How does this expected value translate into a conditional variance? I'm working with a simple local level model in a textbook
\begin{align}
y_t &= \alpha_t + \epsilon_t, \qquad \epsilon_t \sim N(0, \sigma_\epsilon^2) \\
\alpha_{t+1} &= \alpha_t + \eta_t, \qquad \eta_t \sim N(0, \sigma_\eta^2)
\end{align}
The conditional distribution of $\alpha_t$ given $Y_{t-1}$ (the set of all observations $y_j$ where $1 \leq j \leq t-1$) is $N(a_t, P_t)$, where we have $a_t = E(\alpha_t \mid Y_{t-1})$ and $P_t = Var(\alpha_t \mid Y_{t-1})$.
The book includes this calculation:
\begin{equation}
E[\alpha_t(\alpha_t - a_t)] = E[Var(\alpha_t \mid Y_{t-1})] = P_t
\end{equation}
I don't understand where this first equality comes from.
Working backwards from the law of conditional variance, I know that
\begin{align}
Var(\alpha_t \mid Y_{t-1})
&= E[ (\alpha_t - E[\alpha_t \mid Y_{t-1}])^2\mid Y_{t-1}] \\
&= E[ (\alpha_t - a_t)^2 \mid Y_{t-1}] \\
&= E[ (\alpha_t^2 - 2 \alpha_t a_t + a_t^2) \mid Y_{t-1}] \\
&= E[\alpha_t^2 \mid Y_{t-1}] - 2E[\alpha_t a_t \mid Y_{t-1}] + E[a_t^2 \mid Y_{t-1}] \\
\end{align}
but I don't see how to get $\alpha_t (\alpha_t - a_t)$ from this, which would give me the correct value inside the expectation. 
 A: For any two random variables $X$ and $Y$, we always have: $$V(Y)=E[V(Y|X)]+V[E(Y|X)].$$ Now in above formula let $Y=\alpha_t$ and $X=Y_{t-1}$ to have: $$V[\alpha_t]=E[V(\alpha_t|Y_{t-1})]+V[E(\alpha_t|Y_{t-1})]=E[\alpha_t^2]-E^2[\alpha_t]$$ Hence
$$V[\alpha_t]=E[V(\alpha_t|Y_{t-1})]+V[a_t]=E[\alpha^2_t]-E^2[\alpha_t]$$ So $$E[V(\alpha_t|Y_{t-1})]=E[\alpha^2_t]-E^2[\alpha_t]-V[a_t]=E[\alpha^2_t]-E^2[\alpha_t]-E[a^2_t]+E^2[a_t] \quad (1)$$
Now take expected value from both sides of $a_t=E[\alpha_t|Y_{t-1}]$ to have: $$E[a_t]=E\big[E[\alpha_t|Y_{t-1}]\big]=E[\alpha_t].$$ So $$E[\alpha_t]=E[a_t] \quad (2)$$
Next use (2) in (1) to have:
$$E[V(\alpha_t|Y_{t-1})]=E[\alpha^2_t]-E^2[\alpha_t]-E[a^2_t]+E^2[\alpha_t]=E[\alpha^2_t]-E[a^2_t] \quad (3)$$ Note that $$E[\alpha_ta_t]=E\big[E[\alpha_ta_t|Y_{t-1}]\big]=E[a_tE\big[\alpha_t|Y_{t-1}]\big]=E[a^2_t] \quad (4)$$ Finally use (4) in (3) i.e. replace $E[a^2_t]$ with $E[\alpha_ta_t]$  in (3) to have:
$$E[V(\alpha_t|Y_{t-1})]=E[\alpha^2_t]-E[\alpha_ta_t]=E[\alpha_t(\alpha_t-a_t)].$$
