Joint distribution of sum of independent normals

Question

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](http://en.wikipedia.org/wiki/Multivariate_normal_distribution). This is a duplicate question addressed repeatedly on both math.SE and stats.SE. But, the Wikipedia article linked to should suffice. — cardinal, Jan 04 '12 at 01:20
possible duplicate of [What is the distribution of the sum of non i.i.d. gaussian variates?](http://stats.stackexchange.com/questions/19948/what-is-the-distribution-of-the-sum-of-non-i-i-d-gaussian-variates) — cardinal, Jan 04 '12 at 01:24
@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 04 '12 at 02:36
Yes, as the page linked to suggests, the word *normal* got inadvertently dropped when creating the link text. Thanks for catching this. (And, it would be a bit unusual, in general, to refer to $Y$ as a multivariate normal random *variable*.) — cardinal, Jan 04 '12 at 02:49
Why do you say $\vec{Y}$ is multivariate normal? The fact that $Y_0$ and $Y_1$ both are Gaussian does not imply that $\vec{Y}$ is Gaussian. See http://en.wikipedia.org/wiki/Multivariate_normal_distribution#Two_normally_distributed_random_variables_need_not_be_jointly_bivariate_normal — emrea, Jan 04 '12 at 02:51
@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](http://en.wikipedia.org/wiki/Multivariate_normal_distribution#Definition) on the Wiki page.) — cardinal, Jan 04 '12 at 02:55
You're right, they're not. But that's irrelevant. The whole point of a multivariate normal is to have a vector of random variables that are each marginally normal, but that are not independent of one another. Look at the definition on the Wikipedia page and the form of the density. See the $\Sigma$? That's a covariance matrix. It codifies the dependence between the elements of $Y$. Can you find the $\Sigma$ that corresponds to your example? — cardinal, Jan 04 '12 at 03:02
Thanks @cardinal, I missed the "second point under definition" part in your previous answer. Now it's clear. — emrea, Jan 04 '12 at 03:15
**Good**. I'm glad things have clarified for you. Cheers. :) — cardinal, Jan 04 '12 at 03:41

score 1 · Accepted Answer · answered Jul 24 '17 at 01:30

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

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

Joint distribution of sum of independent normals

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