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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!

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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 $$ \Var\sum_i X_i = \sum_i \Var(X_1) + \sum_{i\ne j} \Cov(X_1, X_2) = n \Var(X_1) + n(n-1)\Cov(X_1, X_2) $$

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  • $\begingroup$ Do I need to show that the pairs $(X_i,X_j) i\neq j$ are identically distributed? How do I do that? It does not follow from the fact that $X_1,...,X_n$ are identically distributed? $\endgroup$ – Buzi Jun 26 '14 at 7:31
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    $\begingroup$ @Buzi You cannot show this from your assumptions, you have to assume it additionally if you want (2) to hold. It will not follow from $X_1, \ldots, X_n$ being identically distributed: Let $X_1, X_2$ be independent and $N(0,1)$ distributed (to be concrete). Let $X_3 := X_1$. Then $X_1, X_2, X_3$ are identically distributed, but: ${\rm Cov}(X_1, X_2) = 0$ and ${\rm Cov}(X_1, X_2) = {\rm Var}(X_1) = 1$. $\endgroup$ – martini Jun 26 '14 at 7:44
  • $\begingroup$ Oh, you are right, I even got this very own counter example in class. Thanks for your help! You helped me a lot! $\endgroup$ – Buzi Jun 26 '14 at 7:46

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