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h) 2ha (a + h)2 + o(h). Hence, rearranging and taking the limit as h → 0, we have F(a, v) + 2v a2 (a, v) = − 2 a ∂ F ∂a. (3) (4) (5) (6) (1) (2) Worked Examples and Exercises 377 Integrating (2) using the condition F(a, a) = 1 gives F(a, v) = 2av − v2 a2. (3) The densities of U and W may be found by the same method (ex... |
What is the density of X (k+1) − X (k); 1 ≤ k ≤ n − 1? Exercise (6) Continued What is the joint density of X (k+1) − X (k) and X ( j+1) − X ( j) for j = k? Exercise Two points are picked at random on the perimeter (including its diameter) of a semicircle with radius 1. Show that the expected area of the resulting tria... |
ab)−1. (i) Now consider Figure 8.2. The needle can intersect both A and B only when C = (x, y) lies in the positive quadrant of the circle, radius l, centred at the origin (region I). If the angle it makes with OB lies between ± sin−1(x/l), then it cuts only OA. Likewise the needle cuts only OB if it lies within the an... |
(l 2 − y2) 1 2 ) cos−1 2(a − l) cos−1 y l dy 2θ(cos2 θ − sin2 θ) dθ + l 0 = 2al − l 2, as required. y l dy π 2 0 (2l + 2(a − l))θ sin θ dθ (4) Exercise Show that the probability that the needle intersects no line of the grid is 1 − 2l πb − 2l πa + l 2 πab. (5) (6) (7) (Do this in two ways, one of which is an integral.... |
n ≤ y) = 2n+1 k=1 P(Yn ≤ y ∩ Ak) = (2n + 1)P(Yn ≤ y ∩ A1), = (2n + 1) y f X 1(y) −1 2n n n 1 + y 2 1 − (1 + y) 2 by symmetry, n dy, by conditional probability. Hence, Yn has density fY (y) = ((2n + 1)!/(n!)2)((1 − y2)n/22n+1). Because this is a density, its integral over (−1, 1) is unity, so 1 (1 − y2)n dy = 22n(n!)2 (... |
(X ) E(X ) + E(Y ). (c) What is the density of V? Worked Examples and Exercises 381 (a) We use the change of variables technique. The transformation u = x + Solution y, v = x/(x + y) for x, y > 0, is a one–one map of the positive quadrant onto the strip 0 < v < 1, u > 0, with inverse x = uv and y = u(1 − v). Hence, J =... |
≤ f S(X )) = x −∞ f S(y) dy. (b) Explain how this result may be used to produce realizations of a random variable Z with density f S(z). (a) By conditional probability, Solution P(X ≤ x|aU f X (X ) ≤ f S(X )) = P(X ≤ x, aU f X (X ) ≤ f S(X )) P(aU f X (X ) ≤ f S(X )) x = −∞ ∞ −∞ P(aU f X (x) ≤ f S(x)) f X (x) d x P(aU... |
es (1). (4) (5) (6) Find the mass function and mean of the number N of pairs (Uk, X k) that are rejected Exercise before the first occasion on which A occurs. What does this imply about a? If X is exponential with parameter 1, show that (2) takes the form P(X ≤ x|aU1U2 ≤ Exercise f S(X )) = FS(x), where U1 and U2 are in... |
nonnegative. Hence, (1) (2) Likewise, Hence, P(C(t) > z) = 1 e−λz. P(B(t) > y) = 1 e−λy y < 0 y > 0. E(B + C) = 1 λ + t 0 λte−λt dt + te−λt = 2 λ − 1 λ e−λt. If we suppose N (t) is the number of renewals of (say) light bulbs, then (2) Remark says that the expected life of the light bulb inspected at time t is 2/... |
independent, but (X k, Yk) is independent of (X j, Y j ) for j = k. At time t, let R1 and R2 be the number of aesthetes in the respective rooms. Show that R1 and R2 are independent Poisson random variables. Exercise Find cov (N (s), N (t)), and the correlation ρ(N (s), N (t)) 8.18 Example: von Neumann’s Exponential Va... |
k−1),..., X (1), X (k)), 2 ≤ k ≤ r, the event R = r occurs, and for no others. Hence, P(R = r ) = r − 1 r!. The above remarks also show that P(X R ≤ x) = = = ∞ r =2 ∞ r =2 ∞ r ==2 r P(X (k) ≤ x) r r j k=2 j=k (F(x)) j (1 − F(x))r − j by (8.7.12) (r F(x) − 1 + (1 − F(x))r ) = e(F(x) − 1) + exp (1 − F(x)), on summing the... |
Hence, (4) P(V > v) = P(N ≥ [v] + 1) + P(N = [v]; X 0 > v − [v]) 1 − 1 − e−v+[v] 1 − e−1 = e−[v]−1 + ((1 − e−1)e−[v]) = e−v for 0 < v < ∞. Thus, V is exponentially distributed. Remark This method of generating exponential random variables from uniform ones was devised by von Neumann in 1951. Notice that it is computat... |
