Marginal PMF values of a function

[Longines]

Hello once again!

I've been doing this question and I was wondering if my workings are correct, if they are not correct, can you please correct them?

The question is as follows:
View attachment 3243

My workings are:

$\binom{y}{x} \frac{e^{-1}}{2^y y!}$We can rewrite this:

$\frac{y!}{x!(y-x)!} \times \frac{e^{-1}}{2^y y!}$$\frac{1}{x!(y-x)!} \times \frac{e^{-1}}{2^y}$

So we have:
$f_{x, y}(x,y) = \frac{1}{x!(y-x)!} \times \frac{e^{-1}}{2^y}$

Now using the formula:

$f_{x} (x,y) = \sum_{y}^{\infty} P_{x,y}(x,y)$

and subbing in $P_{x,y}(x,y)$:$\frac{e^{-1}}{x!} \sum_{y}^{\infty} \frac{1}{(y-x)!}$

= $\frac{e^{-1}}{x!}$

And in a same fashion, for y we get:

$f_{y}(x,y) = \frac{e^{-1}}{y!}$

 

[mmwave]

Hello! Your workings look mostly correct, but there are a few minor errors that I wanted to point out.

Firstly, in the first line where you rewrote the expression, the exponent for 2 should be y-x, not just y. This is because the denominator of the binomial coefficient is (y-x)!, not y!.

Secondly, in the line where you use the formula for f_x(x,y), the summation should start at x, not y. This is because for each value of x, the summation should go from x to infinity, not y to infinity.

Finally, in the last line where you calculate f_y(x,y), the exponent for 2 should be y, not y!. This is because the denominator of the binomial coefficient is y!, not (y-1)!.

Other than these minor errors, your workings look good and you have correctly calculated the probability distributions for x and y. Keep up the good work!