Tentamentsskrivning: Stochastic processes 1

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1 Tentamentsskrivning: Stochastic processes 1 Tentamentsskrivning i MSF2/MVE33, 7.5 hp. Tid: fredagen den 31 maj 213 kl Examinator och jour: Serik Sagitov, tel , mob , rum H326 i MV-huset. Hjälpmedel: miniräknare, egen formelsamling på 4 A4 sidor (2 blad). CTH: för 3 fordras 12 poäng, för 4-18 poäng, för 5-24 poäng. GU: för G fordras 12 poäng, för VG - 2 poäng. Inklusive eventuella bonus poäng. 1. (6 points) Define a process Y n recursively by Y n = Z n.3 Y n 1, n 1, where Z n are independent r.v. with zero means and unit variance. (a) Under which conditions on Y this process is weakly stationary? Compute its autocovariance function. (b) Find the best linear predictor Ŷr+3 of Y r+3 given (Y,..., Y r ). (c) What is the mean squared error of this prediction? (d) Give an example when this process is strongly stationary. 2. (4 points) Consider a Poisson process with parameter λ. (a) Describe the Poisson process as a renewal process. Compute its renewal function. (b) Let E(t) be the excess life at time t. Write down a renewal equation for P(E(t) > x). Solve this equation to find the distribution of E(t). (c) What is the stationary distribution of the excess life? 3. (4 points) Let X 1, X 2,... be iid r.v. with finite mean µ, and let M be a stopping time with respect to the sequence X n such that E(M) <. Then E(X X M ) = µe(m). (a) Give a detailed proof of the above formulated Wald s equation lemma. (b) Illustrate by an example that the statement may fail if M is not a stopping time. 4. (6 points) Let {X n } n=1 be a sequence of stochastic variables defined by X n = I 1 sin(φn) + I 2 cos(φn), where I 1, I 2, φ are independent random variables such that P (I 1 = 1) = P (I 1 = 1) = P (I 2 = 1) = P (I 2 = 1) = 1 2 and φ is uniformly distributed over [ π, π]. (a) On the same graph draw three examples of trajectories of the process {X n }. What is random in these trajectories? (b) Verify that {X n } is a stationary process. (c) Find the spectral density function of {X n }. (d) Find the spectral representation for the process {X n } in the following form X n = cos(nλ)du(λ) + sin(nλ)dv (λ).

2 Tentamentsskrivning: Stochastic processes 2 5. (4 points) Let random variables {X n } be iid random variables with P(X n = 2) = P(X n = ) = 1/2. (a) Show that the product S n = X 1 X n is a martingale. (b) Show that S n in probability but not in mean. (c) Does S n converge a.s.? 6. (3 points) For a filtration (F n ) let B n F n. Put X n = 1 B Bn. (a) Show that (X n ) is a submartingale. (b) What is the Doob decomposition for X n? 7. (3 points) Theorem. Let (Y n, F n ) be a submartingale and let T be a stopping time. Then (Y T n, F n ) is a submartingale. If moreover, E Y n <, then (Y n Y T n, F n ) is also a submartingale. (a) Referring to the above theorem carefully deduce the following statement. If (Y n, F n ) is a martingale and T is a stopping time, then (Y T n, F n ) and (Y n Y T n, F n ) are martingales. (b) Illustrate this theorem with a simple example. Partial answers and solutions are also welcome. Good luck!

