I originally posted this question in Mathstackexchange, but since I got no answer I'm posting it also here.

Let $X_1,X_2,...$ be a sequence of identically distributed and $m$-dependent random variables with $\mathbb{E}[X_i]=0$, $0<\operatorname{Var}(X_i)<\infty$ ($m$-dependent means that each $X_i$ is independent of $X_{i+j}$ for $|i-j|\ge m $.

Suppose $Y$ is a random variable with $\mathbb{E}[Y]=0$ and $\operatorname{Var}(Y)<\infty$.

Assume also that $Y$ is independent of $X_m,X_{m+1},...$

We know that $$ \frac{Y+\sum_{i=1}^{n}X_i}{\sqrt{n}}\overset{d}{\longrightarrow} N(0,\sigma^2) $$

from the Hoeffding-Robbins theorem, but I am struggling to show that $\sigma^2>0 $ even though intuitively it seems true.

Do you have any ideas?

  • $\begingroup$ To start, the appearance of $Y$ doesn't affect anything, because $Y/\sqrt{n} \to 0$ a.s., and Slutsky's lemma. $\endgroup$ Mar 7, 2019 at 16:13

1 Answer 1


First, the random variable (r.v.) $Y$ plays no role here, since $Y/\sqrt n\to0$.

Second, $\sigma^2$ may be zero. However, in the abstract of Janson we find this complete answer to your question:

It is well-known that the central limit theorem holds for partial sums of a stationary sequence $(X_i)$ of $m$-dependent random variables with finite variance; however, the limit may be degenerate with variance $0$ even if $\text{Var}\,(X_i)\ne0$. We show that this happens only in the case when $X_i − \text{E}\,X_i = Y_i − Y_{i−1}$ for an $(m − 1)$-dependent stationary sequence $(Y_i)$ with finite variance (a result implicit in earlier results)


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