Random variables $(X,Y)$ with $\text{Var}(X)<\text{Var}(Y)$ and $\mathbb{E}(|X-\mu_X|)>\mathbb{E}(|Y-\mu_Y|)$ I am looking for an example of a pair of random variables $(X,Y)$ with expected values $(\mu_X,\mu_Y)$ satisfying the following relationships:
$$
\text{Var}(X)<\text{Var}(Y)
$$ and
$$
\mathbb{E}(|X-\mu_X|)>\mathbb{E}(|Y-\mu_Y|).
$$
 A: Consider a Laplace distribution with scale parameter equal to $1$. One can compute that the MAD equals $1$, and the variance equals $2$.
Now consider a Normal distribution with arbitrary mean, and with variance equal to $1.9$.  One can compute the MAD to be
$$\sqrt{\frac{1.9 \cdot 2}{\pi}} = \sqrt{\frac{3.8}{\pi}} > \sqrt{\frac{3.8}{3.2}} > 1,$$
using that $0 < \pi < 3.2$. As such,

*

*the variance is less than that of our Laplace distribution, but

*the MAD is greater.

A: Consider a discrete random variable $Y$ that, for some constants $M\in\mathbb{R}$ and $\epsilon\in [0, 1]$, takes the following values:
$$
Y = \left\{\begin{array} ~-M+\mu_Y & \text{w.p.}~\epsilon/2 \\ \mu_Y & \text{w.p.}~1-\epsilon \\ M+\mu_Y & \text{w.p.}~\epsilon/2\end{array}\right.
$$
Simple calculations show that $Y$ has the desired mean $\mu_Y$, $Var(Y) = M^2\epsilon$, and $E[|Y-\mu_Y|] = M\epsilon$. Taking $M = \epsilon^{-0.75}$ and $\epsilon\rightarrow 0$, we have $Var(Y)\rightarrow\infty$ and $E[|Y-\mu_Y|]\rightarrow 0$. So by selecting small enough $\epsilon$, we can obtain a $Y$ with the desired mean, arbitrarily large $Var(Y)$, and arbitrarily small $[|Y-\mu_Y|]$.
For instance, with $\epsilon=10^{-12}$ and $M = 10^9$, we have $Var(Y) = 10^6$ and $E[|Y-\mu_Y|] = 0.001$.
So now we have a wealth of options for $X$ that will cause the desired inequalities to hold. An example would be, a normal distribution with mean $\mu_X$ and variance 1.
