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At a local two year college, \( {60}\% \) of freshmen become sophomores, \( {25}\% \) remain freshmen, and \( {15}\% \) drop out. \( {70}\% \) of sophomores graduate and transfer to a four year college, \( {20}\% \) remain sophomores and \( {10}\% \) 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\nLet \( h\left( x\right) \) be the probability that a student currently in state \( x \) eventually graduates. By considering what happens on one step\... | 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 probabilit... | 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</td><td>1</td><t... | Yes |
Theorem 1.28. Consider a Markov chain with state space \( S \) . Let \( A \) and \( B \) be subsets of \( S \), so that \( C = S - \left( {A \cup B}\right) \) is finite. Suppose \( h\left( a\right) = 1 \) for \( a \in A, h\left( b\right) = 0 \) for \( b \in B \), and that for \( x \in C \) we have\n\n\[ h\left( x\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 \( x \in C \) . (1.18) implies that \( h\left( x\right) = {E}_{x}h\left( {X}_{1}\right) \) when \( x \in C \) . The Markov property implies\n\n\[ h\left( x\right) = {E}_{x}h\left( {X}_{T \lan... | Yes |
Example 1.42 (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 two pennies (for a profit of 1). If the two coins are different, then Charlie gets the two pennies. They quit when someone has all of the pennies.... | 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 1.43 (Wright-Fisher Model with No Mutation). The state space is \( S = \) \( \{ 0,1,\ldots N\} \) and the transition probability is\n\n\[ p\left( {x, y}\right) = \left( \begin{array}{l} N \\ y \end{array}\right) {\left( \frac{x}{N}\right) }^{y}{\left( \frac{N - x}{N}\right) }^{N - y} \] | The right-hand side is the binomial \( \left( {N, x/N}\right) \) distribution, i.e., the number of successes in \( N \) trials when success has probability \( x/N \), so the mean number of successes is \( x \) . From this it follows that if we define \( h\left( x\right) = x/N \), then\n\n\[ h\left( x\right) = \mathop{\... | 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 stop. Let\n\n\[ h\left( x\right) = {P}_{x}\... | 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\[ h\left( {x + 1}\right) - h\left( x\right) = \frac{q}{p} \cdot \left( {h\left( x\right) - h\left( {x - 1}\right) }\right) \]\n\nIf we set \(... | Yes |
If we bet \( \$ 1 \) on red on a roulette wheel with \( {18}\mathrm{{red}},{18} \) black, and 2 green \( \left( {0\text{and 00}}\right) \) 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 hop... | Here \( \theta = q/p = {20}/{18} \), so (1.24) 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 |
Example 1.46. If we rearrange the matrix for the seven state chain in Example 1.14 we get\n\n<table><thead><tr><th></th><th>2</th><th>3</th><th>1</th><th>5</th><th>4</th><th>6</th><th>7</th></tr></thead><tr><td>2</td><td>.2</td><td>.3</td><td>.1</td><td>0</td><td>.4</td><td>0</td><td>0</td></tr><tr><td>3</td><td>0</td>... | The first step is to note that by our formula for two state chains the stationary distribution on the closed irreducible set \( A = \{ 1,5\} \) is \( 2/3,1/3 \) . With a little more work one concludes that the stationary distribution on \( B = \{ 4,6,7\} \) is \( 8/{17},5/{17},4/{17} \) the third row of\n\n\[ \n{\left(... | 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 the \( 1... | 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\n\n\[ g\left( x\right) = 1 + \mathop{\sum }\limits_{y}p\left( {x, y}\right) g\left( y\right) \]\n\nSince \( g\left( 2\right) = g\left( {-2}\right) = 0 \), if we let \( r\left( {x, y}\right) \) be the restriction of the transition probability to \( 1,0, - 1 \) we have\n\n\[ g\left... | Yes |
