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.


1 Answer 1


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 $h=0$.

  • 1
    $\begingroup$ 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. :-) $\endgroup$
    – Mars
    Aug 7, 2019 at 17:16

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