For one Bernoulli variable, direct expansion gives $\operatorname{Var}F(X_1)=p(1-p)(F(1)-F(0))^2$. Applying the law of total variance successively to the coordinates gives the <variance tensorization>
$$
\boxed{\operatorname{Var}F(X)\leq p(1-p)\sum_{i=1}^n
\mathbb E\bigl(F(X)-F(X^i)\bigr)^2.}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2018 / ia / Paper 2 / 9F / b / Solution

For independent coordinates $X_1,\ldots,X_n$ and square-integrable $f$,
$$
\operatorname{Var}f(X)
\leq\sum_i\mathbb E\!\left[\operatorname{Var}(f(X)\mid X_j,\ j\ne i)\right].
$$
For Bernoulli coordinates, each conditional variance is an explicit squared coordinate difference.


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