# Definition of moment generating function

1. May 13, 2010

### donutmax

$$M(t)=E(e^{ty})=\sum_{y=0}^{n} e^{ty}p(y)$$

Is this correct?

2. May 13, 2010

### eachus

No. I'm not going to try to play with TeX here too much, so I'll point you to http://en.wikipedia.org/wiki/Moment-generating_function which is fairly complete. What you really need, in is probably

$$$M(t)=\sum_{n=0}^{\infty}\frac{t^nm_n}{n!}$$$

In other words, differentiating M(x) n times gives you the nth moment of the distribution.

3. May 14, 2010

Added: the upper limit of $n$ only if the distribution takes on finitely many values (example: binomial). if the distribution takes on infinitely many values, the mgf is
$$m_Y(t) = \sum_{y=1}^\infty {e^{ty} p(y)}$$