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

Somebody asked me this question in a job interview and I replied that their joint distribution is always Gaussian. I thought that I can always write a bivariate Normal with their means and variance and covariances. I am wondering if there can be a case for which the join probability of two Gaussians is not Gaussian?

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– user10525 Jun 9 '12 at 23:30
– Stéphane Laurent Jun 10 '12 at 9:01

## 2 Answers

The bivariate normal distribution is the exception, not the rule!

It is important to recognize that "almost all" joint distributions with normal marginals are not the bivariate normal distribution. That is, the common viewpoint that joint distributions with normal marginals that are not the bivariate normal are somehow "pathological", is a bit misguided.

Certainly, the multivariate normal is extremely important due to its stability under linear transformations, and so receives the bulk of attention in applications.

Examples

It is useful to start with some examples. The figure below contains heatmaps of six bivariate distributions, all of which have standard normal marginals. The left and middle ones in the top row are bivariate normals, the remaining ones are not (as should be apparent). They're described further below.

The bare bones of copulas

Properties of dependence are often efficiently analyzed using copulas. A bivariate copula is just a fancy name for a probability distribution on the unit square $[0,1]^2$ with uniform marginals.

Suppose $C(u,v)$ is a bivariate copula. Then, immediately from the above, we know that $C(u,v) \geq 0$, $C(u,1) = u$ and $C(1,v) = v$, for example.

We can construct bivariate random variables on the Euclidean plane with prespecified marginals by a simple transformation of a bivariate copula. Let $F_1$ and $F_2$ be prescribed marginal distributions for a pair of random variables $(X,Y)$. Then, if $C(u,v)$ is a bivariate copula, $$F(x,y) = C(F_1(x), F_2(y))$$ is a bivariate distribution function with marginals $F_1$ and $F_2$. To see this last fact, just note that $$\renewcommand{\Pr}{\mathbb P} \Pr(X \leq x) = \Pr(X \leq x, Y < \infty) = C(F_1(x), F_2(\infty)) = C(F_1(x),1) = F_1(x) \>.$$ The same argument works for $F_2$.

For continuous $F_1$ and $F_2$, Sklar's theorem asserts a converse implying uniqueness. That is, given a bivariate distribution $F(x,y)$ with continuous marginals $F_1$, $F_2$, the corresponding copula is unique (on the appropriate range space).

The bivariate normal is exceptional

Sklar's theorem tells us (essentially) that there is only one copula that produces the bivariate normal distribution. This is, aptly named, the Gaussian copula which has density on $[0,1]^2$ $$c_\rho(u,v) := \frac{\partial^2}{\partial u \partial v} C_\rho(u,v) = \frac{\varphi_{2,\rho}(\Phi^{-1}(u),\Phi^{-1}(v))}{\varphi(\Phi^{-1}(u)) \varphi(\Phi^{-1}(v))} \>,$$ where the numerator is the bivariate normal distribution with correlation $\rho$ evaluated at $\Phi^{-1}(u)$ and $\Phi^{-1}(v)$.

But, there are lots of other copulas and all of them will give a bivariate distribution with normal marginals which is not the bivariate normal by using the transformation described in the previous section.

Some details on the examples

Note that if $C(u,v)$ is am arbitrary copula with density $c(u,v)$, the corresponding bivariate density with standard normal marginals under the transformation $F(x,y) = C(\Phi(x),\Phi(y))$ is $$f(x,y) = \varphi(x) \varphi(y) c(\Phi(x), \Phi(y)) \> .$$

Note that by applying the Gaussian copula in the above equation, we recover the bivariate normal density. But, for any other choice of $c(u,v)$, we will not.