j, X k) > x), and so on for any product. Furthermore, for all x, we see by inspection that I (max I j > x) = 1 − = j n (1 − I j ) j=1 I (X j > x) − j<k I (X j ∧ X k > x) + · · · + (−1)n+1 I (min X j > x). Now taking expectations gives P(max X j > x) = j P(X j > x) − j<k P(X j ∧ X k > x) + · · · Finally, integrating ov... |
all the rest. On failing, it is replaced by one with identical and independent properties. Find an expression for the expected time until every distinct component has failed at least once. Worked Examples and Exercises 387 8.20 Example: Binormal and Trinormal Let U and V be independent random variables having the stan... |
) = b2 + ρab a2 + 2ρab + b2 z b var (Y |a X + bY = z) = a2(1 − ρ2) a2 + 2ab + b2. 388 8 Jointly Continuous Random Variables (5) Exercise that Let X, Y, and Z have the standard trivariate normal density, defined in (8.10) Show P(X > 0, Y > 0, Z > 0) = 1 8 + 1 4π Express X, Y, and Z in terms of three independent N (0, 1) ... |
− 1√ n √, where Fn is the distribution function of (Sn − n)/ we have n, and by the Central Limit Theorem, Fn(x) → (x) as n → ∞. Furthermore, because (x) = φ(x), we have, as n → ∞, √! − 1√ n Because the convergence in (2) is uniform on finite intervals, we may let n → ∞ in (1) and use (3) to yield the result. → φ(0) = 1... |
(s))|N (u); 0 ≤ u ≤ s) W (t) W (s) exp[(−λ(t − s))(1 − e−θ )]|N (u); 0 ≤ u ≤ s = E = exp[−λ(t − s)(1 − e−θ )] because W (t) is a martingale. As this does not depend on N (u), 0 ≤ u ≤ s, it follows that N (t) has independent increments. Furthermore, we recognise the final expression as the moment generating function of ... |
we have E exp[−t log W ] = E exp{t Z {− log X − log Y }} = E{E{exp[t Z (− log X − log Y )]|Z }}, since − log X and − log Y are exponential, 1 (1 − t Z )2 1 = E = 0 1 (1 − t z)2 dz = 1 1 − t. Hence, by Example 7.5.4, − log W has an exponential density, so that W has a uniform density on (0, 1), by the remark above. Rem... |
X ), Problems 391 but | cos t X + i sin t X | = cos2 t X + sin2 t X = 1, and so |φ(t)| ≤ 1. If X is exponential, ∞ Eeit X = 0 λe−λx cos t x d x + i ∞ 0 λe−λx sin t x d x. Integrating by parts gives λe−λx cos t x d x = 1 − ∞ 0 te−λx sin λ2 λe−λx cos t x d x. ∞ 0 Hence, Likewise, Hence, ∞ 0 ∞ 0 λe−λx cos t x d x 1 − t 2... |
? Find cov (X, Y ) for the joint density of Problem 1. 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 392 8 Jointly Continuous Random Variables 2 /(U 1 2 + V 1 Let X and Y have joint density f (x, y) = c exp(−x − y) for x > 0, y > 0. Find (a) c, (b) P(X + Y > 1), and (c) P(X < Y ). Let X and Y have joint density f (x, y) = g... |
U − 1)(2Z −1 log Z −1) Y = (2V − 1)(2Z −1 log Z −1) 1 2 1 2. 2 (x 2 + y2)). Show that the conditional joint density of X and Y given Z < 1, is (2π)−1 exp (− 1 Explain how this provides a method for simulating normal random variables. Let X and Y be independent exponential random variables with respective parameters λ a... |
that for any integer k, the density fn(x) is a polynomial in x for x ∈ [k, k + 1). Let X and Y have the bivariate normal density of Examples 8.4.3 and 8.20. (a) If σ = τ, what is E(X |X + Y )? (b) If σ = τ, what are E(X |X + Y ) and E(Y |X + Y )? 19 20 21 22 23 24 25 26 27 28 29 30 31 32 Problems 393 Let (X n; n ≥ 1) ... |
. (b) Let (Xi ; 1 ≤ i ≤ n) be independent and exponential with parameter 1. Use the lack-of-memory property of the exponential density to show that max {X 1,..., X n} has the same distribution as X 1 + X 2/2 +... + X n/n. 2nx) → (x). Let X 1, X 2, X 3, and X 4 be independent standard normal random variables. Show that ... |
, but he tells you the sum of these two numbers. (a) If the sum is 75, which 11 boxes should you look in? (b) Give an approximation to the probability of finding the ring. 2 Y, |ρ| ≤ 1, has a standard bivariate normal density. ring in one of 1 11 10 0 1√ 2π e− u2 2 du = 0.36. 394 8 Jointly Continuous Random Variables 33... |
36 Molecules A molecule M has velocity v = (v1, v2, v3) in Cartesian corrdinates. Suppose that (iv) X n = U2n(1)/(1 + U2n−1(1)). (v) X n = Un(n). 2 )!. 35 v1, v2, and v3 have joint density: f (x, y, z) = (2πσ 2)− 3 2 exp Show that the density of the magnitude |v| of v is − 1 2σ 2 (x 2 + y2 + z2). 37 38 39 f (w) = 1 2 ... |
