# Moments of Laplace distribution

I am a newbie in stat. I am working on the Laplace distribution for my algorithm.

• Could tell me the first what the four moments of the Laplace distribution are?
• Does it have infinite tail like the Cauchy distribution?
• What is the empirical rule?
• One ring to rule them all: Wikipedia: at the side you have a table holding mean, variance, skewness and kurtosis. – Nick Sabbe May 19 '11 at 15:27
• Try Wikipedia? Moments. You can also use the moment-generating function. :) – cardinal May 19 '11 at 15:29
• @Nick, I did not mean to discourage you from answering, either. Please consider submitting an answer detailing your favorite technique (or two) for deriving the results. It may be instructive for the OP and others. :) – cardinal May 19 '11 at 16:46
• @crucified I would like to know what is meant by "the empirical rule." Could it perhaps refer to unbiased estimators of the moments? – whuber May 19 '11 at 19:00

Here is a quick check using a symbolic algebra package ...

Let $$X \sim \text{Laplace}(\mu, \sigma)$$ with pdf $$f(x)$$:

Then, the first 4 raw moments $$E[X^i]$$ are given by:

where I am using the Expect function from the mathStatica package for Mathematica.

It is worth noting that the $$3^\text{rd}$$ and $$4^\text{th}$$ raw moments are different to those given in the answer above.

• +1 -- but the second moment is the same as given previously. – whuber Feb 1 at 17:13

It's been awhile and looks like this hasn't been answered. I'll provide one and hopefully we can mark this as correct. I'll answer in order the questions asked using the parameterization of the wikipedia page

$$f(x\mid\mu,b)= \frac{1}{2b} \exp \left( -\frac{|x-\mu|}{b} \right), x\in \mathbb{R}.$$

1. For the case $$\mu=0$$, the first four moments are: $$\mathbb{E}(X)=0, \mathbb{E}(X^2)=2b^2 + \mu^2, \mathbb{E}(X^3)=0, and\ \mathbb{E}(X^4) = 24b^4.$$ As whuber indicates in a comment you can related a non-central random variable $$Y$$ via a binomial expansion of $$Y^k=(Xb+\mu)^k$$. The value $$\mu=0$$ is often chosen to simplify the calculation and to build up to the solution.

2. The Laplace have infinite tails like the Cauchy, the support is $$x \in (-\infty, \infty)$$.

3. For the empirical rule, I'm assuming the OP is using the shorthand for the probability of observations within $$\sigma$$ of the mean, $$\mu$$, $$2\sigma$$ of $$\mu$$ and $$2\sigma$$ of $$\mu$$ respectively. These probabilities are: (0.75688, 0.94089, 0.98563) to 5 significant digits, respectively.

A couple of different ways to calculate the expected values are:

a. Direct integration (usually split the integral at the point $$\mu$$ where the sign changes.

b. Differentiate the moment generating function $$m$$ times and set $$t=0$$ to get the $$m$$th moment.

c. Formulate the Laplace random variable (r.v.) as a scale mixture of Normal and Exponential random variable. Then use conditional expectations.

Note approach (c) is only easier than (a) if you know the moments of Normal and Exponential random variables-or can calculate them easier than directly calculating the moments of the Laplace distribution

• Are you sure the value of the fourth moment is correct? – user144410 Jan 30 at 17:32
• @user144410, I think so. How I got that was to calculate the 4th derivative of the moment generating function (MGF) w.r.t. $t$. The MGF is $(1-b^2t^2)^{-1}$ and the fourth derivative of the MGF is: $$\frac{-24b^4(5b^4t^4 +10b^2t^2+1)}{(b^2t^2-1)^5}.$$ Put $t=0$ into this to get $24b^4$ (negative signs in numerator and denominator cancel). The fourth moment (raw) is different from Kurtosis and different from excess kurtosis. – Lucas Roberts Feb 1 at 1:00
• This is not what you have in your answer. – user144410 Feb 1 at 8:18
• ah that is a typo, yeah, thanks for the call out! The power in the fourth moment should be 4. Fixed now. – Lucas Roberts Feb 1 at 16:11
• The higher moments are clearly wrong: after all, when $|\mu|$ is large and $b$ is small, the (non-central) moment of order $k$ will be approximately $\mu^k,$ but $\mu$ doesn't even appear in your expression for $E(X^4)$! You have also introduced "$\sigma$" as an (undefined) synonym for "$b.$" A quick and easy way to check is to derive the non-central moments for the case $\mu=0,b=1$ (the odd ones are obviously zero, making this pretty easy) and then use the Binomial Theorem to expand $(b X+\mu)^k.$ – whuber Feb 1 at 20:11