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Unformatted text preview: Stat 302, Introduction to Probability Jiahua Chen JanuaryApril 2011 Jiahua Chen () Lecture 9 JanuaryApril 2011 1 / 32 Variance of continuous random variables If X has continuous distribution with pdf f ( x ) , then var ( X ) = E ( X ) 2 = integraldisplay ( x ) 2 f ( x ) dx where = E ( X ) and the range of integration is over R . We do require the existence of E ( X ) and this integration to claim the variance of X exists. We rarely directly compute E ( X ) 2 but make use of var ( X ) = E ( X 2 ) { E ( X ) } 2 . Jiahua Chen () Lecture 9 JanuaryApril 2011 2 / 32 Three continuous random variables Uniform distribution on [0, 1]: density function f ( x ) = 1 for x 1; Exponential distribution with rate parameter > 0: density function f ( x ; ) = exp ( x ) for x > 0. Normal distribution with parameters and 2 : density function is f ( x ; , 2 ) = 1 2 exp { ( x ) 2 2 2 } . Jiahua Chen () Lecture 9 JanuaryApril 2011 3 / 32 Variance of Uniform [0, 1] distribution Uniform distribution on [0, 1]: density function f ( x ) = 1 for x 1; It is seen that E ( X 2 ) = integraldisplay 1 x 2 1 dx = 1 3 . Hence, we have var ( X ) = 1 3 parenleftbigg 1 2 parenrightbigg 2 = 1 12 . Jiahua Chen () Lecture 9 JanuaryApril 2011 4 / 32 Variance of Uniform [0, 1] distribution Uniform distribution on [0, 1]: density function f ( x ) = 1 for x 1; It is seen that E ( X 2 ) = integraldisplay 1 x 2 1 dx = 1 3 . Hence, we have var ( X ) = 1 3 parenleftbigg 1 2 parenrightbigg 2 = 1 12 . Remember again that the variance of any random variable is positive (if exists). Jiahua Chen () Lecture 9 JanuaryApril 2011 4 / 32 Variance of exponential distribution Exponential distribution with rate parameter > 0: density function f ( x ; ) = exp ( x ) for x > 0. Its mean is given by = 1 / . It is seen that E ( X 2 ) = integraldisplay x 2 exp ( x ) dx = ( 3 ) 2 = 2 2 Hence, we have var ( X ) = 2 2 parenleftbigg 1 parenrightbigg 2 = 1 2 In terms of = E ( X ) , we have var ( X ) = 2 . Jiahua Chen () Lecture 9 JanuaryApril 2011 5 / 32 Variance of standard normal The standard normal distribution has pdf ( x ) = 1 2 exp { 1 2 x 2 } . Its mean is 0. It is seen that if X has standard normal distribution, E ( X 2 ) = 2 integraldisplay x 2 ( x ) dx = 2 integraldisplay t 3 / 2 1 exp ( t ) dt = 2 ( 3 / 2 ) = 1 ( 1 / 2 ) = 1. Hence, we have var ( X ) = 1 2 = 1. Jiahua Chen () Lecture 9 JanuaryApril 2011 6 / 32 Variance of N( , 2 ) distribution If Y N ( , 2 ) , then E ( Y ) = shown already....
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 Spring '11
 Dr.Chen
 Probability, Variance

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