Patricia Bean

2022-07-21

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. ${p}_{X}\left(n\right)={p}_{Y}\left(n\right)=\left\{\begin{array}{cc}p\left(1-p{\right)}^{n-1}& n=1,2,...\\ 0& otherwise\end{array}$ equas to

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taguetzbo

Expert

Step 1
Since $X,Y\ge 1$, the summation should run over $k=1,2,\dots ,n-1$.
Step 2
$\begin{array}{rcl}P\left(X+Y=n\right)& =& \sum _{k=1}^{n-1}{p}^{2}\left(1-p{\right)}^{n-2}\\ & =& {p}^{2}\left(1-p{\right)}^{n-2}\sum _{k=1}^{n-1}1\\ & =& {p}^{2}\left(1-p{\right)}^{n-2}\left(n-1\right).\end{array}$

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Ruby Briggs

Expert

Step 1
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.
$\mathsf{P}\left(X=n\right)=\left(1-p{\right)}^{n-1}p\phantom{\rule{2em}{0ex}}:n\in \left\{1,2,\dots \right\}$
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.
$\mathsf{P}\left(X+Y=n\right)=\left(n-1\right)\left(1-p{\right)}^{n-2}{p}^{2}\phantom{\rule{2em}{0ex}}:n\in \left\{2,3\dots \right\}$
Step 2
This may also be counted by summing $\begin{array}{rlr}\mathsf{P}\left(X+Y=n\right)& =\sum _{k=1}^{n-1}\mathsf{P}\left(X=k,Y=n-k\right)& \text{note the range}\\ & =\sum _{k=1}^{n-1}\mathsf{P}\left(X=k\right)\mathsf{P}\left(Y=n-k\right)& \text{by independence}\\ & =\sum _{k=1}^{n-1}\left(1-p{\right)}^{k-1}p\cdot \left(1-p{\right)}^{n-k-1}p\\ & =\left(1-p{\right)}^{n-2}{p}^{2}\sum _{k=1}^{n-1}1\\ & =\left(n-1\right)\left(1-p{\right)}^{n-2}{p}^{2}\end{array}$
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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