I have a probability density function as follows:
\begin{equation} P_{X|T}(t, a, b) = \begin{cases} \text{Bernoulli}(p = a(1 - e^{-\frac{-t}{b}})) \text{ where } t\leq T \\ 0 \text{ otherwise} \end{cases} \end{equation}
\begin{equation} f_{T}(c) = \text{Exp}(c) = \frac{e^{-\frac{t}{c}}}{c} \end{equation}
It is trivial to show from the tower property ($E(X) = E(E(X|t))$) that:
\begin{equation} E(X; a, b, c) = \int^{T}_{0} P_{X|T}(t,a,b) \cdot f_{T}(c) = \int^{T}_{0} a(1 - e^{-\frac{-t}{b}}) \frac{e^{-\frac{t}{c}}}{c} \end{equation}
\begin{equation} E(X; a,b,c) = \frac{a}{b + c} \left(c - e^{-\frac{T}{c}}\left( c + b\cdot(1 - e^{-\frac{T}{b}}) \right) \right) \end{equation}
I have noticed that then, this is equivalent to saying:
\begin{equation} P_{X}(t, a, b, c) = \text{Bernoulli}(E(X; a,b,c)) \end{equation}
Is there a trivial way that I can prove that this statement, or is it somehow implicit?
I mistakenly forgot an important feature that trivialises this problem:
\begin{equation} P_{X|t}(t, a, b) = \begin{cases} \text{Bernoulli}(p = a(1 - e^{-\frac{-t}{b}})) \text{ where } t\leq T \\ 0 \text{ otherwise} \end{cases} \end{equation}
The probability of a success is $E_{X}(t; a, b, c)$, as we have calculated. This follows from the Law of Total Probability:
\begin{equation} P_{X}(a, b, c) = \int^{T}_{0} P_{X|T}(t, a, b) \cdot f_{t}(c).dt \end{equation}
Equivalently, the probability of a failure must be:
\begin{equation} P_{\overline{X}}(a, b, c) = \int^{T}_{0} (1 - P_{X|T}(t, a, b))\cdot f_{t}(c).dt + \int^{\infty}_{T} f_{t}(c).dt \end{equation}
The second part of the expression arises from the fact that there is certain failure if $t>T$.
We integrate and obtain: \begin{equation} P_{\overline{X}}(a, b, c) = (1 - e^{-\frac{T}{c}}) - P_{X}(a, b, c) + e^{-\frac{T}{c}} \end{equation}
Which cancels to: \begin{equation} P_{\overline{X}}(a, b, c) = 1 - P_{X}(a, b, c) \end{equation}
Since $X$ cannot take a value other than a success or a failure, we must conclude that:
\begin{equation} X \sim P_{X}(a, b, c)^{s} + (1-P_{X}(a, b, c))^{1-s} \text{ where } s \in (0, 1) \end{equation}
and that therefore:
\begin{equation} X \sim \text{Bernoulli}(P_{X}(a, b, c)) \end{equation}