Finding MGF and PMF given a related MGF?

If E(X^r) =5^r, r = 1, 2, 3, ..., find the moment-generating function M(t) of X and the p.m.f. of X.

The m.g.f. of X is sum((e^tx)*f(x)) of all x in S.

so the mgf of X^r is sum((e^t(X^r))*f(x^r)... this is where I'm lost. Can I get some guidance how to apply the E(X^r) = 5^r to this? Also, do I have to find the pmf first or the mgf? or can I find both from the original statement?

Thank you in advance. (-goes to look up how to use math tags and equations-)

Re: Finding MGF and PMF given a related MGF?

Let me ask a more basic question first.

one of the examples in the book states let the moments of X be defined by E(X^r) = 0.8, r = 1,2,3...

then goes on to find the MGF.

and that P(X=0) = 0.2 and P(X=1) = 0.8.

What is r? If it was just E(X) = 0.8, then the expected value of the random variable X is 0.8. but what is the expected value of a random variable raised to a variable power??

and if E(X^r) = a constant is a bernoulli distribution, is there some similar association I can make for E(X^r) = Constant^r?

edit: just realized/rememberd/associated the notation E(X) finds the first moment, and E(X^2) finds the second etc. so 5^1 is the first moment and 5^2 is the second etc.

Re: Finding MGF and PMF given a related MGF?

Alright, I'm still struggling with this and so are some other people in my class apparently.

We found that the MGF is e^(5t). but I haven't been able to see what function could modify e^(tx) to make it e^(5t) to fit with the Sum(e^(tx) f(x)).

Edit: found out the answer is f(x)=1 x=5 and e^(5t) is indeed the MGF. I kind of understand why e^(5t) is the MGF but i'm at somewhat of a loss why the pmf is 1.

Re: Finding MGF and PMF given an expected value?

Hello,

There's a formula that gives

(this comes from the exponential series)

So the mgf is indeed exp(5t).

Now by recognizing the mgf of an almost surely constant random variable, you can deduce that almost surely. Thus its pmf is 1 at x=5.

Re: Finding MGF and PMF given an expected value?

Quote:

Originally Posted by

**Moo** Hello,

There's a formula that gives

(this comes from the exponential series)

So the mgf is indeed exp(5t).

Now by recognizing the mgf of an almost surely constant random variable, you can deduce that

almost surely. Thus its pmf is 1 at x=5.

Thank you, I think I get it now. I lost sight of the fact that the pmf is a function and is a result of the x in e^(tx) so i was looking for some function to directly turn e^(5t) into e^(tx). thank you again.