# $X+Y \sim N(5,5)$ two random variables normal bi-variate

The exercise is the following. Let $$(X,Y)$$ be a bi-variate normal vector with $$\mu=(2,3)$$ and covariance matrix $$\Sigma=\begin{pmatrix}1 & -2 \\ -2 & 4\end{pmatrix}.$$ What is the distribution of the random variable $$X+Y$$?

My try is: Consider the moment generating functions of $$X$$ and $$Y$$. Since $$\sigma_X^2=1$$, $$\sigma_Y^2=4$$, $$\mu_X=2$$ and $$\mu_Y=3$$, we have $$\varphi_X(t)=\exp(2t+\frac{t^2}{2})$$ $$\varphi_Y(t)=\exp(3t+2t^2).$$

Thus, $$\varphi_{X+Y}(t)=\varphi_X(t)\varphi_Y(t)=exp(5t+\frac{5t^2}{2}).$$

Hence, $$X+Y\sim N(5,5)$$ since the moment generating function of a normal r.v. is given by $$\varphi_X(t)=\exp(\mu_X t + \frac{\sigma_X^2 t^2}{2})$$.

I am asking this because my teacher put as a solution $$X+Y\sim N(5,1)$$. Any feedback?

EDIT:

I think that we can not use $$\varphi_{X+Y}(t)=\varphi_X(t)\varphi_Y(t)$$ since $$X$$, $$Y$$ are not independent.

The variance of $$X+Y$$ is $$\operatorname{Var}(X)+\operatorname{Var}(Y)+2\operatorname{Cov}(X,Y)=1+4+2\times(-2)=1.$$ So $$N(5,1)$$ is correct. Your problem is the assertion $$\varphi_{X+Y}(t)=\varphi_X(t)\varphi_Y(t)$$ which doesn't work in this case because $$X$$ and $$Y$$ are dependent.
$$X+Y \sim N(\mu_x + \mu_y, \sigma_x^2 + \sigma_y^2 +2 \sigma_{x,y})$$