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Let ${X_{n}}$ be a sequence of independent identically distributed random variables with mean $\mu$ and finite variance. So,

$$\operatorname{Var} \left(\sum_{1\leqslant i< j\leqslant n} X_i X_j \right) = \binom{n}{2} \operatorname{Var}(X) \text{ ?} $$

Thank you

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    $\begingroup$ no, it is not true. First you need to consider $Var(X_{i}X_{j})$, then you need to also consider the covariance such as $Cov(X_{i}X_{j},X_{i+1}X_{j})$. $\endgroup$ – lzstat Dec 19 '15 at 20:31
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\begin{align} \operatorname{var}(X_1 X_2) & = \operatorname{E}(X_1^2 X_2^2) - (\operatorname{E}(X_1 X_2))^2 \\[6pt] & = \operatorname{E}(X_1^2)\operatorname{E}(X_2^2) - (\operatorname{E}(X_1) \operatorname{E}(X_2))^2 = (\mu^2+\sigma^2)^2 - \mu^2 \\[10pt] \operatorname{cov}(X_1 X_2,X_2 X_3) & = \operatorname{E}(X_1 X_2^2 X_3) - \operatorname{E}(X_1 X_2) \operatorname{E}(X_2 X_3) \\[6pt] & = \mu(\mu^2+\sigma^2)\mu - (\mu^2)(\mu^2) \\[10pt] \operatorname{cov}(X_1 X_2,X_3 X_4) & = 0. \end{align}

Figure out how many terms of each of these three kinds you need to add up.

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