# Turning a mgf into a pmf

• September 29th 2008, 03:58 PM
wolverine21
Turning a mgf into a pmf
this is given m(t) = (1/3) + (2/3)e^t

m (t) = (.25 + .75e^t)^12

How do these become pmf's and also what is the first and second derivative of the first one. Is it just 2/3 e^t?
• September 29th 2008, 08:10 PM
mr fantastic
Quote:

Originally Posted by wolverine21
this is given m(t) = (1/3) + (2/3)e^t

m (t) = (.25 + .75e^t)^12

How do these become pmf's and also what is the first and second derivative of the first one. Is it just 2/3 e^t? Mr F says: Yes.

Ah yes. The recognition problem.

Big hint: If X ~ Binomial(n, p) then $m_X (t) = [(1 - p) + p e^t]^n$