# Joint distribution of sum of independent normals

Suppose we have three independent normally distributed random variables $$X_0 \sim \mathcal{N}(\mu_0, \sigma_0^2),$$ $$X_1 \sim \mathcal{N}(\mu_1, \sigma_1^2),$$ $$X_2 \sim \mathcal{N}(\mu_2, \sigma_2^2).$$

Now, define two new random variables $Y_0 = X_0+X_1$ and $Y_1 = X_1+X_2$.

Let $\vec{Y} = [Y_0 \;\;\; Y_1]^T$

What can we say about the distribution of $\vec{Y}$? Obviously, $Y_0$ and $Y_1$ are not independent. If they were, then $\vec{Y}$ would have been a multivariate normal variable. Any ideas?

• Yes, $Y$ is a multivariate random vector. This is a duplicate question addressed repeatedly on both math.SE and stats.SE. But, the Wikipedia article linked to should suffice. – cardinal Jan 4 '12 at 1:20
• possible duplicate of What is the distribution of the sum of non i.i.d. gaussian variates? – cardinal Jan 4 '12 at 1:24
• @cardinal I think you meant to say "Yes, $\vec{Y}$ is a multivariate normal random variable". That $\vec{Y}$ is a multivariate random vector is a tautology. But this is an issue that comes up repeatedly on math.SE as well as stats.Se where many assume that marginally normal automatically means jointly normal as well and many assume (as eakbas has done) that joint normality requires independence. The overlap between the two groups of people may be large too. – Dilip Sarwate Jan 4 '12 at 2:36
• @eakbas: You supplied the additional assumption that $X_0, X_1,$ and $X_2$ were independent. This is sufficient to guarantee the multivariate normality of $Y$. (Read the second point under definition on the Wiki page.) – cardinal Jan 4 '12 at 2:55
• Thanks @cardinal, I missed the "second point under definition" part in your previous answer. Now it's clear. – emrea Jan 4 '12 at 3:15

Not entirely clear to me from reading the comments if the OP has solved this but there is no answer so I will write one.

The distribution of each $Y_i$ will be normal with given means and variances:

$\mu_0+\mu_1$ and $\sigma_0^2+\sigma^2_1$ for $Y_0$ and

$\mu_1+\mu_2$ and $\sigma_1^2+\sigma^2_2$ for $Y_1$. Now finally we need to determine if there is a correlation between $Y_0$ and $Y_1$. To do this we can calculate

$$\mathbb{C}ov(Y_0,Y_1)=\mathbb{C}ov(X_0+X_1,X_1+X_2) =\mathbb{C}ov(X_1,X_1) =\mathbb{V}ar(X_1) =\sigma_1^2.$$ Now you can turn this into a correlation by dividing by the square roots of the variances

$$\rho = \frac{\sigma_1^2}{\sqrt{(\sigma_0^2+\sigma^2_1)(\sigma_1^2+\sigma^2_2)} }.$$

Now we know that the sum of two normal random variables is normally distributed so that both $Y_0$ and $Y_1$ have normal distributions with the stated means and variances and the correlation is given by $\rho$ above. So the joint density of $Y_0, Y_1$ is

$$f(y_0,y_1) = N\left(\vec{\mu} = \begin{bmatrix} \mu_0+\mu_1 \\ \mu_1+\mu_2 \\ \end{bmatrix}, \Sigma = \begin{bmatrix} \sigma^2_0+\sigma^2_1 &\sigma_1^2 \\ \sigma_1^2 & \sigma^2_1+\sigma^2_2 \\ \end{bmatrix} \right).$$