Let $X \sim N(4,3)$ and $Y \sim N(2,4)$. Which of the following do we know for sure about $X+Y$? That 


*

*$Var(X+Y)=7$

*$E(X+Y)=6$

*$X+Y$ is normally distributed. 

*$X+Y$ is not normally distributed.


My answer
It follows from the linearitet of $E$ that $E(X+Y) = 6$. I do not know how to determine whether the rest is true or not.
 A: You conclusion about the expected value, and your reason for your conlusion are correct.
Now consider:
$$\operatorname{var}(X+Y) = \operatorname{var}(X) + \operatorname{var}(Y) + 2\operatorname{cov}(X,Y).$$
Now suppose
\begin{align}
X & = Z_1 + Z_2 + Z_3 + 4 \\
Y & = Z_2 + Z_3 + Z_4 + Z_5 + 2 \\
\text{where } & Z_1,Z_2,Z_3,Z_4 \sim \text{i.i.d.} \operatorname N(0,1).
\end{align}
Then the distributions of $X$ and $Y$ are just those that you specified, but $\operatorname{cov}(X,Y)\ne 0.$
There are also cases where $X+Y$ is not normally distributed. For example, suppose $X\sim\operatorname N(4,3)$ and
$$
Y = 2 + 2\times\begin{cases} (X-4)/\sqrt 3 & \text{if “heads''} \\ (4-X)/\sqrt 3 & \text{if “tails''} \end{cases}
$$
where $\text{“heads''}$ or $\text{“tails''}$ is the outcome of the toss of a fair coin that is indpendent of $X$.
In that case, $X$ and $Y$ have the specified distributions, but it is not hard to show that $X+Y$ is not normally distributed. Thus the pair $(X,Y)$ does not have a bivariate normal distribution.
