Suppose $X$ and $Y$ are discrete random variable with joint p.f. Find the marginal p.f. of $X$ and $Y$

Question

The joint probability function of x and y is:

$$f(x,y)=\frac{e^{-2}}{x!(y-x)!}$$ where $$x = 0, 1, ..., y; y= 0, 1, ...$$

My solution to this problem:

The marginal p.f. of x is:

$$f_x(x)=\sum_{y=0}^\infty \frac{e^{-2}}{x!(y-x)!}$$ Then $$f_x(x)=\frac{e^{-2}}{x!}\sum_{y=0}^\infty\frac{1}{(y-x)!}$$

and this is where I got stuck.

In the solution, it writes $$f_x(x)=\sum_{\color{red}{y=x}}^\infty \frac{e^{-2}}{x!(y-x)!}$$

$$=\frac{e^{-2}}{x!}\sum_{t=0}^\infty\frac{1}{t!}$$ $$=\frac{e^{-2}}{x!}e$$ $$=\frac{e^{-1}}{x!}$$

I understand that $y=x$ as the initial condition since when $y=0$, $x=0$. However, when $y$ starts to increase, the series will not always conform the series of Euler's number $e$. For example when $y=2$, $x$ can also be 2 then the third entry in the series would be $\frac{1}{(2-2)!}$ which is 1 again like the first two entries in the series.

How to interpret this solution?

Answer 1

The joint probability only applies when $y \ge x$ (e.g., $(y-x)!$, apart from using the Gamma function, doesn't make sense when $y \lt x \implies y - x \lt 0$). This is why the summation starts at $y = x$ as it's given in the solution. Note the function $f_x(x)$ depends only on $x$, with $y$ being a summation variable. Thus, as $y$ increases on the RHS, the value of $x$ is not changing. So your statement of

For example when $y=2$, $x$ can also be 2 then the third entry in the series would be $\frac{1}{(2-2)!}$ which is 1 again like the first two entries in the series.

is not correct. If $x = 2$, then the first entry in the series would be for $y = 2$ since $y$ starts at $x = 2$, i.e., it's not the third entry.

Also, note that in the summation expression of

$$\sum_{y=x}^\infty \frac{1}{(y-x)!} \tag{1}\label{eq1A}$$

you have for some value of $x$, starting $y$ at $x$ and incrementing it by $1$ each time in the summation up to infinity. If you let $t = y - x$, then when $y = x$, you have $t = 0$, when $y = x + 1$, you have $t = 1$, etc. Thus, you get

$$\sum_{y=x}^\infty \frac{1}{(y-x)!} = \sum_{t=0}^\infty \frac{1}{t!} \tag{2}\label{eq2A}$$

Since $\frac{e^{-2}}{x!}$ doesn't involve $y$ at all, it's a constant factor in all of the terms so you can move it outside of the summation. You thus get

$$\begin{equation}\begin{aligned} \sum_{y=x}^\infty \frac{e^{-2}}{x!(y-x)!} & = \frac{e^{-2}}{x!}\sum_{y=x}^\infty \frac{1}{(y-x)!} \\ & = \frac{e^{-2}}{x!}\sum_{t=0}^\infty \frac{1}{t!} \end{aligned}\end{equation}\tag{1}\label{eq3A}$$

Answer 2

You've written $f_x(x)$ where you should have $f_X(x).$ For example, if $x=4$ then $f_X(x) = f_X(4) = \Pr(X=4).$ The number $4$ replaces $x$ but it doesn't replace $X.$ Without this distinction one cannot even understand something like $\Pr(X\le x).$

You wrote $x=0,1,\ldots,y,$ so the range of values of $x$ depends on $y.$ You always have $x\le y.$ Thus, for example, when $x=4$ then $y \ge 4,$ i.e. $y\in\{4,5,6,\ldots\}.$

Then you have \begin{align} f_X(4) = {} & \sum_{y\,=\,4}^\infty \frac{e^{-2}}{x!(y-x)!} \\[8pt] = {} & \frac {e^{-2}} {x!} \sum_{y\,=\,4}^\infty \frac 1 {(y-4)!} \\ & \text{since as $y$ goes from 4 to $\infty,$} \\ & \text{$e^{-2}/x!$ does not change,} \\[10pt] = {} & \frac{e^{-2}}{x!} \sum_{w\,=\,0}^\infty \frac 1 {w!} \\ & \text{where $w=y-4.$} \\ & \text{When $y=4,$ then $w=0.$} \\[10pt] = {} & \frac{e^{-2}}{x!} \cdot e = \frac{e^{-1}}{x!}. \end{align}