MHB Problem of the Week #124 - August 11th, 2014

[Chris L T521]

Here's this week's problem!

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Problem: Let $X$ be a Poisson random variable with parameter $\lambda$, and let $Y$ be a Geometric random variable with parameter $p$ which is independent of $X$. In simplest terms of $\lambda$ and $p$, what is the value of $\Bbb{P}(Y>X)$?

Recall: The Poisson pmf is given by $f(x) = \dfrac{e^{-\lambda}\lambda^x}{x!}$ (with support $x\geq0$) and the Geometric pmf is given by $f(x) = p(1-p)^{x-1}$ (with support $x\geq 1$).

-----Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[Chris L T521]

This week's problem was correctly answered by laura123. You can find her solution below.

[sp]$X$ is a Poisson random variable with parameter $\lambda$ and $Y$ is a Geometric random variable with parameter $p$ then $\mathbb{P}(X=n)=\dfrac{e^{-\lambda}\lambda^n}{n!}$ and $\mathbb{P}(Y=m)=pq^{m-1}$ where $q=1-p$, with $n\geq0$ and $m\geq 1$.
$\mathbb{P}(Y>X)=1-\mathbb{P}(X\geq Y)$. Let us find the value of $\mathbb{P}(X\geq Y)$:

$\begin{aligned}\displaystyle\mathbb{P}(X\geq Y) &= \sum_{n=1}^{+\infty}\mathbb{P}[X=n\wedge(Y=1\vee...\vee Y=n)]\\ &=\sum_{n=1}^{+\infty}\left[\dfrac{e^{-\lambda}\lambda^n}{n!}\cdot \sum_{m=1}^{n}pq^{m-1}\right]\\ & =\sum_{n=1}^{+\infty}\left[\dfrac{e^{-\lambda}\lambda^n}{n!}\cdot p \sum_{m=1}^{n}q^{m-1}\right]\\ & =\sum_{n=1}^{+\infty}\left[\dfrac{e^{-\lambda}\lambda^n}{n!}\cdot p \dfrac{1-q^n}{1-q}\right]\\ & =\sum_{n=1}^{+\infty}\left[\dfrac{e^{-\lambda}\lambda^n}{n!}\cdot p \dfrac{1-q^n}{1-1+p}\right]\\ & =\sum_{n=1}^{+\infty}\left[\dfrac{e^{-\lambda}\lambda^n}{n!} \cdot(1-q^n)\right] \\ & =
e^{-\lambda}\sum_{n=1}^{+\infty}\left[\dfrac{\lambda^n}{n!}\cdot(1-q^n)\right] \\ & =
e^{-\lambda}\sum_{n=1}^{+\infty}\left[\dfrac{\lambda^n}{n!}-\dfrac{\lambda^n}{n!}q^n\right] \\ & =
e^{-\lambda}\sum_{n=1}^{+\infty}\left[\dfrac{\lambda^n}{n!}-\dfrac{(\lambda q)^n}{n!}\right] \\ & =
e^{-\lambda}\left[\sum_{n=1}^{+\infty}\dfrac{\lambda^n}{n!}-\left(\sum_{n=1}^{+\infty}\dfrac{(\lambda q)^n}{n!}\right)\right]
\end{aligned}$

because $\displaystyle e^x-1=\sum_{n=1}^{+\infty}\dfrac{x^n}{n!}$ then $\displaystyle\sum_{n=1}^{+\infty}\dfrac{\lambda^n}{n!}=e^{\lambda}-1$ and $\displaystyle\sum_{n=1}^{+\infty}\dfrac{(\lambda q)^n}{n!}=e^{\lambda q}-1$.

Therefore
$\displaystyle e^{-\lambda}\left[\sum_{n=1}^{+\infty}\dfrac{\lambda^n}{n!}-\left(\sum_{n=1}^{+\infty}\dfrac{(\lambda q)^n}{n!}\right)\right]=
e^{-\lambda}[e^{\lambda}-1-e^{\lambda q}+1]=
e^{-\lambda}[e^{\lambda}-e^{\lambda q}]=
e^{-\lambda}[e^{\lambda}-e^{\lambda -\lambda p}]=
1-e^{-\lambda p}$.

Then $\mathbb{P}(Y>X)=1-\mathbb{P}(X\geq Y)=1-(1-e^{-\lambda p})=e^{-\lambda p}$[/sp]