elemprob-page15 - If x ∈ [k/2n , (k + 1)/2n ), then x...

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Unformatted text preview: 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 Proposition 6.2. E g (X ) = g (x)f (x)dx. As in the discrete case, Var X = E [X − E X ]2 . As an example of these calculations, let us look at the uniform distribution. We say that a random 1 variable X has a uniform distribution on [a, b] if fX (x) = b−a if a ≤ x ≤ b and 0 otherwise. To calculate the expectation of X , ∞ EX = b xfX (x)dx = −∞ x a b 1 1 dx = b−a b−a x dx = a b2 a2 1 − b−a 2 2 This is what one would expect. To calculate the variance, we first calculate ∞ E X2 = b x2 fX (x)dx = −∞ x2 a 1 a2 + ab + b2 dx = . b−a 3 We then do some algebra to obtain Var X = E X 2 − (E X )2 = (b − a)2 . 12 7. Normal distribution. A r.v. is a standard normal (written N (0, 1)) if it has density 2 1 √ e−x /2 . 2π 15 = a+b . 2 ...
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This note was uploaded on 12/29/2011 for the course MATH 317 taught by Professor Wen during the Spring '09 term at SUNY Stony Brook.

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