# Probability variance

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

• 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 $Cov(X_{i}X_{j},X_{i+1}X_{j})$. – lzstat Dec 19 '15 at 20:31