Laplace Absolute Deviations and Central Mass

Distance From the Laplace Center

For a Laplace distribution, the scale parameter has a direct interpretation in terms of distance from the center. It is the expected absolute deviation from that center. At the same time, the distribution places a particularly simple amount of probability within any symmetric interval around it. These facts make the scale useful for describing both typical deviations and central coverage.

Let $X$ have a Laplace distribution with location $\mu\in\mathbb{R}$ and scale $b>0$. Its density is

f(x)=\frac{1}{2b}\exp\left(-\frac{|x-\mu|}{b}\right),\qquad -\infty<x<\infty.

The location $\mu$ is the center of the density. To study distances from it, define the nonnegative random variable $D=|X-\mu|$. The central-mass question asks for $P(D\le c)$, where $c\ge 0$. The absolute-deviation question asks for $E[D]$.

Definition: The mean absolute deviation about the center of a Laplace variable is $E[|X-\mu|]$. Its median absolute deviation about the center is any median of $|X-\mu|$. These are distances and hence are nonnegative; they are distinct from the location $\mu$, which is a signed value on the original measurement scale.

Central Probability and Absolute-Deviation Distribution

A symmetric interval from $\mu-c$ to $\mu+c$ consists of two pieces, one on each side of the center. Integrating the density across both pieces gives the central probability.

Theorem: If $X$ has density $f(x)=(2b)^{-1}\exp(-|x-\mu|/b)$, with $b>0$, then for every $c\ge 0$, $$ P(|X-\mu|\le c)=1-\exp(-c/b). $$ Equivalently, $P(|X-\mu|>c)=\exp(-c/b)$.

Proof. For $c\ge 0$, integrate the density over the central interval. On the left of $\mu$, $|x-\mu|=\mu-x$; on the right, $|x-\mu|=x-\mu$. Therefore

\begin{aligned} P(|X-\mu|\le c) &=\int_{\mu-c}^{\mu}\frac{1}{2b}e^{-(\mu-x)/b}\,dx +\int_{\mu}^{\mu+c}\frac{1}{2b}e^{-(x-\mu)/b}\,dx\\ &=\frac{1}{2}(1-e^{-c/b})+\frac{1}{2}(1-e^{-c/b})\\ &=1-e^{-c/b}. \end{aligned}

When $c=0$, both integrals are zero and the formula gives zero, as required for a continuous distribution. The probability outside the interval is the complement, $e^{-c/b}$. This proves both statements. $\square$

The result also gives the distribution of the absolute deviation itself. For $d\ge 0$, $P(D\le d)=1-e^{-d/b}$; for $d<0$, that probability is zero. Thus $D$ has density $b^{-1}e^{-d/b}$ for $d>0$. The two sides of the Laplace density together produce this one-sided distribution of distances.

Worked Example: Probability Within a Fixed Distance

Suppose $X\sim\operatorname{Laplace}(12,3)$, so its location is 12 and its scale is 3. Find the probability that $X$ lies within 4 units of its center.

Here $c=4$ and $b=3$. The central-mass formula gives

P(|X-12|\le 4)=1-e^{-4/3}.

Numerically, $e^{-4/3}\approx 0.2636$, so the probability is approximately $1-0.2636=0.7364$. Equivalently, the interval is $[8,16]$, and about $73.64\%$ of the probability lies in it. The probability outside this interval is $e^{-4/3}\approx 0.2636$, split equally between the two tails by symmetry.

Mean, Median, and Mode

The location parameter is simultaneously the mean, median, and mode of this Laplace distribution. The mean absolute deviation about that common center is the scale $b$. The median absolute deviation is slightly different: it is $b\ln 2$. The two distance summaries describe different points in the distribution of $D$, rather than different centers on the original scale.

Theorem: For a Laplace variable with location $\mu$ and scale $b>0$, the mean, median, and mode of $X$ are all $\mu$, while $$ E[|X-\mu|]=b,\qquad \operatorname{median}(|X-\mu|)=b\ln 2. $$

Proof. First, the density is symmetric about $\mu$. The integral defining the mean exists: on either side, the absolute first moment is bounded by an integral of a linear function times an exponentially decaying density, which is finite. By symmetry, the contributions to $E[X-\mu]$ from equal distances on opposite sides cancel, so $E[X]=\mu$.

The density increases up to $\mu$ from the left and decreases after $\mu$ to the right, since it is proportional to $e^{-|x-\mu|/b}$. Its unique maximum is therefore at $\mu$, which is the mode. For the median, symmetry gives $P(X\le\mu)=1/2$. Moreover, the density is positive everywhere, so the cumulative distribution function is strictly increasing; no other point can also have cumulative probability $1/2$. Thus the median is $\mu$.

