# Determine the density of sum of three normal variables.

Setting

$\pmb{X} = (X_1,X_2,X_3)$ is a properly center normal with covariance matrix

$$\begin{pmatrix} a & b & 0\\ b & d & 0\\ 0 & 0 & e \end{pmatrix}$$

Determine the density of $Y = X_1 + 2 X_2 - X_3$

So I see $X_1 \overset{d}{\sim} \mathcal{N}(0,a)$, $X_2 \overset{d}{\sim} \mathcal{N}(0,d)$, and $X_3 \overset{d}{\sim} \mathcal{N}(0,e)$.

But the non-zero covariance between $X_1$ and $X_2$ is really throwing me off, please explain?

• Are you sure you have transcribed all the numbers properly? The covariance matrix that you have written down says that cov$(X_1,X_2)= 4$ while var$(X_1)=3$, var$(X_2)=5$ which makes the correlation coefficient $$\rho_{1,2}=\frac{\operatorname{cov}(X_1,X_2)}{\sqrt{\operatorname{var}(X_1) \operatorname{var}(X_2)}} = \frac{4}{\sqrt{3\times 5}} > 1.$$ – Dilip Sarwate Dec 18 '14 at 21:46
• Alternatively, $$\operatorname{var}(X_1-X_2)=\operatorname{var}(X_1) + \operatorname{var}(X_2) - 2\cdot \operatorname{cov}(X_1,X_2) = 3+5-2\cdot 4 = 0$$ showing that something is awry. – Dilip Sarwate Dec 18 '14 at 23:41
• You're right the numbers are not correct, I made an edit for a generic covariance matrix – chibro2 Dec 18 '14 at 23:42

Hint: \begin{align*} \operatorname{Var}[Y] &= \operatorname{Var}[X_1 + 2X_2 - X_3] \\ &= \operatorname{Var}[X_1] + 4\operatorname{Var}[X_2] + \operatorname{Var}[X_3] + 4 \operatorname{Cov}[X_1, X_2] - 2 \operatorname{Cov}[X_1, X_3] - 4 \operatorname{Cov}[X_2, X_3] \end{align*}