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Unformatted text preview: Probability and Stochastic Processe Problem Set 7 Solutions Spring’2010 Problem Solutions : Yates and Goodman, 7.1.2 7.2.1 7.2.2 7.2.4 7.3.1 7.3.3 7.4.2 and 7.4.3 Problem 7.1.2 Solution X 1 ,X 2 ...X n are independent uniform random variables with mean value μ X = 7 and σ 2 X = 3 (a) Since X 1 is a uniform random variable, it must have a uniform PDF over an interval [ a,b ]. From Appendix A, we can look up that μ X = ( a + b ) / 2 and that Var[ X ] = ( b − a ) 2 / 12. Hence, given the mean and variance, we obtain the following equations for a and b . ( b − a ) 2 / 12 = 3 ( a + b ) / 2 = 7 (1) Solving these equations yields a = 4 and b = 10 from which we can state the distri bution of X . f X ( x ) = braceleftbigg 1 / 6 4 ≤ x ≤ 10 otherwise (2) (b) From Theorem 7.1, we know that Var[ M 16 ( X )] = Var[ X ] 16 = 3 16 (3) (c) P [ X 1 ≥ 9] = integraldisplay ∞ 9 f X 1 ( x ) dx = integraldisplay 10 9 (1 / 6) dx = 1 / 6 (4) (d) The variance of M 16 ( X ) is much less than Var[ X 1 ]. Hence, the PDF of M 16 ( X ) should be much more concentrated about E [ X ] than the PDF of X 1 . Thus we should expect P [ M 16 ( X ) > 9] to be much less than P [ X 1 > 9]. P [ M 16 ( X ) > 9] = 1 − P [ M 16 ( X ) ≤ 9] = 1 − P [( X 1 + ··· + X 16 ) ≤ 144] (5) By a Central Limit Theorem approximation, P [ M 16 ( X ) > 9] ≈ 1 − Φ parenleftbigg 144 − 16 μ X √ 16 σ X parenrightbigg = 1 − Φ(2 . 66) = 0 . 0039 (6) As we predicted, P [ M 16 ( X ) > 9] ≪ P [ X 1 > 9]....
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 Spring '08
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 Standard Deviation, Variance, Probability theory, var, Chebyshev inequality

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