Sonia Rowland

2022-10-01

Geometric sum of geometric random variables

I am looking to find the probability mass function of $Y=\sum _{i=1}^{N}{X}_{i}$ where ${X}_{i}\sim \text{Geometric}(a)$ and $N\sim \text{Geometric}(b)$. I attempted to do this by finding the probability generating function of Y and comparing it to known probability generating functions to take advantage of the uniqueness property. (In my searches online, it sounds like I should find that $Y\sim \text{Geometric}(ab)$.)

I am looking to find the probability mass function of $Y=\sum _{i=1}^{N}{X}_{i}$ where ${X}_{i}\sim \text{Geometric}(a)$ and $N\sim \text{Geometric}(b)$. I attempted to do this by finding the probability generating function of Y and comparing it to known probability generating functions to take advantage of the uniqueness property. (In my searches online, it sounds like I should find that $Y\sim \text{Geometric}(ab)$.)

Cody Petty

Beginner2022-10-02Added 6 answers

Step 1

Assume that $\begin{array}{}\text{(1)}& \mathbb{P}[{X}_{i}=k]=p(1-p{)}^{k-1},\phantom{\rule{2em}{0ex}}k=1,2,\dots \end{array}$

and $\mathbb{P}[N=n]=r(1-r{)}^{n-1}$ for $r=1,2,\dots $. Then: $\begin{array}{rcl}\mathbb{P}[Y=m]& =& \sum _{n\ge 1}\mathbb{P}[N=n]\cdot \mathbb{P}[{X}_{1}+\dots +{X}_{n}=m]\\ \text{(2)}& & =& \frac{r}{1-r}\sum _{n\ge 1}(1-r{)}^{n}{\left(\frac{p}{1-p}\right)}^{n}(1-p{)}^{m}\cdot r(m,n)\end{array}$

Where: $\begin{array}{rcl}r(m,n)& =& \mathrm{\#}\{({a}_{1},\dots ,{a}_{n}):{a}_{i}\in {\mathbb{N}}_{>0},{a}_{1}+\dots +{a}_{n}=m\}\\ & =& [{x}^{m}]{(x+{x}^{2}+{x}^{3}+\dots )}^{n}\\ \text{(3)}& & =& [{x}^{m}]\frac{{x}^{n}}{(1-x{)}^{n}}={\textstyle (}\genfrac{}{}{0ex}{}{m-1}{n-1}{\textstyle )}\end{array}$

Step 2

Hence: $\begin{array}{rcl}\mathbb{P}[Y=m]& =& rp(1-p{)}^{m-1}\sum _{n\ge 1}{\textstyle (}\genfrac{}{}{0ex}{}{m-1}{n-1}{\textstyle )}{\left(\frac{p-pr}{1-p}\right)}^{n-1}\\ & =& rp(1-p{)}^{m-1}{(1+\frac{p-pr}{1-p})}^{m-1}\\ \text{(4)}& & =& pr(1-pr{)}^{m-1}\end{array}$

proving your claim for geometric distributions supported on 1,2,…. There is little to change in order to deal with geometric distributions supported on 0,1,…, too.

Assume that $\begin{array}{}\text{(1)}& \mathbb{P}[{X}_{i}=k]=p(1-p{)}^{k-1},\phantom{\rule{2em}{0ex}}k=1,2,\dots \end{array}$

and $\mathbb{P}[N=n]=r(1-r{)}^{n-1}$ for $r=1,2,\dots $. Then: $\begin{array}{rcl}\mathbb{P}[Y=m]& =& \sum _{n\ge 1}\mathbb{P}[N=n]\cdot \mathbb{P}[{X}_{1}+\dots +{X}_{n}=m]\\ \text{(2)}& & =& \frac{r}{1-r}\sum _{n\ge 1}(1-r{)}^{n}{\left(\frac{p}{1-p}\right)}^{n}(1-p{)}^{m}\cdot r(m,n)\end{array}$

Where: $\begin{array}{rcl}r(m,n)& =& \mathrm{\#}\{({a}_{1},\dots ,{a}_{n}):{a}_{i}\in {\mathbb{N}}_{>0},{a}_{1}+\dots +{a}_{n}=m\}\\ & =& [{x}^{m}]{(x+{x}^{2}+{x}^{3}+\dots )}^{n}\\ \text{(3)}& & =& [{x}^{m}]\frac{{x}^{n}}{(1-x{)}^{n}}={\textstyle (}\genfrac{}{}{0ex}{}{m-1}{n-1}{\textstyle )}\end{array}$

Step 2

Hence: $\begin{array}{rcl}\mathbb{P}[Y=m]& =& rp(1-p{)}^{m-1}\sum _{n\ge 1}{\textstyle (}\genfrac{}{}{0ex}{}{m-1}{n-1}{\textstyle )}{\left(\frac{p-pr}{1-p}\right)}^{n-1}\\ & =& rp(1-p{)}^{m-1}{(1+\frac{p-pr}{1-p})}^{m-1}\\ \text{(4)}& & =& pr(1-pr{)}^{m-1}\end{array}$

proving your claim for geometric distributions supported on 1,2,…. There is little to change in order to deal with geometric distributions supported on 0,1,…, too.

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