\right\} \) we have \[ E\left( {{X}_{n + 1} \mid {A}_{v}}\right) = \frac{{R}_{n} + 1}{n + k + 1} \cdot \frac{{R}_{n}}{n + k} + \frac{{... | Yes |
Subcritical. If \( \mu < 1 \) then \( P\left( {{Z}_{n} > 0}\right) \leq {\mu }^{n}E{Z}_{0} \rightarrow 0 \) as \( n \rightarrow \infty \) | Proof. Since \( {Z}_{n}/{\mu }^{n} \) is a martingale, \( E{Z}_{n} = {\mu }^{n}E{Z}_{0} \) . Using this with \( P\left( {{Z}_{n} \geq 1}\right) \) \( \leq E{Z}_{n} \) gives the desired result. | Yes |
Theorem 5.19. If \( \mathop{\sum }\limits_{k}k{p}_{k} > 1 \) and \( \mathop{\sum }\limits_{k}{k}^{2}{p}_{k} < \infty \) then \( P\left( {W = 0}\right) = \rho \) . | Proof. We begin with the easy part: if \( P\left( {W = 0}\right) < 1 \) the \( P\left( {W = 0}\right) = \rho \) . If we have \( {Z}_{n}/{\mu }^{n} \rightarrow 0 \) then this must be true for the branching processes started by the \( {Z}_{1} \) individuals in generation 1. Breaking things down according to the value of ... | 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\[ \n{V}_{0} = \frac{18}{19} \cdot \frac{4}{9} \cdot {15} = \frac{120}{19} = {6.3158} \n\] | 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\[{P}^{ * }\left( a\right) = {\left( {p}^{ * }\right) }^{H\left( a\right) }{\left( 1 - {p}^{ * }\right) }^{T\left( a\right) }\n\]\nwhere \( H\left( a\right) \) and \( T\left( a\right) \... | Yes |
Theorem 6.5. Assume that the holdings \( {\Delta }_{n}\left( a\right) \) can be determined from the outcomes of the first \( n \) stock movements 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} = ... | 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\[ \n{V}_{3} = \mathop{\max }\limits_{{0 \leq m \leq 3}}{S}_{m} - {S}_{3} \n\] | Our goal is to compute the value \( {V}_{n}\left( a\right) \) and the replicating strategy \( {\Delta }_{n}\left( a\right) \) for this option in the binomial model given in (6.16) with \( {S}_{0} = 4 \) . Here the numbers above the nodes are the stock price, while those below are the values of \( {V}_{n}\left( a\right)... | 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\nConsider the two cases \( {S}_{N} \geq K \) and \( {S}_{N} \leq K \) . Dividing by \( {\left( 1 + r\right) }^{N} \), taking \( {E}^{ * } \) expected value and using the fact that \( {... | 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( ... | \[ {V}_{n}\left( a\right) = \frac{6}{7}\left\lbrack {{0.6}{V}_{n}\left( {aH}\right) + {0.4}{V}_{n}\left( {aT}\right) }\right\rbrack \] with the extra boundary condition that if the price is \( \leq {20} \) the value is 0 . To check the answer note that the knockout feature eliminates one of the paths to 36 so \[ {V}_{0... | Yes |
Lemma 6.7. If \( U \) is smooth then \( U \) is concave if \( {U}^{\prime \prime } \leq 0 \) . | Proof. \( {U}^{\prime \prime } \leq 0 \) implies that the \( {U}^{\prime } \) is decreasing, so if \( x > y \)\n\n\[ \n\frac{1}{\lambda \left( {x - y}\right) }{\int }_{y}^{{\lambda x} + \left( {1 - \lambda }\right) y}{U}^{\prime }\left( z\right) {dz} \geq \frac{1}{x - y}{\int }_{y}^{x}{U}^{\prime }\left( z\right) {dz} ... | 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 6.8. 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 to exercise. | Proof. Since \( {S}_{n}/{\left( 1 + r\right) }^{n} \) is a martingale under \( {P}^{ * } \n\n\[ \ng\left( {S}_{n}\right) = g\left( {{E}_{n}^{ * }\left( \frac{{S}_{n + 1}}{1 + r}\right) }\right) \n\]\n\nUnder the risk neutral probability\n\n\[ \n{S}_{n}\left( a\right) = {p}_{n}^{ * }\left( a\right) \frac{{S}_{n + 1}\lef... | Yes |
