Expectation of a Random Variable

[Firepanda]

I know the E[X] = Integral between [-inf,inf] of X*f(x) dx

Where X is normally distributed and f(x) is the PDF

How do I find the expectation of X⁴?

Bare with me because I'm useless in Latex

So far what I've done is written the integral as Integral between [-inf,inf] of X⁴*f(x) dx

and I started to subsitute u = exp{-x²/2t}

So now I have Integral between [-inf,inf] of -tX³*u dx

I really don't think this is correct.. I'm trying to follow the same way my lecturer did it for the expectation of X², but at that statge he started to integrate by parts, yet mine doesn't look like that and it's essentially the same!

Can anyone help?

Thanks

 

[Ray Vickson]

I know the E[X] = Integral between [-inf,inf] of X*f(x) dx

Where X is normally distributed and f(x) is the PDF

How do I find the expectation of X⁴?

Bare with me because I'm useless in Latex

So far what I've done is written the integral as Integral between [-inf,inf] of X⁴*f(x) dx

and I started to subsitute u = exp{-x²/2t}

So now I have Integral between [-inf,inf] of -tX³*u dx

I really don't think this is correct.. I'm trying to follow the same way my lecturer did it for the expectation of X², but at that statge he started to integrate by parts, yet mine doesn't look like that and it's essentially the same!

Can anyone help?

Thanks

The standard way of dealing with this type of problem is to note that $d\left(e^{-x^2/2}\right) = - x e^{-x^2/2} \, dx,$ and use that in integration by parts.

RGV