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

Moreover, what about this special case :

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.

  • $\begingroup$ In your special case the joint distribution has a 1 dimensional line as it's support $\endgroup$ – Calvin Khor Jan 22 '18 at 10:05
  • $\begingroup$ Yes, their joint distribution is a multivariate gaussian. $\endgroup$ – drhab Jan 22 '18 at 10:29
  • $\begingroup$ Ok thanks, but why? And @CalvinKhor, what do you mean by "a 1 dimensional line"? $\endgroup$ – Ségo Jan 22 '18 at 10:54
  • $\begingroup$ @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. $\endgroup$ – 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.

  • $\begingroup$ 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! $\endgroup$ – Ségo Jan 22 '18 at 12:43
  • $\begingroup$ 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 $\endgroup$ – drhab Jan 22 '18 at 12:54
  • $\begingroup$ But then how would I actually compute a probability of the joint variable? Some like, for instance, $P((X-a_1,...,X-a_m) > (1,...,1))$ $\endgroup$ – Ségo Jan 22 '18 at 13:24
  • $\begingroup$ I guess I can just put the $a_i$ on the other side of the inequality, actually $\endgroup$ – Ségo Jan 22 '18 at 13:25
  • $\begingroup$ By reducing it to $P(X>\max(a_1+1,\dots,a_m+1))$ $\endgroup$ – drhab Jan 22 '18 at 13:26

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