On the other hand suppose that pj 0 for innitely many

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Unformatted text preview: , ￿￿ ￿￿ ￿￿ n ￿￿ j−1 j−1 j 2n − 1 2 − 2 2n − 1 E( Xn ) = P , + P , 2n 2n 2n 2n 2n 2n j =1 ￿ 2n − 1 ￿ j−1￿ j j−1 = −n 2n 2n 2 j =1 2n − 2 ￿ n 2 −1 1￿ = 2n (j − 1) 2 j =1 1 (2n − 2)(2n − 1) 2 2n 2 1 1 1 1 = − n − n+1 + 2n 22 2 2 = implying E(X ) = lim E(Xn ) = lim n→∞ n→∞ ￿ 1 1 1 1 − n − n+1 + 2n 22 2 2 ￿ 1 =. 2 Example 21.4. Consider ([0, 1], B1 , P) where B1 are the Borel sets of [0, 1] and P is the uniform probability. Let Q1 = [0, 1] ∩ Q and consider the random variable Y : Ω → R given by ￿ ω , if ω ∈ [0, 1] \ Q1 , Y (ω ) = 0, if ω ∈ Q1 . Note that Y is not a simple random variable, although it is non-negative. In order to compute E(Y ) it is easiest to use Theorem 20.3. That is, let X (ω ) = ω for ω ∈ [0, 1] and note that {ω : X (ω ) ￿= Y (ω )} = Q1 \ {0}. (The technical point here is that X (0) = Y (0) = 0. However, if ω ∈ Q1 with ω ￿= 0, then X (ω ) ￿= Y (ω ).) Since P {Q1 \ {0}} = 0, we see that X and Y differ on a set of probability 0 so that...
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This document was uploaded on 04/07/2014.

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