elemprob-fall2010-page27

elemprob-fall2010-page27 - We give another definition of...

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Unformatted text preview: We give another definition of the expectation in the continuous case. First suppose X is nonnegative and bounded above by a constant M . Define Xn (ω ) to be k/2n if k/2n ≤ X (ω ) < (k +1)/2n . We are approximating X from below by the largest multiple of 2−n . Each Xn is discrete and the Xn increase to X . We define E X = limn→∞ E Xn . Let us argue that this agrees with the first definition in this case. We have E Xn = k /2 n = k P(Xn = k/2n ) = 2n k 2n k /2 n k P(k/2n ≤ X < (k + 1)/2n ) 2n (k+1)/2n (k+1)/2n f (x)dx = k/2n k/2n k f (x)dx. 2n If x ∈ [k/2n , (k + 1)/2n ), then x differs from k/2n by at most 1/2n . So the last integral differs from (k+1)/2n xf (x)dx k/2n by at most (1/2n )P(k/2n ≤ X < (k + 1)/2n ) ≤ 1/2n , which goes to 0 as n → ∞. On the other hand, (k+1)/2n M xf (x)dx = k/2n xf (x)dx, 0 which is how we defined the expectation of X . We will not prove the following, but it is an interesting exercise: if Xm is any sequence of discrete random variables that increase up to X , then limm→∞ E Xm will have the same value E X . To show linearity, if X and Y are bounded positive random variables, then take Xm discrete increasing up to X and Ym discrete increasing up to Y . Then Xm + Ym is discrete and increases up to X + Y , so we have E (X + Y ) = lim E (Xm + Ym ) = lim E Xm + lim E Ym = E X + E Y. m→∞ m→∞ m→∞ If X is not bounded or not necessarily positive, we have a similar definition; we will not do the details. This second definition of expectation is mostly useful for theoretical purposes and much less so for calculations. Similarly to the discrete case, we have 27 ...
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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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