Conditional Variance of the sum of two variables $\newcommand{\v}{\operatorname{Var}}\newcommand{\c}{\operatorname{Cov}}$I have a simple question that, after thinking for a while, got me confused and I cannot figure it out.
Does the following statements hold?
1
$$ \v(A + B + C \mid \theta ) = \v(A \mid \theta) + \v(B + C \mid \theta) + 2 \c(A, B + C \mid \theta)$$
2
$$ \c(A, B + C \mid \theta ) = \c(A, B \mid \theta) + \c(A, C \mid \theta)$$
I am not sure whether I can do these two steps above.
Thanks
 A: The linearity of conditional expectation implies
$$\operatorname{Cov}(A, B + C \mid \theta ) = \mathbb E\left[A(B+C)\mid\theta\right]-\mathbb E\left[A\mid\theta\right]\mathbb E\left[B+C\mid\theta\right]$$
$$
=\mathbb E\left[AB\mid\theta\right]+\mathbb E\left[AC \mid \theta \right] - \mathbb E\left[A\mid\theta\right]\mathbb E \left[B \mid \theta \right] -\mathbb E\left[A\mid\theta\right]\mathbb E \left[ C \mid \theta\right]$$
$$=\operatorname{Cov}(A, B \mid \theta) + \operatorname{Cov}(A, C \mid \theta).$$
The same for the variance:
$$
\operatorname{Var}(X+Y\mid Z) = \mathbb E\left[(X+Y)^2\mid Z\right] - \bigl(\mathbb E[X+Y\mid Z]\bigr)^2 $$ 
$$ = \mathbb E\left[X^2+Y^2+2XY\mid Z\right] - \bigl(\mathbb E[X\mid Z]+\mathbb E[Y\mid Z]\bigr)^2$$
$$
=\color{red}{\mathbb E\left[X^2\mid Z\right]}+\color{blue}{\mathbb E\left[Y^2\mid Z\right]}+\color{green}{2\mathbb E\left[XY\mid Z\right]} - \color{red}{\bigl(\mathbb E[X\mid Z]\bigr)^2} - \color{blue}{\bigl(\mathbb E[Y\mid Z]\bigr)^2} - \color{green}{2\mathbb E[X\mid Z]\mathbb E[Y\mid Z]} 
$$
$$
=\color{red}{\operatorname{Var}(X\mid Z)}+\color{blue}{\operatorname{Var}(Y\mid Z)} + \color{green}{2\operatorname{Cov}(X,Y\mid Z)}.
$$
