# Variance inequality of identically distributed random variables

Let $X_1,...,X_n$ be (not necessarily independent) identically distributed random variables with $E[X_1^2]<\infty$. Assume that there is a fixed $h\ge1$ so that $Cov(X_i,X_j)=0$ for $|i-j|\ge h$. Proof that with $S_n:=\sum_{i=1}^nX_i$ $$Var(S_n)\le 2nhVar(X_1)$$ Proof.

We have $$Var(S_n)=\sum_{i=1}^nVar(X_i)+2\sum_{\substack{1\le i < j \le n}} Cov(X_i,X_j)$$ $=$(assumption) $$nVar(X_1)+2\sum_{\substack{1\le i < j \le n \\ h > |i-j|}} Cov(X_i,X_j)$$$\le$ (Cauchy-Schwarz-ineq.)$$nVar(X_1)+2Var(X_1)\sum_{\substack{1\le i < j \le n \\ h > |i-j|}}1$$

and that has to be $\le 2nhVar(X_1)$.

Can anyone explain the last inequality to me?

## 1 Answer

Since $i<j$, $h > |i-j|$ means $h > j-i$ and hence $i<j<i+h$, hence fixing $i$, $j$ can take at most $h-1$ values.

Hence $$\sum_{\substack{1\le i < j \le n \\ h > |i-j|}}1 \leq n(h-1)$$

Hence

$$n+2\sum_{\substack{1\le i < j \le n \\ h > |i-j|}}1 \leq n+2n(h-1) \leq 2n + 2n(h-1)=2nh$$