Probability: Discrete Random Variable

1. Jul 12, 2012

stosw

1. The problem statement, all variables and given/known data
Suppose X is a discrete random variable whose probability generating function is
G(z) = z^2 * exp(4z-4)

2. Relevant equations
No idea

3. The attempt at a solution
I'm thinking that due to the exponent on the z term, that the exp(4z-4) would be the
P[X=3] = exp(4z-4), but i'm not even sure of this.

I honestly have no idea where to even start on a problem like this. Any sort of guidance would be great.

2. Jul 12, 2012

Simon Bridge

3. Jul 12, 2012

stosw

All the info given in [1] is what was given for the problem, I forgot to say that I am suppose to find the expected value, var[x], and the distribution on x.

I honestly have no idea what I am doing. Any hints would be great.

4. Jul 12, 2012

Simon Bridge

Well, that link has the information. Have a look, have a go, and then show us where you get stuck.

Start with $$m_X(t) = \sum_{k=0}^n {P(X=k) e^{kt}}$$... to get the distribution, then use the definitions for expectation and variance.

5. Jul 12, 2012

Ray Vickson

What you have written is the "moment-generating function", rather that the probability generating function. For a discrete random variable $X \in \{0,1,2,\ldots \}$ the probability generating function is
$$G(z) \equiv E z^X = \sum_{k=0}^{\infty} P(X=k) z^k.$$
See, eg., http://en.wikipedia.org/wiki/Probability-generating_function .

RGV

6. Jul 13, 2012

Simon Bridge

Why yes I did, and it is indeed - please see post #2, and the link from that post, for the reasoning behind that :)

7. Jul 13, 2012

Ray Vickson

Of course one can use the moment-generating function for discrete, integer-valued random variables, but it is not very convenient; the moment-generating function (or Laplace transform) works better for continuous random variables. In the OP's example, the mgf would be
$$M_X(t) = G(e^t) = e^{2t - 4 + 4e^t},$$
which is not particularly nice to work with.

RGV

8. Jul 13, 2012

Simon Bridge

Well... either way OP has a place to start.