# Variance of sum of $m$ dependent random variables

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?

• To start, the appearance of $Y$ doesn't affect anything, because $Y/\sqrt{n} \to 0$ a.s., and Slutsky's lemma. – Nate Eldredge Mar 7 '19 at 16:13

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)