Theorem 6.9 (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 (6.25) 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 \ | Yes |
Theorem 6.10. Write \( {E}^{ * } \) for expected values when \( \mu = r - {\sigma }^{2}/2 \) in (6.26). 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(6.28)\n\nUsing the formulas for \( u \) and \( d \) in (6.23) and recalling that \( {e}^{x} = 1 + x + {x}^{2}/2 + \cdots \) ,... | Yes |
Theorem 6.11. 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 6.9 (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 = {0.0... | With the help of a little spreadsheet we find that the formula predicts a price of \$32.93. | Yes |
Example 6.10 (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: | \[ {V}_{P} - {V}_{C} = {e}^{-{rT}}K - {S}_{0} \] | Yes |
Example 1.1 (Gambler's Ruin). Consider a gambling game in which on any turn you win \( \$ 1 \) with probability \( p = {0.4} \) or lose \( \$ 1 \) with probability \( 1 - p = {0.6} \) . Suppose further that you adopt the rule that you quit playing if your fortune reaches \( \$ N \) . Of course, if your fortune reaches ... | Let \( {X}_{n} \) be the amount of money you have after \( n \) plays. Your fortune, \( {X}_{n} \) has the \ | No |
Example 1.7 (Repair Chain). A machine has three critical parts that are subject to failure, but can function as long as two of these parts are working. When two are broken, they are replaced and the machine is back to working order the next day. To formulate a Markov chain model we declare its state space to be the par... | \[ \begin{array}{rrrrrrrr} 0 & 1 & 2 & 3 & {12} & {13} & {23} & \\ 0 & {.93} & {.01} & {.02} & {.04} & 0 & 0 & 0 \\ 1 & 0 & {.94} & 0 & 0 & {.02} & {.04} & 0 \\ 2 & 0 & 0 & {.95} & 0 & {.01} & 0 & {.04} \\ 3 & 0 & 0 & 0 & {.97} & 0 & {.01} & {.02} \\ {12} & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ {13} & 1 & 0 & 0 & 0 & 0 & 0 & 0 ... | Yes |
Theorem 1.1. The \( m \) step transition probability \( P\left( {{X}_{n + m} = j \mid {X}_{n} = i}\right) \) is the mth power of the transition matrix \( p \) . | The key ingredient in proving this is the Chapman-Kolmogorov equation\n\n\[ \n{p}^{m + n}\left( {i, j}\right) = \mathop{\sum }\limits_{k}{p}^{m}\left( {i, k}\right) {p}^{n}\left( {k, j}\right) \]\n\n(1.2)\n\nOnce this is proved, Theorem 1.1 follows, since taking \( n = 1 \) in (1.2), we see that\n\n\[ \n{p}^{m + 1}\lef... | Yes |
Example 1.11 (Gambler’s Ruin). Suppose for simplicity that \( N = 4 \) in Example 1.1 , so that the transition probability is\n\n\[\n\\begin{matrix} 0 & 1 & 2 & 3 & 4 & \\\\ 0 & {1.0} & 0 & 0 & 0 & 0 \\\\ 1 & {0.6} & 0 & {0.4} & 0 & 0 \\\\ 2 & 0 & {0.6} & 0 & {0.4} & 0 \\\\ 3 & 0 & 0 & {0.6} & 0 & {0.4} \\\\ 4 & 0 & 0 ... | To compute \( {p}^{2} \) one row at a time we note:\n\n\( {p}^{2}\left( {0,0}\right) = 1 \) and \( {p}^{2}\left( {4,4}\right) = 1 \), since these are absorbing states.\n\n\( {p}^{2}\left( {1,3}\right) = {\left( {.4}\right) }^{2} = {0.16} \), since the chain has to go up twice.\n\n\( {p}^{2}\left( {1,1}\right) = \left( ... | No |
Theorem 1.2 (Strong Markov Property). Suppose \( T \) is a stopping time. Given that \( T = n \) and \( {X}_{T} = y \), any other information about \( {X}_{0},\ldots {X}_{T} \) is irrelevant for predicting the future, and \( {X}_{T + k}, k \geq 0 \) behaves like the Markov chain with initial state \( y \) . | Why is this true? To keep things as simple as possible we will show only that\n\n\[ P\left( {{X}_{T + 1} = z \mid {X}_{T} = y, T = n}\right) = p\left( {y, z}\right) \]\n\nLet \( {V}_{n} \) be the set of vectors \( \left( {{x}_{0},\ldots ,{x}_{n}}\right) \) so that if \( {X}_{0} = {x}_{0},\ldots ,{X}_{n} = {x}_{n} \), t... | Yes |
