5. Homework 7 (Chap 5)5.2. 5.5. 5.7. 5.15.5. Homework 7 (Chap 5)p. 125 #5.2,5,7,155.2. If Xtis a Poisson process with λ = 1 then find E(X1| X2), E(X2| X1).By the definition of a Poisson proces s, E(X2− X1) = E(X1) = λ = 1. Also, X2− X1isindependent of X1. So,E(X2| X1) = E(X2− X1| X1) + E(X1| X1) = X1+ 1If the total X1+ (X2− X1) is given then, by symmetry,E(X1| X2) =12X2.5.5. Suppose that Xnis the numb er of individuals in the nth generation in a branchingprocess. If the mean number of offspring is µ then show thatMn= µ−nXnis a martingale wrt X0, X1, · · ·By definition of µ we haveE(Xn+1| Xn) = µXnSo,E(Mn+1| Fn) = µ−n−1E(Xn+1| Xn) = µ−n−1µXn= Mnand we see that Mnis a martingale.5.7. Take the random walk on Z where the probability of going right at each step is p < 1/2and the probability of going left is 1 − p. Take Sn= a + X1+ · · · + Xnwhere 0 < a < N.(a) Show thatMn=1 − ppSnis a martingale.First note thatMn+1=1 − ppSn+Xn+1= Mn1 − ppXn+1AndE 1 − ppXn+1!=1 − ppp +1 − pp−1(1 − p) = (1 − p) + p = 1So,E(Mn+1| Fn) = MnE 1 − ppXn+1!= MnSo, Mnis a martingale.14(b) Suppose that T is the first time that Snreaches 0 or N. Compute P(ST= 0).By the OST we haveE(MT) = M0=1 − ppaBut this expected value is also given byE(MT) = P(ST= 0) +1 − ppN(1 − P(ST= 0))So,P(ST= 0) =h1−ppia−h1−ppiN1 −h1−ppiN=(1 − p)N− (1 − p)apN−a(1 − p)N− pN5.15. Suppose that Mnis a martingale. Suppose there exists Y ≥ 0 so that E(Y ) < ∞ and|Mn| < Y for all n. Then show that Mnare uniformly integrable.Since E(Y ) < ∞, the size of the tail of Y goes to zero. I.e., for any > 0,E(Y IY >K) < for K sufficiently large. But |Mn| > K implies Y > |Mn| > K. So, the indicator function of|Mn| > K is less than or equal to the indicator f unction for Y > K. So,|Mn|I|Mn|>K≤ Y IY >KandE(|Mn|I|Mn|>K) ≤ E(Y IY >K) < Since the same K works for all n we have uniform
View Full Document
Unlocking...