Let $X_1,…,X_n$ be random variables with identical distribution, and for all $i=1,…,n$ $\mathrm{Var}(X_i)$ exist.
1. Show that the covariance between each two random variables exist.
2. Show that :
$\mathrm{Var} \sum _{i=1}^{n} X_i = n \mathrm{Var} (X_1) + n(n-1)\mathrm{Cov}(X_1,X_2)$
Well, at the first question I tried to split it to two cases: case 1, that $\mathrm{Cov}(X_i,X_j)$ when $i=j$, and in this case it is simply $\mathrm{Var}(X_i)$ which exists.
but then I tried to show the other case, when $i\neq j$, and I dont really know how. Is it correct to use Cauchy-Schwarz inequality $|\mathrm{Cov}(X_i,X_j|^2 \leq \mathrm{Var}(X_i)\mathrm{Var}(X_j)$?
Other then that, I don't know how to show the second question, can someone give me a hint?
Thanks!
Best Answer
Cauchy Schwarz is fine. For the second equation $\def\Var{\mathop{\rm Var}}\def\Cov{\mathop{\rm Cov}}$(which will hold only if not only the $X_i$, but the pairs $(X_i,X_j)$, $i \ne j$ are identically distributed) recall that \begin{align*} \Var\sum_i X_i &= \Cov\left(\sum_i X_i, \sum_j X_j\right)\\ &= \sum_{i,j} \Cov(X_i, X_j)\\ &= \sum_i \Var(X_i) + \sum_{i\ne j} \Cov(X_i, X_j) \end{align*} and the fact the the $X_i$ are identical distributed. This gives $\Var(X_i) = \Var(X_1)$ for all $i$. If we have that the pairs are identically distributed, we have $\Cov(X_i, X_j) =\Cov(X_1, X_2)$ for all $i \ne j$, hence $$ \Var\sum_i X_i = \sum_i \Var(X_1) + \sum_{i\ne j} \Cov(X_1, X_2) = n \Var(X_1) + n(n-1)\Cov(X_1, X_2) $$