Are its marginal distributions Laplace?

Yes, for the symmetric family described here.

Are uncorrelated coordinates independent?

Not in the nondegenerate example.

Are asymmetric variants available?

Yes; their parameterization must be specified separately.

AI Glossary / Statistics and probability / Advanced

Multivariate Laplace distribution

Definition

A multivariate Laplace distribution extends Laplace-type probability modeling to several variables. This entry focuses on the symmetric, elliptically contoured family, whose linear projections have univariate Laplace distributions.

What it means

The model describes joint variation with more pronounced tails than a comparable normal model. Its coordinates can have zero correlation while remaining dependent. Specifying Laplace marginal distributions alone does not identify this joint distribution; a vector of independent Laplace variables is a different construction.

How it works

One representation is X = mu + sqrt(W) Z, where Z is a zero-mean multivariate normal vector with covariance Sigma, W is independent exponential with mean one, and mu is a fixed location vector. The same W scales every coordinate in a draw.
Under this convention, X has mean mu and covariance Sigma. Conditional on W, the vector is normal; averaging over W produces the Laplace family. Asymmetric extensions require additional parameters and should not be confused with simply shifting mu.

Three practical examples

A fictional sensor study compares a shared-scale Laplace error model with a normal model for two simultaneous measurements.
A fictional simulation exercise generates dependent variables with zero correlation to test an analyst's independence assumptions.
A fictional signal-processing team checks whether linear combinations of residuals resemble the proposed model before using it for simulation.

Think of it this way

Imagine two independent noise sources connected to one randomly changing volume control. Their signs need not move together, but a high-volume setting makes larger magnitudes more plausible for both. Sharing the volume control creates a relationship that ordinary correlation can miss.

Worked example

Take an original two-dimensional model with mu=(0, 0) and Sigma equal to the identity matrix. Generate independent standard normal variables Z1 and Z2, and independently generate W from an exponential distribution with mean one. Set X1=sqrt(W) Z1 and X2=sqrt(W) Z2.
For an illustrative single draw, suppose W=4 and the normal values are Z1=0.5 and Z2=-1. Then the result is X=(1, -2). If the same normal values were paired with W=0.25 instead, the result would be (0.25, -0.5). These are arithmetic illustrations of the transformation, not representative frequencies.
The exponential variable has E(W)=1 and E(W squared)=2. Each standard normal has mean zero, second moment one, and fourth moment three. Independence between W and the normal vector makes the following calculations straightforward.
E(X1 squared)=E(W) E(Z1 squared)=1. The same holds for X2. Also E(X1 X2)=E(W) E(Z1 Z2)=0, so the covariance and correlation of the two coordinates are zero.
Now examine their squared magnitudes. E(X1 squared times X2 squared)=E(W squared) E(Z1 squared) E(Z2 squared)=2. If X1 and X2 were independent, this expectation would instead equal the product of their second moments, which is one. The disagreement proves dependence in this example.
More precisely, the covariance of their squared magnitudes is two minus one, or one. Each coordinate's fourth moment is E(W squared) E(Z1 to the fourth power)=2 times 3=6. Its squared magnitude therefore has variance six minus one, or five. The correlation between X1 squared and X2 squared is consequently 1/5=0.2.
This distinguishes two statements: the signed measurements have correlation zero, while their squared magnitudes have correlation 0.2. Neither number should be substituted for the other.
Consider S=X1+X2. Conditional on W, S is normal with mean zero and variance 2W. Its unconditional variance is two. The shared exponential mixture gives a univariate Laplace distribution with scale b=1, using the density exp(-absolute(s)/b)/(2b).
For this sum, the probability that its absolute value exceeds three is exp(-3), approximately 0.04979, or 4.98 percent. Each individual coordinate instead has scale 1/sqrt(2), since its variance is one and a Laplace variable in this convention has variance 2b squared.
A matched normal sum with variance two has probability about 0.03389 of exceeding three in absolute value. The two models agree on the sum's mean and variance but disagree on that event's probability. This comparison is specific to the stated threshold and parameters.
For another check, the joint characteristic function of the example is 1 divided by (one plus (t1 squared plus t2 squared)/2). This follows by averaging exp(-W(t1 squared plus t2 squared)/2) against the exponential density.
At t1=t2=1, its value is one half. Multiplying the two marginal characteristic functions instead gives (2/3) times (2/3)=4/9. Their inequality is another way to see that the coordinates are not independent.
All these calculations describe the chosen model. They do not establish that a real dataset follows it.

Illustrative case study

Fictional company Willow Signals considers this model for paired sensor residuals. The analysts first check signed correlation, then examine squared residuals and several linear combinations.
They compare simulated summaries with held-out observations and retain the shared-scale interpretation only if the evidence supports it. Their report explicitly states the symmetric model and its parameter convention. This fictional case illustrates model checking, not a claim that sensor errors generally follow a Laplace law.

Common misconceptions and mistakes

Zero correlation is insufficient for independence in the worked example.
Independent Laplace coordinates do not reproduce the shared-scale construction.
A location shift changes the center without introducing asymmetry.
Matching covariance does not guarantee matching tail probabilities.
The symbol used for scale must be defined: b in the example is not a standard deviation.

Business use

A joint error model can influence simulated combinations of outcomes. Before using the fictional sensor model operationally, compare its predictions with observed joint behavior and document where it fails, especially for combinations that matter to the decision.

Check your understanding

What creates dependence in the example?

Answer: The shared random W.

What is the correlation of the coordinates?

Answer: Zero.

What is the correlation of their squares?

Answer: 0.2.

Related terms

Multivariate normal distribution
Elliptical distribution
Characteristic function
Covariance
Laplace distribution
scale mixture
statistical independence

Learn next

Compare shared-scale and independent-coordinate simulations, then examine multivariate asymmetric Laplace models.

Sources

Kotz, Kozubowski, and Podgorski, Symmetric Multivariate Laplace Distribution, publisher preview pages 231-232: https://page-one.springer.com/pdf/preview/10.1007/978-1-4612-0173-1_6 . Geraci and Cortina Borja, Notebook: The Laplace Distribution: https://academic.oup.com/jrssig/article/15/5/10/7029313 . Calculations and company scenario are original.

Previous: Multivariate kernel density estimation Next: Multivariate model