Inequality in probability theory.

Question

Let X be a nonnegative random variable and let $(\mathcal{H}_i)$ be a sequence of increasing $\sigma$ algebras. Let $(A_i)_{0 \leqslant i \leqslant N}$ be a sequence of pairwise disjoint events. Do we have $$ \sum_{i=0}^N \mathbb{E} \big( \mathbb{1}_{A_i} \mathbb{E}\big(X | \mathcal{H}_i \big) \big) \leqslant \mathbb{E}(X) ? $$

Answer 1

In $\Omega=[0,1]$ with Lebesgue measure, let $H_0=\{\emptyset,\Omega\}$ and let $H_1$ be the $\sigma$-field generated by $X={\bf 1}_{[0,1/2]}$ (or we could take $H_1$ to be the full Lebesgue $\sigma$-field.) Then $$ {\bf 1}_{( 1/2,1]}E(X|H_0)=\frac12 \cdot {\bf 1}_{( 1/2,1]} \quad \text{a.e.} $$ and
$$ {\bf 1}_{[0,1/2 ]}E(X|H_1)= {\bf 1}_{[0,1/2 ]}\quad \text{a.e.} $$ Adding the expectations of the two displays yields $1/4+1/2>E(X)$.