Consider $X_1, ..., X_n,...$ independent random variables for which $\text{Var}(X_i) = \sqrt{i}$. Let $Y_i=f(X_i, X_{i+1}), f:R^2\to R$. Show that $cov(Y_i,Y_j)=0$ for $|i-j|>2$ and then show that the variance of $\frac{1}{n}\sum\limits_{i=1}^{n}Y_i$ converges to $0$ as $n\to\infty$.
I have: $Cov[Y_i,Y_j]=E[Y_i,Y_j]-E[Y_i]E[Y_j]$. As $Y_i$ is a transformation of the random variables $X_i, X_{i+1}$, the covariance can be rewritten as (omitting constants):
$=E[(aX_i+bX_{i+1} ),(cX_j+dX_{j+1} )]-E[(aX_i+bX_{i+1} )]E[cX_j+dX_{j+1}]$.
Take $|i-j|>2$ and use the independence of $X_1,...,X_n$, it follows that the covariance can be rewritten as:
$=E[(aX_i+bX_{i+1})]E[cX_j+dX_{j+1} ]-E[(aX_i+bX_{i+1} )]E[cX_j+dX_{j+1}]\\=E[Y_i]E[Y_j]-E[Y_i]E[Y_j]=0$
I guess this is correct, but I suppose this is not the kind of proof they are asking for. Can anyone give me a hint to give a better proof?
For the second part, I suppose that $\lim_{n \to \infty}\frac{1}{n^2}\sum\limits_{i=1}^{n}Y_i= \lim_{n \to \infty}\frac{1}{n^2}\sum\limits_{i=1}^{n}\sum\limits_{j=1}^{n}Cov[Y_i,Y_j]\neq \lim_{n \to \infty}\frac{1}{n^2}\sum\limits_{i=1}^{n}\sum\limits_{j=1}^{n}0$, as the second part also assumes the case that $|i-j|\leq2$. Could anyone give me a hint there too?
Part 1: When $|i -j|\geq 2$ the Random Variables $Y_i,Y_j$ depend on are independent of each other. For example, $$Y_1 = f(X_1, X_2), Y_3 = f(X_3, X_4)$$ This implies that $Y_i$ and $Y_j$ are independent and therefore have no correlation.
Part 2: \begin{eqnarray*} \text{Var}\left(\frac{1}{n}\sum_{i=1}^n Y_i\right) &=& \frac{1}{n^2} \sum_{i=1}^n \sum_{j=1}^n \text{Cov}(Y_i, Y_j) \\ &=& \frac{1}{n^2} \left(\sum_{i=1}^n \text{Var}(Y_i) + 2\sum_{i=1}^{n-1} \text{Cov}(Y_i, Y_{i+1})\right) \\ &\leq& \frac{1}{n^2} \left(\sum_{i=1}^n \text{Var}(Y_i) + 2\sum_{i=1}^{n-1} \sqrt{\text{Var}(Y_i)\text{Var}(Y_{i+1}})\right) \\ \end{eqnarray*}
The third line follows because $\text{Cov}(X, Y) \leq \sqrt{\text{Var}(X)\text{Var}(Y)}$.