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

Question

$X = \{X(t): t \geq 0\}$ 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$[X(T)]$.

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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.

Answer 1

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 $$ \mathbb{E}[X(T)]=\int_0^{\infty}\mathbb{E}[X(T)|T=t]f_T(t)\;dt=\int_0^{\infty}\mathbb{E}[X(t)]f_T(t)\;dt=\int_0^{\infty}\lambda t\cdot\theta e^{-\theta t}\;dt=\frac{\lambda}{\theta} $$