Consider the following distributions:
$Y \sim \text{Exp}(-\lambda_{0})$
$Z \sim \text{Binom}(N, (1 - e^{-\lambda_{1}y}))$
$a = (1 - e^{-\lambda_{2}y})$
I would like to calculate the probability distribution of $a$, given $Z>0$.
Attempt at a solution
We may calculate the distribution of $Y$ given $Z>0$ by considering Bayes Theorem:
\begin{equation} P_{Z|Y}(z>0|y)f_{Y}(y) = f_{Y|Z}(y|z>0)P(z>0) \end{equation}
So my tactic is to substitute $y$ for $a$, by rearranging:
\begin{equation} y = \frac{-\log(1 - a)}{\lambda_{2}} \end{equation}
Then, we solve for $P_{Z|A}(z>0|a)$:
\begin{equation} P_{Z|A}(z>0|a) = 1 - P_{Z|A}(z=0|a) \end{equation}
Using the fact that: $\text{Binom}(N, 0, p) = {{N}\choose{0}}p^{0}(1-p)^{N}$
\begin{equation} P_{Z|Y}(z>0|y) = 1 - (1 - e^{-\lambda_{1}y})^{N} \end{equation}
So, substituting $y$ for $a$: \begin{equation} P_{Z|A}(z>0|a) = 1 - (1 - e^{\frac{\lambda_{1}\log(1 - a)}{\lambda_{2}} })^{N} \rightarrow 1 - (1 - (1-a)^{\frac{\lambda_{1}}{\lambda_{2}}})^{N} \end{equation}
I also know that $P_{Z}(z>0) = N\frac{\lambda_{1}}{\lambda_{0} + \lambda_{1}}$, which is proved in this question I previously asked here.
The problem I have now is the $f_{A}(a)$ term. I do not know what this should be.
I would assume that one can use Bayes Theorem again. I can obtain at least one of the terms, but I am a bit perplexed as to what the others are.
\begin{equation} f_{Y|A}(y|a) = \lambda_{0} e^{-\lambda_{0} \frac{-\log(1 - a)}{\lambda_{2}}} = \lambda_{0}(1 - a)^{\frac{\lambda_{0}}{\lambda_{2}}} \end{equation}
Am I on the right track here? Is there something I am missing?
Any tips on how to proceed would be appreciated.