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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).

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2 Answers 2

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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*}

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  • $\begingroup$ Could you please clarify more what happened here with the summation? Based on what was it converted in $\sum_{k=1}^{n-1} 1$ and then further into (n-1). I don't get those steps. $\endgroup$
    – Michal
    Sep 14, 2015 at 21:31
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    $\begingroup$ @Michal First move all factors that aren't terms of $k$ to the outside of the sum, as $\sum_{k=1}^{n-1} a = a \sum_{k=1}^{n-1} 1$ Then: $\sum_{k=1}^{n-1} 1 = \underbrace{1+ 1+ \ldots +1}_{\text{how many?}}$ $\endgroup$ Sep 14, 2015 at 21:51
  • $\begingroup$ ok, clear now. Thanks guys! $\endgroup$
    – Michal
    Sep 14, 2015 at 21:54
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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.

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  • $\begingroup$ Thank you Graham for the detailed explanation! $\endgroup$
    – Michal
    Sep 14, 2015 at 22:19
  • $\begingroup$ Could you tell me what happens when there are different parameters for $X$ and $Y$? $\endgroup$ Jun 17, 2017 at 13:25
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    $\begingroup$ @StubbornAtom Then you do not get the nice neat cancelation and:... $$\sum\limits_{k=1}^{n-1} \Big(\big(p(1 - p)^{k - 1}\big) \big(q(1 - q)^{n - k - 1}\big)\Big) = \dfrac{p q \big((1 - p)^{n-1}-(1-q)^{n-1} \big)}{q-p}\text{ for }n\in\{2,...\}$$ $\endgroup$ Jun 17, 2017 at 13:54
  • $\begingroup$ Thank you. And this is not negative binomial, is it? $\endgroup$ Jun 17, 2017 at 16:52
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    $\begingroup$ @StubbornAtom When $X,Y$ are independent geometric distributions with the same rate, their sum has a negative binomial distribution. Not when their rates differ. $\endgroup$ Jun 18, 2017 at 12:23

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