Definition of moment generating function

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
2 replies · 2K views
donutmax
Messages
7
Reaction score
0
[tex]M(t)=E(e^{ty})=\sum_{y=0}^{n} e^{ty}p(y)[/tex]

Is this correct?
 
Physics news on Phys.org
donutmax said:
[tex]M(t)=E(e^{ty})=\sum_{y=0}^{n} e^{ty}p(y)[/tex]

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

[tex]\[M(t)=\sum_{n=0}^{\infty}\frac{t^nm_n}{n!}\][/tex]

In other words, differentiating M(x) n times gives you the nth moment of the distribution.
 
donutmax said:
[tex]M(t)=E(e^{ty})=\sum_{y=0}^{n} e^{ty}p(y)[/tex]

Is this correct?

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

Added: the upper limit of [itex]n[/itex] only if the distribution takes on finitely many values (example: binomial). if the distribution takes on infinitely many values, the mgf is

[tex] m_Y(t) = \sum_{y=1}^\infty {e^{ty} p(y)}[/tex]