Expected absolute deviation greater than standard Laplace

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 $$a \in \mathbb{R}$$?

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 $$c > 1$$:

Entering the pdf as:

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

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 $$(c=1)$$:

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.

• 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? Commented Aug 19, 2023 at 20:21
• @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 $c=1$ Commented Aug 19, 2023 at 20:49
• 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 $c \to 0^+$ Commented Aug 19, 2023 at 22:11
• @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. Commented Aug 19, 2023 at 22:43
• @SushantVijayan I have added a formal proof. Commented Aug 20, 2023 at 20:04