# What are the exact conditions to get a multivariate gaussian distribution from multiple single gaussian variables?

I would like to get a precise answer to this question, I can't seem to find a clear answer anywhere.

I have $m$ gaussian random variables $X_i$ ($i=1,...,m$), which are dependent because they are defined by

$$X_1 = X - a_1\\ X_2 = X - a_2\\ \vdots\\ X_m = X - a_m,$$

where the $a_i$ are real positive constants and $X$ is a random variable following $N(\mu, \sigma^2)$.

So they all have the same variance, but not the same mean. Is their joint distribution a multivariate gaussian? And if not, what can I say about their joint distribution?

Thanks a lot.

• In your special case the joint distribution has a 1 dimensional line as it's support – Calvin Khor Jan 22 '18 at 10:05
• Yes, their joint distribution is a multivariate gaussian. – drhab Jan 22 '18 at 10:29
• Ok thanks, but why? And @CalvinKhor, what do you mean by "a 1 dimensional line"? – Ségo Jan 22 '18 at 10:54
• @Ségo The image of the joint variable $(X_1,\dots,X_m)$ forms a line. Hence the pushforward measure is a (singular wrt lebesgue) measure supported on this line. – Calvin Khor Jan 22 '18 at 22:21

In general if $X:\Omega\to\mathbb R^n$ is a random vector with normal distribution where $\mathsf EX=\mu\in\mathbb R^n$ and $\mathsf{Covar}X=\Sigma$ then $AX+v$ where $A^{m\times n}$ is a matrix with entries in $\mathbb R$ and $v\in\mathbb R^m$ also has normal distribution.

This with $\mathsf E(AX+v)=A\mu+v$ and $\mathsf{Cov}(AX+v)=A\Sigma A^T$.

Now apply this for $n=1$ and $A^{m\times1}=(1,\dots,1)^T$ and $v=(-a_1,\dots,-a_m)^T$ and you are ready.

• Ok I got it, thanks a lot! I just have one last question. In my case, the variance matrix $\Sigma$ has all of its terms equal to $\sigma^2$ (variance of $X$), right? This makes it singular and I can't write the joint density. Do you know how to treat that case? How could I get an expression for my density? Even an approximated one would be nice! – Ségo Jan 22 '18 at 12:43
• If $m>1$ then you will just have to accept that no density exists. But who cares? The distribution of the vector is completely determined by the distribution of random variable $X$ which is known – drhab Jan 22 '18 at 12:54
• But then how would I actually compute a probability of the joint variable? Some like, for instance, $P((X-a_1,...,X-a_m) > (1,...,1))$ – Ségo Jan 22 '18 at 13:24
• I guess I can just put the $a_i$ on the other side of the inequality, actually – Ségo Jan 22 '18 at 13:25
• By reducing it to $P(X>\max(a_1+1,\dots,a_m+1))$ – drhab Jan 22 '18 at 13:26