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Page 1 of 2 �� Massachusetts Institute of Technology Department of Electrical Engineering & Computer Science 6.041/6.431: Probabilistic Systems Analysis (Spring 2006) Tutorial 9 Answers April 20-21, 2006 1. P (D > α ) = P ( | ( X μ ) | > α ) = P ( | X μ | > αμ ) Using Chebyshev Inequality, σ 2 1 P ( | X μ | > αμ ) α 2 μ 2 = r 2 α 2 Therefore, 1 P ( D > α ) r 2 α 2 1 P ( D α ) 1 r 2 α 2 2. (a) Let X i be random variables indicating the quality of the i th bulb (“1” for good bulbs, “0” for bad ones). Then X i are independent Bernoulli random variables. Let Z n be X 1 + X 2 + ... + X n Z n = . n We apply the Chebyshev inquality and obtain σ 2 P ( | Z n p | ≥ ǫ ) 2 , where σ 2 is the variance of the Bernoulli random variable. Hence, we obtain lim P ( | Z n p | ≥ ǫ ) = 0 , n →∞ σ 2 by noticing lim n →∞ 2 = 0. This means that Z n converges to p in probability. (b) For any number greater than 500, we know the number of bulbs would be enough for the test by using Chebyshev. Since the variance of a Bernoulli random variable is p (1 p ) 4 , we have σ 2 1 which is less than or equal to 1 4 . Hence, for
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