Variance and covariance inequality

Question

Given a real-valued random variable $X$, is

$$2\mathbb E[X] \mathrm{Var}(X) \geq \mathrm{Cov}(X, X^2)$$

true?

Any pointers for how to tackle this problem would be immensely helpful.

Answer 1

Use the variance and covariance identities $$\text{Var}(X)=\mathbb E[X^2] − \mathbb E[X]^2$$ and $$\text{Cov}(X,Y) = \mathbb E[XY] − \mathbb E[X] \mathbb E[Y]$$

Then the given inequality is equivalent to

$$ 2 \mathbb E[X] \Big( \mathbb E[X^2] - \mathbb E[X]^2 \Big) \overset{?}{\geq} \mathbb E[X^3] - \mathbb E[X] \mathbb E[X^2] \tag{1}$$

For a counterexample, let $X \sim \text{Exp}(\lambda)$ be the exponential distribution with parameter $\lambda \in (0, \infty)$ and probability mass function

$$ f_X(x) = \begin{cases} \lambda e^{-\lambda x} & \text{if } x \geq 0 \\ 0 & \text{if } x < 0 \end{cases} $$

Its $n$-th moment is given by $\mathbb E[X^n]=\frac{n!}{\lambda^n}$, so the inequality $(1)$ becomes

$$2 \cdot \frac{1}{\lambda} \Big( \frac{2}{\lambda^2} - \frac{1}{\lambda^2} \Big) \geq \frac{6}{\lambda^3} - \frac{1}{\lambda} \cdot \frac{2}{\lambda^2}$$

which reduces to

$$\frac{2}{\lambda^3} \geq \frac{4}{\lambda^3}$$

This is false for all $\lambda \in (0, \infty)$, hence we found a counterexample.

Answer 2

Alternate approach / observation...

Assume $2 E[X] Var(X) \ge Cov(X, X^2)$ for any real-valued r.v. $X.$ Then in particular it would also be true for $-X$, so that:

$$ \begin{align} 2E[-X] Var(-X) &\ge Cov(-X, (-X)^2)\\ -2E[X] Var(X) &\ge -Cov(X, X^2)\\ 2E[X] Var(X) &\le Cov(X, X^2) \end{align} $$

which, combined with the original inequality, means: $2E[X]Var(X) = Cov(X, X^2)$ for all $X$.

At this point, it would seem one should be able to find a counter-example easily, without using e.g. $Cov(X, X^2) = E[X^3] - E[X]E[X^2]$ etc, but I just couldn't think of a quick one. :(