Moment-Generating Function

**Yan** · March 1st 2009, 01:03 PM

Find the moment-generating function of the negative binomial distribution by making use of the fact that if k independent random variable have geometric distributions with the same parameter Θ, their sum is random variable having the negative binomial distribution with the parameters Θ and k.

**matheagle** · March 1st 2009, 05:40 PM

Let $Y=\sum_{i=1}^kX_i$ where the X's are i.i.d. Geometrics, which parameter theta. NOW, be careful, not all Geo's are defined the same, it depends if you count the trivial in which that success happens or not. If $P(X_1=x)=\theta(1-\theta)^{x-1}$ , then X's MGF is
$M(t)={\theta e^t\over 1-(1-\theta)e^t}$ .
Then by independence, the MGF of Y is
$E(e^{Yt})=E(e^{(X_1+\cdots +X_k)t})=E(e^{X_1t})\cdots <br /> E(e^{X_kt})={\theta e^t\over 1-(1-\theta)e^t}\cdots {\theta e^t\over 1-(1-\theta)e^t}$
$=\biggl({\theta e^t\over 1-(1-\theta)e^t}\biggr)^k$ .

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