It is apparently the case that if $X_i \sim N(0,1)$, then

$X_1 X_2 + X_3 X_4 \sim \mathrm{Laplace(0,1)}$

I've seen papers on arbitrary quadratic forms, which always results in horrible non-central chi-squared expressions.

The above simple relationship does not seem at all obvious to me, so (if it is true!) does anyone have a simple proof of the above?


2 Answers 2


An elementary sequence of steps using well-known relationships among distributions and a simple algebraic polarization identity provide an elementary and intuitive demonstration.

I have found this polarization identity generally useful for reasoning about, and computing with, products of random variables, because it reduces them to linear combinations of squares. It is a bit like working with matrices by diagonalizing them first. (There's more than a superficial connection here.)

A Laplace distribution is a difference of two Exponentials (which intuitively makes some sense, because an Exponential is a "half-Laplace" distribution). (The link demonstrates this by manipulating characteristic functions, but the relation can be proven using an elementary integration following from the definition of a difference as a convolution.)

An Exponential distribution (which itself is a $\Gamma(1)$ distribution) is also a (scaled version of a) $\chi^2(2)$ distribution. The scale factor is $1/2$. This can easily be seen by comparing the PDFs of the two distributions.

$\chi^2$ distributions are obtained naturally as sums of squares of iid Normal distributions (having zero means). The degrees of freedom, $2$, count the number of Normal distributions in the sum.

The algebraic relation

$$X_1X_2 + X_3X_4 = \left[\left(\frac{X_1+X_2}{2}\right)^2 + \left(\frac{X_3+X_4}{2}\right)^2\right] - \left[\left(\frac{X_1-X_2}{2}\right)^2 + \left(\frac{X_3-X_4}{2}\right)^2\right]$$

exhibits $X_1X_2 + X_3X_4$ in terms of squares of four distributions, each of which is a linear combination of standard Normals. It is easy to check that all four linear combinations are linearly independent (and each follows a Normal$(0,\sqrt{1/2})$ distribution). Thus the first two terms, which sum the squares of two identically distributed Normal distributions of mean zero, form a scaled $\chi^2(2)$ distribution (and its scale factor of $\sqrt{1/2}\ ^2=1/2$ is exactly what is needed to make it an Exponential distribution) and the second two terms independently have an Exponential distribution, too, for the same reason.

Therefore $X_1X_2+X_3X_4$, being the difference of two independent Exponential distributions, has a (standard) Laplace distribution.

  • 4
    $\begingroup$ That is absolutely delightful! $\endgroup$
    – Corvus
    Sep 26, 2013 at 15:35
  • 2
    $\begingroup$ I just noticed that another answer, based on moment generating functions, appears at stats.stackexchange.com/a/51717/919: see the paragraph in the middle beginning "incidentally" (another name for the Laplace distribution is "bi-exponential"). That thread concerns the MGF of a generalization of the present question. $\endgroup$
    – whuber
    Sep 26, 2013 at 17:25
  • $\begingroup$ Nice derivation, but how do you know that the difference of two independent exponential distributed variables has a Laplacian distribution? $\endgroup$ Sep 24, 2016 at 14:14
  • $\begingroup$ @Hello Please follow the link: it goes to a Wikipedia article that includes a brief demonstration. $\endgroup$
    – whuber
    Sep 24, 2016 at 14:50

$X\sim \mathrm{Laplace}(0,1)$ has characteristic function $$ \phi_X(t) = \frac{1}{1+t^2} $$ which is the square of the characteristic function of a product standard normal (see https://math.stackexchange.com/questions/74013/characteristic-function-of-product-of-normal-random-variables). The claim follows by the fact that sums of independent random variables relate to products of characteristic functions.


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