Calculating Var(X) Let $\Theta$ be an unknown random variable with mean $1$ and variance $2$. Let $W$ be another unknown random variable with mean $3$ and variance $5$. $\Theta$ and $W$ are independent.
Let: $X_1=\Theta+W$ and $X_2=2\Theta+3W$. We pick measurement $X$ at random, each having probability $\frac{1}{2}$ of being chosen. This choice is independent of everything else.
How does one calculate $Var(X)$ in this case?
Is 
$$
Var(X)\;\; = \;\; \frac{1}{2}(Var(\Theta)+Var(W))+\frac{1}{2}(Var(2\Theta)+Var(3W)) \;\; =\;\; \frac{1}{2}(5Var(\Theta)+10Var(W))?
$$
 A: Hint:
Denoting the random index by $I$ we have:
$$\mathbb EX=\mathbb E(X\mid I=1)P(I=1)+\mathbb E(X\mid I=1)P(I=1)=\mathbb EX_1.\frac12+\mathbb EX_2.\frac12$$and:
$$\mathbb EX^2=\mathbb E(X^2\mid I=1)P(I=1)+\mathbb E(X^2\mid I=1)P(I=1)=\mathbb EX_1^2.\frac12+\mathbb EX^2_2.\frac12$$
Now use the well known identity:
$$\text{Var}(X)=\mathbb EX^2-(\mathbb EX)^2$$
The equalities $X_1=\Theta+W$ and $X_2=2\Theta+3W$ can be used to find $\mathbb EX_i$ and $\mathbb EX_i^2$ for $i=1,2$.
A: The choice between $X_1$ and $X_2$ is a coin toss independent of everything else. Let $Y$ equal 1 if $X_1$ is chosen, and 0 if $X_2$ is chosen. Then
$$
X = X_1 Y + X_2(1-Y)\;.
$$
Calculate
$$E(X) = E(X_1Y)+ E[X_2(1-Y)] = 0.5  (EX_1+EX_2)
$$
and
$$E(X^2) = E(X_1^2Y^2) + 2E(X_1X_2Y(1-Y)) + E[X_2^2(1-Y)^2] = 0.5(EX_1^2+EX_2^2)
$$
using independence between $Y$ and everything else, and the facts $Y^2=Y$, $Y(1-Y)=0$, $(1-Y)^2=1-Y$, and $EY=E(1-Y)=0.5$.
Finally calculate the variance using
$$\text{var}(X)=E(X^2)-[E(X)]^2\;.$$
