Let us assume we have two continuous random variables X1 and X2 and the correlation coefficient rho.
The first question is about the following assumptions: If we know that the marginals of each RV are a normal distribution and we also know that the correlation coefficient is rho can we conclude that the joint density function is the bivariate normal density function?
The second question is the reverse of the first one: if we assume that the joint density function of X1 and X2 is the bivariate normal density function with some coefficient rho then can we conclude that the marginals of X1 and X2 are normal distributions?
Thanks,
Bogdan.
Best Answer
Short answer: (1) No, (2) Yes (refer to Wikipedia: Multivariate normal distribution)
For (1) all you need is a counterexample. There are many different possibilities. Say, suppose you already have a normal $X_1$. Then you flip a coin and if it lands head you take $X_2 = X_1$, whereas if it lands tails you take $X_2 = -X_1$. It's not hard to show that this $X_2$ is normal: all it does is randomly flips the sign of another normal random variable $X_1$. What about the join distribution? The pair $(X_1,X_2)$ takes values on two crossing lines $y=\pm x$ -- this is definitely not a bivariate normal! As for the correlation coefficient $\rho$, in this example it will be $0$ if the coin is fair. However if the coin is not fair, then the correlation can be anything from $1$ (if the coin lands only on head) to $-1$ (if it lands only tails). That is, picking a right coin will give you whatever coefficient $\rho$ you need.
As for the second question, the answer is Yes, almost by definition. If you have a bivariate normal $(X_1,X_2)$, then its single component, say $X_1$, can be obtained as a scalar product of $(X_1,X_2)$ and the unit vector $(1,0)$. By definition of a multivariate normal, this scalar product must be normally distributed (as a reminder, the definition says that a variable is multivariate normal if its scalar product with any vector has univariate normal distribution).