derived from n independent random variables, each uniformly distributed on [0, 1]. Show that (a) EX (k) = k n + 1 (b) var X (k) = k(n − k + 1) (n + 1)2(n + 2) Let (X (k); 1 ≤ k ≤ n) be the order statistics derived from n independent random variables each uniformly distributed on [0, 1]. Show that they have the same di... |
Consider, for example, the air temperature outside your window on successive days or the sequence of morning fixes of the price of gold. It is desirable and necessary to consider more general types of sequences of random variables. After some thought, you may agree that for many such systems it is reasonable to suppose... |
integers, and the states relabelled accordingly. (3) Example: Simple Random Walk Let (Sn; n ≥ 0) be a simple random walk. Because the steps (Sn+1 − Sn; n ≥ 0) are independent, the sequence Sn clearly has the Markov property, and the transition probabilities are given by pik = P(Sn+1 = k|Sn = i) = The state space S i... |
ov Chains best suits the context of the problem and your own psyche. For example: (i) A particle performs a random walk on the vertices of a graph. The distribution of its next step depends on where it is, but not on how it got there. (ii) A system may be in any one of d states. The distribution of its next state depen... |
Definition 1 and (6), (7), and (8) in Problem 6. Notice that (8) expresses in a precise form our previously expressed rough idea that given the present state of a Markov chain, its future is independent of its past. Finally, it should be noted that the Markov property is preserved by some operations, but not by others,... |
0 otherwise (12) Example Let X be a Markov chain. Show that Yn = |X n|; n ≥ 0, is not necessarily a Markov chain. Solution cept for Then Let X have state space S = {−1, 0, 1} and transition probabilities zero, ex- p−1,0 = 1 2, p−1,1 = 1 2, p0,−1 = 1, p1,0 = 1. P(Yn+1 = 1|Yn = 1, Yn−1 = 1) = P(X n+1 = 1|X n = 1, X n−1 ... |
n+1 = s0|X n = sr,...) = 1, and so on. It is easy to see that X n now is a Markov chain. 9.2 Transition Probabilities Recall that X is a Markov chain with state space S, where |S| = d. The transition probabilities pik are given by pik = P(X n+1 = k|X n = i) for n ≥ 0. The d × d matrix ( pi j ) of transition probabilit... |
(Sn; n ≥ 0) are the successive values of a simple random walk. If Sn = k and S0 = i, then from Theorem 5.6.4 we have (8) pik(n) = n 2 (n + k − i) 1 0 p 1 2 (n+k−i)(1 − p) 1 2 (n−k+i) if n + k − i is even otherwise. (Note that this chain has infinite state space.) (9) Example: Survival A traffic sign stands in a vulnerab... |
... j1∈S jn ∈S pi j 1 p j 1 j 2... p jn k. Proof Recall that if (A j ; j ≤ d) is a collection of disjoint events such that ∪d then for any events B and C 1 A j =, P(B|C) = d j=1 P(B ∩ A j |C). Hence, setting A j = {X m = j}, we have pik(m + n) = = = = j∈S j∈S j∈S j∈S P(X m+n = k, X m = j|X 0 = i) P(X m+n = k|X m = j, X... |
(n); 1 ≤ i ≤ d, 1 ≤ k ≤ d) ; 1 ≤ i ≤ d) as a row can be regarded as a matrix Pn, and the absolute probabilities (α(n) vector αn. It follows from Theorem 12 and (18) that i and αn = αPn, where α = (α1,..., αd ). Pm+n = PmPn = Pm+n (16) (17) (18) (19) (20) Example: Two State Chain The following simple but important examp... |
Let X and Y be independent regular Markov chains with transition matrices P = ( pik) and Q = (qik), respectively. Show that Zn = (X n, Yn); n ≥ 0, is a regular Markov chain. Solution Using the independence of X and Y, P(Zn = (k, l)|Zn−1 = (i, j), Zn−2,..., Z0) = P(X n = k|X n−1 = i, X n−2,..., X 0) × P(Yn = l|Yn−1 = j... |
r =1 E[log( pXr −1 Xr )]. Solution Hence, Because X is a Markov chain, f (x0, x1,..., xn) = αx0 px0x1... pxn−1xn. E[log( f (X 0,..., X n))]... = αx0 px0x1 xn ∈S αx0 log αx0 + = x0∈S x0 x0,x1 n... pxn−1xn (log αx0 + log px0x1 + · · · + log pxn−1xn ) αx0 px0x1 log px0x1 + · · · + αxn−1 pxn−1xn log pxn−1xn xn−1,xn = E[lo... |
