I'm working on the following problem, which I quote verbatim from the assignment.
Suppose that in the regression equation $$Y_i=a+bX_i+u_i$$ the coefficients are $a=1$ and $b=2$. Suppose also the probability distribution of $u$ is discrete, equaling $-5$ with probability $1/4$, $0$ with probability $1/2$, and $5$ with probability $1/4$. Calculate the expected value and the variance of $Y_i$ when $X_i=4$. How does the variance of $Y$ compare with the variance of $u$?
Computing $E[Y_i | X_i=4]$ is easy:
$$E[Y_i | X_i=4] = E[1+2X_i+u_i | X_i=4] = E[1|X_i=4] + 2E[X_i|X_i=4]+E[u_i|X_i=4]$$
The first term is $1$, and the second term is $2(4)$. For the last term, I'm not sure whether I'm supposed to argue that by mean independence between $u$ and $X$ we have $E[u_i|X_i=4]=E[u_i]$, or whether I should just pretend the problem gives me not the unconditional distribution of $u$ but the conditional distribution of $u$. Either way, the last term is zero. Thus the expected value of $Y_i$ given $X_i=4$ is $9$.
I also know that $$\text{Var}(Y_i|X_i=4)=\text{Var}(1+2X_i+u_i|X_i=4)=\text{Var}(u_i|X_i=4)$$ because $1+2X_i$ is a constant, conditional on $X_i=4$, and adding a constant to a random variable makes no difference to the variance. Again assuming $\text{Var}(u_i|X_i=4)=\text{Var}(u_i)$, we just calculate $E[u_i^2]-E[u_i]^2=12.5-0=12.5$.
My problem is with the last question: "how does the variance of $Y$ compare with the variance of $u$?" For "the variance of $Y$," should I use the conditional variance I just calculated? Or does this mean I should now calculate the unconditional variance of $Y$?
I can try to find the unconditional variance using the law of total variance:
$$\text{Var}(Y)=E[\text{Var}(Y|X)]+\text{Var}(E[Y|X])$$
The first term is $E[12.5]=12.5$, as I calculated above. But the second term is $\text{Var}(1+2X)=4\text{Var}(X)$, and I don't know enough about $X$ to calculate this variance.