probability mass function of sum of two independent geometric random variables How could it be proved that the probability mass function of X + Y, where X and Y are independent random variables each geometrically distributed with parameter p; i.e.
$p_X(n)=p_Y(n)=\left\{\begin{matrix}
p(1-p)^{n-1} & n=1,2,...\\ 
0 & otherwise
\end{matrix}\right.$
equals to $\mathbf{P_{X+Y}(n)= \color{Red}{(n-1)}\ p^2(1-p)^{n-2}}$
 
 
Using convolution I get
 
$\mathbf{P(X+Y=n)=\sum_{n}^{k=0} Pr(X=k)*Pr(Y=n-k) =\sum_{k=1}^n p_X (1-p_x)^{k-1} p_Y(1-p_Y)^{n-k-1}}$ 
 
as $p=p_X=p_Y$  it reduces to
 
$\mathbf{P(X+Y=n)=\sum_{k=1}^n p^2(1-p)^{n-2}}$
  
is this a correct way? I am stuck here, I don't know how to get the final formula.I miss some transition in order to get the (n-1).
 A: Since $X, Y \geq 1$, the summation should run over $k = 1,2, \dots, n-1$.  Using this your convolution becomes
\begin{eqnarray*} P(X+Y = n) &=& \sum_{k=1}^{n-1} p^2(1-p)^{n-2} \\
& = & p^2(1-p)^{n-2}\sum_{k=1}^{n-1} 1 \\
& = & p^2(1-p)^{n-2}(n-1) .
\end{eqnarray*}
A: A geometric random variable is the count of Bernouli trial until a success. We measure the probability of obtaining $n-1$ failures and then $1$ success.$$\mathsf P(X=n) = (1-p)^{n-1} p\qquad :n\in\{1,2,\ldots\}$$ 
The sum of two such is the count of Bernouli trials until the second success. We measure the probability of obtaining $1$ success and $n-2$ failures, in any arrangement of those $n-1$ trials, followed by the second success.
$$\mathsf P(X+Y=n) = (n-1) (1-p)^{n-2} p^2\qquad :n\in\{2,3\ldots\}$$
This may also be counted by summing $$\begin{align}\mathsf P(X+Y=n) & = \sum_{k=1}^{n-1} \mathsf P(X=k, Y=n-k) & \text{note the range} \\[1ex]  & = \sum_{k=1}^{n-1} \mathsf P(X=k)\mathsf P(Y=n-k) & \text{by independence} \\[1ex] & = \sum_{k=1}^{n-1} (1-p)^{k-1} p \cdot (1-p)^{n-k-1} p \\[1ex] & = (1-p)^{n-2} p^2 \sum_{k=1}^{n-1} 1 \\[1ex] & = (n-1) (1-p)^{n-2}p^2\end{align}$$ 
Since $X+Y$ must equal $n$ and neither can be less than $1$, then neither can be more than $n-1$.   Hence this the range of $X$ values we must sum over.