y0,..., X n = yn) pyn x log pyn x P(X n = yn) = H(X n+1|X n). 9.3 First Passage Times For any two states i and k of X, we are often interested in the time it takes for the chain to travel from i to k. This is not merely a natural interest, these quantities are also of theoretical and practical importance. For example,... |
). But, by the Markov property, E(T12|X 1 = 1) = 1 + E(T12), and obviously E(T12|X 1 = 2) = 1. Hence, µ12 = 1 + 1 3 find µ21 = 4, and using conditional expectation again yields µ1 = 1 + 2 3 before. µ12 as above. Likewise, we 3 as µ21 = 11 For a rather different type of behaviour consider the following. (6) Example Let X... |
7) (8) (9) Theorem For i = d = k, (10) rik(n) = or in matrix form Rn = Rn 1.... j1=d j2=d jn−1=d pi j1 p j1 j2... p jn−1k, Proof We use the idea of paths. Every distinct path of the chain that goes from i to k in n steps and does not enter d is of the form i, j1, j2,..., jn−1, k, where jr ∈ S\d for... p jn−1k and one o... |
because n k proving (12). rik(nm0) < t n 0 < 1. k=d k rik(n) is nondecreasing (11) follows. Finally, rik(n) ≤ m0 1 + rik(m0n) ≤ m0 n,k 1 1 − t n 0 < ∞ With these preliminaries completed, we can get on with proving the main claim of the paragraph following Example 6. (15) Theorem For a regular chain, Tid is finite with ... |
(18) P(X T +m = k|Xr = xr for 1 ≤ r ≤ T, X T = d) = pdk(m). Solution conditional probability, the left-hand side of (18) may be written as Let us denote the event {Xr = xr = d for1 ≤ r < T } by A(T ). Then, using (19) P(X T +m = k, A(T ), X T = d) P(A(T ), X T = d) Now the numerator can be expanded as P(X T +m = k, A(... |
is much the same as that of Example 17. For each k in S, let us define the event Am = {X m = k}, and let Bm be the event that the first visit to k after time 0 takes place at time m. That is, Bm = {Xr = k for 1 ≤ r < m, X m = k}. Then following a now familiar route, we write (23) pik(m) = P(Am|X 0 = i) = m r =1 P(Am ∩ B... |
on the 2nth step (not necessarily for the first time), is 2n 2k = 1 2 (α + (1 − α))2n + 1 2 (α − (1 − α))2n n p0(2n) = α2k(1 − α)2n−2k k=0 = 1 2 ((1 − 2α)2n + 1). 412 Hence, 9 Markov Chains P0(z) = ∞ 0 p0(2n)z2n = 1 2 1 1 − (1 − 2α)2 z2 + 1 1 − z2. Hence, by (22), we have F(z) = P0(z) − 1 P0(z) = z2 1 + (1 − 2z2)(1 − 2... |
Suppose that π is the mass function of X 0, then X 1 has mass function αk(1) = πi pik = πk i because π is a solution of (1). Hence, X 1 has mass function π, and by a trivial induction so does X n for all n: (5) P(X n = k) = πk; n ≥ 0. Remark The chain is sometimes said to be in equilibrium. In formal terms, (1) says t... |
�i pi j pklδ jk = j π j pklδ jk = πk pkl = ηkl. Furthermore, i, j ηi j = i, j πi pi j = j = 1 by (3). π j by (1) Hence, η is the stationary distribution of Y. (10) Example: Nonhomogeneous Random Walk Let (Sn; n ≥ 0) be a Markov chain with transition matrix given by pi,i+1 = λi, pi,i−1 = µi, pi,k = 0, if |i − k| = 1, wh... |
a famous result with many algebraic and analytical proofs. Most of these are neither elementary nor probabilistic. We prefer to give a proof that uses the ideas of probability theory and is elementary. (11) Theorem tribution π. A regular Markov chain with transition matrix P has a stationary dis- Proof Let s be an arb... |
∞ P(X n−1 = j, Ts ≥ n − 1|X 0 = s) j=s = ρs(s) psk + n=2 p jkρ j (s) = j ρ j (s) p jk. p jk j=s Dividing throughout by µs yields the result (12), as required. In view of the appearance of mean recurrence times in the above proof, it is perhaps not surprising to discover another intimate link between π and µ. Theorem s... |
edges that meet there, to a neighbouring vertex. Find the mean recurrence time of each vertex. Solution abilities are The state space can be chosen as S = {i: 1 ≤ i ≤ 8} and the transition prob- pi j = if i and j are joined by an edge 1 3 0 otherwise. Hence, 15, µi = 8; 1 ≤ i ≤ 8. i∈s pi j = 1, and so the stationary d... |
(i) = pi,i+1; i ≥ 0. by conditional probability = P(X n+1 = i + 1|X n = i) Otherwise, (25) P(X n+1 = 0|X n = i, R) = 1 − pi,i+1 = f (i) 1 − F(i). Hence, X is a Markov chain with transition probabilities given by (24) and (25). If Wr is uniform on {1,..., d}, then pi,i+ and pd−1,0 = 1. Hence, any stationary distribution... |
