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).
Best Answer
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*}