Theorem 1.29. Let \( {V}_{A} = \inf \left\{ {n \geq 0 : {X}_{n} \in A}\right\} \) . Suppose \( C = S - A \) is finite, and that \( {P}_{x}\left( {{V}_{A} < \infty }\right) > 0 \) for any \( x \in C \) . If \( g\left( a\right) = 0 \) for all \( a \in A \), and for \( x \in C \) we have\n\n\[ g\left( x\right) = 1 + \math... | Proof. It follows from Lemma 1.3 that \( {E}_{x}{V}_{A} < \infty \) for all \( x \in C \) . (1.27) 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( {{V}_{A} \land n}\right) + {E}_{x}g\left( {X}_{{V}_{A} \la... | 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 happens after we... | 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. | 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\[ E{T}_{HT} = 2 + \frac{1}{4}\left( {0 + 2 + 2 + 4}\right) = 4 \] | Yes |
On the average how many times do we need to roll one die in order to see a run of six rolls with all different numbers? | To formulate the problem we need some notation. Let \( {X}_{n}, n \geq 1 \) be independent and equal to \( 1,2,3,4,5,6 \), with probability \( 1/6 \) each. Let \( {K}_{n} = \max \left\{ {k : {X}_{n},{X}_{n - 1},\ldots {X}_{n - k + 1}}\right\} \) are all different. To explain the last definition note that if the last fi... | 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 \( \tau = \min \left\{ {n : {X}_{n} \notin \left( {0, N}\right) }\right\} \) . We claim that\n\n\[ \n{E}_{x}\tau = \frac{x}{q - p} - \frac{N}{q - p} \cdot \frac{1 - {\left( q/... | 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 s... | \\[ \n{p\\pi }\\left( i\\right) = \\left( {1 - p}\\right) \\pi \\left( {i + 1}\\right) \\;\\text{ when }i \\geq 0 \n\\]\n\nRewriting this as \\( \\pi \\left( {i + 1}\\right) = \\pi \\left( i\\right) \\cdot p/\\left( {1 - p}\\right) \\) and setting \\( \\pi \\left( 0\\right) = c \\), we have\n\n\\[ \n\\pi \\left( i\\rig... | Yes |
Theorem 1.30. 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 1.24 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 |
The extinction probability \( \rho \) is the smallest solution of the equation \( \phi \left( x\right) = x \) with \( 0 \leq x \leq 1 \) . | Proof. Extending the reasoning for (1.36) 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 |
II. If \( \mu > 1 \), then there is positive probability of avoiding extinction. | Proof. In view of Lemma 1.31, 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 \( \phi \left( 0\right) = {p}_{0} > 0 \) . Differentiatin... | 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 ti... | 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 <... | 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\... | No |
Theorem 1.28. Consider a Markov chain with finite state space \( S \) . Let \( A, B \subset S \), so that \( C = S - \{ a, b\} \) is finite, and \( {P}_{x}\left( {{V}_{A} \land {V}_{B} < \infty }\right) > 0 \) for all \( x \in C \) . If \( h\left( a\right) = 1 \) for \( a \in A, h\left( b\right) = 0 \) for \( b \in B \... | Let \( r\left( {x, y}\right) \) be the part of the matrix \( p\left( {x, y}\right) \) with \( x, y \in C \) and for \( x \in C \) let \( v\left( x\right) = \mathop{\sum }\limits_{{y \in A}}p\left( {x, y}\right) \) which we think of as a column vector. Since \( h\left( a\right) = 1 \) for \( a \in A \) and \( h\left( b\... | Yes |
Theorem 1.29. Consider a Markov chain with finite state space \( S \) . Let \( A \subset S \) so that \( C = S - A \) is finite and \( {P}_{x}\left( {{V}_{A} < \infty }\right) > 0 \) for \( x \in C \) . If \( g\left( a\right) = 0 \) for all \( a \in A \) , and\n\n\[ g\left( x\right) = 1 + \mathop{\sum }\limits_{y}p\lef... | 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 |
Example 2.1. Anne and Betty enter a beauty parlor simultaneously, Anne to get a manicure and Betty to get a haircut. Suppose the time for a manicure (haircut) is exponentially distributed with mean 20 (30) minutes. (a) What is the probability Anne gets done first? (b) What is the expected amount of time until Anne and ... | (a) The rates are 1/20 and 1/30 per hour so Anne finishes first with probability\n\n\[ \n\frac{1/{20}}{1/{20} + 1/{30}} = \frac{30}{{30} + {20}} = \frac{3}{5} \n\]\n\n(b) The total service rate is \( 1/{30} + 1/{20} = 5/{60} \), so the time until the first customer completes service is exponential with mean 12 minutes.... | Yes |
