Notes7 - For a discrete random variable X and event A define E(X | A = x P(X = x | A x∈X E(g(X | A = g(x)P(X = x | A x∈X whenever the sums

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Unformatted text preview: For a discrete random variable X and event A, define E (X | A) = x P (X = x | A) x∈X E (g (X ) | A) = g (x)P (X = x | A) x∈X whenever the sums converge absolutely. Analog of “Law of Total Probability” for expectations: If the events B1 , B2 , . . . , Bk form a partition, then k E (g (X ) | B i )P (B i ) . Eg (X ) = i=1 Note: It is also true when k = ∞. Proof: k E (g (X ) | B i )P (B i ) i=1 k g (x)P (X = x | B i ) = i=1 P (B i ) x∈X k g (x)P (X = x | B i )P (B i ) = x∈X i=1 k = P ({ X = x} ∩ B i ) g (x) x∈X = i=1 g (x)P (X = x) = Eg (X ) . x∈X Conditional expectations have the same properties as ordinary expectations. For example: E (aX + bY + c | A) = aE (X | A) + bE (Y | A) + c Application: Mean and Variance of Geometric Distribution: Suppose X ∼ Geometric(p). The Discrete Memoryless Property can be written: For all y > 0, P (X − y > z | X > y ) = P (X > z ) for z = 1, 2, 3, . . . Or in other words: L(X − y | X > y ) = L(X ) which implies E (g (X − y ) | X > y ) = E (g (X )) for any g (“L” means “distribution” or “law”) Mean of X : EX pEX = = = = = E (X | X = 1)P (X = 1) + E (X | X > 1)P (X > 1) 1 · p + E (1 + (X − 1) | X > 1) · (1 − p) p + E (1 + X ) · (1 − p) p + (1 + E (X ))(1 − p) so that 1 leading to EX = 1/p. Variance of X : Similarly, EX 2 = E (X 2 | X = 1)P (X = 1) + E (X 2 | X > 1)P (X > 1) = 1 · p + E (1 + (X − 1))2 | X > 1 · (1 − p) = 1 · p + E (1 + X )2 · (1 − p) pEX 2 = p + (1 + 2EX + EX 2 )(1 − p) 2 = p + 1 + + EX 2 (1 − p) so that p 2 2−p = − 1 and EX 2 = p p2 Thus Var(X ) = EX 2 − (EX )2 = 1 1−p 2−p − 2= . p2 p p2 Additional Exercise: Compute EX 3 by this approach. ...
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This note was uploaded on 12/15/2011 for the course STAT 5326 taught by Professor Frade during the Fall '10 term at FSU.

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Notes7 - For a discrete random variable X and event A define E(X | A = x P(X = x | A x∈X E(g(X | A = g(x)P(X = x | A x∈X whenever the sums

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