i have a stopping time $T$ of an Poisson Process $N$ with rate $\lambda$. Somehow this stopping time is exponential distributed. So we have $ T \sim exp(\lambda)$.
I want to know the distribution of the following term: \begin{align} \exp(-\kappa \cdot T) \mathbb{1}_{\{ T < \tau \} } \end{align}
where $\tau$ is fixed and $\kappa$ is actually a intensity rate of another Poisson Process. But unter the circumstances we can hold $\kappa$ and find the fitting distrbution. Thanks for your help.
Best Regards
frakChris
Assuming you mean that $\kappa$ and $\tau$ are positive constants. Let $X = \exp(-\kappa T)1_{\{T < \tau\}}$ be the random variable of interest. Since there is an indicator, we can first easily see that it is a mixture of continuous and discrete random variable, with a point mass at $0$:
$$ \Pr\{X = 0\} = \Pr\{T \geq \tau\} = e^{-\lambda\tau}$$
and it has a continuous support on the interval $[e^{-\kappa\tau},1]$. For $x \in [e^{-\kappa\tau},1],$ we can calculate the CDF as
$$ \begin{align} \Pr\{X \leq x\} & = \Pr\{X = 0\} + \Pr\{e^{-\kappa\tau} < X \leq x\} \\ & = e^{-\lambda\tau} + \Pr\left\{-\frac {\ln x} {\kappa} \leq T < \tau \right\}\\ & = \exp\left\{\frac {\lambda \ln x} {\kappa}\right\} \\ & = x^{\lambda/\kappa} \end{align}$$
As for the expectation, we can directly use the CDF obtained here, as $X$ is a non-negative random variable:
$$ \begin{align} E[X] &= \int_0^{+\infty} 1 - F_X(x)dx \\ &= (1-e^{-\lambda\tau})e^{-\kappa\tau} + \int_{e^{-\kappa\tau}}^1 1 - x^{\frac {\lambda} {\kappa}}dx \\ & = (1-e^{-\lambda\tau})e^{-\kappa\tau} + \left.x - \frac {\kappa} {\lambda + \kappa} x^{\frac {\lambda + \kappa} {\kappa}} \right|_{x=e^{-\kappa\tau}}^{x=1} \\ &= -e^{-(\lambda+\kappa)\tau} + 1 - \frac {\kappa} {\lambda + \kappa} + \frac {\kappa} {\lambda + \kappa}e^{-(\lambda+\kappa)\tau} \\ &= \frac {\lambda} {\lambda + \kappa}(1 - e^{-(\lambda+\kappa)\tau}) \end{align}$$