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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