Let $X$ and $Y$ be independent random variables taking only integer values. Let $Z=X+Y$, which also takes only integer values. Its PMF can be computed by the convolution formula: for any integer $z$,
\begin{align}
\ P_Z(z) & = P(Z=z) =P(X+Y=z) \\
& = \sum_{x=-\infty}^{\infty} P(X=x,X+Y=z)\\
& = \sum_{x=-\infty}^{\infty} P(X=x,Y=z-x) \\
& = \sum_{x=-\infty}^{\infty} P(X=x)P(Y=z-x) \\
& = \sum_{x=-\infty}^{\infty} P_X(x) P_Y(z-x)
\end{align}
Question: Let $X$ and $Y$ be independent and have geometric distribution with parameter $p$. By computing the PMF, show $X+Y$ has a negative binomial distribution with parameters $r=2$ and $p$.
Geometric distribution is given by $f(x)=(1-p)^{x-1} p$ and negative binomial distribution is $P(X=x|r,p)={x-1 \choose r-1}p^r (1-p)^y$. I would like to know how to calculate the convolution formula for PMF. Any help is appreciated.
Best Answer
You have your geometric distribution supported on $\{1,2,3,\ldots\}$ and negative binomial supported on $\{r,r+1,r+2,\ldots\}$
So $$\mathbb P(Z=z) = \sum\limits_{x=-\infty}^{\infty} P_X(x) P_Y(z-x) \\= \sum\limits_{x=1}^{z-1} P_X(x) P_Y(z-x) \\= \sum\limits_{x=1}^{z-1}(1-p)^{x-1} p \,(1-p)^{z-x-1} p \\= \sum\limits_{x=1}^{z-1}p^2 (1-p)^{z-2}\\= (z-1) p^2 (1-p)^{z-2} \\={z-1 \choose 1}p^2 (1-p)^{z-2} \\= {z-1 \choose r-1}p^r (1-p)^{z-r}$$ with $r=2$