 Average conditional variance decreases with more information

For square-integrable $X$ and nested information $\mathcal{H}\subseteq \mathcal{G}$, let $m_{\mathcal{H}}=\mathbb{E}[X\mid \mathcal{H}]$ and $m_{\mathcal{G}}=\mathbb{E}[X\mid \mathcal{G}]$. The predictor $m_{\mathcal{H}}$ is also $\mathcal{G}$-measurable, so the conditional variance error identity givesMore information cannot increase the average best squared prediction error. Equality holds exactly when the two conditional means agree almost surely; additional information can change the conditional distribution without changing its mean.

This is an averaged statement. Take $X=-4,0,4$ with equal probabilities, and learn only whether $X$ is zero. The unconditional variance is $32/3$. In the nonzero group, the conditional variance is $16$, which is larger; in the zero group it is zero. Their weighted average remains $(2/3)16+(1/3)0=32/3$. Both groups have conditional mean zero, so the extra information has not improved squared-error prediction, even though it has revealed whether the outcome is extreme or central. An individual group's conditional variance need not decrease.