It remains to calculate the absolute deviations. Using the density and symmetry,

\begin{aligned} E[|X-\mu|] &=2\int_0^\infty t\frac{1}{2b}e^{-t/b}\,dt\\ &=\frac{1}{b}\int_0^\infty t e^{-t/b}\,dt\\ &=b\int_0^\infty u e^{-u}\,du\\ &=b. \end{aligned}

In the third line, $u=t/b$, so $t=bu$ and $dt=b\,du$. The remaining integral equals 1 by integration by parts: $\int_0^\infty u e^{-u}\,du=[-(u+1)e^{-u}]_0^\infty=1$.

Finally, the distribution function of $D=|X-\mu|$, for $d\ge 0$, is $1-e^{-d/b}$. Its median $m_D$ solves $P(D\le m_D)=1/2$, so

1-e^{-m_D/b}=\frac12 \quad\Longrightarrow\quad e^{-m_D/b}=\frac12 \quad\Longrightarrow\quad m_D=b\ln 2.

The distribution of $D$ is continuous and strictly increasing on positive distances, so this median is unique. The mean, median, mode, and both deviation formulas are established. $\square$

Worked Example: Recovering the Scale From Central Coverage

Let $X\sim\operatorname{Laplace}(-2,5)$. Find a symmetric interval about its center that contains $80\%$ of the probability.

Write the interval as $[-2-c,-2+c]$. Set the central probability equal to $0.80$ and solve for $c$:

1-e^{-c/5}=0.80 \quad\Longrightarrow\quad e^{-c/5}=0.20 \quad\Longrightarrow\quad c=-5\ln(0.20)=5\ln 5\approx 8.047.

Thus the interval is approximately $[-10.047,6.047]$. Substitution checks the result: $1-e^{-8.047/5}\approx 1-0.20=0.80$. The scale also tells us that the mean absolute deviation from $-2$ is 5, while the median absolute deviation is $5\ln 2\approx 3.466$.

Using the Scale to Describe Deviations

The central-mass formula can be inverted. For a desired central probability $q$, where $0\le q<1$, the radius $c_q$ satisfying $P(|X-\mu|\le c_q)=q$ is

c_q=-b\ln(1-q).

In particular, multiplying the scale by a fixed constant always gives the same central probability, regardless of the location and scale of the original measurement. For example, radius $b$ captures $1-e^{-1}$, or about $63.21\%$, of the probability. Radius $b\ln 2$ captures $1/2$, which is why it is the median absolute deviation. As $q$ approaches 1, the required radius grows without bound; no finite interval captures all the probability.

Worked Example: Mean and Median Absolute Deviations

A Laplace measurement has location $\mu=30$ and scale $b=7$. Find its mean absolute deviation about the center, its median absolute deviation, and the radius of a central interval containing $95\%$ of the probability.

The mean absolute deviation is exactly the scale:

$$ E[|X-30|]=7. $$

The median absolute deviation is

m_D=7\ln 2\approx 7(0.693147)=4.852.

For $95\%$ central coverage, set $q=0.95$:

c_{0.95}=-7\ln(1-0.95)=7\ln 20\approx 20.970.

The corresponding interval is approximately $[30-20.970,30+20.970]=[9.030,50.970]$. Its probability is $1-e^{-20.970/7}$, and since $20.970/7$ is approximately $\ln 20$, this is $1-e^{-\ln 20}=1-1/20=0.95$. This verifies the coverage.

Common Mistakes and Interpretation

A frequent error is to treat $e^{-c/b}$ as the probability inside the central interval. It is the probability outside that interval. The central probability is its complement, $1-e^{-c/b}$. Another error is to identify the location $\mu$ with the scale $b$: $\mu$ locates the distribution on the measurement scale, while $b$ controls distances from that location.

The three summaries should also not be confused. The mean absolute deviation $b$ is an average distance; the median absolute deviation $b\ln 2$ is a median distance; and the median and mode of $X$ itself are both $\mu$. These distinctions are useful when describing the center and spread of a Laplace model or translating a desired central probability into an interval width.

Key takeaway: For a Laplace variable, $P(|X-\mu|\le c)=1-e^{-c/b}$. The center $\mu$ is the mean, median, and mode; the mean absolute deviation is $b$, and the median absolute deviation is $b\ln 2$.

Check Your Understanding

Use the definitions and results above to answer these questions.

If $X$ has location $\mu$ and scale $b$, what is $P(|X-\mu|>2b)$?
For a Laplace variable with scale 4, find the mean absolute deviation and the median absolute deviation about its center.
A symmetric interval about the center contains $90\%$ of a Laplace distribution. Express its radius in terms of $b$.
Explain why the mean, median, and mode of $X$ are location parameters, whereas $E[|X-\mu|]$ is a scale summary.
Someone claims that $e^{-c/b}$ is the probability within distance $c$ of the center. State the correction and the corresponding outside probability.