Exam Stochastic Processes 2WB08 - March 10, 2009,

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1 Exam Stochastic Processes WB8 - March, 9, The number of points that can be obtained per exercise is mentioned between square brackets. The maximum number of points is 4. Good luck!!. (a) [ pts.] Let X, X,... be random variables such that the partial sums S n = n i= X i determine a martingale. Show that E(X i X j ) = if i j. Solution: Suppose i < j and remark that X j = S j S j. Then E(X ix j) = E(E(X i(s j S j ) S, S,..., S j )) = E(X ie((s j S j ) S, S,..., S j )) = E(X i(s j S j )) =. (b) [3 pts.] Let (Z n ) n be the size of the nth generation of an ordinary branching process with Z = and with family size Z having mean E(Z ) = µ and variance Var(Z ) = σ. Using the properties of conditional expectation, compute the mean E(Z n ) and show that the variance is given by { nσ if µ = Var(Z n ) = σ (µ n )µ n µ if µ. Hint: Recall that Z n = Z n i= Y i, where (Y i ) i is a sequence of i.i.d. random variable with E(Y ) = µ and Var(Y ) = σ. Solution: We have Z n E(Z n) = E i= Y i Z n = E E Y i Z n i= = E(Z n E(Y )) = µe(z n ). By iteration it is found E(Z n) = µ n. A similar computation for the second moment E(Z n) yields the recursion relation Var(Z n) = µ Var(Z n ) + σ µ n which is solved by the claimed expression for Var(Z n). (c) [3 pts.] Let η be the probability that the branching process ultimately becomes extinct. Define M n () = µ n Z n and M n () = η Zn. Check that both the processes (M n () ) n and (M n () ) n are martingales. Hint: for M n () use that η is given by the smallest non-negative root of the equation s = G(s), where G(s) = E(s Z ) is the probability generating function of the family size. Solution: This was done during the lecture. (d) [ pts.] Using the martingale M n (), show that E(Z n Z m ) = µ n m E(Zm) for m n. Hence find the correlation coefficient ρ(z m, Z n ) in term of µ (you might use the result of item (b)). We recall the definition ρ(z m, Z n ) = cov(z m, Z n ) var(zm )var(z n )

2 Solution: Using the martingale M () n we immediately have E(Z n Z m) = Z mµ n m. Hence E(Z mz n Z m) = Z mµ n m and E(Z mz n) = E(E(Z mz n Z m)) = E(Z m)µ n m. Thus we have and, by the result of item (b), cov(z m, Z n) = E(Z m)µ n m E(Z n)e(z m) = µ n m Var(Z m), ρ(z m, Z n) = { µ n m ( µ m )/( µ n ) if µ m/n if µ =. Let (B t ) t be a standard Wiener process and let (X t ) t be a standard Brownian motion process with drift µ >. ( ) (a) [ pt.] For θ R show that the process (M t ) t defined by M t = exp θb t θ t is a martingale. Solution: This has been proved during the lecture. (b) [3 pts.] Let a > and T a = inf{t : B t = a}. By making use of the martingale in the previous item, compute the Laplace transform E(e λta ) where λ >. Solution: T a is a stopping time and the martingale stopping theorem can be applied (why?) to the martingale M t. We have ( ( )) = E exp θb Ta θ T a Remark that B Ta = a. Thus, choosing θ = λ, we find E(e λta ) = e λ a (c) [ pts.] Conclude from (b) that E((T a ) ) = a. Hint: Use the identity x = e λx dλ for x >. Solution: ( ) E = E(e λta )dλ = T a e λ a dλ = a. (d) [ pts.] Let a > and τ a = inf{t : X t = a}. Use again the above martingale to show that the following expression holds where λ >. E(e λτa ) = e a( µ +λ µ) Solution: Notice that X t = B t + µt and use the stopping theorem as in item (b). The final result is obtained by choosing θ = µ + λ µ. (e) [ pts.] Compute the mean E(τ a ) and the variance Var(τ a ). Solution: Define g(λ) = E(e λτa ). Then E(τ a) = lim λ + dg(λ) dλ Starting from the g(λ) of the previous item, an immediate computation yields E(τ a) = x. A µ similar computation with the second derivative of g(λ) allows to compute the second moment, from which it is found Var(τ a) = x. µ 3

