Exam1___Solutions

# Exam1___Solutions - Midterm 1 Solutions 1 Let X and Y be...

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Midterm 1 Solutions 1. Let X and Y be random variables with the following joint distribution. x y -1 0 1 -1 0.1 0.2 0.2 0 0.3 0 0.1 1 0 0 0.1 (a) Find P ( X 0 and Y 0) . Solution. 0 + 0.1 + 0.1 + 0 = 0.2. (b) Compute P ( X = x ) for all possible values of x (i.e., - 1 , 0 , and 1 ). Solution. x -1 0 1 P X ( x ) 0.4 0.2 0.4 Each of the above values is the sum of the entries in the corresponding column in the original table. (E.g., 0.2 = 0.2 + 0 + 0.) (c) What is the variance of X ? Solution. First we compute the mean: E ( X ) = x xP X ( x ) = - 1 · 0 . 4+0 · 0 . 2+1 · 0 . 4 = 0. Then Var( X ) = x x 2 P X ( x ) - E ( X ) 2 = ( - 1) 2 · 0 . 4+0 2 · 0 . 2+1 2 · 0 . 4 - 0 2 = 0 . 8 . (d) Are X and Y independent? Solution. No, because, for example, P XY (0 , 0) = 0, while P X (0) = 0 . 2 and P Y (0) = 0 . 3 + 0 + 0 . 1 = 0 . 4. 2. Let X be a random variable with probability density f X ( x ) = 1 2 (1 + 3 x 2 ) , where 0 < x < 1 . (a) What is the mean of X ? Solution. R 1 0 xf X ( x ) dx = R 1 0 x · 1 2 (1 + 3 x 2 ) dx = 1 2 ( x 2 2 + 3 x 4 4 ) fl fl 1 0 = 5 8 = 0 . 625 . (b) What is the standard deviation of X ? Solution. First we compute the variance: Var( X ) = R 1 0 x 2 f X ( x ) dx - E ( X ) 2 = R 1 0 x 2 · 1 2 (1+3 x 2 ) dx - ( 5 8 ) 2 = 1 2 ( x 3 3 + 3 x 5 5 ) fl fl 1 0 = 14 30 - 25 64 = 73 960 .

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