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 $$\mathbb{E}(|X-\mu_X|)>\mathbb{E}(|Y-\mu_Y|).$$

• Hints: (1) simplify, with no loss in generality, by taking $\mu_X=\mu_Y=0.$ (2) Variance calculations emphasize more extreme results compared to mean absolute values. Thus, give $Y$ some outliers compared to $X.$
– whuber
Jun 16, 2020 at 14:03

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
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.