We will show that eventually the chain gets stuck in either the bankrupt (0) or happy winner (4) state. In the terms of our recent definitions, we will show that states \( 0 < y < 4 \) are transient, while the states 0 and 4 are recurrent. | It is easy to check that 0 and 4 are recurrent. Since \( p\left( {0,0}\right) = 1 \), the chain comes back on the next step with probability one, i.e., \[ {P}_{0}\left( {{T}_{0} = 1}\right) = 1 \] and hence \( {\rho }_{00} = 1 \) . A similar argument shows that 4 is recurrent. In general if \( y \) is an absorbing stat... | Yes |
Example 1.13 (Social Mobility). Recall that the transition probability is\n\n12 3\n\n1.7 .2 .1\n\n\[ \text{2.3.5.2} \]\n\n\[ \text{3.2 .4 .4} \] | To begin we note that no matter where \( {X}_{n} \) is, there is a probability of at least 0.1 of hitting 3 on the next step so\n\n\[ {P}_{3}\left( {{T}_{3} > n}\right) \leq {\left( {0.9}\right) }^{n} \rightarrow 0\text{ as }n \rightarrow \infty \]\n\ni.e., we will return to 3 with probability 1 . The last argument app... | No |
Lemma 1.3. Suppose \( {P}_{x}\left( {{T}_{y} \leq k}\right) \geq \alpha > 0 \) for all \( x \) in the state space \( S \) . Then | \[ {P}_{x}\left( {{T}_{y} > {nk}}\right) \leq {\left( 1 - \alpha \right) }^{n} \] | Yes |
Lemma 1.4. If \( x \rightarrow y \) and \( y \rightarrow z \), then \( x \rightarrow z \) . | Proof. Since \( x \rightarrow y \) there is an \( m \) so that \( {p}^{m}\left( {x, y}\right) > 0 \) . Similarly there is an \( n \) so that \( {p}^{n}\left( {y, z}\right) > 0 \) . Since \( {p}^{m + n}\left( {x, z}\right) \geq {p}^{m}\left( {x, y}\right) {p}^{n}\left( {y, z}\right) \) it follows that \( x \rightarrow z... | Yes |
Theorem 1.5. If \( {\rho }_{xy} > 0 \), but \( {\rho }_{yx} < 1 \), then \( x \) is transient. | Proof. Let \( K = \min \left\{ {k : {p}^{k}\left( {x, y}\right) > 0}\right\} \) be the smallest number of steps we can take to get from \( x \) to \( y \) . Since \( {p}^{K}\left( {x, y}\right) > 0 \) there must be a sequence \( {y}_{1},\ldots {y}_{K - 1} \) so that\n\n\[ p\left( {x,{y}_{1}}\right) p\left( {{y}_{1},{y}... | Yes |
Lemma 1.6. If \( x \) is recurrent and \( {\rho }_{xy} > 0 \), then \( {\rho }_{yx} = 1 \) . | Proof. If \( {\rho }_{yx} < 1 \), then Lemma 1.5 would imply \( x \) is transient. | No |
Example 1.14 (A Seven-State Chain). Consider the transition probability: | To identify the states that are recurrent and those that are transient, we begin by drawing a graph that will contain an arc from \( i \) to \( j \) if \( p\left( {i, j}\right) > 0 \) and \( i \neq j \) . We do not worry about drawing the self-loops corresponding to states with \( p\left( {i, i}\right) > 0 \) since suc... | No |
Theorem 1.8. If the state space \( S \) is finite, then \( S \) can be written as a disjoint union \( T \cup {R}_{1} \cup \cdots \cup {R}_{k} \), where \( T \) is a set of transient states and the \( {R}_{i},1 \leq i \leq k \), are closed irreducible sets of recurrent states. | Proof. Let \( T \) be the set of \( x \) for which there is a \( y \) so that \( x \rightarrow y \) but \( y \nrightarrow x \) . The states in \( T \) are transient by Theorem 1.5. Our next step is to show that all the remaining states, \( S - T \), are recurrent.\n\nPick an \( x \in S - T \) and let \( {C}_{x} = \{ y ... | Yes |
