This is easy to show, keeping in mind the linearity of the expectation value. Consider the stochastic variables \( X_i \) and \( X_j \), (\( i\neq j \)) $$ \begin{align*} \mathrm{Cov}(X_i,\,X_j) &= \langle (x_i-\langle x_i\rangle)(x_j-\langle x_j\rangle)\rangle\\ &=\langle x_i x_j - x_i\langle x_j\rangle - \langle x_i\rangle x_j + \langle x_i\rangle\langle x_j\rangle\rangle\\ &=\langle x_i x_j\rangle - \langle x_i\langle x_j\rangle\rangle - \langle \langle x_i\rangle x_j \rangle + \langle \langle x_i\rangle\langle x_j\rangle\rangle\\ &=\langle x_i x_j\rangle - \langle x_i\rangle\langle x_j\rangle - \langle x_i\rangle\langle x_j\rangle + \langle x_i\rangle\langle x_j\rangle\\ &=\langle x_i x_j\rangle - \langle x_i\rangle\langle x_j\rangle \end{align*} $$ If \( X_i \) and \( X_j \) are independent, we get $$ \langle x_i x_j\rangle = \langle x_i\rangle\langle x_j\rangle=\mathrm{Cov}(X_i, X_j) = 0\ \ (i\neq j). $$