Continuity of poisson process Is a Poisson process continuous in probability? Or more generally, when is a counting process continuous in probability?
 A: More generally, if $N=\{N_t,t\ge 0\}$ is a counting process with $N_0=0$, $T_n=\inf\{t>0: N_t\ge n\}$ is the $n$-th time of jumps of $N$, then $N$ is continuous in probability if and only if the distribution functions, $\{F_{T_n}(x)=\mathsf{P}(T_n\le x),n\ge 1\}$, all are continuous functions.
A: Yes, the Poisson process is continuous in probability. A simple way to check this is to visualize the distribution.
A: Yes it is. In fact, any Levy process is continuous in probability. (Note that this is very different from being almost surely continuous, which is kind of counter intuitive!)
Here is an intuition for the Poisson process: Suppose that $s \to t$, from below. The probability that there is an arrival between $s$ and $t$ is going to zero as $s$ get's closer to $t$. So the probability that $X_s$ is not $X_t$ is also going to zero.
You can compute this explicitly for the Poisson process.
For a general Levy process $X_t$, the stationary increments property means that you only have to argue that as $\delta \to 0$, $X_{\delta} \to 0$ in distribution. Note that convergence in distribution would enough to show that $P ( |X_{\delta}| > \epsilon) \to 0$ as $\delta \to 0$, so we will show that in the next paragraphs. (To show it is enough: use a continuous, non-negative test function which is $0$ around $0$, and $1$ on $\{ x \in \mathbb{R} : \epsilon > 0\}$ to conclude that if $Z_n \to c$ in distribution, where $c$ is a constant, then $Z_n \to c$ in probability.)
There is hopefully an easier way to show this, but convergence in distribution follows from the Lévy–Khintchine theorem. 
Lévy–Khintchine says that some distribution $\mu$ is an infinitely divisible distribution iff it has characteristic function of the form $\hat{\mu}( \xi ) = exp( - \Psi(\xi))$, where $\Psi(\xi)$ is a function you can see here: https://en.wikipedia.org/wiki/L%C3%A9vy_process#L%C3%A9vy%E2%80%93Khintchine_representation
This implies (using stationary independent increments, and right continuity of Levy process paths) that if $X_t$ is a Levy process, then for all $t$, $\hat{X}_t = \exp ( - t \Psi(\xi))$.
(I was a little confused about branches and $n$th roots and stuff -- I think the point is that there are canonical $n$th roots of $\hat{X}_1$, given by $\hat{X}_{1/n}$. Writing $(\hat{X}_1)^{1/n}$ is a little sloppy, since this doesn't uniquely define a function. The point is that $\hat{\mu}_{1/n}$ satisfies the correct equation making it an $n$th root of $\hat{\mu}$. Maybe I'm still missing something... will ask and update.)
This implies that $\hat{ X}_{\delta}  = exp (- \delta \Psi(\xi)) \to 1$ pointwise, which by Levy's continuity theorem implies that $X_{\delta}$ converges to $\delta_0$ in distribution as $\delta \to 0$. As discussed above, this implies that $X_{\delta} \to \delta_0$ in probability.  
Overall this shows that $X_{\delta} \to 0$ in probability as $\delta \to 0$. Now, $P ( | X_t - X_s | > \epsilon) = P ( |X_{t - s} | > \epsilon)$ by stationary increments, so the continuity of the Levy process in probability follows.
