Let $X_{i}$, $i=1,2,\dots, n$, be independent random variables of geometric distribution, that is, $P(X_{i}=m)=p(1-p)^{m-1}$. How to compute the PDF of their sum $\sum_{i=1}^{n}X_{i}$?
I know intuitively it's a negative binomial distribution $$P\left(\sum_{i=1}^{n}X_{i}=m\right)=\binom{m-1}{n-1}p^{n}(1-p)^{m-n}$$ but how to do this deduction?
Let $X_{1},X_{2},\ldots$ be independent rvs having the geometric distribution with parameter $p$, i.e. $P\left[X_{i}=m\right]=pq^{m-1}$ for $m=1,2.\ldots$ (here $p+q=1$).
Define $S_{n}:=X_{1}+\cdots+X_{n}$.
With induction on $n$ it can be shown that $S_{n}$ has a negative binomial distribution with parameters $p$ and $n$, i.e. $P\left\{ S_{n}=m\right\} =\binom{m-1}{n-1}p^{n}q^{m-n}$ for $m=n,n+1,\ldots$.
It is obvious that this is true for $n=1$ and for $S_{n+1}$ we find for $m=n+1,n+2,\ldots$:
$P\left[S_{n+1}=m\right]=\sum_{k=n}^{m-1}P\left[S_{n}=k\wedge X_{n+1}=m-k\right]=\sum_{k=n}^{m-1}P\left[S_{n}=k\right]\times P\left[X_{n+1}=m-k\right]$
Working this out leads to $P\left[S_{n+1}=m\right]=p^{n+1}q^{m-n-1}\sum_{k=n}^{m-1}\binom{k-1}{n-1}$ so it remains to be shown that $\sum_{k=n}^{m-1}\binom{k-1}{n-1}=\binom{m-1}{n}$.
This can be done with induction on $m$:
$\sum_{k=n}^{m}\binom{k-1}{n-1}=\sum_{k=n}^{m-1}\binom{k-1}{n-1}+\binom{m-1}{n-1}=\binom{m-1}{n}+\binom{m-1}{n-1}=\binom{m}{n}$
Another way to do this is by using moment-generating functions. In particular, we use the theorem, a probability distribution is unique to a given MGF(moment-generating functions).
Calculation of MGF for negative binomial distribution:
$$X\sim \text{NegBin}(r,p),\ P(X=x) = p^rq^x\binom {x+r-1}{r-1}.$$
Then, using the definition of MGF:
$$E[e^{tX}]=\sum_{x=0}^{\infty}p^rq^x\binom {x+r-1}{r-1}\cdot e^{tx} = p^r(1-qe^t)^{-r}=M(t)^r,$$
where $M(t)$ denotes the moment generating function of a random variable $Y \sim \text{Geo}(p)$. As
$$E[e^{t(X_1+X_2+\dots+X_n)}]=\prod_{i=1}^nE[e^{tX_i}]$$
since they are independent, and we are done.
This is an old question, but a probabilistic interpetation was requested, but not given. So, I give it here.
Consider continually tossing a $p$-weighted coin. Let $X_i$ be the number of tosses between the $(i-1)$-th and $i$-th head. Then, $X_1, X_2, ... \sim^\mathsf{iid} \operatorname{Geometric}(p)$. The sum $Y := \sum_{i=1}^n X_i$ is just the time until the $n$-th head.
How do we calculate this? Well, if it takes $m \ge n$ tosses, then precisely $n$ of the tosses must be heads and the remaining $m-n$ tails. These particular events have probability $p^n$ and $(1-p)^{m-n}$, respectively.
Now, we just need to count how many ways they can be ordered. One might think it's $\binom mn$: after all, there are $n$ heads to place amongst $m$ tosses. However, the last toss is always a head: that's when you stop. So, there are actually $n-1$ heads to places amongst $m-1$ tosses. Hence, there are $\binom{m-1}{n-1}$ orderings.
Multiplying these three numbers gives the required PDF: $$ \mathbb P(Y = m) = \binom{m-1}{n-1} p^n (1-p)^{m-n}. $$
