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The joint probability of the random variables $X$ and $Y$ is given by $$P(X=n, Y=m) =\frac{\alpha ^{m}\beta ^{n-m}}{m!(n-m)!}e^{-(\alpha +\beta )} $$ $$m=0,1,2...$$ $$n=m, m+1, m+2, ...$$

a) Evaluate $P(X=n)$ and $P(Y=m)$.
b) Are $X$ and $Y$ independent?

My solution so far:

a) $$f_{XY}(n,m) = P(X=n, Y=m)$$ $$P(X=n) = f_{X}(n) = \sum_{m}f_{XY}(n,m) = \sum_{m=0}^{n}\frac{\alpha ^{m}\beta ^{n-m}}{m!(n-m)!}e^{-(\alpha +\beta )}$$ $$=e^{-(\alpha +\beta )}\frac{1}{n!} \sum_{m=0}^{n}\frac{n!}{m!(n-m)!}\beta^{n-m}\alpha^{m}$$ $$=\frac{(\beta+\alpha)^{n}}{e^{(\alpha+\beta)}n!}.$$

So far, am I doing this correctly?

For $P(Y=m)$:

$$P(Y=m) = f_{Y}(n) = \sum_{n}f_{XY}(n,m) = \sum_{n=m}^{\infty}\frac{\alpha ^{m}\beta ^{n-m}}{m!(n-m)!}e^{-(\alpha +\beta )}$$ $$=e^{-(\alpha +\beta )}\frac{1}{n!} \sum_{n=m}^{\infty}\frac{n!}{m!(n-m)!}\beta^{n-m}\alpha^{m}$$ $$\mbox{ from here-> }=\frac{e^{\beta }\alpha ^{m}}{m!}e^{-(\alpha +\beta )}=\frac{e^{-\alpha }\alpha ^{m}}{m!}=\frac{\alpha ^{m}}{e^{\alpha }m!}.$$ Here, I have no idea how to the get the summation "from here". I used an online summation calculator because I don't know how to get there. Any help as to how to get there would be appreciated!!

b)

$X$ and $Y$ are independent if $f_{X,Y} = f_{X}(x)f_{Y}(y)$.
From a, $$f_{X}(x)f_{Y}(y) = \frac{(\beta+\alpha)^{n}}{e^{(\alpha+\beta)}n!}\frac{\alpha ^{m}}{e^{\alpha }m!}. $$
From given, $$f_{X,Y}= \frac{\alpha ^{m}\beta ^{n-m}}{m!(n-m)!}e^{-(\alpha +\beta )}.$$
So, since $f_{X,Y} \neq f_{X}(x)f_{Y}(y)$, $X$ and $Y$ are NOT independent.

Here, I'm not sure if $f_{X}(x)f_{Y}(y)$ actually reduces to $f_{X,Y}$ somehow.

Thanks for answering!

beepboopbeepboop
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1 Answers1

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Your solution to (a) appears correct.

Regarding (b), we simplify and obtain the series expansion for $e^\beta$ as highlighted in blue below:- $$\begin{align}P(Y=m)&=e^{-(\alpha +\beta )}\frac{1}{n!} \sum_{n=m}^{\infty}\frac{n!}{m!(n-m)!}\beta^{n-m}\alpha^{m}\\&=\frac{e^{-(\alpha+\beta)}\alpha^m}{m!}\sum_{n=m}^{\infty}\frac{1}{(n-m)!}\beta^{n-m}&\\&=\frac{e^{-(\alpha+\beta)}\alpha^m}{m!}\color{blue}{\sum_{k=0}^{\infty}\frac{1}{k!}\beta^k}\\&=\frac{e^{-(\alpha+\beta)}\alpha^m}{m!}\color{blue}{e^\beta}\\&=\frac{e^{-\alpha}\alpha^m}{m!}\end{align}$$

Finally, from the expressions obtained from (a) and (b), $X$ and $Y$ are indeed NOT independent.

Alijah Ahmed
  • 11,609
  • Beautiful, thank you very much!! – beepboopbeepboop May 27 '18 at 19:51
  • Quick question - does it not matter that k starts at 0 whereas n starts at m? – beepboopbeepboop May 27 '18 at 19:53
  • You are welcome. The condition of k starting at 0 is the same as n starting at m - the index k results from taking a fixed offset of m from index n, i.e. k=n-m – Alijah Ahmed May 27 '18 at 20:05
  • Understood, thanks again! – beepboopbeepboop May 27 '18 at 20:19
  • I also wanted to clarify something else: Should there actually be limits for the answers as such? $P(X=n)= \begin{cases} \frac{(\beta+\alpha)^{n}}{e^{(\alpha+\beta)}n!} & \text{, } n=m, m+1, m+2, ... \ 0 & \text{, otherwise} \end{cases}$. $P(Y=m)= \begin{cases} \frac{\alpha ^{m}}{e^{\alpha }m!} & \text{, } m\geq0 \ 0 & \text{, otherwise} \end{cases}$. – beepboopbeepboop May 28 '18 at 00:14
  • From the wording of the question, I think they are only looking for the expression for $P(X=n)$ and $P(Y=m)$, but there's no harm in including the limits. For $P(Y=m)$ the limit is clear ($m\geq 0$), but looking at $P(X=n)$ in isolation, we do not have a context for $m$, so the limits would need to be $n\geq 0$ as well. – Alijah Ahmed May 28 '18 at 11:23