Theorem 2.1. Let \( V = \min \left( {{T}_{1},\ldots ,{T}_{n}}\right) \) and \( I \) be the (random) index of the \( {T}_{i} \) that is smallest.\n\n\[ P\left( {V > t}\right) = \exp \left( {-\left( {{\lambda }_{1} + \cdots + {\lambda }_{n}}\right) t}\right) \]\n\n\[ P\left( {I = i}\right) = \frac{{\lambda }_{i}}{{\lambd... | Proof. Arguing as in the case of two random variables:\n\n\[ P\left( {\min \left( {{T}_{1},\ldots ,{T}_{n}}\right) > t}\right) = P\left( {{T}_{1} > t,\ldots {T}_{n} > t}\right) \]\n\n\[ = \mathop{\prod }\limits_{{i = 1}}^{n}P\left( {{T}_{i} > t}\right) = \mathop{\prod }\limits_{{i = 1}}^{n}{e}^{-{\lambda }_{i}t} = {e}^... | Yes |
A submarine has three navigational devices but can remain at sea if at least two are working. Suppose that the failure times are exponential with means 1 year, 1.5 years, and 3 years. (a) What is the average length of time the boat can remain at sea? | (a) The rates for the three exponentials are 1, 2/3, and 1/3, per year. Thus the time to the first failure is exponential with rate \( 2 = 1 + 2/3 + 1/3 \), so the mean time to first failure is \( 1/2 \). \( 1/2 \) of the time part 1 is the first to fail. In this case the time to the next failure has rate \( 2/3 + 1/3 ... | Yes |
Theorem 2.2. 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 |
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(2.10)\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 Poisson \( \left( {\lambda }_{i}\right) \), then\n\n\[ \n{X}_{1} + \cdots + {X}_{k} = \operatorname{Poisson}\left( {{\lambda }_{1} + \cdots + {\lambda }_{n}}\right) .\n\] | Proof. It suffices to prove the result for \( k = 2 \), for then the general result follows by induction.\n\n\[ \nP\left( {{X}_{1} + {X}_{2} = n}\right) = \mathop{\sum }\limits_{{m = 0}}^{n}P\left( {{X}_{1} = m}\right) P\left( {{X}_{2} = n - m}\right)\n\]\n\n\[ \n= \mathop{\sum }\limits_{{m = 0}}^{n}{e}^{-{\lambda }_{1... | Yes |
Theorem 2.5. 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, (2.11) 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 |
Lemma 2.6. \( N\left( s\right) \) 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 |
Lemma 2.7. \( N\left( {t + s}\right) - N\left( s\right), t \geq 0 \) 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 reuse the last picture) 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 must have \( {\tau }_{5} > \)... | Yes |
Lemma 2.8. \( N\left( t\right) \) has independent increments. | Why is this true? Lemma 2.7 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 n... | 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 cal... | 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 \( N = 0 \) .\n\n(i) If \( E\left| {Y}_{i}\right| ,{EN} < \infty \), then \( {ES} ... | 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... | \[ {81} \cdot \left\{ {{\left( \$ 6\right) }^{2} + {\left( \$ 8\right) }^{2}}\right\} = \$ {8100} \] Taking 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 rate \( {\lambda P}\left( {{Y}_{i} = j}\right) \) Poisson processes. | 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 \( {N}_{2}\left( t\right) \) . It should be clear that the independent increments property of the Poisson process i... | Yes |
Ellen catches fish at times of a Poisson process with rate 2 per hour. \( {40}\% \) of the fish are salmon, while \( {60}\% \) of the fish are trout. What is the probability she will catch exactly 1 salmon and 2 trout if she fishes for 2.5 hours? | The total number of fish she catches in 2.5 hours is Poisson with mean 5, so the number of salmon and the number of trout are independent Poissons with means 2 and 3. Thus the probability of interest is\n\n\[ \n{e}^{-2}\frac{{2}^{1}}{1!} \cdot {e}^{-3}\frac{{3}^{2}}{2!} \n\] | Yes |
Two copy editors read a 300-page manuscript. The first found 100 typos, the second found 120, and their lists contain 80 errors in common. Suppose that the author’s typos follow a Poisson process with some unknown rate \( \lambda \) per page, while the two copy editors catch errors with unknown probabilities of success... | Let \( {X}_{0} \) be the number of typos that neither found. Let \( {X}_{1} \) and \( {X}_{2} \) be the number of typos found only by 1 or only by 2, and let \( {X}_{3} \) be the number of typos found by both. If we let \( \mu = {300\lambda } \), the \( {X}_{i} \) are independent Poisson with means\n\n\[ \mu \left( {1 ... | Yes |
