# Why are processes with stationary independent increments nonstationary?

This answer states that any process with stationary independent increments is nonstationary.

Why? Specifically:

Let $$X(t-s) = N(t) - N(s)$$ have distribution $$F(t-s)$$ for all $$s\leq t$$ [increments are stationary] and $$\{X(t_i-s_i) : i\in{1,\dots,n}\}$$ are independent whenever the $$[s_i,t_i]$$ overlap only at endpoints if at all.

The claim then is that it's impossible that $$N(t_1)=y_1, N(t_2)=y_2,\ldots,N(t_m)=y_m$$ and $$N(t_1+h)=y_1, N(t_2+h)=y_2,\ldots,N(t_m+h)=y_m$$ must have the same distribution. I don't see why that must be the case.

I suspect that this has been answered previously, but I haven't managed to find it. Perhaps there is something obvious that I'm just not seeing.

If $$(N_t)_{t \geq 0}$$ is a Lévy process (i.e. a stochastic process with stationary and independent increments) and $$h \neq 0$$, then the random variables $$N_t$$ and $$N_{t+h}$$ cannot have the same distribution unless $$(N_t)_{t \geq 0}$$ is trivial, i.e. $$N_t=0$$ for all $$t \geq 0$$. This implies, in particular, that the only stationary Lévy process is the trivial process.
To prove this, assume that $$(N_t)_{t \geq 0}$$ is a non-trivial Lévy process and fix $$t,h \geq 0$$ such that $$N_t$$ and $$N_{t+h}$$ have the same distribution. Since $$N_{t+h} = (N_{t+h}-N_t) + N_t$$ it follows from the independence and stationarity of the increments that $$\mathbb{E}\exp(i \xi N_{t+h}) = \mathbb{E}\exp(i \xi N_t) \exp(i \xi N_h)$$ for all $$\xi$$. By assumption, $$N_t$$ and $$N_{t+h}$$ have the same distribution, and so $$\mathbb{E}\exp(i \xi N_{t}) = \mathbb{E}\exp(i \xi N_t) \exp(i \xi N_h)$$ for all $$\xi$$. Since $$\mathbb{E}\exp(i \xi N_t) \neq 0$$ for all $$\xi$$ (that's a property of Lévy processes) it follows that $$1= \mathbb{E}\exp(i \xi N_h)$$ for all $$\xi$$. Note that $$\chi(\xi)=1$$ is the Fouier transform of the random variable $$X=0$$. By the uniqueness of the Fourier transform, we therefore get $$N_h=0$$ almost surely. This is, however, only possible if $$h=0$$; this can be, for instance, proved using the Lévy-Khintchine formula. Hence, we conclude that $$N_t$$ and $$N_{t+h}$$ can only have the same distribution in the trivial cases that $$h=0$$.
• Thanks very much saz. I had to work through some of the steps and notation, but I understand now, and the argument is very clear. I will take your word for it that $N_h=0$ a.s. implies $h=0$ (it feels right, intuitively, anyway), and I see that therefore $N$ can't be stationary except in the trivial case. Even without that implication, we would have that $N_h=0$ if $N$ is stationary, which in itself would be an extreme restriction on the idea of a stationary Lévy process. And I guess I'm happy to know that reason wasn't something that I would consider obvious. :-)