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

Question

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.

Answer 1

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.