Suppose that a Poisson number of Duke students with mean 2263 will show up to watch the next women's basketball game. What is the probability that for all of the 365 days there is at least one person in the crowd who has that birthday. (Pretend February 29th does not exist.) | By thinning if we let \( {N}_{j} \) be the number of people who have birthdays on the \( j \) th day of the year then the \( {N}_{j} \) are independent Poisson mean \( {2263}/{365} = {6.2} \) . The probability that all of \( {N}_{j} > 0 \) is\n\n\[ \n{\left( 1 - {e}^{-{6.2}}\right) }^{365} = {0.4764} \n\] | Yes |
Example 2.9 (M/G/ \( \infty \) Queue). As one walks around the Duke campus it seems that every student is talking on their smartphone. The argument for arrivals at the ATM 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... | \[ {\int }_{s = 0}^{t}\lambda \left( {1 - G\left( {t - s}\right) }\right) {ds} = \lambda {\int }_{r = 0}^{t}\left( {1 - G\left( r\right) }\right) {dr} \] | Yes |
What is the probability in equilibrium that there are four men and two women in the store? | By Poisson thinning the arrivals of men and women are independent Poisson process with rate 6 and 4 . Since the mean time in the store is \( 1/2 \) for men and \( 1/4 \) for women, by Theorem 2.13 the number of men \( M \) and women \( W \) in equilibrium are independent Poissons with means 3 and 1. Thus\n\n\[ P\left( ... | Yes |
Example 2.11. People arrive at a puzzle exhibit according to a Poisson process with rate 2 per minute. The exhibit has enough copies of the puzzle so everyone at the exhibit can have one to play with. Suppose the puzzle takes an amount of time to solve that is uniform on \( \\left( {0,{10}}\\right) \) minutes. (a) What... | (a) The probability a customer who arrived \( x \) minutes ago is still working on the puzzle is \( x/{10} \), so by Poisson thinning the number is Poisson with mean \( 2\\int_{0}^{10} x/{10} dx = 10 \). (b) The number that has been working more than four minutes is Poisson with mean \( 2\\int_{0}^{6} x/{10} dx = 36/{1... | Yes |
Theorem 2.14. 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. 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 6 red arrivals before a total of 4 green ones? | The first step is to note that the event in question is equivalent to having at least 6 red arrivals in the first 9 . If this happens, then we have at most 3 green arrivals before the 6th red one. On the other hand, if there are 5 or fewer red arrivals in the first 9, then we have had at least 4 red arrivals and at mos... | Yes |
Theorem 2.15. 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 3 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( t\ri... | Yes |
Theorem 2.16. 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 \( \left( {n, s/t}\right) \) .\n\nOne can also prove this directl... | Yes |
For a concrete example, suppose \( N\left( 3\right) = 4 \) . | \[ P\left( {N\left( 1\right) = 1 \mid N\left( 3\right) = 4}\right) = 4 \cdot {\left( 1/3\right) }^{1}{\left( 2/3\right) }^{3} = {32}/{81} \] | Yes |
Example 2.14. Trucks and cars on highway US 421 are Poisson processes with rate 40 and 100 per hour, respectively. 1/8 of the trucks and 1/10 of the cars get off on exit 257 to go to the Bojangle's in Yadkinville. (a) Find the probability that exactly six trucks arrive at Bojangle's between noon and 1 p.m. | (a) By thinning trucks are Poisson with rate 5, so \( {e}^{-5}{5}^{6}/6! = {0.1462} \) . | Yes |
Theorem 3.2 (Strong Law of Large Numbers). Let \( {x}_{1},{x}_{2},{x}_{3},\ldots \) be i.i.d. with \( E{x}_{i} = \mu \), and let \( {S}_{n} = {x}_{1} + \cdots + {x}_{n} \) . Then with probability one,\n\n\[ \n{S}_{n}/n \rightarrow \mu \;\text{ as }n \rightarrow \infty \n\] | Taking \( {x}_{i} = {t}_{i} \), we have \( {S}_{n} = {T}_{n} \), so Theorem 3.2 implies that with probability one, \( {T}_{n}/n \rightarrow \mu \) as \( n \rightarrow \infty \) . Now by definition,\n\n\[ \n{T}_{N\left( t\right) } \leq t < {T}_{N\left( t\right) + 1} \n\]\n\nDividing by \( N\left( t\right) \), we have\n\... | Yes |
