Moment Generating Function

1. Aug 9, 2009

cse63146

1. The problem statement, all variables and given/known data

Let X be uniformly distributed over the unit interval (0,1). Determine the moment generating function of X, and using this, determine all moments around the origin.

2. Relevant equations

3. The attempt at a solution

I know that the MGT is M(x) = E[ext]

I'm just not sure how to start this problem. Could someone give me a hint, or point me in the right direction?

2. Aug 9, 2009

Dick

That makes it the integral from 0 to 1 of exp(x*t)*1*dx, doesn't it?

3. Aug 9, 2009

cse63146

What does the "1" exp(x*t)*1*dx supposed to represent?

4. Aug 9, 2009

Dick

If the distribution is uniform over [0,1], then the pdf is f(x)=1, isn't it? The moment generating function is integral f(x)*exp(x*t)*dx. The 1 is the f(x).

5. Aug 9, 2009

cse63146

Can't believe I didn't realise that. Thanks again.

6. Aug 10, 2009

cse63146

So I did the integral and I got $$\frac{1}{t}e^t - \frac{1}{t}$$

so the nth derivative of e^t is $$\frac{e^t}{t!}$$ (from taylor series)

but not sure what to do about the 1/t.

7. Aug 10, 2009

Dick

The series expansion of e^t is 1+t+t^2/2!+t^3/3!+... Which is quite unlike what you said. The 1/t part cancels.

Last edited: Aug 10, 2009
8. Aug 10, 2009

cse63146

so all moments around the origin is just e^t (evaluated at t = 0)?

9. Aug 11, 2009

Dick

No! Not at all! I was suggesting you expand the infinite series for e^t and use some algebra to write it in a way that's easier to handle for derivatives. You can work directly with e^t/t-1/t as well, but you can't just put t=0 after you take the derivative since that would be undefined - you would have to take the limit as t->0.

10. Aug 11, 2009

cse63146

I see what you mean. I get tn - 1/n!

11. Aug 11, 2009

Dick

Good! You mean the sum of that for n=1 to infinity, right? I.e. 1+t/2!+t^2/3!+..., right? Now it's easy to find the nth derivative of that, also right?

12. Aug 11, 2009

cse63146

the nth derivative would be [(1-n)^n]*(tn - 1/n!)

13. Aug 11, 2009

Dick

Why on earth would it be that? Look, take 1+t/2!+t^2/3!+t^3/4!+... What's the first derivative evaluated at t=0? Now do the second. Try the third. Do you see a pattern? Those are the moments. They are also pretty easy to compute from the pdf without even using a generating function. You should probably do the zero moment as well.

14. Aug 11, 2009

cse63146

probably wrong, but it look's like tn/(n+1)! but if I substitute t = 0, then I get 0.

15. Aug 11, 2009

Dick

Yes, it's wrong. Can you help me out here? m(t)=1+t/2!+t^2/3!+t^3/4!+... That is your moment generating function isn't it? It's a infinite SUM of terms like you have suggested before. Not just a naked t^n/(n+1)!. It's a summation. What's the first derivative at t=0? Please, help me. Exactly ONE of the terms in that series has a nonzero first derivative at t=0. NONE of the others do.

Last edited: Aug 11, 2009
16. Aug 13, 2009

cse63146

So I was looking through my notes and found this:

$$M(0) + \sum M^{(n)}(0)\frac{t^{n-1}}{n!} =1 + \sum^{\infty}_{n=2} E[X^n] \frac{t^{n-1}}{n!}$$

where E[Xn] = 1/(n+1) since it's uniformly distributed on (0,1)

17. Aug 13, 2009

Dick

Yes. E[X^n]=1/(n+1). Because i) the integral from 0 to 1 of x^n is 1/(n+1). And ii) you can also read that off from the generating faction. If you take the nth derivative and set t=0 the only term that contributes is t^n/(n+1)!. All of the others vanish. And differentiating that term n times gives 1/(n+1).

18. Aug 13, 2009

cse63146

So it's supposed to be:

$$M(0) + \sum M^{(n)}(0)\frac{t^{n-1}}{n!} =1 + \sum^{\infty}_{n=2} \frac{1}{n+1} \frac{t^{n-1}}{n!}$$

19. Aug 13, 2009

Dick

Forget the n. Just deal with say the first 3 or 4 moments. Try and prove that the first few moments of your distribution are the same as what you would get from the generating function. Or even the first 1 or 2. Any of them. I'm on the verge of giving up here. Just do numbers, forget the symbols.

Last edited: Aug 13, 2009
20. Aug 13, 2009

cse63146

can't believe it took me so long to get it. Thanks for being so patient, and sorry for all the trouble.

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