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lecture13 - Stat 302 Introduction to Probability Jiahua...

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Stat 302, Introduction to Probability Jiahua Chen January-April 2011 Jiahua Chen () Lecture 13 January-April 2011 1 / 24
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Moments and covariance formulas Let X be a random variable. If X is discrete, E ( X ) = x xP ( X = x ) , where P ( X = x ) = p ( x ) is the pmf of X . If X is (absolutely) continuous, E ( X ) = integraltext - xf ( x ) , where f ( x ) is the pdf of X . Jiahua Chen () Lecture 13 January-April 2011 2 / 24
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Moments and covariance formulas Let X be a random variable. If X is discrete, E ( X ) = x xP ( X = x ) , where P ( X = x ) = p ( x ) is the pmf of X . If X is (absolutely) continuous, E ( X ) = integraltext - xf ( x ) , where f ( x ) is the pdf of X . The above formulas have their own range of applicability . Be sure to check whether X is discrete or continuous before one of them is applied. Jiahua Chen () Lecture 13 January-April 2011 2 / 24
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Moments and covariance formulas Whether or not X is discrete or continuous, E ( aX + b ) = aE ( X ) + b where a , b are two non-random constants. A trivial example is E ( 2 X - 4 ) = 2 E ( X ) - 4 in which a = 2 and b = - 4. Jiahua Chen () Lecture 13 January-April 2011 3 / 24
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Moments and covariance formulas Let X and Y be two random variables (on the same Ω ). Apparently, aX + bY + c is another random variable (on the same Ω ), where a , b , c are two non-random constants. There is a formula that E ( aX + bY + c ) = aE ( X ) + bE ( Y ) + c . More concretely, E ( 2 X - 3 Y + 4 ) = 2 E ( X ) - 3 E ( Y ) + 4; E ( 3 X + 5 Y ) = 3 E ( X ) + 5 E ( Y ) . Jiahua Chen () Lecture 13 January-April 2011 4 / 24
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Moments and covariance formulas Let X and Y be two random variables (on the same Ω ). Apparently, aX + bY + c is another random variable (on the same Ω ), where a , b , c are two non-random constants. There is a formula that E ( aX + bY + c ) = aE ( X ) + bE ( Y ) + c . More concretely, E ( 2 X - 3 Y + 4 ) = 2 E ( X ) - 3 E ( Y ) + 4; E ( 3 X + 5 Y ) = 3 E ( X ) + 5 E ( Y ) . Technically, the formulas are applicable when E ( X ) and E ( Y ) both exist. Jiahua Chen () Lecture 13 January-April 2011 4 / 24
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Negative binomial Example Let T be the number of failures before the 2nd success in a sequence of Bernoulli trials. Using either probabilistic or mathematical tool, we find T has negative binomial distribution and its pmf is P ( T = k ) = ( k + 1 )( 1 - p ) k p 2 for k = 0, 1, . . . and p is the probability of success. How much is E ( T ) ? A direct approach is to compute E ( T ) = k = 0 k ( k + 1 )( 1 - p ) k p 2 . Jiahua Chen () Lecture 13 January-April 2011 5 / 24
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Negative binomial Example Let T be the number of failures before the 2nd success in a sequence of Bernoulli trials. Let X be the number of failures before the 1st success, and Y be the number of failures between the first and second success.
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