3 3. A staircase is constructed such that the kth stair has width W k and height H k, where W, H, W, H,... are nonnegative i.i.d. random variables with distribution function F (x) = P(H x) = P(W x) = ( + x), x. (a) [ pts.] Show that E[H ] =. Solution: We have E[H ] = ( F (u)) du = du = dw =. ( + u) w (b) [3 pts.] Let D(h) be the number of stairs needed to reach at least a total height h. Calculate the limit lim h E[D(h)]/h. Solution: The successive heights H, H,... form a renewal process with renewal function E[D(H)], hence E[D(h)] lim = h h E[H = ]. (c) [3 pts.] Let L(h) = D(h) k= W k denote the total length of the construction. Derive the limit of the steepness E[L(h)]/h of the stairway as the height h tends to. Solution: Consider the renewal process from (a). Taking W, W,... as rewards, we see that L(h) is the total reward. We have from the renewal reward theorem E[W] lim E[L(h)]/h = t E[H =. ] (d) [3 pts.] Find the equilibrium distribution F e (x) of F (x). Assuming that h is very large, give an estimate for the probability P(Z(h) 3) for the overshoot Z(h) over the height h (see figure). Solution: F e(x) = x P(H > u) du E[H ] = x du x+ (+u) = u du = x + x. Z(h) is equivalent to the time to the next event in a renewal process with interarrival times H, H,.... We know from the lecture, that the distribution converges to the equilibrium distribution F e(x). Hence, if h is large, P(Z(h) > ) 3 4 = 4. Hint: Note that H, H,... form a renewal process. 4. Buses arrive at a bus stop according to a renewal process with interarrival times X, X,... (measured in minutes), so that the first bus arrives at time X. We assume that the X, X,... are i.i.d. with E[X ] < and E[X ] <.

4 (a) [ pts.] Some busses are green, the others are red. The probability that an arbitrary bus is red is given by p (, ]. The color of a bus is independent from the color of the other busses and independent of X, X,.... Let τ be the time of the first arrival of a red bus. Show that E[τ] = E[X ]/p. Hint: What is the distribution of the number of green busses arriving before τ? Solution: We have K {,,...} and has a geometric distribution with success probability p and E[K] = ( p)/p. Since K is independent of X, X,... it follows from Wald s equation that K E[τ] = E[ X k ] + E[X ] = E[K]E[X ] + E[X ] = E[X]. p k= (b) [ pts.] Let M t denote the number of red busses minus the number of green busses that arrived during the time interval (, t]. Show that E[M t ] lim = p t t E[X ]. Solution: Let R t denote the number of red busses. Then E[R t] lim = t t E[τ] = p E[X ] and similarly for G t, the number of green busses, E[G t] lim = p t t E[X. ] Then E[R t G t] lim = t t E[τ] = p E[X. ] (c) [3 pts.] Assume that people arrive with constant rate person/min to the bus stop. At the time a bus arrives, all waiting passengers receives /p Euros if the bus is red and /( p) Euros if the bus is green. Let C t denote the total amount of money spend after time t by the bus company. Show that E[C t ] lim =. t t Solution: Let C k denote the money that the company has to pay when bus k arrives. Then E[C k ] = ( p + ( p))e[x] = E[X] and by the renewal reward theorem p p E[C t] lim = E[C k] t t E[X =. ] (d) [3 pts.] At time t let A t denote the time that went by since the last bus arrived and B t the time until the next bus arrives. Sketch the process J t = t min{a s, B s } ds. By using an appropriate renewal reward process, show that E[J t ] lim = E[X ] t t 4E[X ]. Solution:

5 Let J Tk be the reward at time T k, where T k = X + X X k. Then (draw the process!) E[J Tk ] = E[ 4 X k], since the area under the process min{a s, B s} between T k and T k (one cycle) is equal to one quarter of the square with area X k X k. The result follows from renewal theory for reward processes.