Let $X$ be a discrete random variable, where $X\in\{1,\ldots, n\}$. Suppose $P(X=i)=p_i$ for all $i=1,\ldots, n$, such that $p_1+\ldots+p_n = 1$. I am interested to know the value of $$ \mathbb{E}\left[|1_{X=i}-\mathbb{E}(1_{X=i})|\cdot |1_{X=j}-\mathbb{E}(1_{X=j})|\right], \quad\text{for}\quad i\neq j. $$
Is this value $0$, due to independence? I can't see why. Can someone explain this explicitly? Thank you!
P.s. $1_{X=i}$ represents the indicator function.
Split the expectation over $X=i$, $X=j$ and $X\neq i, X \neq j$.
$$ \mathbb{E}\left[|1_{X=i}-\mathbb{E}(1_{X=i})|\cdot |1_{X=j}-\mathbb{E}(1_{X=j})|\right]$$ $$=\mathbb{E}1_{X=i} (1-p_i)(p_j)+\mathbb{E}1_{X=j} (p_i)(1-p_j)+p_ip_j \mathbb{E}1_{X\neq i, X \neq j}.$$
Using the fact that $P((X=i)\cup (X=j)=p_i+p_j$ we get $P(X\neq i, X \neq j)=1-p_i-p_j$. Hence,
$$ \mathbb{E}\left[|1_{X=i}-\mathbb{E}(1_{X=i})|\cdot |1_{X=j}-\mathbb{E}(1_{X=j})|\right]$$ $$=p_i(1-p_i)(p_j)+p_j(p_i)(1-p_j)+p_ip_j(1-p_i-p_j)$$
I will let you simplify this expression.