Expectation of a Random Variable

  • Thread starter Firepanda
  • Start date
  • #1
430
0
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
 

Answers and Replies

  • #2
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
10,706
1,728
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 [itex]d\left(e^{-x^2/2}\right) = - x e^{-x^2/2} \, dx, [/itex] and use that in integration by parts.

RGV
 

Related Threads on Expectation of a Random Variable

  • Last Post
Replies
5
Views
1K
Replies
3
Views
7K
  • Last Post
Replies
3
Views
1K
Replies
5
Views
548
Replies
6
Views
3K
  • Last Post
Replies
4
Views
1K
Replies
2
Views
3K
Replies
13
Views
3K
Replies
4
Views
1K
Replies
12
Views
4K
Top