Lemma 1.10. In a finite closed set there has to be at least one recurrent state. | To prove these results we need to introduce a little more theory. Recall the time of the \( k \) th visit to \( y \) defined by\n\n\[ \n{T}_{y}^{k} = \min \left\{ {n > {T}_{y}^{k - 1} : {X}_{n} = y}\right\} \n\]\n\nand \( {\rho }_{xy} = {P}_{x}\left( {{T}_{y} < \infty }\right) \) the probability we ever visit \( y \) a... | No |
Lemma 1.11. \( {E}_{x}N\left( y\right) = {\rho }_{xy}/\left( {1 - {\rho }_{yy}}\right) \) | Proof. Accept for the moment the fact that for any nonnegative integer valued random variable \( X \), the expected value of \( X \) can be computed by\n\n\[ \n{EX} = \mathop{\sum }\limits_{{k = 1}}^{\infty }P\left( {X \geq k}\right) \]\n\n(1.6)\n\nWe will prove this after we complete the proof of Lemma 1.11. Now the p... | No |
Lemma 1.12. \( {E}_{x}N\left( y\right) = \mathop{\sum }\limits_{{n = 1}}^{\infty }{p}^{n}\left( {x, y}\right) \) . | Proof. Let \( {1}_{\left\{ {X}_{n} = y\right\} } \) denote the random variable that is 1 if \( {X}_{n} = y,0 \) otherwise. Clearly\n\n\[ N\left( y\right) = \mathop{\sum }\limits_{{n = 1}}^{\infty }{1}_{\left\{ {X}_{n} = y\right\} } \]\n\nTaking expected values now gives\n\n\[ {E}_{x}N\left( y\right) = \mathop{\sum }\li... | Yes |
Theorem 1.13. \( y \) is recurrent if and only if\n\n\[ \mathop{\sum }\limits_{{n = 1}}^{\infty }{p}^{n}\left( {y, y}\right) = {E}_{y}N\left( y\right) = \infty \] | Proof. The first equality is Lemma 1.12. From Lemma 1.11 we see that \( {E}_{y}N\left( y\right) = \) \( \infty \) if and only if \( {\rho }_{yy} = 1 \), which is the definition of recurrence. | Yes |
Consider the weather chain (Example 1.3) and suppose that the initial distribution is \( q\left( 1\right) = {0.3} \) and \( q\left( 2\right) = {0.7} \) . | In this case\n\n\[ \left( \begin{array}{ll} {.3} & {.7} \end{array}\right) \left( \begin{array}{ll} {.6} & {.4} \\ {.2} & {.8} \end{array}\right) = \left( \begin{array}{ll} {.32} & {.68} \end{array}\right) \]\n\n\[ \text{since}\;{.3}\left( {.6}\right) + {.7}\left( {.2}\right) = {.32} \]\n\n\[ {.3}\left( {.4}\right) + {... | Yes |
Consider the social mobility chain (Example 1.4) and suppose that the initial distribution: \( q\left( 1\right) = {.5}, q\left( 2\right) = {.2} \), and \( q\left( 3\right) = {.3} \) . Multiplying the vector \( q \) by the transition probability gives the vector of probabilities at time 1 . | \[ \left( \begin{array}{lll} {.5} & {.2} & {.3} \end{array}\right) \left( \begin{array}{lll} {.7} & {.2} & {.1} \\ {.3} & {.5} & {.2} \\ {.2} & {.4} & {.4} \end{array}\right) = \left( \begin{array}{lll} {.47} & {.32} & {.21} \end{array}\right) \] To check the arithmetic note that the three entries on the right-hand sid... | Yes |
Example 1.17 (Weather Chain). To compute the stationary distribution we want to solve\n\n\[ \left( \begin{array}{ll} {\pi }_{1} & {\pi }_{2} \end{array}\right) \left( \begin{array}{ll} {.6} & {.4} \\ {.2} & {.8} \end{array}\right) = \left( \begin{array}{ll} {\pi }_{1} & {\pi }_{2} \end{array}\right) \] | Multiplying gives two equations:\n\n\[ {.6}{\pi }_{1} + {.2}{\pi }_{2} = {\pi }_{1} \]\n\n\[ {.4}{\pi }_{1} + {.8}{\pi }_{2} = {\pi }_{2} \]\n\nBoth equations reduce to \( {.4}{\pi }_{1} = {.2}{\pi }_{2} \) . Since we want \( {\pi }_{1} + {\pi }_{2} = 1 \), we must have \( {.4}{\pi }_{1} = {.2} - {.2}{\pi }_{1} \), and... | Yes |
