# Bayes Theorem with Conditional Variables

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:

$$$$P_{Z|Y}(z>0|y)f_{Y}(y) = f_{Y|Z}(y|z>0)P(z>0)$$$$

So my tactic is to substitute $$y$$ for $$a$$, by rearranging:

$$$$y = \frac{-\log(1 - a)}{\lambda_{2}}$$$$

Then, we solve for $$P_{Z|A}(z>0|a)$$:

$$$$P_{Z|A}(z>0|a) = 1 - P_{Z|A}(z=0|a)$$$$

Using the fact that: $$\text{Binom}(N, 0, p) = {{N}\choose{0}}p^{0}(1-p)^{N}$$

$$$$P_{Z|Y}(z>0|y) = 1 - (1 - e^{-\lambda_{1}y})^{N}$$$$

So, substituting $$y$$ for $$a$$: $$$$P_{Z|A}(z>0|a) = 1 - (1 - e^{\frac{\lambda_{1}\log(1 - a)}{\lambda_{2}} })^{N} \rightarrow 1 - (1 - (1-a)^{\frac{\lambda_{1}}{\lambda_{2}}})^{N}$$$$

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.

$$$$f_{Y|A}(y|a) = \lambda_{0} e^{-\lambda_{0} \frac{-\log(1 - a)}{\lambda_{2}}} = \lambda_{0}(1 - a)^{\frac{\lambda_{0}}{\lambda_{2}}}$$$$

Am I on the right track here? Is there something I am missing?

Any tips on how to proceed would be appreciated.

• $A$ is exactly determined by $Y$, so $f_{A\mid Y}$ is infinite Oct 26, 2021 at 0:51

$$f_A(a)=f_Y({-\tfrac 1{\lambda_2}\ln(1-a)})\cdot\left\lvert\dfrac{\mathrm d (-\tfrac 1{\lambda_2}\ln(1-a))}{\mathrm d a\hspace{14ex}}\right\rvert\\=\dfrac{f_Y({-\tfrac 1{\lambda_2}\ln(1-a)})}{\lvert1-a\rvert}$$