Finding probability involving Poisson process. Let $N(t)$ for $t\geq{0}$ be a Poisson process with intensity $\lambda > 0$. Now, let $X(t)$ be a process defined such that the arrival process of $X$ is every even numbered arrival of $N$. Then for $t>0$:


*

*What is $P(X(t)=k)$ for $k=0,1,2,3,...$?

*Is $X(t)$ a Poisson process? If yes what is the intensity parameter?


I am thinking this could be the thinning of the Poisson process with probability $0.5$ of choosing the even arrival time of the process $N$ but then I am not sure how to proceed then!
Is the following correct?
$P(X(t)=k)=P(N(t)=2k)$
 A: $X(t)$ is not a Poisson process. Note that the inter-arrival times of a Poisson process are i.i.d. and distributed with $\exp(\lambda)$. It is easy to see that the inter-arrival times of $X(t)$ has the distribution of the sum of 2 exponential random variables. In general, the sum of $n$ i.i.d. exponential random variables is called an Erlang random variable, with the probability density
$$
f_X(x) = \frac{\lambda^n x^{n-1} e^{-\lambda x}}{(n-1)!}
$$
where $\lambda$ is the parameter of the exponential distribution, and $n$ is the number of exponential random variables used to generate the Erlang variable. For the special case of $n=2$, you have
$$
f_X(x) = \lambda^2 x e^{-\lambda x}
$$
For a fixed time interval $[0,T]$, you can compute the probability that $X(t)$ has $k$ arrivals, $P(X(T)-X(0)=k)$, as follows:
$$
\begin{align*}
P(X(T)-X(0)=k) &= P(N(T)-N(0)=2k) + P(N(T)-N(0)=2k+1)\\
& = e^{-\lambda T} \left( \frac{(\lambda T)^{2k}}{(2k)!}+\frac{(\lambda T)^{2k+1}}{(2k+1)!}\right) \\
&= e^{-\lambda T}\frac{(\lambda T)^{2k}}{(2k)!} \left( 1 + \frac{\lambda T}{2k+1}\right)
\end{align*}
$$
The reason for adding up the probabilities of $N(t)=2k$ and $N(t)=2k+1$ is that even if $N(t)$ has $2k+1$ arrivals, $X(t)$ will not have received its $(k+1)$th arrival until $N(t)$ has its $(2k+2)$th arrival.
As a final note, this is not a thinning of the Poisson process, since it is a deterministic mapping. Thinning is done by erasing each arrival of the process with probability $p$, which is a probabilistic way of generating $X(t)$. Note that in that case, the inter-arrival distribution would be a mixture of an infinite sequence of Erlang-K distributions, where $K$ is distributed geometrically with parameter $p$. It can be shown that this mixture distribution is in fact exponential distribution with parameter $\lambda p$.
