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)/µ| > α) = P(|X − µ| > αµ) Using Chebyshev Inequality, σ2 1 P(|X − µ| > αµ) ≤ α2µ2 = r2α2 Therefore, 1 P(D > α) ≤ r2α2 1 P(D ≤ α) ≥ 1 − r2α2 2. (a) Let Xi be random variables indicating the quality of the ith bulb (“1” for good bulbs, “0” for bad ones). Then Xi are independent Bernoulli random variables. Let Zn be X1 + X2 + ... + XnZn = . n We apply the Chebyshev inquality and obtain σ2 P (|Zn − p| ≥ ǫ) ≤ nǫ2 , where σ2 is the variance of the Bernoulli random variable. Hence, we obtain lim P (|Zn − p| ≥ ǫ) = 0, n→∞ σ2 by noticing limn→∞ nǫ2 = 0. This means that Zn converges to p in probability. (b) For any number greater than 500, we know the number of bulbs would be en ough for the test by using Chebyshev. Since the variance of a Bernoulli random variable is p(1 − p) 4 , we have σ2 ≤1which is less than or equal to 1 4 . Hence, for n ≥ 500, � X1 + X2 + ... + Xn � σ2 P � − p� ≥ 0.1 ≤ � n � n0.12 1 ≤ 4 n × 0.12 ≤ 1 − 0.95 = 0.05. However, for a number less than 500, we can not tell if the number of bulbs is enough for the test because we don’t know the variance. If the variance is very small, w hich is possible when p is quite small, 27 bulb s could be enough actually. Thus, the ans wer is “cannot be decided”. In reality, we need to estimate the variance first.Page 2 of 2Massachusetts Institute of Technology Department of Electrical Engineering & Computer Science 6.041/6.431: Probabilistic Systems Analysis (Spring 2006) 13. (a) E[Xn] = n var(Xn) = nn−2 1 E[Yn] = 1 var(Yn) = n − 1 (b) Xn is convergent in probability (c) In this case, Chebyshev tells us nothing. (d) Yes, to zero. (e) Yes, to
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