that the probability of winning from i is 1 − pi. Hence, for 0 ≤ i ≤ K, as n → ∞, pi0(n) → pi pi K (n) → 1 − pi pi j (n) → 0 for 0 < j < K. The pair { pi, 1 − pi } is a stationary distribution, but it depends on the initial state of the chain. These examples illustrate the possibilities and agree with our intuition. R... |
(X 0 = i) = 1, and Y0 have the stationary distribution of pi j so that P(Y0 = i) = πi. Define the Markov chain W = (X, Y ), and let T be the first passage time of W to the set D = {(x, y): x = y}, namely, T = min {n: X n = Yn}. Now, by (9.2.11), because X and Y are regular, so is W. Hence, T is finite with probability 1 (... |
20, we see that when 0 < α + β < 2, as n → ∞ 9.5 The Long Run 421 p11(n) → β, p21(n) → β p12(n) → α, p22(n) → α. And, of course, so β, αβ, α), is the stationary distribution as it must be. When α + β = 0, the chain is not irreducible, and when α + β = 2, the chain is not regular (being periodic). (6) Example: Entropy o... |
visiting k should converge to πk. The following theorem shows that a more precise version of this vague statement is indeed true. It may be thought of as a type of weak law of large numbers for Markov chains. (10) (11) (12) (13) (14) (15) 422 9 Markov Chains (9) Theorem Let X be regular with transition matrix P and st... |
( pkk(t) − πk) − (αk(m) − πk) − (αk(r ) − πk)]) ((αk(s) − πk)( pkk(t) − πk) r =0 m,r ≤ 1 (n + 1)22 → 0 m,r 2c2(λs + λt ) by (14) and (15) as n → ∞, establishing (10). by (11) and (12) 9.5 The Long Run 423 (16) Corollary For any bounded function g(x), and any > (Xr ) − πk g(k) > → 0, r =0 k∈S $ $ $ $ $ (17) P as n → ∞.... |
HX + 1 $ n P log f (X 0,..., X n) $ $ $ $ > δ → 0. Solution Markov chain with stationary distribution (πi pik; i ∈ S, k S). Second, we have First, from Example 9.4.8, the sequence Yn = {X n, X n+1}; n ≥ 0, is a − 1 n log f (X 0,..., X n) = − 1 n = − 1 n log( pX 0 X 1 pX 1 X 2 n−1 log pXr Xr +1 r =0.,..., pX n−1 X n ) F... |
you are seeking efficient transmission of messages, it therefore makes sense to concentrate on the typical sequences. It follows that a natural question is, how many typical sequences are there? At this point, we recall that by convention the logarithms in Example 18 are taken to base 2. Hence, from (22), But also, fro... |
space S can be uniquely partitioned as S = T ∪ C1 ∪ C2 ∪..., where T is the set of transient states and each Ci is an irreducible closed set of recurrent states. This means that eventually the chain ends up in some one of the Ci and never leaves it, or it remains forever in the transient states. Of course, if there is... |
��nition, analogous to Definition 9.1.1. As usual, X (t) ∈ S, where S is a subset of the integers called the state space. (1) Definition The process X = (X (t); t ≥ 0) taking values in S is a Markov process (or has the Markov property), if P(X (t) = k|X (t1) = i1,..., X (tn) = in) = P(X (t) = k|X (tn) = in) for all possi... |
a counter or particle moving around the vertices of a graph according to some specified distributions, and if necessary we could actually do it. Here, we have started with a collection of probabilities, with no description of how we might actually produce a sequence X (t) having these transition probabilities and joint... |
memory property of the exponential distribution that is basically responsible for this essential role in the theory of Markov processes. However, we can do no more here than state the fact baldly; exploring its ramifications is beyond our scope. One example will suffice to give some trivial insight into these remarks. Le... |
) = o(t). N (t) is nondecreasing. (5) (6) The probability of an event in [s, s + t] is proportional to t, for small t, and does not depend on previous events. (7) The probability of two or more events in [s, s + t], for small t, is o(t). What we are going to do now is to seek a Markov process X (t) with transition prob... |
. A simple induction now yields (10). A second method relies on the generating function G(z, t) = ∞ k=i pik(t)zk = E(z N (t)|N (0) = i). Multiply (8) by zk and sum over k to obtain From (9), we have ∂G ∂t = λ(z − 1)G. G(z, 0) = zi. The solution of (11) that satisfies (12) is and the coefficient of zk in this expression i... |
