# N.B. Expectatation[ (x)] is the expectation of the function (x), given some probability density function p(x)

1. consider the pdf p(x), p(x)=kx^2 exp (-lambda*x^2), x between minus infinity and infinity, where lambda (>0) and k are both constants. show that k= 2*lambda^3/sqrt of pi.

without explicitly integrating any function, exlain why the odd moments of p(x) are all zero, ie,

E[x^2n+1] = 0, n=0, 1, 2,...

Relevance
• Anonymous

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,......

.