Lower bound of an expectation

Question

Suppose a random variable $X$ has unit variance i.e. $\sigma^{2} = 1$. Is there a positive constant $c > 0$ such that $$\mathbb{E}[\ | X - \mathbb{E}[X] | \ ] \ge c $$

My attempt of a solution is as follows:

Let $Y = X-\mathbb{E}(X)$. Then by Chebyshev's inequality, for any $k > 0$, $$P(|Y-\mathbb{E}[Y] | \geq k) \leq 1/k^{2} \ \ (1) $$

Also by the triangle inequality,

$$P(|Y-\mathbb{E}[Y]| \geq k) \geq P(|Y| \geq k) - P(|\mathbb{E}[Y]| \geq k) $$

which implies

$$P(|Y-\mathbb{E}[Y]| \geq k) + P(|\mathbb{E}[Y]| \geq k) \geq P(|Y| \geq k) $$

or $$1/k^2 + \epsilon \geq P(|Y| \geq k) $$

where $\epsilon \in [0, 1]$.

From this statement, can it be concluded that $0$ is the greatest lower bound of $\mathbb{E}[|Y|]$?

Answer 1

I am not sure if I understood your question correctly, but for each random variable $X$ with positive variance, there of course is such a constant $c(X)$ depending on $X$.

However, there is no uniform constant that works for all random variables with unit variance, as the following counterexample shows.

For $n \in \mathbb{N}$ consider a random variable $X_n$ with $\mathbb{P}[X_n = n] = \mathbb{P}[X_n = -n] = \frac{1}{2n^2}$ and $\mathbb{P}[X_n = 0] = 1 - \frac{1}{n^2}$. Then we have $\mathbb{E}[X_n]=0$ and $\mathbb{E}[X_n^2]=1$ for all $n$, but also $\mathbb{E}[\lvert X_n \rvert] = \frac{1}{n}$. Therefore there cannot be a uniform lower bound $c$ without any additional assumptions.