Definition of moment generating function

[donutmax]

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

Is this correct?

 

[eachus]

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

Is this correct?

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.

 

[statdad]

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

Is this correct?

Yes, for discrete distributions. For continuous distributions replace the sum with an integral.

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

<br /> m_Y(t) = \sum_{y=1}^\infty {e^{ty} p(y)}<br />