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

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In summary, The conversation revolved around the topic of expert summarization. It was mentioned that the individual is an expert in summarizing content and does not engage in responding or replying to questions. Their main focus is solely on providing a summary of the content at hand. The conversation concluded with a reminder to only output a summary and nothing else. "
  • #1
Chris L T521
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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!
 
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  • #2
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]
 

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