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ECE 541 Probability and Stochastic Processes; Fall 2008 Assignment 8; Due date: Tuesday, Nov. 18, 2008
Exercise 1. Show that − limn→∞ Xn = limn→∞(−Xn).
This follows from the observation that supk≥n Xk = − infk≥n(−Xk).
Exercise 2. Prove that (supk≥n Wk)−^1 (a, ∞) = ∪∞ k=nWk(a, ∞). Where have we used this fact in the class?
Suppose that ω ∈ (supk≥n Wk)−^1 (a, ∞). Note that if Wk(ω) ≤ a for every k ≥ n, then supk≥n Wk(ω) ≤ a, a contradiction. Hence, for some i ≥ n, Wi(ω) > a and hence ω ∈ ∪∞ k=nWk(a, ∞). On the other hand, suppose ω ∈ ∪∞ k=nWk(a, ∞). Then for some k ≥ n, Wk(ω) > a and hence ω ∈ (supk≥n Wk)−^1 (a, ∞). We have used this simple fact in proving that the lim and lim of a sequence of D-measurable random variables are D-measurable.
Problem 1. Show that if f is a uniformly continuous function, then Xn → X in probability implies f (Xn) → f (X) in probability. Hint: Use Theorem 16 in the notes.
Pick ≤ > 0 and consider P{|f (Xn) − f (X)| > ≤}. By uniform continuity of f , there exists δ > 0 (possibly dependent on ≤) such that |f (x) − f (xn)| ≤ ≤ whenever |x − xn| ≤ δ. Equivalently, |x − xn| > δ whenever |f (x) − f (xn)| > ≤. Hence, {|f (X) − f (Xn)| > ≤} ⊂ {|X − Xn| > δ}, and therefore P{|f (X) − f (Xn)| > ≤} ≤ P{|X − Xn| > δ}. Since Xn converges to X in probability, P{|X − Xn| > δ} converges to 0 as n → ∞, and we obtain P{|f (X) − f (Xn)| > ≤} → 0 as n → ∞.
Problem 2. Referring to the proof of Theorem 8 in the notes: where did we use the fact that P is a finite measure?
We used the fact that if E 1 ⊃ E 2 ⊃... then ( ¯
E) = limN →∞ P(EN ). This statement is not true in general if P is not a finite measure. Consider the example given in the notes, just after the proof of the theorem.
Problem 3. Refer to the notes in Section 9. Consider a probability space (Ω, F, P) and let X ∈ L 1 (Ω, F, P). Let D be a sub σ-algebra. For each n ≥ 1, we define Xn as X when |X| ≤ n and n other- wise. Note that Xn ∈ L 2 (Ω, F, P) (since it is bounded), so we can talk about E[Xn|D] as a projection of Xn onto L 2 (Ω, D, P ), which we call Zn.
a) Show that Zn is Cauchy in L 1.
Since Zn = E[Xn|D], we know that E[(Zn −Zm)Y ] = E[(Xn −Xm)Y ] for any Y ∈ L 2 (Ω, D, P ). In par- ticular, pick Y = I{Zn−Zm> 0 } − I{Zn−Zm≤ 0 }, which is clearly in L 2 (Ω, D, P ). In this case, (Zn − Zm)Y = |Zn − Zm|. Thus, we conclude that E[|Zn − Zm|] = E[(Xn − Xm)(I{Zn−Zm> 0 } − I{Zn−Zm≤ 0 })] ≤ E[(Xn − Xm)(I{Xn−Xm> 0 } − I{Xn−Xm≤ 0 })]= E[|Xn − Xm|]. Hence, we have E[|Zn − Zm|] ≤ E[|Xn − Xm|]. However, E[|Xn − Xm|] ≤ E[|X|I{|X|≥max(m,n)}] → 0 by the dominated convergence theorem. This is because |X|I{|X|≥max(m,n)} ≤ |X|, E[|X|] < ∞ and |X|I{|X|≥max(m,n)} → 0 a.s. We therefore conclude that limm,n→∞ E[|Zn − Zm|] = 0.
b) We argued in the notes, using the completeness of L 1 , that there exists Z ∈ L 1 (Ω, D, P) such that limn→∞ E[|Zn − Z|] = 0. Further, Zn has a subsequence, Znk , that converges almost surely to Z. We took this Z as a candidate for E[X|D]. Show that Z satisfies E[ZY ] = E[XY ] for any bounded, D-measurable Y.
Suppose that |Y | < M , for some M. We already know that E[ZnY ] = E[XnY ] (why?), and by the dominated convergence theorem, E[XnY ] → E[XY ] (since Xn → X a.s., |Xn| ≤ |X| and |XnY | ≤ |XY |, and E[|XY |] ≤ M E[|X|] < ∞). Also, |E[ZnY ] − E[ZY ]| ≤ E[|Zn − Z||Y |] ≤ E[|Zn − Z|M ] → 0 (since limn→∞ E[|Zn − Z|] = 0). Hence, E[ZnY ] → E[ZY ].
Problem 4. Consider a sequence Nn = X 1 +... Xn, where the Xi’s are iid, Poisson distributed, and E[X 1 ] = λ.
a) What can you say about the quantity n−^1 Nn when n is large? In other words, does limn→∞ n−^1 Nn exist? If so, precisely describe in what stochastic sense it exists and characterize the limit.
By the strong law of large numbers, limn→∞ n−^1
∑n i=1 Xn^ =^ λ^ with probability one.
b) Show that limn→∞ P{Nn > nλ} = 1/2. (Hint: Think of the central limit theorem.) Interpret this result.
P{Nn > λn} = P{Nn − λn > 0 } = P{(Nn − λn)/
λn > 0 }. But by the central limit theorem, (Nn − λn)/
λn converges in distribution to a zero-mean unit-variance Gaussian rv (verify the hypothesis). Thus, limn→∞ P{(Nn − λn)/
λn > 0 } = 0.5 since the median of a zero-mean Gaussian rv is 0. Hence limn→∞ P{Nn > λn} = 0.5. This result indicates that for large counting times, the median and the mean of the Poisson process are approximately equal.
c) Use the result in (b) to show that limλ→∞ P{Nn > λn} = 1/2. Interpret this result. Is P{Nn > λn} = 1/2 for any λ? Support your answer by an example. Hint: Think of the central-limit theorem. Assume that λ is a non-negative integer. Fix n > 0 and consider the random variable Mλ =^4 Y 1 +... + Yλ, where the Yi’s are iid, Poisson distributed, and E[Y 1 ] = n. Note that just as Nn, Mλ is also a Poisson rv with mean λn. Thus, P{Nn > λn} = P{Mλ > λn}. But from part (b) and by interchanging n and λ, we obtain limλ→∞ P{Mλ > λn} = 0.5. Hence, limλ→∞ P{Nn > λn} = 1/2. This result shows that at high mean values, the median and the mean of the Poisson random variable are approximately equal. This is an asymptotic result only, for if we take n = λ = 1, then P{N 1 > 1 } = 1 − P{N 1 = 0} − P{N 1 = 1} = 1 − exp(−1) − (1) exp(−1)/1 = 1 − 2 exp(−1) 6 = 1/2.
