Clearly this translated version of ni is a poisson

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Unformatted text preview: on splitting Poisson processes, these will be independent, and process Ni will be a Poisson process on R of rate λpi Now shift all the points split into the ith process, Ni , by the same value vi . Clearly this translated version of Ni is a Poisson process, and it is independent of all the other Poisson processes, and hence independent of the translated versions as well. Therefore, once we merge the shifted processes, again we have a Poisson process of the same rate. Problem 5: (10 points) Let {Xn } be a sequence of nonnegative random variables such that limn→∞ Xn = 0, almost surely. For the following statements, answer (together with a brief justification) whether it is: (i) always true, (ii) always false; (iii) sometimes true and sometimes false, (a) (5 pts.) limn→∞ P(Xn > 0) = 0. 4 (b) (5 pts.) For every � > 0, limn→∞ P(Xn > �) = 0. Solution: part (a) is sometimes true, sometimes false. Consider Xn = 1/n with prob­ ability 1, in which case (a) is false; and consider Xn = 0 with probability 1, in which case (a) is true. Part (b) is always true because P (Xn > �) ≤ P (|Xn − 0| > �), and since con­ vergence almost everywhere implies convergence in probability, the last expression ap­ proaches 0 as n goes to infinity. Problem 6: (17 points) Let {Xn } be a sequence of random variables defined on the same probability space. (a) (4 pts.) Suppose that limn→∞ E[|Xn |] = 0. Show that Xn converges to zero, in probability. (b) (5 pts.) Suppose that Xn converges to zero, in probability, and that for some constant c, we have |Xn | ≤ c, for all n, with probability 1. Show that lim E[|Xn |] = 0. n→∞ (c) Suppose that each Xn can only take the values 0 and 1 and, that P(Xn = 1) = 1/n. (i) (4 pts.) Given an example in which we have almost sure convergence of Xn to 0. (ii) (4 pts.) Given an example in which we do not have almost sure convergence of Xn to 0. Solution: For part (a), observe that Markov’s inequality implies P (|Xn − 0| ≥ �) ≤ E[|Xn |] , � so that if E[|Xn |] approaches 0, we have that Xn approaches 0 in probability. � For part (b), fix � > 0 and define a new random variable Xn as follows. We have � � Xn = � whenever |Xn | ≤ �, and Xn = c whenever |Xn | > �. Then, it is always true � that |Xn | ≤ Xn and therefore � E[|Xn |] ≤ E[Xn ] = �P (|Xn | ≤ �) + cP (|Xn | > �) Taking limits as n goes to infinity, we get lim E[|Xn |] ≤ �, n and since this holds for all � > 0, we get limn E[|Xn |] = 0. For part (c).i, we can generate a uniform random variable on [0, 1] and declare that Xn = 1 if the outcome is at most 1/n, and 0 otherwise. It immediately follows Xn 5 is binary with P (Xn = 1) = 1/n. Now suppose that the uniform random variable generated the value x with x > 0. Then eventually 1/n is smaller than x, and Xn = 0 after this point. Since the outcome is positive with probability 1 (the probability of getting 0 is 0), it follows that Xn approaches 0 almost surely. For part (c).ii, we can take Xn to be independent. Since ∞ � P (Xn = 1) = i=1 ∞ �1 = ∞, n n=1 and the events {Xn = 1} are independent, the Borel-Cantelli lemma implies that Xn = 1 occurs infinitely often with probability 1. It follows that Xn does not converge to 0 almost surely. 6...
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This document was uploaded on 03/19/2014 for the course EECS 6.436 at MIT.

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