Yahoo Answers: Answers and Comments for N.B. Expectatation[ (x)] is the expectation of the function (x), given some probability density function p(x) [Homework Help]
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From agbo
enUS
Mon, 06 Aug 2007 12:04:12 +0000
3
Yahoo Answers: Answers and Comments for N.B. Expectatation[ (x)] is the expectation of the function (x), given some probability density function p(x) [Homework Help]
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From Anonymous: 1)
you know
∫p(x) dx = 1 by definition of a ...
https://answers.yahoo.com/question/index?qid=20070806120412AAoJY0H
https://answers.yahoo.com/question/index?qid=20070806120412AAoJY0H
Tue, 07 Aug 2007 15:16:30 +0000
1)
you know
∫p(x) dx = 1 by definition of a pdf
so you need to find
∫kx^2 exp (lambda*x^2) dx
using integration by parts twice and then equate it to 1 will give you the required value of k, it is much too long and tedious for me to do for you here
2.
E[x^2n+1] = ∫x^(2n+1)p(x) dx
let
f(x)
= x^(2n+1)p(x)
= x^(2n+1)kx^2 exp (lambda*x^2)
= kx^(2n+3) exp (lambda*x^2)
= kx^(2n) x^3 exp (lambda*x^2)
then
f(x)
= k(x)^(2n) (x)^3 exp (lambda*(x)^2)
= kx^(2n) x^3 exp (lambda*x^2)
= f(x)
so f(x) is an odd function and you are probably aware that if you integrate (ie find the area under) an odd function over a symmetric interval of the x axis that it equals zero
hence
E[x^2n+1] = ∫f(x)dx = 0 for n = 0,1,2,......
.