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 last step we have us... | 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 ... | Yes |
Example 3.5 (Poisson Janitor). A light bulb burns for an amount of time having distribution \( F \) with mean \( {\mu }_{F} \) then burns out. A janitor comes at times of a rate \( \lambda \) Poisson process to check the bulb and will replace the bulb if it is burnt out. (a) At what rate are bulbs replaced? (b) What is... | Solution. Suppose that a new bulb is put in at time 0 . It will last for an amount of time \( {s}_{1} \) . Using the lack of memory property of the exponential distribution, it follows that the amount of time until the next inspection, \( {u}_{1} \), will have an exponential distribution with rate \( \lambda \) . The b... | 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 \( \leq \lambda /\mu \) . | 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... | Yes |
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... | Yes |
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 | \[ {W}_{Q} = W - E{s}_{i} \] | Yes |
Theorem 3.7. If \( \lambda < \mu \), then \( {X}_{n} \) is positive recurrent and \( {E}_{0}{T}_{0} = \mu /\left( {\mu - \lambda }\right) \) . | Proof. Turning to the second conclusion, note that setting \( \lambda = \mu \) in the first result gives \( {E}_{0}{T}_{0} = \infty \) . To begin the proof of recurrence note that \( E{\zeta }_{i} = 0 \) so if \( x > 0 \) then \( {E}_{x}{X}_{1} = x \) . Let \( \tau \left( M\right) = \min \left\{ {n : {X}_{n} = 0}\right... | Yes |
Theorem 3.8. \( {\alpha }_{n} = \pi \left( n\right) \) . | Why is this true? Suppose the process has been running since time \( - \infty \) and hence the system is in equilibrium. If we condition on there being arrival at time \( t \), then the times of the previous arrivals are a Poisson process with rate \( \lambda \) . Thus knowing that there is an arrival at time \( t \) d... | No |
Theorem 3.9 (Pollaczek-Khintchine Formula). The long run average waiting time in queue\n\n\\[ \n{W}_{Q} = \\frac{{\\lambda E}\\left( {{s}_{i}^{2}/2}\\right) }{1 - {\\lambda E}{s}_{i}} \n\\]\n\n(3.8) | Proof. 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\\[ \nZ = \\mathop{\\lim }\\limits_{{t \\rightarrow \\infty }}\\frac{1}{t}{\\int }_{0}^{t}{Z}_{s}{ds} \n\\]\n\nAs in the p... | Yes |
(a) In the long run what is the fraction of time, \( \pi \left( 0\right) \), that the server is idle? | \[ \lambda = 1/6, E{s}_{i} = 5 = 1/\mu \text{, so by (3.5)}\pi \left( 0\right) = 1 - \left( {1/6}\right) /\left( {1/5}\right) = 1/6\text{.} \] | Yes |
Example 3.8 (Visits to Go). In Monopoly one rolls two dice and then moves that number of squares. As in Example 1.23 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... | <table><tr><td>0</td><td>1</td><td></td><td>3</td><td></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>\( \frac... | Yes |
Theorem 3.11. As \( t \rightarrow \infty \)\n\n\[ \frac{1}{t}{\int }_{0}^{t}{1}_{\left\{ {A}_{s} > x,{Z}_{s} > y\right\} }{ds} \rightarrow \frac{1}{E{t}_{1}}{\int }_{x + y}^{\infty }P\left( {{t}_{i} > z}\right) {dz} \] | 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 3.9 (Exponential). In this case the limiting density given in (3.11) is | \[ \frac{\lambda {e}^{-\lambda \left( {a + z}\right) }}{1/\lambda } = \lambda {e}^{-{\lambda a}} \cdot \lambda {e}^{-{\lambda z}} \] | Yes |
