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I already tried to search an answer for this question, but none of the other posts were satisfactory.

If we have n independent random variables $X_1,\dots,X_n$ where each $X_i$ is distributed according to $q_i(1 - q_i)^k, k \in \mathbb{Z}_+$, is the sum $S_n = \sum_{i = 1}^{n}X_i$ a geometric random variable?

I tried to see how this works in the case of $n = 2$ by computing the probability generating functions for $X_1$ and $X_2$ and then comparing whether or not $\phi_{S_2}(s) = \phi_{X_1}(s)\phi_{X_2}(s)$ has the form of $\frac{p}{1 - s(1 - p)}$ for some $p$. Opening the expression for $\phi_{S_2}(s)$ gives $\phi_{X_1}(s)\phi_{X_2}(s) = \frac{q_1}{1 - s(1 - q_1)}\frac{q_2}{1 - s(1 - q_2)} = \frac{q_1q_2}{1 - s((1 - q_1) + (1 - q_2) + s(1-q_1)(1-q_2))}$. To me this does not have the desired shape, but I do not really know how to perform more deeper analysis. So what gives?

Epsilon Away
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1 Answers1

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The MGF of your geometric distribution is the following

$$\frac{p}{1-(1-p)e^t}$$

Being $X_i$ iid, MGF of $Y=\Sigma_i X_i$ is

$$\frac{p^n}{[1-(1-p)e^t]^n}$$

Which is the MGF of a Negative Binomial

tommik
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  • I see (now) that the rvs are independent but not iid. So your answer needs some modification. Maybe there was an edit, but I cannot see that from here (iphone). – drhab Dec 05 '20 at 15:15