The equation \( {\pi p} = \pi \) says\n\n\[ \left( \begin{array}{lll} {\pi }_{1} & {\pi }_{2} & {\pi }_{3} \end{array}\right) \left( \begin{array}{lll} {.7} & {.2} & {.1} \\ {.3} & {.5} & {.2} \\ {.2} & {.4} & {.4} \end{array}\right) = \left( \begin{array}{lll} {\pi }_{1} & {\pi }_{2} & {\pi }_{3} \end{array}\right) \] | which translates into three equations\n\n\[ {.7}{\pi }_{1} + {.3}{\pi }_{2} + {.2}{\pi }_{3} = {\pi }_{1} \]\n\n\[ {.2}{\pi }_{1} + {.5}{\pi }_{2} + {.4}{\pi }_{3} = {\pi }_{2} \]\n\n\[ {.1}{\pi }_{1} + {.2}{\pi }_{2} + {.4}{\pi }_{3} = {\pi }_{3} \]\n\nNote that the columns of the matrix give the numbers in the rows o... | Yes |
Using the first two equations and the fact that the sum of the \( \pi \) ’s is 1\n\n\[{.8}{\pi }_{1} + {.2}{\pi }_{2} + {.3}{\pi }_{3} = {\pi }_{1}\]\n\n\[{.1}{\pi }_{1} + {.6}{\pi }_{2} + {.3}{\pi }_{3} = {\pi }_{2}\]\n\n\[{\pi }_{1} + {\pi }_{2} + {\pi }_{3} = 1\] | Subtracting \( {\pi }_{1} \) from both sides of the first equation and \( {\pi }_{2} \) from both sides of the second, this translates into \( {\pi A} = \left( {0,0,1}\right) \) with\n\n\[A = \left( \begin{matrix} - {.2} & {.1} & 1 \\ {.2} & - {.4} & 1 \\ {.3} & {.3} & 1 \end{matrix}\right)\]\n\nNote that here and in t... | Yes |
Example 1.20 (Hot Hand (Continuation of 1.10)). To find the stationary matrix in this case we can follow the same procedure. \( A \) consists of the first three columns of the transition matrix with 1 subtracted from the diagonal, and a final column of all 1's. | The answer is given by the fourth row of \( {A}^{-1} \) :\n\n\[ \left( {{0.5},{0.1875},{0.1875},{0.125}}\right) = \left( {1/2,3/{16},3/{16},1/8}\right) \]\n\nThus the long run fraction of time the player hits a shot is\n\n\[ \pi \left( \mathrm{{HH}}\right) + \pi \left( \mathrm{{MH}}\right) = {0.6875} = {11}/{36}. \] | Yes |
Theorem 1.14. If \( p \) is a doubly stochastic transition probability for a Markov chain with \( N \) states, then the uniform distribution, \( \pi \left( x\right) = 1/N \) for all \( x \), is a stationary distribution. | Proof. To check this claim we note that if \( \pi \left( x\right) = 1/N \) then\n\n\[ \mathop{\sum }\limits_{x}\pi \left( x\right) p\left( {x, y}\right) = \frac{1}{N}\mathop{\sum }\limits_{x}p\left( {x, y}\right) = \frac{1}{N} = \pi \left( y\right) \]\n\nLooking at the second equality we see that conversely, if \( \pi ... | Yes |
Example 1.21 (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 1.22 (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 + ... | It is clear that the columns add to one, so the stationary distribution is uniform. To check the hypothesis of the convergence theorem, we note that after 3 turns we will have moved between three and nine spaces so \( {p}^{3}\left( {i, j}\right) > 0 \) for all \( i \) and \( j \) . | Yes |
Example 1.23 (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 shifts one 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 th... | Yes |
\( \pi \) is said to satisfy the detailed balance condition if\n\n\[ \pi \left( x\right) p\left( {x, y}\right) = \pi \left( y\right) p\left( {y, x}\right) \] | To see that this is a stronger condition than \( {\pi p} = \pi \), we sum over \( x \) on each side to get\n\n\[ \mathop{\sum }\limits_{x}\pi \left( x\right) p\left( {x, y}\right) = \pi \left( y\right) \mathop{\sum }\limits_{x}p\left( {y, x}\right) = \pi \left( y\right) \]\n\nAs in our earlier discussion of stationary ... | 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(... | \[ \pi \left( {x + 1}\right) = \frac{{p}_{x}}{{q}_{x + 1}} \cdot \pi \left( x\right) \]\n\n(1.12)\n\nUsing this with \( x = \ell \) gives \( \pi \left( {\ell + 1}\right) = \pi \left( \ell \right) {p}_{\ell }/{q}_{\ell + 1} \) . Taking \( x = \ell + 1 \)\n\n\[ \pi \left( {\ell + 2}\right) = \frac{{p}_{\ell + 1}}{{q}_{\e... | Yes |