3 Tentamentsskrivning: Stochastic processes 3 Solutions 1. This is an example of AR(1) model Y n = αy n 1 + Z n. Observe that for n, m, Y n+m = Z n+m + αz n+m α n 1 Z m+1 + α n Y m. 1a. Let Y has mean µ and variance σ 2. In the stationary case Y n should also have mean µ and variance σ 2. This leads to equations µ = α n µ, σ 2 = 1 + α α 2n 2 + α 2n σ 2, which imply µ = and σ 2 = 1 1 α Hence, using these values we compute for n, c(n) = Cov(Y n+m, Y m ) = E(Y n+m Y m ) = α n E(Y 2 m) = αn (.3)n = ( 1)n 1 α b. The best linear predictor Ŷr+k = r j= a jy r j is found from the equations or r a j c( j m ) = c(k + m), m r. j= r a j α j m = α k+m, m r. j= Intuition suggests seeking a solution of the form Ŷr+k = a Y r. Indeed, plugging a 1 =... = a r = we easily obtain a = α k. Thus Ŷr+k = (.3) k Y r, and Ŷr+3 =.27 Y r. 1c. The mean square error is E ( (Y r+k Ŷr+k) 2) = E ( (Y r+k α k Y r ) 2) = E ( (Z r+k + αz r+k α k 1 Z r+1 ) 2) = 1 α2k 1 α 2 = 1 (.9)k.91 = for k = 3. 1d. If Y has a normal distribution with mean zero and variance σ 2 = 1 1 α 1.989, then the 2 process becomes Gaussian. For the Gaussian processes weak and strong stationarity are equivalent. 2a. Let T =, T n = X X n, where X i are iid exponential inter-arrival times with parameter λ. The Poisson process N(t) is the renewal process giving the number of renewal events during the time interval (, t], so that {N(t) n} = {T n t}, T N(t) t < T N(t)+1. Its renewal function m(t) := E(N(t)) satisfies the renewal equation m(t) = F (t) + m(t u)df (u), F (t) = 1 e λt. In terms of the Laplace-Stieltjes transforms ˆm(θ) := e θt dm(t) we get Since ˆF (θ) = ˆm(θ) = ˆF (θ) + ˆm(θ) ˆF (θ). λ θ+λ, we obtain ˆm(θ) = λ θ, implying m(t) = λt.

4 Tentamentsskrivning: Stochastic processes 4 2b. The excess life time E(t) := T N(t)+1 t. The distribution of E(t) is given by the following recursion for b(t) := P(E(t) > y) ( ) ( ) b(t) = E E(1 {E(t)>y} X 1 ) = E 1 {E(t X1)>y}1 {X1 t} + 1 {X1>t+y} = 1 F (t + y) + This renewal equation gives P(E(t) > y) = 1 F (t + y) + = e λ(t+y) + λ b(t x)df (x). (1 F (t + y u))dm(u) e λ(t+y u) du = e λ(t+y) + λ Thus E(t) is exponentially distributed with parameter λ. y+t y e λx dx = e λy. 2c. From the previous calculation we see that the stationary distribution of E(t) is geometric with parameter λ. This is in agreement with the general result, with µ standing for the mean inter-arrival time, y lim P(E(t) y) = t µ 1 (1 F (x))dx = λ y e λx dx = 1 e λy. 3a. Let X 1, X 2,... be iid r.v. with finite mean µ, and let M be a stopping time with respect to the sequence X n such that E(M) <. Then Proof. By dominated convergence ( E(X X M ) = E = E(X X M ) = µe(m). lim n i=1 n ) ( X i 1 {M i} = lim E n ) X i 1 {M i} n E(X i )P(M i) = µe(m). i=1 Here we used independence between {M i} and X i, which follows from the fact that {M i} is the complimentary event to {M i 1} σ{x 1,..., X i 1 }. 3c. Turn to the Poisson process and observe that M = N(t) is not a stopping time and in this case E(T N(t) ) µm(t). Indeed, for the Poisson process µm(t) = t while T N(t) = t C(t), where C(t) > is the current lifetime. i=

