A question was recently asked by a new user, SomeoneHAHA, and then deleted by the user, after receiving an answer. I think the question and the answer (QA) to it may be of interest to some users. Therefore, I am resurrecting the QA. Here is the question, in a bit brushed-up form:

Let $(X_t)_{t\in\mathbb R}$ be a normalized wide-sense stationary stochastic process in $\mathbb R$, so that $g(t):=\text{Cov}(X_s,X_{s+t})$ does not depend on $s$ and $g(0)=1$. By Bochner's theorem, $g$ must then be the characteristic function of a probability distribution.

The question is: Is it possible that $\frac{1-g(t)}{t^2}\underset{t\to0}\longrightarrow0$?


1 Answer 1


$\newcommand{\R}{\mathbb{R}}$ This is only possible when $g=1$ identically. Indeed, otherwise, by Hoeffding's Lemma 3.2, page 217, one of the following two alternatives must take place:

(i) $g$ is the characteristic function of a degenerate distribution, so that $g(t)=e^{itx}$ for some real $x\ne0$ and all real $t$, and hence $\frac{1-g(t)}{t^2}\underset{t\to0}\longrightarrow0$ is false;

(ii) there are positive real numbers $b$ and $c$ such that $1-|g(t)|\ge ct^2$ for $|t|\le b$, which again precludes $\frac{1-g(t)}{t^2}\underset{t\to0}\longrightarrow0$, because $|1-g(t)|\ge1-|g(t)|$.


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