elemprob-fall2010-page7

# elemprob-fall2010-page7 - P X = n =(1-p n-1 p Such a random...

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Proposition 3.2 If Y = f ( X ) , then E Y = E f ( Y ) = X f ( x ) P ( X = x ) . Proof. P ( Y = y ) = X { x : f ( x )= y } P ( X = x ) , so E Y = X y P ( Y = y ) = X y X { x : f ( x )= y } y P ( X = x ) = X y X { x : f ( x )= y } f ( x ) P ( X = x ) = X x f ( x ) P ( X = x ) . Theorem 3.3 E ( X 1 + ··· + X n ) = E X 1 + ··· + E X n . Proof. We will do the case when n = 2, the general case being done similarly. Using the alternate deﬁnition, E ( X 1 + X 2 ) = X ω ( X 1 + X 2 )( ω ) P ( ω ) = X ω X 1 ( ω ) P ( ω ) + X ω X 2 ( ω ) P ( ω ) = E X 1 + E X 2 . Suppose X is such that
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Unformatted text preview: P ( X = n ) = (1-p ) n-1 p . Such a random variable is said to have a geometric distribution. Let’s ﬁnd the expectation of X . We need to evaluate ∞ X n =1 n (1-p ) n-1 p. 7...
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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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