Getting E[Y|X] and E[Y] of joint PMF ${x \choose y}\frac{4}{9}^y \frac{5}{9}^{x-y}$

Question

Let X be number of rolls till a 6. Next we choose with replacement X balls from an urn with 5 red and 4 green, Y is defined as the number of green balls sampled.

conditional PMF = ${x \choose y}\frac{4}{9}^y \frac{5}{9}^{x-y}$

Get E[Y|X] and E[Y]

For E[Y|X] I got $\int_0^{\infty}yf_{Y|X}(y,x)dy=\sum_0^{\infty}y {x \choose y}\frac{4}{9}^y \frac{5}{9}^{x-y}$, which can't be integrated

Then for E[Y]
$E[Y]=\int_0^{\infty}E[Y|X]*P_{Y|X}(y,x)dx= [\sum_0^{\infty}y {x \choose y}\frac{4}{9}^y \frac{5}{9}^{x-y}]*[\int_0^{\infty}{x \choose y}\frac{4}{9}^y \frac{5}{9}^{x-y}]$

Is there a way to integrate these?

Answer 1

First notice that, $$Y|X \sim Binom(x,\frac{4}{9})$$

Now for any $x$ the value of $E[Y|X]$ should be easy.

Second part you only need to use law of total expectation.

$$E[Y] = \sum_x E[Y|X=x]\cdot P(X=x)$$

I leave the rest for you to work on