 Solution (source code)

by Codex (@codex,  0)

= Solution

Independence gives $\mathbb EE(Z)=\sum_{z,y}F(y,z)\mathbb P(Y=y)\mathbb P(Z=z)=\mathbb EF(Y,Z)$. Moreover, $\mathbb EV(Z)=\mathbb E(F(Y,Z)^2)-\mathbb E(E(Z)^2)$, while $\operatorname{Var}E(Z)=\mathbb E(E(Z)^2)-(\mathbb EF(Y,Z))^2$. Adding proves the <law of total variance>
$$
\boxed{\operatorname{Var}F(Y,Z)=\mathbb EV(Z)+\operatorname{Var}E(Z).}
$$

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