MATH360 Applied Stochastic Models 2022 Solutions Final Exam
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Solutions Final Exam
Applied Stochastic Models (MATH360)
2022
1 Solution Exercise A1
Part A1 (a). The Kolmogorov backward equations are given by
dtp1,1 (t) = -αp1,1 (t) + αp2,1 (t),
d
dtp2,2 (t) = βp1,2 (t) - βp2,2 (t),
d
for t > 0, p1,1 (0) = p2,2 (0) = 1 and p1,2 (0) = p2,1 (0) = 0.
The Kolmogorov forward equations are given by
dtp1,1 (t) = -αp1,1 (t) + βp1,2 (t),
d
dtp2,2 (t) = αp2,1 (t) - βp2,2 (t),
d
for t > 0, p1,1 (0) = p2,2 (0) = 1 and p1,2 (0) = p2,1 (0) = 0.
Part A1 (b). Observe that p1,2 (t) = 1 -p1,1 (t) and p2,1 (t) = 1 -p2,2 (t). From the Forward equations, we see that
p1,1 (t) = β - (α + β)p1,1 (t) and p2,2 (t) = α - (α + β)p2,2 (t),
for t > 0.
The previous equations are of the form given in the hint and we obtain that
p1,1 (t) = c1 e-(α+β)t + and p2,2 (t) = c2 e-(α+β)t + ,
for t > 0, and some unknown constants c1, c2 e Ⅸ. Since p1,1 (0) = p2,2 (0) = 1, we see that c1 = α/(α + β) and c2= β/(α + β). Therefore,
P (t) = = α(α) , for t > 0.
Part A1 (c). The invariant distribution ξ = (ξ(1), ξ(2)) of (Xt, t > 0) must satisfy the equation ξQ = 0 (i.e. component-wise (ξQ)1 = -αξ(1) + βξ(2) = 0 and (ξQ)2 = αξ(1) - βξ(2) = 0) together with ξ(1) + ξ(2) = 1. Therefore, ξ(1) = β/(α + β) and ξ(2) = α/(α + β).
Part A1 (d). It follows from the explicit expressions of the transition probabilities in Part A1 (b) by letting t 二 o.
Alternatively, this exercise can be proved as follow. Since α > 0 and β > 0, we observe that the continuous-time Markov chain (Xt, t > 0) is irreducible. In Part A1 (c), we have proved that (Xt, t > 0) has an invariant distribution ξ = (ξ(1), ξ(2)). By Proposition 12 in the lecture notes, we
conclude that (Xt, t > 0) is positive recurrent. Therefore, Theorem 3 in the lectures notes implies the result.
Part A1 (e). Since α > 0 and β > 0, we observe that the continuous-time Markov chain (Xt, t > 0) is irreducible. In Part A1 (c), we have proved that (Xt, t > 0) has an invariant distribution ξ = (ξ(1), ξ(2)). By Proposition 12 in the lecture notes, we conclude that (Xt, t > 0) is positive recur- rent. Therefore, the ergodic theorem (Theorem 4 in the lecture notes) implies that the asymptotic proportion of time that the machine is down is equal to ξ(2) = α/(α + β) with probability one, i.e.,
T『 t(l)im二o (0t 1(Xs=2】ds = ξ(2)│ = 1.
2 Solution Exercise B1
Part B1 (a). For the continuous-time Markov chain (Xt, t > 0), we have three communicating classes(0】,(1】and(2】.
We see that(1】and(2】are not closed communicating classes and thus, they are transient. We also see that(0】is closed, absorbing and recurrent.
Part B1 (b). For i > 0 and t > 0, we want to compute p2,i(t) = T(Xt = i|X0 = 2). We use the Kolmogorov forward equations to compute such probabilities. More precisely, we have that
p2,0 (t) = λ2p2,1 (t), p2,1 (t) = -λ2p2,1 (t) + λ1p2,2 (t), p2,2 (t) = -λ 1p2,2 (t),
for t > 0, p2,0 (0) = p2,1 (0) = 0 and p2,2 (0) = 1.
We see that the solution of the previous system is given by
p2,0 (t) = 1 + ‘λ1 e-λ2 t - λ2 e-λ 1 t \, p2,1 (t) = ‘e-λ 1 t - e-λ2 t \, p2,2 (t) = e-λ 1 t,
for t > 0. Note that the previous solution also includes the case λ 1 = λ2 by viewing this case as the limit when λ2 二 λ 1 .
Part B1 (c). We see that
T(T s t) = T(Xt = 0|X0 = 2) = p2,0 (t) = 1 + ‘λ1 e-λ2 t - λ2 e-λ 1 t \, for t > 0,
and T(T s t) = 0, for t < 0, where we have used the expression found in Part B1 (b). Note that the previous solution also includes the case λ 1 = λ2 by viewing this case as the limit when λ2 二 λ 1 .
3 Solution Exercise C1
We observe that
正[VT] = 正[max(0,K - XT)] = (0o T(max(0,K - XT) > u)du.
On the other hand, for u > 0,
T(max(0,K - XT) > u) = T(max(0,K - XT) > u,K - XT > 0) + T(max(0,K - XT) > u,K - XT s 0) = T(K - XT > u) +T(0 > u,K - XT s 0) = T(K - XT > u).
Then,
正[VT] = (0o T(K - XT > u)du = (0o T(K - yeBT > u)du = (0K T(yeBT < K - u)du = (0K T『BT < log『 ││ du.
Recall that BT ~ N(0,T ). Therefore,
正[VT] = (0K ( e- dxdu.
4 Solution Exercise C2
Part C2 (a). Consider the function f (s,x) = xn+1, for s > 0, x e Ⅸ and n > 1. Observe that
f (s,x) = 0, f (s,x) = xn and f (s,x) = nxn-1 .
By Itô’s formula, we see that
B+1 =t(n) (0t Bs(n)dBs + (0t nBs(n)-1ds, for t > 0,
which implies the result.
Part C2 (b). Consider the function f (s,x) = sx, for s > 0, x e Ⅸ. Observe that
f (s,x) = x, f (s,x) = s and f (s,x) = 0.
By Itô’s formula, we see that
Xt = tBt = (0t Bsds + (0t sdBs = (0t ds + (0t sdBs, for t > 0.
2022-05-19