h + o(h))(1 − λh + o(h))i−1 = iλh + o(h) pik(h) = o(h); pik(h) = 0; k > i + 1 k < i. Following the by now familiar routine, we find pk(t + h) = (1 − λkh) pk(t) + λ(k − 1)hpk−1(t) + o(h) and so ∂ ∂t pk(t) = −λkpk(t) + λ(k − 1) pk−1(t). Now we set G X (z, t) = E(z X (t)), and notice that because probability generating fun... |
1, t) = eλt. However, we could have obtained E(X (t)) without solving (18). If we assume E[(X (t))2] exists, then differentiating (18) with respect to z and setting z = 1 yields ∂ ∂t ∂G X ∂z (1, t) = λ ∂G X ∂z (1, t). This has solution given by (21). Or we could simply have noted that X has a geometric distribution wit... |
(h), independently of its past, or it stays down with probability 1 − βh + o(h). (a) If it is up at t = 0, find the probability that it is down at time t > 0. (b) Let N (t) be the number of occasions on which it has gone down during [0, t]. Find E(N (t)). (c) Find the probability generating function E(z N (t)). (a) Let ... |
k, k − 1)) = βh + o(h) P(Y(t + h) = (k, k − 1)|Y(t) = (k, k − 1)) = 1 − βh + o(h). 9.8 Forward Equations: Equilibrium 433 Hence, if fk j (t) = P(Y(t) = (k, j)), the forward equations may be derived routinely as (9) (10) d dt where f0,−1(t) = 0. d dt fkk(t) = −α fkk(t) + β fk,k−1(t); k ≥ 0. fk,k−1(t) = −β fk,k−1(t) + α ... |
and so, using the result of (a), E(N (t)) = t α 0 β α + β + αe−(α+β)ν α + β dν. It follows that as t → ∞, t −1E(N (t)) → αβ/(α + β). (c) Let x(t, z) = ∞ k=0 fkk(t)zk and y(t, z) = ∞ k=1 fk,k−1(t)zk. 434 9 Markov Chains Then E(z N (t)) = x(t, z) + y(t, z), and multiplying each of (9) and (10) by zk and summing over k g... |
Markov process with transition matrix pi j (t). Then If X (t) is irreducible then the limit is independent of i, we write Furthermore, π j satisfies j pi j (t) = π j. lim t→∞ π j = 1 and π j = πi pi j (t); t ≥ 0, i π is the stationary distribution of X (t). [A chain is irreducible if for each i, j > 0.] there is some fi... |
) Example: Queue Let X (t) be the length of queue formed before a single service point at time t. The times between arrivals are exponentially distributed with parameter λ; each individual is served on reaching the head of the queue; each service time is exponentially distributed with parameter µ; interarrival times an... |
the integers (or some other countable set). They have thus been “jump processes,” in the sense that transitions between discrete states take place instantaneously at times that may be fixed (as in the simple random walk), or they may be random times indexed by a continuous parameter (as in the Poisson process). But it ... |
nition, so our first task must be to justify it. First, of course, we must check that it is indeed a Markov process satisfying Definition (9.6.1). We write, for any t0 < t1 <... < tn < tn + s, P(W (tn + s) ≤ w|W (tn) = wn,..., W (t0) = w0) = P(W (tn + s) − wn ≤ w − wn|W (tn), W (tn) − W (tn−1),..., W (t0) = w0) = P(W (tn... |
�ciently small particle, inanimate as well as animate. He began his observations on pollen, which is in general too large to show the effect, but he observed that the smaller particles were in motion. Further, the smaller the particle the more vigorous the motion, and he obtained the same result using “every mineral wh... |
to execute a random walk, whose steps (on any scale) are equally likely to be in any direction. Clearly, the steps must be independent, and an application of the central limit theorem tells us that steps in any given direction must be normally distributed. These are just the properties we set out in Definition (1). The... |
(x). Also, recall from the end of 8.10, the review of Chapter 8, that the joint distribution of a collection of multivariate normal random variables is determined by their means, variances, and covariances. Hence, we have for the Wiener process (2) Example: Joint Distribution Because increments are independent and norm... |
