5
$\begingroup$

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?

$\endgroup$
1
  • $\begingroup$ 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

6
$\begingroup$

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)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.