1

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.

Marc van Leeuwen
  • 115,048
Bogdan
  • 33
  • No. And yes. // Which texts are your references on gaussian random variables? – Did Mar 19 '13 at 08:40
  • What do you mean by this question ? – Bogdan Mar 19 '13 at 08:54
  • The question is clear, let me repeat it: your post mentions bivariate normal densities, which texts are your references on this subject? This piece of information could help to understand why you have a problem at all with the second question (note the two answers already posted, basically saying they do not see the problem you have with (2)). – Did Mar 19 '13 at 12:01

2 Answers2

1

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).

Pasha
  • 271
0

As a counterexample for the first part, consider a joint distribution where the conditional distribution for $X_2$ is the upper half of a normal distribution when $X_1$ is above the mean and the lower part of that normal distribution when $X_1$ is below the mean.

The answer to the second part is affirmative; you get this directly by integrating out one of the variables.

joriki
  • 238,052
  • Thanks for your answer but I didnt understood your example. Can you please give me a numerical one? – Bogdan Mar 19 '13 at 08:50
  • @Bogdan: What do you mean by a numerical example? This is a concrete joint distribution function that's not a bivariate normal distribution but has normal marginal distributions. In case by "numerical example" you mean formulas instead of words, you can take the density to be proportional to

    $$ n(x_1)n(x_2)(\Theta(x_1)\Theta(x_2)+(1-\Theta(x_1))(1-\Theta(x_2));, $$

    where $n$ is a normal distribution with zero mean and $\Theta$ is the Heaviside step function.

     – joriki Mar 19 '13 at 08:55