Tell me more ×
Cross Validated is a question and answer site for statisticians, data analysts, data miners and data visualization experts. It's 100% free, no registration required.

Possible Duplicate:
Is it possible to have a pair of Gaussian random variables for which the joint distribution is not Gaussian?

For our upcoming exam we had to calculate the joint density of two normally distributed random variables a few times.

Say $X \sim N(0,1)$ and $Y \sim N(2,3)$.

We just assumed that $Z=(X,Y)^\top $ have a joint bivariate normal distribution $ Z \sim N\left( \left(\begin{array}{c} 0\\ 2 \end{array}\right) , \left(\begin{array}{cc} 1 & a\\ a & 3 \end{array}\right) \right) $ and calculated the only missing parameter, the covariance $a$ of the two random variables.

Now, wikipedia says that this is not always true and brings a counter example. Here is the image of the example's resulting plot.

My question: Under which conditions are they jointly normally distributed?

share|improve this question
1  
Related: stats.stackexchange.com/q/30159/2970 – cardinal Jul 30 '12 at 18:17
1  
The condition is that every linear combination of $X$ and $Y$ is normally distributed. The wikipedia page includes this characterization under the Definition subsection. – cardinal Jul 30 '12 at 18:20
1  
Given $X\sim N(0,1)$ and $Y \sim N(2,3)$ and that the random variables are jointly normal, how did you calculate the covariance $a$? There must have been some other information provided to you that enabled you to calculate $a$ because given just that $X\sim N(0,1)$ and $Y \sim N(2,3)$ (whether jointly normal or not), $a$ can have any value in $[\sqrt{3}, -\sqrt{3}]$. – Dilip Sarwate Jul 30 '12 at 18:31
3  
Also of interest: surprising characterizations of the Gaussian distribution – whuber Jul 30 '12 at 18:47
Dilip, I didn't calculate $a$, and there normally is more information. I just made up a few quick numbers :) – Alexx Hardt Jul 30 '12 at 18:49
show 3 more comments

marked as duplicate by cardinal, gung, Andy W, whuber Dec 8 '12 at 15:27

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

1 Answer

You should read up on copulas! That will give you a way to construct as many counterexamples as you wish. You can start with http://en.wikipedia.org/wiki/Copula_(probability_theory)

share|improve this answer
Indeed, this is taken up, to a certain degree, in the answer at the first link in the comments. :-) – cardinal Aug 8 '12 at 23:40

Not the answer you're looking for? Browse other questions tagged or ask your own question.