Example 3.10 (Uniform on \( \\left( {0, b}\\right) \) ). Plugging into (3.11) gives for \( a, z > 0, a + z < b \) : | \[ \n\\frac{1/b}{b/2} = \\frac{2}{{b}^{2}} \n\] \n\ni.e., the limiting joint density is uniform on the triangle of possible values. The marginal densities given in (3.10) are \n\n\[ \n\\frac{\\left( {b - x}\\right) /b}{b/2} = \\frac{2}{b} \\cdot \\left( {1 - \\frac{x}{b}}\\right) \n\] \n\nIn words, the limiting density... | 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 accordin... | 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\[ \mathop{\sum }\limits_{k}{p}_{s}\left( {i, k}\right) {p}_{t}\left( {k, j}\right) = {p}_{s + t}\left( {i, j}\right) \] | Why is this true? In order for the chain to go from \( i \) to \( j \) in time \( s + t \), it must be in some state \( k \) at time \( s \), and the Markov property implies that the two parts of the journey are independent.\n\nProof. Breaking things down according to the state at time \( s \), we have\n\n\[ P\left( {{... | Yes |
Given the rates, how do you construct the chain? | Let \( {\lambda }_{i} = \mathop{\sum }\limits_{{j \neq i}}q\left( {i, j}\right) \) be the rate at which \( {X}_{t} \) leaves \( i \) . If \( {\lambda }_{i} = \infty \), then the process will want to leave \( i \) immediately, so we will always suppose that each state \( i \) has \( {\lambda }_{i} < \infty \) . If \( {\... | No |
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 (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 this we have to differentiate the formula in (4.11).\n... | 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 chain the states are the four nucleotides \( A, C, G, T \) . Jumps, which correspond to nucleotide substitutions, occur according at rate \( q\left( {x, y}\right) = \mu \) if \( x \neq y \) . To find the transition probability, we modify the chain so that jumps occur at rate \( {4\mu } \) and the new state is u... | \[ {p}_{t}\left( {x, y}\right) = \left( {1/4}\right) \left( {1 - {e}^{-{4\mu t}}}\right) \;y \neq x \] \[ {p}_{t}\left( {x, x}\right) = {e}^{-{4\mu t}} + \left( {1/4}\right) \left( {1 - {e}^{-{4\mu t}}}\right) \] | Yes |
Example 4.10 (Kimura Two Parameter Model). The nucleotides \( A \) and \( G \) are purines while \( C \) ’s and \( T \) ’s are pyrimidines. Kimura’s model takes into account that mutations that do not change the type of base (called transitions) happen at a different rate than those that do (called transversions) so th... | \[ \n{p}_{1}\left( t\right) + {p}_{2}\left( t\right) = {p}_{t}\left( {B, B}\right) = \frac{1}{2}\left( {1 + {e}^{-{4\beta t}}}\right) \n\]\n\n(4.13)\n\n\[ \n{p}_{3}\left( t\right) = {p}_{t}\left( {B, D}\right) /2 = \frac{1}{4}\left( {1 - {e}^{-{4\beta t}}}\right) \n\]\n\n(4.14)\n\n\nso it remains to compute \( {p}_{1}\... | Yes |
The transition probability of the Yule process is given by\n\n\[ \n{p}_{t}\left( {1, j}\right) = {e}^{-{\beta t}}{\left( 1 - {e}^{-{\beta t}}\right) }^{j - 1}\;\text{ for }j \geq 1 \n\]\n\n\[ \n{p}_{t}\left( {i, j}\right) = \left( \begin{matrix} j - 1 \\ i - 1 \end{matrix}\right) {\left( {e}^{-{\beta t}}\right) }^{i}{\... | Proof. To check (4.15), 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... | Yes |
Consider the process \( Z\\left( t\\right) \) in which each individual gives birth at rate \( \\lambda \) and dies at rate \( \\mu \) . That is, the transition rates are\n\n\[ \nq\\left( {i, i + 1}\\right) = {\\lambda i},\\;q\\left( {i, i - 1}\\right) = {\\mu i}, \n\]\n\nand \( q\\left( {i, j}\\right) = 0 \) otherwise.... | Since each individual gives birth at rate \( \\lambda \) and dies as rate \( \\mu \)\n\n\[ \n\\frac{d}{dt}{EZ}\\left( t\\right) = \\left( {\\lambda - \\mu }\\right) {EZ}\\left( t\\right) \n\]\n\nso \( {E}_{1}Z\\left( t\\right) = \\exp (\\left( {\\lambda - \\mu }\\right) t \) . Because of this we will restrict our atten... | No |
Lemma 4.3. \( \partial F/\partial t = - \left( {\lambda + \mu }\right) F + \lambda {F}^{2} + \mu = \left( {1 - F}\right) \left( {\mu - {\lambda F}}\right) \) | Proof. The second equality is just algebra. If \( h \) is small, then the probability of more than one event in \( \left\lbrack {0, h}\right\rbrack \) is \( O\left( {h}^{2}\right) \), the probability of a birth is \( \approx {\lambda h} \), of a death is \( \approx {\mu h} \). In the second case we have no particles, s... | Yes |
