.

Now make the substitution and use a well known result.

It's simple to confirm the answer, see here: Chi-Squared Distribution -- from Wolfram MathWorld

Results 1 to 10 of 10

- May 6th 2010, 10:33 PM #1

- Joined
- May 2010
- Posts
- 6

## MGF Help Please

Suppose X ~ N(0,1), i.e. X is normally distributed with mean 0 and variance 1.

Let Y = X^2

Find the MGF of Y DIRECTLY...

Can't seem to get a good answer out. Any help would be appreciated.

Also...

If f(z) = e^(-z/2)/sqrt(2*pi*z) for z > 0

and f(z) = 0 for z <= 0

Find the MGF of Z.

THANKS ALOT guys!

- May 7th 2010, 04:26 AM #2
.

Now make the substitution and use a well known result.

It's simple to confirm the answer, see here: Chi-Squared Distribution -- from Wolfram MathWorld

- May 7th 2010, 05:29 PM #3

- Joined
- May 2010
- Posts
- 6

- May 7th 2010, 05:30 PM #4

- Joined
- May 2010
- Posts
- 6

- May 8th 2010, 03:00 AM #5

- May 8th 2010, 07:55 PM #6

- Joined
- May 2010
- Posts
- 6

Well firstly, considering

Can't x take on any value between negative infinity and infinity? Since X ~ N(0,1)?

So the new limits from the substitution would be negative infinity to positive infinity?

And also, when I take the function of t out, i get...

I've researched a bit, and i've tried to integrate the function directly (cause its only an exponential) but the limits give me weird answers...

I'm really stuck mr fantastic...

- May 9th 2010, 02:33 AM #7
Your integral limits are correct. As far as the final integral, read this for example: Gaussian Integral -- from Wolfram MathWorld

(You really should know this at this level).

- May 9th 2010, 04:42 PM #8

- Joined
- May 2010
- Posts
- 6

I'm in 1st year university, studying an actuarial science degree.

I haven't seen this "Gaussian Integral" before mr fantastic.

So I couldn't really understand where you were coming from...

I thought I could just integrate the integral directly considering it wasn't a tricky integral, but the limits don't give a suitable answer.

Is there any other method to integrate, say by recognition of a statistical distribution?

P.S you've been really helpful, and I appreciate it very much.

Thank you

- May 9th 2010, 05:12 PM #9

- May 10th 2010, 04:49 AM #10