hw7-sol

# hw7-sol - EE 278 Statistical Signal Processing Handout#19...

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Unformatted text preview: EE 278 November 20, 2009 Statistical Signal Processing Handout #19 Homework #7 Solutions 1. Convergence examples. Consider the following sequences of random variables defined on the probability space (Ω , F , P), where Ω = { , 1 , . . ., m- 1 } , F is the collection of all subsets of Ω, and P is the uniform distribution over Ω. X n ( ω ) = braceleftBigg 1 n ω = n mod m otherwise Y n ( ω ) = braceleftBigg 2 n ω = 1 otherwise Z n ( ω ) = braceleftBigg 1 ω = 1 0 otherwise Which of these sequences converges to zero (a) with probability one, (b) in mean square, and/or (c) in probability? Justify your answers. Solution (15 points) First consider X n . Since X n ( ω ) is either 0 or 1 n , ≤ X n ( ω ) ≤ 1 n for every ω . Therefore lim n →∞ X n ( ω ) = 0 for every ω ∈ Ω. Thus P { ω : lim n →∞ X n ( ω ) = 0 } = P(Ω) = 1 , which shows that X n → 0 with probability one. Now consider convergence in mean square. E(( X n- 0) 2 ) = summationdisplay x ∈X x 2 p X ( x ) = 0 · P { X n = 0 } + 1 n 2 P { X n = 1 n } = 1 n 2 P { ω = ( n mod m ) } = 1 n 2 · 1 m . Obviously, lim n →∞ 1 mn 2 = 0 , so X n → 0 in mean square. This implies that X n → 0 in probability also. Next consider Y n . For any epsilon1 > 0, lim n →∞ P {| Y n- | > epsilon1 } = lim n →∞ P { Y n > epsilon1 } = lim n →∞ P { Y n = 2 n } = P { ω = 1 } = 1 m negationslash = 0 . Thus Y n negationslash→ 0 in probability. Since convergence with probability 1 implies convergence in probability, Y n negationslash→ 0 with probability 1. Similarly, Y n negationslash→ 0 in mean square. Finally consider Z n , which is independent of n . For any epsilon1 such that 0 < epsilon1 < 1, lim n →∞ P {| Z n- | > epsilon1 } = lim n →∞ P { Z n > epsilon1 } = P { Z i = 1 } = P { ω = 1 } = 1 m negationslash = 0 . Thus Z n negationslash→ 0 in probability, hence Z n negationslash→ 0 either with probability 1 or in mean square. Comments: • X n also converges to 0 in distribution since it converges in probability. • Y n does not converge in any sense. To show this it suffices to show that it does not converge in distribution. But P { Y n ≤ y } → 1- 1 m < 1 for every y < ∞ , so the limit of F Y n is not a cdf at all. • Z n converges in every sense to the nonzero random variable Z defined by Z ( ω ) = braceleftBigg 1 ω = 1 0 otherwise This convergence is immediate since Z n = Z for every n . 2. Convergence with probability 1. Let X 1 , X 2 , X 3 , . . . be i.i.d. random variables with X i ∼ Exp( λ ). Show that the sequence of random variables Y n = min { X 1 , . . ., X n } converges with probabil- ity 1. What is the limit? Solution (10 points) For any values of { X n } , the sequence of Y n values is monotonically decreasing in n . Since the random variables are ≥ 0, we know that the limit of Y n is ≥ 0. We suspect that Y n → 0. To prove that Y n converges w.p.1 to 0, we show that for every epsilon1 > 0, lim m →∞ P {| Y n- | < epsilon1...
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## This note was uploaded on 11/28/2009 for the course EE 278 taught by Professor Balajiprabhakar during the Fall '09 term at Stanford.

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hw7-sol - EE 278 Statistical Signal Processing Handout#19...

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