X,Y are random discrete variables with joint probability function of
$f(x,y) = \dfrac {1}{e^2\cdot y!\cdot (x-y)!} $ with $x \in\{ 0,1,2,…\}$ and $y\in\{0,1,2,…x\}$
Find marginals $f_X(x)$ , $f_Y(y)$.
I tried to compute firstly $f_Y(y)$ but i have a problem with the summation of the series:
$$f_Y(y) = \sum_{x=0}^\infty \dfrac {1}{e^2\cdot y!\cdot (x-y)!} = \frac {1}{e^2\cdot y!}\cdot \sum_{x=0}^\infty \frac {1}{(x-y)!} $$
How can I compute $f_Y(y)$ and $f_X(x)$ ?
Thank you very much!
Best Answer
The marginal $f_X(x)$ is $$f_X(x) =\frac{1}{e^2} \sum\limits_y \frac{1}{y!(x-y)!} = \frac{1}{x!e^2} \sum\limits_y \frac{x!}{y!(x-y)!} = \frac{1}{x!e^2} \sum\limits_y \binom{x}{y}(1)^y(1)^{x- y} = \frac{1}{x!e^2} (1+1)^{x} $$ So $$f_X(x) = \frac{1}{x!e^2} 2^x$$
Similarly $$f_Y(y) =\frac{1}{y!e^2} \sum\limits_{x=0}^{\infty} \frac{1}{(x-y)!}=\frac{1}{y!e^2} \sum\limits_{x=y}^{\infty} \frac{1}{(x-y)!}$$ where the last equation is because $ y \leq x$. Let $k = x - y$, you'd get $$f_Y(y) = \frac{1}{y!e^2}\sum\limits_{k=0}^{\infty} \frac{1}{k!} =\frac{1}{y!e^2}\sum\limits_{k=0}^{\infty} \frac{1^k}{k!} = \frac{1}{y!e^2}e^1 = \frac{1}{ey!}$$ The sum of $\frac{1}{k!}$ becomes $e$ due to Taylor series.