Sum of two normal distributions, $Z=X+Y$

Question

Having trouble with this probability question:

IF $X\sim\mathcal{N}(1,1)$ and $Y\sim\mathcal{N}(1,2)$ are two normally distributed random variables with means and standard deviations as indicated, and $Cov(X,Y)=-\frac{1}{2}$ and $Z=X+Y$, calculate $Var(Z)$.

Right since the $Cov(X,Y)\neq 0$, $X$ and $Y$ are dependent and correlated. Therefore $E(XY)\neq E(X)E(Y)$. From the equation for covariance I can calculate $E(XY)$. $$Cov(X,Y)=E(XY)-E(X)E(Y)$$ Therefore $$-\frac{1}{2}=E(XY)-1$$ And $$E(XY)=\frac{1}{2}$$

Now $$Var(Z)=Var(X+Y)=E((X+Y)^{2})-E(X+Y)^{2}$$ $$Var(X+Y)=E(X^2+2XY+Y^2)-E(X+Y)^2$$ $$E(X^2)=2, E(Y^2)=3$$ Calculated from the moment generating function $$m(t)=e^{t\mu+\frac{1}{2}t^2\sigma^2}$$ Now this is as far as I've got: $$Var(X+Y)=8-E(X+Y)^2$$ How would I go about finding $E(X+Y)$?

Thanks

Answer 1

$$ \mathrm{Var}(X+Y) = \mathrm{Var(X)}+\mathrm{Var}(Y)+2\mathrm{Cov}(X,Y) \\ \mathrm{Var}(Z) = 1+2+2\cdot\frac{-1}{2} = 2 $$ Also there is an error in your calculation of $E[Y^2]$. $$ E[Y^2] = \mu^2+\sigma^2 = 1+2 = 3 $$ (You probably squared the $2$ again confusing it with $\sigma$, but it's actually $\sigma^2$)

Answer 2

$E(X+Y) = E(X)+E(Y) = 1+1=2$.

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