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,...

1 Answer

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  • Anonymous
    1 decade ago
    Best Answer

    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,......

    .

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