05 Discrete random variables examples - solutions

05 Discrete random variables examples - solutions -...

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University of California, Los Angeles Department of Statistics Statistics 100A Instructor: Nicolas Christou Examples - solutions Example 1: E ( N ) = E ( Cπr 2 ) = CπEr 2 = ( σ 2 + μ 2 ) . The mean and variance of the distribution of r are: E ( r ) = 23 . 5, var ( r ) = 1 . 54. Therefore, E ( N ) = 8 π (1 . 54 + 23 . 4 2 ) = 13800 . 39 13801 . Or simply compute E ( N ) as: E ( N ) = E ( Cπr 2 ) = X r r 2 P ( r ) = 8 π ± 21 2 (0 . 05) + 22 2 (0 . 20) + ... + 26 2 (0 . 05) ² = 13800 . 39 13801 . Example 2: It is given that P ( X = i ) = cP ( X = i - 1) for i = 1 , 2. Let P(X=0)=p. P ( X = 1) = cP ( X = 0) = cp , and P ( X = 2) = cP ( X = 1) = c 2 p . Also, P ( X = 0) + P ( X = 1) + P ( X = 2) = 1. Or p + cp + c 2 p = 1 p = 1 1+ c + c 2 . The expected value of X is: E ( X ) = 0( p ) + 1( cp ) + 2( c 2 p ) = c 1 + c + c 2 + 2 c 2 1 + c + c 2 = c (1 + 2 c ) 1 + c + c 2 . Example 3:
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This note was uploaded on 01/14/2011 for the course STATS 100A taught by Professor Wu during the Fall '07 term at UCLA.

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