Lemma 4.4. If we let\n\n\\[ \n\\alpha = \\frac{\\mu {e}^{\\rho t} - \\mu }{\\lambda {e}^{\\rho t} - \\mu }\\;\\text{ and }\\;\\beta = \\frac{\\lambda {e}^{\\rho t} - \\lambda }{\\lambda {e}^{\\rho t} - \\mu },\n\\]\n\n(4.22)\n\nthen \\( {Z}_{t} \\) has a generalized geometric distribution\n\n\\[ \n{p}_{0} = \\alpha \\;... | Proof. To check this claim, note that taking \\( x = 0 \\) in (4.18) and multiplying top and bottom by \\( - {e}^{\\rho t} \\) confirms the size of the atom at 0 and suggests that we write\n\n\\[ \nF\\left( {x, t}\\right) = \\frac{\\left( {\\mu {e}^{\\rho t} - \\mu }\\right) - x\\left( {\\mu {e}^{\\rho t} - \\lambda }\... | Yes |
Theorem 4.5. Suppose \( \lambda > \mu \) . As \( t \rightarrow \infty ,{e}^{-{\rho t}}{Z}_{0}\left( t\right) \) converges in distribution to \( W \) with \( P\left( {W = 0}\right) = \mu /\lambda \) and\n\n\[ P\left( {W > x \mid W > 0}\right) = \exp \left( {-{x\rho }/\lambda }\right) \]\n\n(4.25)\n\nIn words, \( \left( ... | Proof. From (4.23), the atom at 0\n\n\[ \alpha \left( t\right) = \frac{\mu - \mu {e}^{-{\rho t}}}{\lambda - \mu {e}^{-{\rho t}}} \rightarrow \frac{\lambda }{\mu }. \]\n\nThe geometric distribution \( \left( {1 - \beta }\right) {\beta }^{n - 1} \) has mean\n\n\[ \frac{1}{1 - \beta } = \frac{\lambda {e}^{\rho t} - \mu }{... | Yes |
Lemma 4.6. If \( {X}_{t} \) is irreducible and \( t > 0 \), then \( {p}_{t}\left( {i, j}\right) > 0 \) for all \( i, j \) . | 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 4.7. \( \pi \) 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 |
Theorem 4.8. If a continuous time Markov chain \( {X}_{t} \) is irreducible and has a stationary distribution \( \pi \), then\n\n\[ \mathop{\lim }\limits_{{t \rightarrow \infty }}{p}_{t}\left( {i, j}\right) = \pi \left( j\right) \] | Proof. When \( {nh} \leq t \leq \left( {n + 1}\right) h,{p}_{t}\left( {i, j}\right) \geq {e}^{-{\lambda }_{j}h}{p}_{nh}\left( {i, j}\right) \) . From this it follows that \( \mathop{\liminf }\limits_{{t \rightarrow \infty }}{p}_{t}\left( {i, j}\right) \geq \pi \left( j\right) \) . If \( A \) is a finite set of states w... | Yes |
Example 4.13 (L.A. Weather Chain). There are three states: \( 1 = \) sunny, \( 2 = \) smoggy, and \( 3 = \) 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... | 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\[ \n\left( {{\pi }_{1}{\pi }_{2}{\pi }_{3}}\right) A = \left( \begin{array}{lll} 0 & 0 & 1 \end{array}\right) \;\text{ where }\... | Yes |
Theorem 4.9. If (4.28) holds, then \( \pi \) is a stationary distribution. | Proof. Summing (4.28) 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 ... | Yes |
Example 4.15 (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 4.16 (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( ... | 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 4.17 (M/M/ \( \infty \) 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 |
Example 4.18 (Two Barbers). Suppose that a shop has two barbers that can each cut hair at rate 3 people per hour customers arrive at times of a rate 2 Poisson process, but will leave if there are two people getting their haircut and two waiting. Find the stationary distribution for the number of customers in the shop. | The transition rate matrix is\n\n\[ \begin{matrix} & 0 & 1 & 2 & 3 & 4 \\ 0 & - 2 & 2 & 0 & 0 & 0 \\ 1 & 3 & - 5 & 2 & 0 & 0 \\ 2 & 0 & 6 & - 8 & 2 & 0 \\ 3 & 0 & 0 & 6 & - 8 & 2 \\ 4 & 0 & 0 & 0 & 6 & - 6 \end{matrix} \]\n\nThe detailed balance conditions say\n\n\[ {2\pi }\left( 0\right) = {3\pi }\left( 1\right) ,\;{2... | Yes |
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 are working? | 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 \( i = 1,2,3 \) . Setting \( \pi \l... | 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\[ \rho = \frac{\mu }{\lambda + \mu } \cdot 1 + \frac{\lambda }{\lambda + \mu } \cdot {\rho }^{2} \]\nsince starting from state 2 extinction occurs if and only if each indivi... | Yes |