5 Tentamentsskrivning: Stochastic processes 5 4a. What looks on the figure as a cloud of stationary allocated points is produced by three sinusoids: pluses: sin(πn/5) + cos(πn/5), circles: sin(πn/5) cos(πn/5), stars: sin(πn/3) + cos(πn/3). The randomness is created by the random choice of the frequency and phase. 4b. The process X n = I 1 sin(φn) + I 2 cos(φn) has zero means by independence and E(I 1 ) = E(I 2 ) =. Its autocovariancies are Cov(X n, X m ) = E(X n X m ) = E(sin(φn) sin(φm) + cos(φn) cos(φm)) = E(cos(φ(n m)) = π cos(φ(n m))dφ = 1 {n=m}. Thus (X n ) is a weakly stationary sequence with mean zero, variance 1, and zero autocorrelation ρ(n) = for n. 4c. The spectral density g must satisfy ρ(n) = cos(λn)g(λ)dλ. Then it is the uniform density g(λ) = π 1 1 {λ [,π]}. where 4d. In this case the spectral representation for the process {X n } is straightforward X n = I 1 sin(φn) + I 2 cos(φn) = cos(nλ)du(λ) + U(λ) = I 1 1 {λ φ}, V (λ) = I 2 1 {λ φ}. sin(nλ)dv (λ), Let us check the key properties of the random process (U(λ), V (λ)): (i) U(λ) and V (λ) have zero means, (ii) U(λ 1 ) and V (λ 2 ) are uncorrelated, (iii) Var(U(λ)) = Var(V (λ)) = E(1 {λ φ} ) = λ = G(λ), (iv) increments of U(λ) are uncorrelated, and increments of V (λ) are uncorrelated. The last property is obtained as follows: for λ 1 < λ 2 < λ 3, E(U(λ 3 ) U(λ 2 ))(U(λ 2 ) U(λ 1 )) = E(1 {λ3 φ} 1 {λ2 φ})(1 {λ2 φ} 1 {λ1 φ}) = E(1 {λ2<φ λ 3}1 {λ1<φ λ 2}) =. 5a. We have S n = 2 n with probability 2 n and S n = with probability 1 2 n, so that E(S n ) = 1. Clearly, E(S n+1 S n = 2 n ) = 2 n and E(S n+1 S n = ) =, so that E(S n+1 S n ) = S n. 5b. On one hand, P( S n > ɛ) = 2 n, so that S n P. On the other hand, E(Sn ) = 1, so that S n does not converge in mean. 5c. Put A n = {S n = }. We have A n A n+1 and A = n A n = {ω : S n (ω) }. Since P(A) = lim P(A n ) = 1, we conclude that S n almost surely. 6a. We verify three properties. (i) Since B i F i F n, we conclude that X n is measureable with respect to F n. (ii) The random variables X n have finite means a n := E(X n ) = P(B 1 ) P(B n ). (iii) E(X n+1 F n ) = X n + P(B n+1 ) X n.

6 Tentamentsskrivning: Stochastic processes 6 6b. Doob s decomposition: X n = M n + S n, where (M n, F n ) is a martingale and (S n, F n ) is an increasing predictable process (called the compensator of the submartingale). Here M and S are computed as: M =, S =, and for n Thus M n = X n a n and S n = a n. M n+1 M n = X n+1 E(X n+1 F n ) = 1 {Bn+1} P(B n+1 ), S n+1 S n = E(X n+1 F n ) X n = P(B n+1 ). 7a. Theorem 1. Let (Y n, F n ) be a submartingale and let T be a stopping time. Then (Y T n, F n ) is a submartingale. If moreover, E Y n <, then (Y n Y T n, F n ) is also a submartingale. Changing the sign we easily get the next sister theorem. Theorem 2. Let (Y n, F n ) be a supermartingale and let T be a stopping time. Then (Y T n, F n ) is a supermartingale. If moreover, E Y n <, then (Y n Y T n, F n ) is also a supermartingale. Since a martingale is both a submartingale and a supermartingale, these two theorems imply the following desired result. Theorem 3. Let (Y n, F n ) be a martingale and let T be a stopping time. Then (Y T n, F n ) is a martingale. If moreover, E Y n <, then (Y n Y T n, F n ) is also a martingale. 7b. Consider a symmetric simple random walk Y n = S n and take T = the first hitting time of the level a. On the figure we show the trajectories of the three martingales in question. Т Т Т