= Laplace distribution
{c}
{title2=$\operatorname{Laplace}(\mu,b),\quad b>0$}
{wiki}
= Double exponential distribution
{synonym}
The <Laplace distribution>, also called the <double exponential distribution>, is a <location-scale family> with <probability density function>
$$
f(x;\mu,b)=\frac1{2b}\exp\!\left(-\frac{|x-\mu|}{b}\right),\qquad x\in\mathbb R,\quad\mu\in\mathbb R,\quad b>0.
$$
The location $\mu$ is its <mean>, <median> and <mode>; $b$ controls its spread. In the centred rate convention, the same law has $\mu=0$ and $\theta=1/b>0$, giving $f(x)=\theta e^{-\theta|x|}/2$. Keeping track of whether a parameter is a scale or its reciprocal prevents reversing the effect of increasing it. <NIST and SEMATECH e-Handbook of Statistical Methods/1.3.6.6.12. Double Exponential Distribution>[NIST and SEMATECH e-Handbook of Statistical Methods; NIST/SEMATECH e-Handbook, Section 1.3.6.6.12, Probability Density Function and Common Statistics; NIST denotes the scale by beta]{ref}
The two halves of the <probability density function> are reflected <exponential decay> curves, each with total <probability> $1/2$. Their meeting point has a sharp corner: the left and right <derivatives> are $1/(2b^2)$ and $-1/(2b^2)$. This finite peak is not a positive <probability mass> at $\mu$. Indeed
$$
\mathbb P(|X-\mu|\leq\varepsilon)=1-e^{-\varepsilon/b}\longrightarrow0\qquad(\varepsilon\downarrow0),
$$
so $\mathbb P(X=\mu)=0$, just as for any <absolutely continuous probability distribution>.
\Image[/laplace-density-and-log-tail-comparison.png]
{title=Laplace and normal densities and two-sided tails with the same mean and variance}
The figure compares a <Laplace distribution> with $\mu=0$, $b=1$ against a <normal distribution> with <mean> zero and <standard deviation> $\sigma=\sqrt2$: both have <variance> 2. The right panel uses a <logarithmic scale> for the two-sided tail <probability> $\mathbb P(|X|>r)$. Its straight <Laplace distribution> curve comes from $e^{-r}$; the <normal distribution> tail falls faster. Thus values far from the center can be more plausible under a <Laplace distribution> even after matching the <variance>. These are theoretical curves, not measured frequencies. The tails still have <exponential decay>; this is not a <heavy-tailed distribution> in the convention requiring every positive <exponential moment> to diverge.
Integrating either half of the <probability density function> gives the <cumulative distribution function>
$$
F(x)=\begin{cases}
\tfrac12e^{(x-\mu)/b},&x<\mu,\\
1-\tfrac12e^{-(x-\mu)/b},&x\geq\mu.
\end{cases}
$$
In particular $F(\mu)=1/2$. The basic <moments> are
$$
\mathbb E[X]=\mu,\qquad \mathbb E|X-\mu|=b,\qquad \operatorname{Var}(X)=2b^2,\qquad \operatorname{SD}(X)=\sqrt2\,b.
$$
Thus the scale is the <expected value> of the absolute distance from the center, and is not the <standard deviation>. <NIST and SEMATECH e-Handbook of Statistical Methods/1.3.6.6.12. Double Exponential Distribution>[NIST and SEMATECH e-Handbook of Statistical Methods; NIST/SEMATECH e-Handbook, Section 1.3.6.6.12, Cumulative Distribution Function and Common Statistics; rescale the standard form by x=mu+bz]{ref} A short way to obtain the <moments> is to note that $R=|X-\mu|$ has an <exponential distribution> with rate $1/b$: $\mathbb P(R>r)=e^{-r/b}$ for $r\geq0$. Integrating gives $\mathbb E R=b$ and $\mathbb E R^2=2b^2$; the <symmetric distribution> gives the centered first <moment> zero.
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