A shop has two barbers that can cut hair at rate 3, people per hour customers arrive at times of a rate 2 Poisson process, but will leave if there are two people getting their haircut and two waiting. The state of the system to be the number of people in the shop. Find \( {P}_{i}\left( {{V}_{0} < {V}_{4}}\right) \) for... | The transition rate matrix is\n\n\n\nThe matrix \( R \) is the \( 3 \times 3 \) matrix inside the box, while \( Q\left( {i,0}\right) \) is the column vector to its left. Using (4.31) now\n\n\[ h = {\left( -R\right) }... | Yes |
In Exercise 2.10. Ron, Sue, and Ted arrive at the beginning of a professor's office hours. The amount of time they will stay is exponentially distributed with means of \( 1,1/2 \), and \( 1/3 \) hour. Part (b) of the question is to compute the probability each student is the last to leave. | 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 |
If we want to compute \( {E}_{x}{T}_{0} \), then \( A = \{ 0\} \), so the matrix \( R \) is\n\n\[ \begin{matrix} 1 & 2 & 3 & 4 & \\ 1 & - 5 & 2 & 0 & 0 \\ 2 & 6 & - 8 & 2 & 0 \\ 3 & 0 & 6 & - 8 & 2 \\ 4 & 0 & 0 & 6 & - 6 \end{matrix} \] | which has\n\n\[ - {R}^{-1} = \left( \begin{matrix} 1/3 & 1/9 & 1/{27} & 1/{81} \\ 1/3 & 5/{18} & 5/{54} & 5/{16} \\ 1/3 & 5/{18} & 7/{27} & 7/{81} \\ 1/3 & 5/{18} & 7/{27} & {41}/{162} \end{matrix}\right) \]\n\nMultiplying by 1 so we have\n\n\[ g\left( i\right) = {\left( {40}/{81},{119}/{162},{155}/{162},/{91}/{81}\rig... | Yes |
Part (c) of Problem 2.10 is to compute the expected time until all students are gone. Letting \( R \) be the matrix in (4.32) with the last column deleted, the first row of \( - {R}^{-1} \) is | \[ \begin{array}{lllllll} 1/6 & 1/6 & 1/{12} & 1/{30} & 7/{12} & 2/{15} & 1/{20} \end{array} \] The sum is \( {63}/{60} \), or one hour and three minutes. We have already explained the first four entries in the row. Similarly the last three entries in the row are \[ \frac{35}{60} \cdot 1\;\frac{16}{60} \cdot \frac{1}{2... | Yes |
We now take a different approach to analyzing the Duke Basketball chain, Example 4.14. The transition rate matrix with rates per minute is\n\n\[ \n\\begin{array}{lllll} & 0 & 1 & 2 & 3 \\ 0 & - 4 & {2.8} & {1.2} & 0 \\ 1 & 0 & - {7.5} & {7.5} & 0 \\ 2 & 1 & 0 & - 3 & 2 \\ 3 & {10} & 0 & 0 & - {10} \\end{array} \n\] | Our first step is to compute \( g\\left( i\\right) = {E}_{i}\\left( {V}_{1}\\right) \) for \( i = 0,2,3 \) . Removing the row and column for 1:\n\n\[ \nR = \\begin{matrix} 0 & 2 & 3 & \\\\ 0 & - 4 & {1.2} & 0 \\\\ 2 & 1 & - 3 & 2 \\\\ 3 & {10} & 0 & - {10} \\end{matrix} \n\]\n\nso using (4.33)\n\n\[ \ng = - {R}^{-1}\\m... | 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 + ... | Plugging into our formula for the stationary distribution, (4.30), we have\n\n\[ \pi \left( n\right) = \frac{{\lambda }_{n - 1}\cdots {\lambda }_{0}}{{\mu }_{n}\cdots {\mu }_{1}} \cdot \pi \left( 0\right) = {\left( \frac{\lambda }{\mu }\right) }^{n}\pi \left( 0\right) \]\n\n(4.34)\n\nTo have the sum 1, we pick \( \pi \... | Yes |
Lemma 4.10. 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 a... | 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 d... | Yes |
Example 4.25 (One Barber). Suppose now we have one barber who cuts hair at rate 3, customers arrive at rate 2 and there are two waiting chairs. In this case \( N = 3 \) , \( \lambda = 2 \), and \( \mu = 3 \), so plugging into (4.37) and multiplying numerator and denominator by \( {3}^{4} = {81} \), we have | \[ \pi \left( 0\right) = \frac{1 - 2/3}{1 - {\left( 2/3\right) }^{4}} = \frac{{81} - {54}}{{81} - {16}} = {27}/{65} \] \[ \pi \left( 1\right) = \frac{2}{3}\pi \left( 0\right) = {18}/{65} \] \[ \pi \left( 2\right) = \frac{2}{3}\pi \left( 1\right) = {12}/{65} \] \[ \pi \left( 3\right) = \frac{2}{3}\pi \left( 2\right) = 8... | Yes |
Theorem 4.11. If \( {a}_{n} \rightarrow 0 \) as \( n \rightarrow infty \), then there is a stationary distribution. | Proof. It follows from (4.29) that\n\n\[ \pi \left( {n + 1}\right) = \frac{{\lambda }_{n}}{{\mu }_{n + 1}} \cdot \pi \left( n\right) = \frac{{a}_{n}\lambda }{\mu } \cdot \pi \left( n\right) \]\n\nIf \( N \) is large enough and \( n \geq N \), then \( {a}_{n}\lambda /\mu \leq 1/2 \) and it follows that\n\n\[ \pi \left( ... | Yes |
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