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Unformatted text preview: Massachusetts Institute of Technology Department of Electrical Engineering & Computer Science 6.041/6.431: Probabilistic Systems Analysis (Spring 2006) Recitation 15 April 20, 2006 1. See textbook pg. 399 2. (a) N = 200, 000. (b) N = 100, 000. 3. Let us fix some ǫ > 0. We will show that P (Yn ≥ 0.5 + ǫ) converges to 0. By symmetry, this will imply that P (Yn ≤ 0.5 − ǫ) also converges to zero, and it will follow that Yn converges to 0.5, in probability. For the event {Yn ≥ 0.5 + ǫ} to occur, we must have at least n + 1 of the random variables X1 , X2 , . . . , X2n+1 to have a value of 0.5 + ǫ or larger. Let Zi b e a Bernoulli random variable which is equal to 1 if and only if Xi ≥ 0.5 + ǫ: Zi = � 1 if Xi ≥ 0.5 + ǫ 0 otherwise {Z1 , Z2 , ....} are i.i.d random variables and E [Zi ] = P (Zi = 1) = P (Xi ≥ 0.5 + ǫ) = 0.5 − ǫ. Hence, for the event {Yn ≥ 0.5 + ǫ} to occur, we must have at least n + 1 of the {Zi } to take value 1, P (Yn ≥ 0.5 + ǫ) = P ( 2� n+1 2n + 1 Zi 1 = P ( i=1 ≥ 0.5 + ) 2n + 1 2(2n + 1) ≤ P( �2n+1 �2n+1 i=1 Zi ≥ 0.5) 2n + 1 = P( Zi ≥ i=1 �2n+1 i=1 Zi n + 1) ≥ n+1 ) 2n + 1 Note that P (Zi = 1) = 0.5 − ǫ. By the weak law of large � numbers, the sequence (Z1 + � · · · + Z2n+1 )/(2n + 1) converges to 0.5 − ǫ. To show that P Z1 +···+Z2n+1 ≥ 0.5 converges 2n+1 to �zero, we need to show that for any given ǫ > 0, there exists N such that for all n > N , � P Z1 +···+Z2n+1 ≥ 0.5 < ǫ. The fact that the sequence (Z1 + · · · + Z2n+1 )/(2n + 1) converges to 2n+1 0.5 − ǫ ensures the existence of such N . Since P (Yn ≥ 0.5 + ǫ) is bounded by P ( it also converges to zero. �2n+1 i=1 Zi 2n+1 ≥ 0.5), Page 1 of 1 ...
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