W (t) = y, as fW (s)|W (t)(x|W (t) = y) = f (s, x; t, y) (y − x)2 t − s x − sy t − 1 2 − 1 2 By inspection, this is a normal density with mean sy/t and variance φt (y)! y2 t t s(t − s) + 1 2! − 1 2 ∝ exp ∝ exp x 2 s. 2 √ s(t − s)/t (where, for simplicity, we have omitted the normalizing constant). That is to say, for ... |
Diffusions 441 A famous and important special case is the so-called “Brownian Bridge,” which (perhaps surprisingly) turns out to be useful in making statistical inferences about empirical distribution functions. (9) Example: Brownian Bridge This is the process B(t), 0 ≤ t ≤ 1, defined to be W (t), conditional on the ev... |
) − sW (t)W (1) − t W (s)W (1) + st W 2(1)) = s − st − st + st = s ∧ t − st = cov(B(s), B(t)). The proof is complete, when we recall from (8.10) that multinormal distributions are determined by their first and second joint moments. Besides conditional processes, there are several other operations on the Wiener process t... |
You can show that U (t) is a Markov process (exercise). (13) Example: Drifting Wiener Process This is obtained from the standard Wiener pro- cess W (t) by setting D(t) = σ W (t) + µt. If σ W (t) represents the position of a particle enjoying Brownian motion in some fluid, then µt may be interpreted as a superimposed gl... |
was most important. This is for two main reasons. First, such first passage times are often naturally important in the real world that our processes describe. The second reason is that it is often useful to condition some event or expectation on the value of T, thus yielding immediate results or at least tractable equa... |
the Brownian Bridge. (21) Example: Maximum points: first, for c > 0, Let M(t) = max{W (s), 0 ≤ s ≤ t}. We note these two useful (22) (23) {M(t) ≥ c} ⊇ {W (t) ≥ c}. Second, for c > 0, denoting the first passage time to c by Tc, {M(t) ≥ c} ≡ {Tc ≤ t}, and after Tc the process has a symmetric distribution about c that is i... |
) = e−2c2. Solution distributed about c, and independent of the process before Tc. Hence, Once again, we use the fact that after Tc, the Wiener process is symmetrically P(Tc < 1; 0 ≤ W (1) < ε) = P(Tc < 1, 2c − ε < W (1) ≤ 2c) = P(2c − ε < W (1) ≤ 2c) = 1√ 2π e−2c2ε + o(ε) But, by definition of B(t), for 0 ≤ t ≤ 1, P(ma... |
∈ R. 446 9 Markov Chains Proof Because W (t) = W (s) + [W (t) − W (s)] and W (t) has independent increments: (a) E(W (t)|W (u), 0 ≤ u ≤ s) = E(W (s) + W (t) − W (s)|W (u), 0 ≤ u ≤ s) = W (s). (b) See Example (9.2.3). (c) E{exp(aW (s) + aW (t) − aW (s))|W (u), 0 ≤ u ≤ s} = eaW (s) exp(+ 1 2 a2(t − s)), because a(W (t) ... |
) it is a relief to know that the formal operation of differentiating with respect to θ and setting θ = 0 can be proved to yield these martingales. Just as we found in Chapter 8, such martingales are particularly useful when taken with a stopping time T. Because we are working with continuous martingales having continu... |
important application of martin- gales and the Wiener process. (36) Example: The Option-Pricing Martingale A popular model for a simple market comprises two available assets: a bond whose value B(t) grows at a continuously compounded constant interest rate r, so that B(t) = B(0)er t, and a stock whose price per unit i... |
the sale should equal the purchase price so that the expectation E0 taken with respect to the pay-off odds of this “fair game” must satisfy E0(e−r t S(t)|S(u); 0 ≤ u ≤ s) = e−r s S(s). That is to say, e−r t S(t) is a martingale. It turns out that if we set µ = r − σ 2/2, then e−r t S(t) is indeed a martingale. It can ... |
� But if the bookie expectation of her outcome is 1 wanted to take risks, she could simply gamble, and this is not why she is a bookie. 2 5000 − 1 9.10 Review and Checklist for Chapter 9 449 In fact, suppose she sets the pay-off odds on A to be 2 : 1, and those on B to be 1 : 2. Then whatever the outcome of the race, s... |
: pik(m + n) = pi j (m) p jk(n). j∈S The first passage time from i to k = i is Tik, with mean first passage time µik = ETik. The recurrence time of i is Ti = min{n > 0 : X n = i|X 0 = i], with mean recurrence time µi = ETi. If µi < ∞, then i is nonnull. A stationary measure is a nonnegative solution (xi ; i ∈ S) of xk = ... |
