# $X$ is a Poisson process and $T$ is exponentially distributed?

$$X = \{X(t): t \geq 0\}$$ is a Poisson process (with intensity $$\lambda$$). Independent of $$X$$, $$T$$ is a random variable that is exponentially distributed with intensity $$\theta$$.

I need to find the PMF for $$Y=X(T)$$ and also Var$$[X(T)]$$.

There is an example in my book that talks about finding the PMF recursively. I'll denote $$p_n=P[X(T)]$$. If I let $$p_0=e^{-\lambda}$$, then supposedly $$p_n=\frac{\lambda}{n}p_{n-1}$$ So $$p_1=\frac{\lambda}{n}e^{-\lambda}$$ $$p_2=\frac{\lambda^2}{n^2}e^{-\lambda}$$ Then, this just becomes $$p_n=\frac{\lambda^n}{n^n}e^{-\lambda}$$

Is that right? I don't understand the first recursive part; I just copied that from a similar example from the book. I don't actually know where that comes from.

There's another example that looks exactly the same but has the answer $$\frac{(\lambda t)^n e^{(-\lambda t)}}{n!}$$

Can someone explain how I should be doing this? Would really appreciate it.

Since $X(t)$ is a Poisson process and $T$ is independent of $\{X(t):t\geq 0\}$, it follows that $$\mathbb{P}(X(T)=n|T=t)=\mathbb{P}(X(t)=n)=\frac{(\lambda t)^ne^{-\lambda t}}{n!}$$ for $n=0,1,2,\dots$.
Therefore if $f_T(t)=\theta e^{-\theta t}1_{t>0}$ is the pdf of $T$, then $$\mathbb{P}(X(T)=n)=\int_{\mathbb{R}}\mathbb{P}(X(T)=n|T=t)f_T(t)\;dt=\int_{0}^{\infty}\frac{(\lambda t)^ne^{-\lambda t}}{n!}\theta e^{-\theta t}\;dt$$ $$= \frac{\theta\lambda^n}{n!}\int_0^{\infty}t^ne^{-(\theta+\lambda)t}\;dt=\frac{\theta\lambda^n}{(\lambda+\theta)^{n+1}n!}\int_0^{\infty}u^ne^{-u}\;du=\frac{\theta\lambda^n}{(\lambda+\theta)^{n+1}}$$ since the last integral is $\Gamma(n+1)=n!$
The expected value and variance of $X(T)$ can be computed in a similar way.
In particular, since $\mathbb{E}[X(t)]=\lambda t$, it follows that $$\mathbb{E}[X(T)]=\int_0^{\infty}\mathbb{E}[X(T)|T=t]f_T(t)\;dt=\int_0^{\infty}\mathbb{E}[X(t)]f_T(t)\;dt=\int_0^{\infty}\lambda t\cdot\theta e^{-\theta t}\;dt=\frac{\lambda}{\theta}$$
• Ah I sort of see where this all comes from. Would you mind showing the integral I need to evaluate for $E[X]$? Commented Mar 21, 2016 at 21:39