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 $\displaystyle M_X(t)=E[e^{tX}]=\sum_{k=0}^\infty t^k \cdot\frac{E[X^k]}{k!}$

(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 $\displaystyle X=5$ 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 $\displaystyle M_X(t)=E[e^{tX}]=\sum_{k=0}^\infty t^k \cdot\frac{E[X^k]}{k!}$

(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 $\displaystyle X=5$ 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.