The examples in the figure were constructed as follows (going across each row, one column at a time):

1. Bivariate normal with independent components.
2. Bivariate normal with $\rho = -0.4$.
3. The example given in this answer of Dilip Sarwate. It can easily be seen to be induced by the copula $C(u,v)$ with density $c(u,v) = 2 (\mathbf 1_{(0 \leq u \leq 1/2, 0 \leq v \leq 1/2)} + \mathbf 1_{(1/2 < u \leq 1, 1/2 < v \leq 1)})$.
4. Generated from the Frank copula with parameter $\theta = 2$.
5. Generated from the Clayton copula with parameter $\theta = 1$.
6. Generated from an asymmetric modification of the Clayton copula with parameter $\theta = 3$.
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+1 for the remark that the bivariate normal density is the exceptional case! – Dilip Sarwate Jun 10 '12 at 23:40
+1, great answer. – NRH Jun 11 '12 at 14:47

It is true that each element of a multivariate normal vector is itself normally distributed, and you can deduce their means and variances. However, it is not true that any two Guassian random variables are jointly normally distributed. Here is an example:

Edit: In response to the consensus that a random variable that is a point mass can be thought of as a normally distributed variable with $\sigma^2=0$, I'm changing my example.

Let $X \sim N(0,1)$ and let $Y = X \cdot (2B-1)$ where $B$ is a ${\rm Bernoulli}(1/2)$ random variable. That is, $Y = \pm X$ each with probability $1/2$.

We first show that $Y$ has a standard normal distribution. By the law of total probability,

$$P(Y \leq y) = \frac{1}{2} \Big( P(Y \leq y | B = 1) + P(Y \leq y | B = 0) \Big)$$

Next,

$$P(Y \leq y | B = 0) = P(-X \leq y) = 1-P(X \leq -y) = 1-\Phi(-y) = \Phi(y)$$

where $\Phi$ is the standard normal CDF. Similarly,

$$P(Y \leq y | B = 1) = P(X \leq y) = \Phi(y)$$

Therefore,

$$P(Y \leq y) = \frac{1}{2} \Big( \Phi(y) + \Phi(y) \Big) = \Phi(y)$$

so, the CDF of $Y$ is $\Phi(\cdot)$, thus $Y \sim N(0,1)$.

Now we show that $X,Y$ are not jointly normally distributed. As @cardinal points out, one characterization of the multivariate normal is that every linear combination of its elements is normally distributed. $X,Y$ do not have this property, since

$$Y+X = \begin{cases} 2X &\mbox{if } B = 1 \\ 0 & \mbox{if } B = 0. \end{cases}$$

Therefore $Y+X$ is a $50/50$ mixture of a $N(0,4)$ random variable and a point mass at 0, therefore it cannot be normally distributed.

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I don't agree with this answer. A degenerate point mass of $1$ at $\mu$ is usually considered to be a degenerate Gaussian random variable with zero variance. Also,$(X, -X)$ are not jointly continuous though they are marginally continuous. For an example of two jointly continuous random variables that are marginally Gaussian but not jointly Gaussian, see, for example, the latter half of this answer. – Dilip Sarwate Jun 10 '12 at 0:02

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– Macro Jun 10 '12 at 0:04

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A standard characterization of the multivariate Gaussian is that $X \in \mathbb R^{n}$ is multivariate Gaussian if and only if $a^T X$ is Gaussian for all $a \in \mathbb R^n$. As @Dilip hints at, it is worth considering if this true for your example. – cardinal Jun 10 '12 at 0:12

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Since you apparently don't like appeals to rationality ;-), how about appeals to authority? (That's a joke, if it's not apparent.) I just happened upon this purely by accident as I was looking something else up: Example 2.4, page 22 of G. A. F. Seber and A. J. Lee, Linear Regression Analysis, 2nd. ed., Wiley. It quoteth: "Let $Y \sim \mathcal N(\mu,\sigma^2)$ and put $\mathbf Y' = (Y, -Y)$...Thus $\mathbf Y$ has a multivariate normal distribution." – cardinal Jun 10 '12 at 0:51

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The discussion is about definitions. Clearly, if the covariance matrix by definition is required to be non-singular Macro provides an example, but this is not a example according to the more liberal definition that @cardinal refers too. One good reason to prefer the more liberal definition is that then all linear transformations of normal variables are normal. In particular, in linear regression with normal errors the residuals have a joint normal distribution but the covariance matrix is singular. – NRH Jun 10 '12 at 20:10
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