he5_soln

# he5_soln - Homework 5 Solutions Problem Solutions Yates and...

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Unformatted text preview: Homework 5 Solutions Problem Solutions : Yates and Goodman, 3.4.2 3.4.5 3.5.3 3.5.5 3.6.1 and 3.6.4 Problem 3.4.2 Solution From Appendix A, we observe that an exponential PDF Y with parameter λ > 0 has PDF f Y ( y ) = λe- λy y ≥ otherwise (1) In addition, the mean and variance of Y are E [ Y ] = 1 λ Var[ Y ] = 1 λ 2 (2) (a) Since Var[ Y ] = 25, we must have λ = 1 / 5. (b) The expected value of Y is E [ Y ] = 1 /λ = 5. (c) P [ Y > 5] = Z ∞ 5 f Y ( y ) dy =- e- y/ 5 ∞ 5 = e- 1 (3) Problem 3.4.5 Solution (a) The PDF of a continuous uniform (- 5 , 5) random variable is f X ( x ) = 1 / 10- 5 ≤ x ≤ 5 otherwise (1) (b) For x <- 5, F X ( x ) = 0. For x ≥ 5, F X ( x ) = 1. For- 5 ≤ x ≤ 5, the CDF is F X ( x ) = Z x- 5 f X ( τ ) dτ = x + 5 10 (2) The complete expression for the CDF of X is F X ( x ) = x <- 5 ( x + 5) / 10 5 ≤ x ≤ 5 1 x > 5 (3) (c) The expected value of X is Z 5- 5 x 10 dx = x 2 20 5- 5 = 0 (4) Another way to obtain this answer is to use Theorem 3.6 which says the expectedAnother way to obtain this answer is to use Theorem 3....
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he5_soln - Homework 5 Solutions Problem Solutions Yates and...

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