Step in proof $E_\theta T^*=E_\theta T$ of Rao-Blackwell Say we have $T$ an estimator for $g(\theta)$, and $T^*=T^*(V)$ an estimator that only depends on the sufficient statistic $V$. My book claims the following:
$$
E_\theta TT^*=\sum_{v}E(TT^*\mid V=v)P_\theta(V=v)=\sum_vT^*(v)E(T\mid V=v)P_\theta(V=v).
$$
Now I don’t see why this holds. How did they factorise the expectation? (as in, which rule did they apply). I know the linearity of expectations, but I don’t think that holds here. And I also know that I can write $E(X+Y)=E(X)+E(Y)$ for $X,Y$ independent. But I don’t understand what happened here.
 A: Write $\mathcal X$ for the range of $X$ and $\mathcal Y$ for the range of $Y$. The definition of conditional expectiation for discrete random variables is
$$
E(X\mid Y=y) = \sum_{x\in\mathcal X} x\cdot P(X=x\mid Y=y)
$$
and this implies
$$
\sum_{y\in\mathcal Y} E(X\mid Y=y)P(Y=y) = \sum_{y\in\mathcal Y} \sum_{x\in\mathcal X} x\cdot P(X=x\mid Y=y)P(Y=y)
$$
The term on the right hand side can be simplified since
$$
P(X=x\mid Y=y)P(Y=y) = P(X=x,Y=y).
$$
Therefore
$$\begin{aligned}
\sum_{y\in\mathcal Y} E(X\mid Y=y)P(Y=y) &= \sum_{y\in\mathcal Y} \sum_{x\in\mathcal X} x\cdot P(X=x,Y=y)\\
&=  \sum_{x\in\mathcal X} x\cdot \sum_{y\in\mathcal Y} P(X=x,Y=y)\\
&= \sum_{x\in\mathcal X} x\cdot P(X=x) \\
&= E(X).
\end{aligned}$$In the second step I used that $\sum_{y\in\mathcal Y} P(X=x,Y=y) = P(X=x, Y\in \mathcal{Y}) = P(X=x)$ because these events are disjoint.
In your case, $X$ is $TT^*$ and $Y$ is called $V$.
In the second step they use
$$
E(TT^* \mid V=v) = E(T  \cdot T^*(V) \mid V=v) = T^*(v) E(T\mid V=v)
$$
since $T^*$ is completely determined by $V$.
