The only strictly stationary random walk in $\mathbb{R}$ is degenerate

Solution 1:

Since $(X_n)$ is stationary, the marginal distributions of $X_n$ form a tight family of distributions, namely, $$ \lim_{M\rightarrow\infty}\sup_{n\ge 1} P(|X_n|>M)=\lim_{M\rightarrow\infty}P(|X_n|>M)=0. $$ On the other hand, the marginal distributions of the random walk $W_n=\sum_{i=1}^n \epsilon_i$ are not tight unless $\epsilon_i$ is zero almost surely. Indeed, if $P(\epsilon_i=0)<1$, then for any $M>0$, we have $$ \lim_{n\rightarrow\infty} P(|W_n|>M)=1. $$ One can check this following Chapter 9 Exercise 2 of Kallenberg's Foundation of Modern Probability 2nd edition using the following inequality of characteristic function \begin{equation} P(|W_n|\le M)\le 2M \int_{-1/M}^{1/M} |\varphi(t)|^n dt, \end{equation} where $\varphi(t)= E[e^{it \epsilon_1}]$.

Lemma If $\epsilon_1$ is not almost surely a constant, then there exists a interval $[-\delta,\delta]$, $\delta>0$, such that $|\varphi(t)|<1$ for all $t\in [-\delta,\delta]$.

Proof of the lemma: this follows from [Post][1] and the continuity of a characteristic function.

So if $\epsilon_1$ is not a constant almost surely, then one can choose $M$ large enough in the inequality above, and conclude by the Dominated Convergence Theorem that $$ \lim_n P(|W_n|\le M)=0. $$ If $\epsilon_1$ is almost surely a nonzero constant, the same conclusion holds trivially as well. So we have shown that the marginal distributions of a non-zero random walk (including deterministic drift) is not tight.

Lemma Suppose random variables $|W_n|\rightarrow \infty$ in probability (in the sense of the displayed line above for arbitrary $M$) and $X$ is a fixed finite random variable. Then $|W_n+X| \rightarrow \infty$ in probability as well.

Proof of the lemma: immediate if using a sub-sub sequence argument to relate to almost sure divergence to infinity: a sequence of variables $Y_n \rightarrow \infty$ in probability if and only if for any subsequence of $(Y_n)$ there exists a further subsequence which diverges to $\infty$ almost surely.

Apply the previous lemma to $X_n=X_0+W_n$ to obtain a contradiction.

We can have all $X_n$ identical to each other (not necessarily a zero random variable), which is a stationary process.

[1]: https://math.stackexchange.com/questions/1410331/characteristic-function-with-modulus-1-implies-degenerate-distribution#:~:text=Characteristic%20function%20with%20modulus%201%20implies%20degenerate%20distribution,-probability%20characteristic%2Dfunctions&text=Let%20X%20be%20a%20random,X%3Dc)%3D1.