Example 1.27 (Ehrenfest Chain). For concreteness, suppose there are three balls. In this case the transition probability is \( \begin{array}{llll} 0 & 1 & 2 & 3 \end{array} \) \( \begin{array}{lllll} 0 & 0 & 3/3 & 0 & 0 \end{array} \) \( {11}/{30}\;2/{30} \) \[ \begin{array}{lllll} 2 & 0 & 2/3 & 0 & 1/3 \end{array} \] ... | Setting \( \pi \left( 0\right) = c \) and using (1.12) we have \[ \pi \left( 1\right) = {3c},\;\pi \left( 2\right) = \pi \left( 1\right) = {3c}\;\pi \left( 3\right) = \pi \left( 2\right) /3 = c. \] The sum of the \( \pi \) ’s is \( {8c} \), so we pick \( c = 1/8 \) to get \[ \pi \left( 0\right) = 1/8,\;\pi \left( 1\rig... | No |
Example 1.28 (Three Machines, One Repairman). Suppose that an office has three machines that each break with probability .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 same... | 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 (.1)(.5), while the state can only increase by 1 if he succeeds and there is no new failure, an event of probability .5... | Yes |
Example 1.29 (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 1.30 (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 ... | Yes |
Theorem 1.15. 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\[ \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\[ P\left( {{Y}_{m + 1} = {i}_{m + 1} \mid {Y}_{m} = {i}_{m},{Y}_{m - 1} = {i}_{m - 1}\ldots {Y}_{0} = {i}_{0}}\right) \]\n\n\[ = \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}_{n} = ... | Yes |
Example 1.31 (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 1.32 (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 |
Consider the chain with transition probability: \n\nand suppose that all entries in the matrix are positive. To satisfy detailed balance we must have\n\n\\[ \n{e\pi }\\left( 2\\right) = {a\pi }\\left( 1\\right) \\;{f\p... | To construct a measure that satisfies detailed balance we take \\( \\pi \\left( 1\\right) = k \\) . From this it follows that\n\n\\[ \n\\pi \\left( 2\\right) = k\\frac{a}{e}\\;\\pi \\left( 3\\right) = k\\frac{ab}{ef}\\;k = \\pi \\left( 1\\right) = k\\frac{abc}{def} \n\\]\n\nFrom this we see that there is a stationary d... | Yes |
Example 1.36 (Ehrenfest Chain (Continuation of 1.2)). For concreteness, suppose there are three balls. In this case the transition probability is \( \begin{array}{llll} 0 & 1 & 2 & 3 \end{array} \) \( \begin{array}{lllll} 0 & 0 & 3/3 & 0 & 0 \end{array} \) | To see that the zeros will persist, note that if we have an odd number of balls in the left urn, then no matter whether we add or subtract one the result will be an even number. Likewise, if the number is even, then it will be odd on the next one step. This alternation between even and odd means that it is impossible t... | No |
Example 4.4 (Triangle and Square). Consider the transition matrix:\n\n\[ \n- 2 - {10}\;{123} \n\]\n\n\[ \n\text{00.500.500} \n\]\n\n\[ \n\begin{array}{lllllll} 1 & 0 & 0 & 0 & 0 & 1 & 0 \end{array} \n\]\n\n\[ \n\begin{array}{lllllll} 3 & 0 & 0 & 1 & 0 & 0 & 0 \end{array} \n\]\n\nIn words, from 0 we are equally likely t... | Clearly, \( {p}^{3}\left( {0,0}\right) > 0 \) and \( {p}^{4}\left( {0,0}\right) > 0 \) so \( 3,4 \in {I}_{0} \) and hence \( {d}_{0} = 1 \) . | No |
