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 Gaussian with their means and variance and covariances. I am wondering if there can be a case for which the joint probability of two Gaussians is not Gaussian?


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




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.

Examples of bivariate distribution with standard normal marginals.

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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    $\begingroup$ +1 for the remark that the bivariate normal density is the exceptional case! $\endgroup$ – Dilip Sarwate Jun 10 '12 at 23:40
  • $\begingroup$ Maybe I am missing something, but if we start from $X_1, X_2\sim\mathcal N(0,1)$, the joint distribution $(X_1, X_2)$ is automatically defined, independently of any copula construction, and if we apply a non-Gaussian copula construction to their CDFs, it is true that we shall obtain a non-Gaussian CDF $F(x_1,x_2)$, but this function in general will not be the CDF of the pair of random variables $X_, X_2$ we started with, right? $\endgroup$ – RandomGuy Jan 6 '17 at 11:06
  • $\begingroup$ Example of how to simulate as in lower right panel: library(copula) kcf <- khoudrajiCopula(copula2 = claytonCopula(6), shapes = fixParam(c(.4, 1), c(FALSE, TRUE))) # force normal margins evil <- mvdc(kcf, c("norm", "norm"), list(list(mean = 0, sd =1), list(mean = 0, sd = 1))) contour(evil, dMvdc, xlim = c(-3, 3), ylim=c(-3, 3)) $\endgroup$ – half-pass Dec 29 '18 at 18:29
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    $\begingroup$ @RandomGuy, You’re missing an unstated assumption that $X_1, X_2 \sim independent N(0, 1)$. If you assume they’re independent, then yes, you know the joint distribution already. Without the independence assumption, knowing the marginal distributions does not give enough information to specify the joint distribution. $\endgroup$ – MentatOfDune Feb 19 '19 at 15:41

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,

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


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

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

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


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

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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    $\begingroup$ 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. $\endgroup$ – Dilip Sarwate Jun 10 '12 at 0:02
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    $\begingroup$ @DilipSarwate, the question was to give an example (if one exists) of two variables that are normally distributed but their joint distribution is not multivariate normal. This is an example. Most standard definitions of the normal distribution (e.g. wikipedia en.wikipedia.org/wiki/Normal_distribution) require the variance the be strictly positive, thus not including a point mass as part of the family of normal distributions. $\endgroup$ – Macro Jun 10 '12 at 0:04
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    $\begingroup$ 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. $\endgroup$ – cardinal Jun 10 '12 at 0:12
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    $\begingroup$ 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." $\endgroup$ – cardinal Jun 10 '12 at 0:51
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    $\begingroup$ 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. $\endgroup$ – NRH Jun 10 '12 at 20:10

The following post contains an outline of a proof, just to give the main ideas and get you started.

Let $z = (Z_1, Z_2)$ be two independent Gaussian random variables and let $x = (X_1, X_2)$ be $$ x = \begin{pmatrix} X_1 \\ X_2 \end{pmatrix} = \begin{pmatrix} \alpha_{11} Z_1 + \alpha_{12} Z_2\\ \alpha_{21} Z_1 + \alpha_{22} Z_2 \end{pmatrix} = \begin{pmatrix} \alpha_{11} & \alpha_{12}\\ \alpha_{21} & \alpha_{22} \end{pmatrix} \begin{pmatrix} Z_1 \\ Z_2 \end{pmatrix} = A z. $$

Each $X_i \sim N(\mu_i, \sigma_i^2)$, but as they are both linear combinations of the same independent r.vs, they are jointly dependent.

Definition A pair of r.vs $x = (X_1, X_2)$ are said to be bivariate normally distributed iff it can be written as a linear combination $x = Az$ of independent normal r.vs $z = (Z_1, Z_2)$.

Lemma If $ x = (X_1, X_2)$ is a bivariate Gaussian, then any other linear combination of them is again a normal random variable.

Proof. Trivial, skipped to not offend anyone.

Property If $X_1, X_2$ are uncorrelated, then they are independent and vice-versa.

Distribution of $X_1 | X_2$

Assume $X_1, X_2$ are the same Gaussian r.vs as before but let's suppose they have positive variance and zero mean for simplicity.

If $\mathbf S$ is the subspace spanned by $X_2$, let $ X_1^{\mathbf S} = \frac{\rho \sigma_{X_1}}{\sigma_{X_2}} X_2 $ and $ X_1^{\mathbf S^\perp} = X_1 - X_1^{\mathbf S} $.

$X_1$ and $X_2$ are linear combinations of $z$, so $ X_2, X_1^{\mathbf S^\perp}$ are too. They are jointly Gaussian, uncorrelated (prove it) and independent.

The decomposition $$ X_1 = X_1^{\mathbf S} + X_1^{\mathbf S^\perp} $$ holds with $\mathbf{E}[X_1 | X_2] = \frac{\rho \sigma_{X_1}}{\sigma_{X_2}} X_2 = X_1^{\mathbf S}$

$$ \begin{split} \mathbf{V}[X_1 | X_2] &= \mathbf{V}[X_1^{\mathbf S^\perp}] \\ &= \mathbf{E} \left[ X_1 - \frac{\rho \sigma_{X_1}}{\sigma_{X_2}} X_2 \right]^2 \\ &= (1 - \rho)^2 \sigma^2_{X_1}. \end{split} $$

Then $$ X_1 | X_2 \sim N\left( X_1^{\mathbf S}, (1 - \rho)^2 \sigma^2_{X_1} \right).$$

Two univariate Gaussian random variables $X, Y$ are jointly Gaussian if the conditionals $X | Y$ and $Y|X$ are Gaussian too.

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    $\begingroup$ It isn't apparent how this observation answers the question. Since the product rule is practically the definition of conditional distribution, it's not special to binormal distributions. The subsequent statement "then in order..." doesn't provide any reason: exactly why must the conditional distributions also be normal? $\endgroup$ – whuber Jan 11 '16 at 0:12
  • $\begingroup$ whuber, I am answering to the main question: "I am wondering if there can be a case for which the joint probability of two Gaussians is not Gaussian?". So, the answer is: when the conditional is not normal. - Ancillary $\endgroup$ – ancillary Jan 11 '16 at 13:02
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    $\begingroup$ Could you complete that demonstration? Right now it's only an assertion on your part, with no proof. It's not at all evident that it's correct. It's also incomplete, because you need to establish existence: that is, you have to demonstrate it's actually possible for a joint distribution to have normal marginals but for which at least one conditional is non-normal. Now in fact that's trivially true, because you may freely alter each conditional distribution of a binormal on a set of measure zero without changing its marginals--but that possibility would seem to contradict your assertions. $\endgroup$ – whuber Jan 11 '16 at 13:24
  • $\begingroup$ Hi @whuber, I hope this helps more. Do you have any suggestions or edits to do? I wrote this very quickly as at the moment I haven't got much spare time :-) but I would value any suggestion or improvement you can make. Best $\endgroup$ – ancillary Aug 30 '16 at 11:57
  • $\begingroup$ (1) What are you trying to prove? (2) Because the question asks when a distribution with Gaussian marginals is not jointly Gaussian, I don't see how this argument is leading to anything relevant. $\endgroup$ – whuber Aug 30 '16 at 13:39

Jointly gaussian implies marginally gaussian but not vice versa.

Independent and marginally gaussian does imply jointly gaussian.

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    $\begingroup$ This does not add anything to the already existing answers! $\endgroup$ – kjetil b halvorsen Dec 29 '19 at 15:35
