Consider $Y$ with a geometric distribution with a parameter $\theta$ such that $0 < \theta < 1$ and let also $P(X=x|Y=y)$ has a binomial distribution with parameters "y" and "1/2". So:
$$ P(X=x|Y=y) = {y \choose x} \left(\frac{1}{2}\right)^x \left(\frac{1}{2}\right)^{y-x} = {y \choose x} \left(\frac{1}{2}\right)^y $$
I am trying to compute the probability of the event $P(X=1)$, but I cannot see a closed form for the solution.
I know $P(X=1) = \sum_{y=1}^{\infty} P(X=1|Y=y) P(Y=y)$. So I evaluated:
$$ P(X=1) =\sum_{i=1}^{\infty} {y \choose 1} \frac{1}{2^y} \theta(1-\theta)^{y-1} = \sum_{i=1}^{\infty} \frac{y}{2^y} \frac{\theta}{1-\theta} (1-\theta)^y$$
We have then the probability as an infinite sum:
$$ P(X=1) = \left(\frac{\theta}{1-\theta}\right) \sum_{i=1}^{\infty} y \left(\frac{1-\theta}{2}\right)^y $$
And I can't find an explicit solution for the sum which has a closed form and not as a series. I applied the ratio test and it will converge, for $ 0 < \theta < 1$. Do you people think there's an explicit solution? I'd appreciate any help and tips!
Note that we have
\begin{align} \sum_{y=1}^\infty y q^y &= \frac{q}{1-q} \sum_{y=1}^\infty yq^{y-1}(1-q)\\ &= \frac{q}{1-q}\cdot \frac1{1-q}\\ &= \frac{q}{(1-q)^2} \end{align}
Here $q=\frac{1-\theta}{2}$. In my working, I have used the property of the mean of the geometric distribution with success probability $1-q$.