elemprob-fall2010-page37

elemprob-fall2010-page37 - + . Note that it is not always...

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gamma with parameters s + t and λ . In particular, the sum of n independent exponentials with parameter λ is a gamma with parameters n and λ . (2) If Z is a N (0 , 1), then F Z 2 ( y ) = P ( Z 2 y ) = P ( - y Z y ) = F Z ( y ) - F Z ( - y ). Differentiating shows that f Z 2 ( y ) = ce - y/ 2 ( y/ 2) (1 / 2) - 1 , or Z 2 is a gamma with parameters 1 2 and 1 2 . So using (1) above, if Z i are independent N (0 , 1)’s, then n i =1 Z 2 i is a gamma with parameters n/ 2 and 1 2 , i.e., a χ 2 n . (3) If X i is a N ( μ i 2 i ) and the X i are independent, then some lengthy calculations show that n i =1 X i is a N ( μ i , σ 2 i ). (4) The analogue for discrete random variables is easier. If X and Y takes only nonnegative integer values, we have P ( X + Y = r ) = r X k =0 P ( X = k,Y = r - k ) = r X k =0 P ( X = k ) P ( Y = r - k ) . In the case where X is a Poisson with parameter λ and Y is a Poisson with parameter μ , we see that X + Y is a Poisson with parameter
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Unformatted text preview: + . Note that it is not always the case that the sum of two independent random variables will be a random variable of the same type. If X and Y are independent normals, then-Y is also a normal (with E (-Y ) =-E Y and Var (-Y ) = (-1) 2 Var Y = Var Y ), and so X-Y is also normal. To dene a conditional density in the discrete case, we write p X | Y = y ( x | y ) = P ( X = x | Y = y ) . This is equal to P ( X = x,Y = y ) P ( Y = y ) = p ( x,y ) p Y ( y ) . Analogously, we dene in the continuous case f X | Y = y ( x | y ) = f ( x,y ) f Y ( y ) . 37...
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This note was uploaded on 12/29/2011 for the course MATH 316 taught by Professor Ansan during the Spring '10 term at SUNY Stony Brook.

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