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.

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

1 Answer 1


Just use the Jacobian transformation:

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

  • $\begingroup$ Thank you very much! This seems to do the trick. $\endgroup$
    – Jack Rolph
    Oct 26, 2021 at 14:39

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