# Finding the parameter of an exponential distribution given certain conditions

Suppose that $T$, $X$ and $C$ are random variables with $X\sim \mathrm{bin}(1,r)$ and $T\mid X=x\sim \exp(\lambda_x)$ for $x=0,1$ and $C\sim\exp(\lambda_C)$ such that $C$ is independent of $T$ and $X$. Let $p,q\in (0,1)$ such that $p+q\in (0,1)$ be given probabilities and define $t=-\log(1-p)$. Assume furthermore that $$P(T\leq t\mid X=x)=p+qx,\quad x=0,1.$$ Then the parameter $\lambda_x$ is given by $$\lambda_x=-\frac{\log(1-p-qx)}{t}$$ and in particular $\lambda_0=1$. In the above setting everything but $\lambda_C$ are known parameters.

My problem is the following: Given a particular value of $P(T\leq C\wedge t)$, how do I find the value of $\lambda_C>0$ that realizes this probability (and for what values of the probability is it possible to find such $\lambda_C$?)

My attempt at this problem, is to note that $$P(T\leq C\wedge t)=(1-r)P(T\leq C\wedge t\mid X=0)+rP(T\leq C\wedge t\mid X=1)$$ and so we only have to find an expression of $P(T\leq C\wedge t\mid X=x)$ for $x=0,1$. This can be done in the following way: \begin{align*} P(T\leq &C\wedge t\mid X=x)=\int_0^\infty\left(1-\exp(-\lambda_x\cdot y\wedge t)\right)\,\lambda_C\,e^{-\lambda_c y}\;\mathrm d y\\ &=1-\int_0^t \lambda_C \exp(-(\lambda_x+\lambda_C)y)\,\mathrm dy-\int_t^\infty\lambda_C\exp(-\lambda_xt)\exp(-\lambda_c y)\,\mathrm dy\\ &=1+\frac{\lambda_C}{\lambda_x+\lambda_C}\exp(-(\lambda_C+\lambda_x)t)-\frac{\lambda_C}{\lambda_x+\lambda_C}-\exp(-(\lambda_C+\lambda_x)t). \end{align*} However, this is where I am stuck. I tried using that $\exp(-\lambda_x t)=(1-p-qx)$ but without success. Does anyone have an idea on how to move on from here? Also I should probably mention that I do not know if the problem is possible to solve analytical.

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The condition that $P(T\leqslant t\mid X=1)=1$ is impossible for an exponential random variable and the formula for $\lambda_1$ diverges. Is this what you mean? – Did Nov 22 '12 at 20:33
That is why I'm assuming $p+q\in (0,1)$ so that $P(T\leq t\mid X=1)=p+q\in (0,1)$ and so that $\lambda_1$ is well-defined. Is my question not clear enough? If not, let me know, then I'll try to rephrase it. – Stefan Hansen Nov 22 '12 at 20:45
Reading it again, the question is quite clear and I misread $p+q\in(0,1)$ for $p+q=1$. Sorry about the noise. – Did Nov 23 '12 at 11:04
That's alright :) – Stefan Hansen Nov 23 '12 at 11:06

As you explain, calling $r_1=r$ and $r_0=1-r$, $$P(T\leqslant\min\{C,t\})=\sum_xr_x\frac{\lambda_x}{\lambda_x+\lambda_C}(1-\mathrm e^{-(\lambda_x+\lambda_C)t})=\sum_xr_x\lambda_x\int_0^t\mathrm e^{-(\lambda_x+\lambda_C)s}\mathrm ds.$$ The RHS is a decreasing function of $\lambda_C$. When $\lambda_C\to\infty$, the RHS goes to zero. When $\lambda_C\to0$, the RHS goes to $$p_0=\sum_xr_x(1-\mathrm e^{-\lambda_xt})\lt1.$$ To sum up, the equation $P(T\leqslant\min\{C,t\})=p$ has indeed a unique solution $\lambda_C$ for every $p$ in $(0,p_0]$ (and no solution if $p\gt p_0$).
This equation can be rewritten as $\mathcal L\varphi(\lambda_C)=p$ where $\mathcal L$ is the Laplace transform and $$\varphi(s)=\sum_xr_x\lambda_x\mathrm e^{-\lambda_xs}\,\mathbb 1_{s\leqslant t},$$ but, at this moment, I fail to see how this remark could lead to an analytical solution, in particular because inversion formulas for Laplace transforms are usually restricted to regular functions while $\varphi$ is discontinuous at $t$.