Could there exist a distribution, other than standard Laplace (probability density of the form $1/2e^{-|x|}$), on $\mathbb{R}$ such that $E[x]=0,E[|x|]=1$ and that \begin{equation*} E[|x-a|] \geq |a|+e^{-|a|} \end{equation*} for all $a \in \mathbb{R}$?


1 Answer 1


Consider a ReflectedGamma($b$, $c$) distribution (also known as a double Gamma distribution) with pdf $f(x)$:

$$f(x) = \frac{ |x|^{c-1} \space e^{-|x|/b}}{ 2 \space \Gamma (c) \space \space b^c} \quad \quad \text{for } x \in \mathbb{R} $$

... but where the usual Gamma parameters $b$ and $c$ are chosen such that $b = \frac{1}{c}$. By this construction, the first two conditions are satisfied for all values of parameter $c$, namely that:

  • $E\big[X \big] = 0$

  • $E\big[\space|X| \space\big] = 1$

When $c = 1$, we obtain the special case of the standard Laplace distribution that is the reference subject of this question.

The following diagram plots our constructed ReflectedGamma pdf for different values of parameter $c$ (including the reference case of the Laplace when $c = 1$). Of note is the changing shape when $c < 1$ or $c > 1$:

enter image description here

Entering the pdf as:

enter image description here

... we can find $E\big[ |X-a| \big]$ as:

enter image description here

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

The following plot compares:

  • the derived expectation $E\big[ |X-a| \big]$ for our ReflectedGamma($c$) distribution (ORANGE curve) when $c <1$ (here with $c = \frac{1}{10}$)

  • with the required lower bound $|a|+e^{-|a|}$ (BLUE curve) relating to the standard Laplace $(c=1)$:

enter image description here

As per the diagram (and numerically checked), $E\big[ |X-a| \big]$ for the ReflectedGamma (orange) EXCEEDS the Laplace lower bound for all real values of $a$, when parameter $c < 1$.

The answer is thus: YES - there do exist distributions (other than standard Laplace) that satisfy the required properties.

Formal Proof

The simplest counterexample I have found is from the family of Reflected Lomax distributions:

$$f(x) = \frac{1}{(1+|x|)^3} \quad \quad \text{for } x \in \mathbb{R} $$

for which $E\big[ |X-a| \big] = |a|+\frac{1}{1+|a|}$. Assuming $a > 0$ without loss of generality (due to symmetry), we then have to prove that:

$$a+\frac{1}{1+a} \geq a + e^{-a}$$

or equivalently that $e^a \geq 1+a$ where the latter is a well-known exponential inequality. All done.

  • $\begingroup$ Thanks for the answer. Could you write out the double gamma density and the expected absolute deviation from a in Latex. I haven't used mathematica so am unable to quite get the expression. Do you think you could turn this into a formal proof? $\endgroup$ Commented Aug 19, 2023 at 20:21
  • 2
    $\begingroup$ @SushantVijayan The density in In[1]:= is $\dfrac{c^c}{2 \Gamma(c)}|x|^{c-1} e^{-c|x|}$ with $0< c \le 1$, and is your Laplace distribution when $c=1$ $\endgroup$
    – Henry
    Commented Aug 19, 2023 at 20:49
  • $\begingroup$ It has a mean of $0$ and average absolute deviation of $1$ as requested, but has a variance of $1+\frac1c$ and becomes extremely leptokurtic as $c \to 0^+$ $\endgroup$
    – Henry
    Commented Aug 19, 2023 at 22:11
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    $\begingroup$ @SushantVijayan Strictly speaking leptokursis is really about the tails of the distribution. Even so, in your Laplace distribution (with variance $2$) almost $1\%$ of the distribution is in the interval $[-0.01,0.01]$. With $c=\frac1{10}$ (despite variance increasing to $11$) more than half the distribution is in the interval $[-0.01,0.01]$, so it becomes more concentrated around the mean. Meanwhile in your Laplace distribution about $0.67\%$ of the distribution is above $5$ or below $-5$ while with $c=\frac{1}{10}$ about $5.86\%$ is that extreme, showing the heavier tails. $\endgroup$
    – Henry
    Commented Aug 19, 2023 at 22:43
  • 1
    $\begingroup$ @SushantVijayan I have added a formal proof. $\endgroup$
    – wolfies
    Commented Aug 20, 2023 at 20:04

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