) → π j. If the chain is not irreducible or has infinite state space, a much wider range of behaviour is possible. Checklist of Terms 9.1 Markov property Markov chain imbedding 9.2 transition matrix doubly stochastic matrix Chapman–Kolmogorov n-step transition probabilities regular chain irreducible aperiodic absorbing ... |
) = (α + β + γ )2n + (β + γ − α)2n + (α + γ − β)2n + (α + β − γ )2n = 1 + a2n + b2n + c2n. Hence, U (s) = ∞ n=0 s2nu(2n) = 1 4 1 1 − s2 + 1 1 − a2s2 + 1 1 − b2s2 + 1 1 − c2s2. Similarly, starting from O, the walk visits V whenever an odd number of steps has been taken in all three possible directions. Hence, u V (2n + ... |
� k=1 kρk−1(1 − ρ)2 = 1. (1) (2) Suppose that at every step the walk may remain at its current vertex with probability Exercise δ, where now α + β + γ + δ = 1. Find: (c) E(T ). (a) The mean recurrence time of O; Let W be the vertex (1, 1, 0), and define ˆT to be the number of steps until the walk Exercise first visits V ... |
i pi j = π j, as required. i πi pi j (n) = π j p ji (n). To see this, consider any n-step path from i to j, and then using (3) gives πi pii1 pi1i2... pin−1 j = pi1i πi1 pi1i2... pin−1 j = pi1i pi2i1... p jin−1 π j after repeated applications of (3). Now summing over all paths from i to j gives (5). Now applying (5) sho... |
,..., Ynr −1 ) = P(Ynr Show that if X has stationary distribution π and transition probabilities P, then, in equilibrium, Y is a Markov chain with transition probabilities qi j = π j π −1 = k|Ynr −1 ). = k|Yn1 P(Ynr i p ji. Exercise 6 Continued Let X be a Markov chain, with transition probabilities pi j. Show that if t... |
. m i (a) Given X 0,..., X n = j, the probability that a particle in C1 is selected Solution for transfer is j/m, and the probability that a particle in C2 is selected for transfer is (m − j)/m. Hence, X is a Markov chain and p j, j+1 = (m − j)/m, p j, j−1 = j/m, p jk = 0 when |k − j| = 1. Worked Examples and Exercises... |
− m 2 + m 2 m )n (b) What is E(X n) when α = 1 = β? Two adjacent containers C1 and C2 each contain m Exercise: Bernoulli Diffusion Model particles. Of these 2m particles, m are of type A and m are of type B. At t = 1, 2,... one particle is selected at random from each container, and these two particles are each transf... |
transition probabilities pi0 = fi+1 1 − Fi pi,i+1 = 1 − pi0 = 1 − Fi+1 1 − Fi ; i ≥ 0. Now let us calculate the first return probability f00(n), that is the probability that the chain first returns to 0 at the nth step. At each stage, the chain either does so return or increases by 1. Hence, f00(n) = p01 p12... pn−2,n−1... |
renewal processes defined in Section 6.7. Events may occur at integer times and the intervals between successive events are independent and identically distributed random variables (Xi ; i ≥ 1), where f X (r ) = fr. An event occurs at n = 0. Now the construction of the chain U in (3) and (4) allows us to identify visit... |
= ∞), 458 (a) Show that (b) Show that 9 Markov Chains ηii = 1 if i is persistent 0 if i is transient. ηi j = P(Ti j < ∞) 0 if j is persistent if j is transient. (c) Show that if i → j and i is persistent then ηi j = η ji = 1. Solution Next, (a) First note that Vii ≥ 1 if and only if Ti < ∞, so P(Vii ≥ 1) = P(Ti < ∞). ... |
, P(V ji = ∞) ≥ 1, and therefore η ji = 1. Hence, P(Ti j < ∞) = 1 and so j → i. It follows that ηi j = 1. (1) (2) (3) (4) (5) Exercise Exercise Exercise Exercise Exercise Show that if i is persistent and i → j, then j is persistent. Show that ηi j = 1 if and only if P(Ti j < ∞) = P(T j < ∞) = 1. Show that if i → j and ... |
s /∈ D into D, with mean µs D = E(Ts D). Also, let φs j be the probability that the chain first enters D at the state j. Show that for i ∈ D µsi = µs D + j∈D φs j µ ji. (c) Hence, show that in an unlimited sequence of tosses of a fair coin, the probability that the consecutive sequence T H T occurs before H H H is 7 12... |
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