Lemma 1.17. If \( {\rho }_{xy} > 0 \) and \( {\rho }_{yx} > 0 \), then \( x \) and \( y \) have the same period. | Proof. Suppose that the period of \( x \) is \( c \), while the period of \( y \) is \( d < c \) . Let \( k \) be such that \( {p}^{k}\left( {x, y}\right) > 0 \) and let \( m \) be such that \( {p}^{m}\left( {y, x}\right) > 0 \) . Since\n\n\[ \n{p}^{k + m}\left( {x, x}\right) \geq {p}^{k}\left( {x, y}\right) {p}^{m}\le... | Yes |
Lemma 1.18. If \( p\left( {x, x}\right) > 0 \), then \( x \) has period 1 . | Proof. If \( p\left( {x, x}\right) > 0 \), then \( 1 \in {I}_{x} \), so the greatest common divisor is 1 . | No |
Theorem 1.23. Suppose I, S.\n\n\[ \frac{1}{n}\mathop{\sum }\limits_{{m = 1}}^{n}{p}^{m}\left( {x, y}\right) \rightarrow \pi \left( y\right) \] | Thus while the sequence \( {p}^{m}\left( {x, y}\right) \) will not converge in the periodic case, the average of the first \( n \) values will. | No |
Example 1.38 (Repair Chain (Continuation of 1.7)). A machine has three critical parts that are subject to failure, but can function as long as two of these parts are working. When two are broken, they are replaced and the machine is back to working order the next day. Declaring the state space to be the parts that are ... | To find the stationary distribution we look at the last row of\n\n\[ {\left( \begin{matrix} - {.07} & {.01} & {.02} & {.04} & 0 & 0 & 1 \\ 0 & - {.06} & 0 & 0 & {.02} & {.04} & 1 \\ 0 & 0 & - {.05} & 0 & {.01} & 0 & 1 \\ 0 & 0 & 0 & - {.03} & 0 & {.01} & 1 \\ 1 & 0 & 0 & 0 & - 1 & 0 & 1 \\ 1 & 0 & 0 & 0 & 0 & - 1 & 1 \... | Yes |
Q. Suppose we make \$12 profit on each unit sold but it costs \$2 a day to store items. What is the long-run profit per day of this inventory policy? | The first thing we have to do is to compute the stationary distribution. The last row of\n\n\[ \n{\left( \begin{matrix} - 1 & 0 & {.1} & {.2} & {.4} & 1 \\ 0 & - 1 & {.1} & {.2} & {.4} & 1 \\ {.3} & {.4} & - {.7} & 0 & 0 & 1 \\ {.1} & {.2} & {.4} & - {.7} & 0 & 1 \\ 0 & {.1} & {.2} & {.4} & - {.7} & 1 \\ 0 & 0 & {.1} &... | Yes |
Theorem 1.20. 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 (1.16). 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 1.21. 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 1.20 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 1.24. Suppose \( p \) is irreducible and recurrent. Let \( x \in S \) and let \( {T}_{x} = \) \( \inf \left\{ {n \geq 1 : {X}_{n} = x}\right\} \)\n\n\[{\mu }_{x}\left( y\right) = \mathop{\sum }\limits_{{n = 0}}^{\infty }{P}_{x}\left( {{X}_{n} = y,{T}_{x} > n}\right)\]\n\ndefines 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\n\n\[ \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}\le... | Yes |
Theorem 1.22. 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}... | No |
Lemma 1.25. 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 |
Lemma 1.26. If \( x \) has period 1, i.e., the greatest common divisor \( {I}_{x} \) is 1, then there is a number \( {n}_{0} \) so that if \( n \geq {n}_{0} \), then \( n \in {I}_{x} \) . In words, \( {I}_{x} \) contains all of the integers after some value \( {n}_{0} \) . | To prove this we begin by proving\n\nLemma 1.27. \( {I}_{x} | No |
Lemma 1.27. \( {I}_{x} \) is closed under addition. That is, if \( i, j \in {I}_{x} \), then \( i + j \in {I}_{x} \) . | Proof. If \( i, j \in {I}_{x} \), then \( {p}^{i}\left( {x, x}\right) > 0 \) and \( {p}^{j}\left( {x, x}\right) > 0 \) so\n\n\[ \n{p}^{i + j}\left( {x, x}\right) \geq {p}^{i}\left( {x, x}\right) {p}^{j}\left( {x, x}\right) > 0 \n\]\n\nand hence \( i + j \in {I}_{x} \) . | Yes |
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