Chebyshev inequality for $n=1$?

Question

Wikipedia suggests that Chebyhev's inequality is only true for $n \ge 2$, but I don't see why we have to exclude the case $n=1$? Is wikipedia right? Chebyshev

Answer 1

There's no reason n can't be 1. It would just make the inequality a little meaningless, though. Basically you're saying at least $1 - \frac{1}{n^2} = 0$ of the values lie within a single standard deviation of the mean. That means there could be anything of zero and upward, which is an all-inclusive, and hence basically meaningless, statement.

Values of n greater than 1 (even those less than 2) are perfectly meaningful.

Answer 2

The way that I would write Chebyshev's inequality is

$$\Pr(|X|>t) \geq \frac{E|X|^2}{t^2},$$

which can be generalized to hold for any $r>0$; it then becomes the Markov inequality.

As for the problem with respect to Wikipedia, it is still true that for $r<2$,

$$\Pr(|X-\mu|>k\sigma) \geq \frac{E|X-\mu|^r}{k^r\sigma^r},$$

but $E|X-\mu|^r$ is not generally greater or equal than $\sigma^r$ because $t^{2/r}$ is not concave.

Answer 3

In fact the Chebyshev inequality is much broader and it states that if $\phi$ is any positive function ($\phi \ge 0$) and $i_A = \inf\{\phi(y) : y\in A\}$ then

$$i_AP(X\in A)\le E\phi(X) $$

To see this, just observe that

$$i_A1_{\{X\in A\}} \le \phi(X)1_{\{X\in A\}}$$ Taking the expected value in both sides and forgetting the conditioned expected value

$$i_AP(X\in A)\le E\phi(X) $$

Now, if you let $\phi(y) = |y|$ and $A = (t,\infty)$ ($i_A = t)$ applying this to the random variable $|X|$ you have

$$tP(|X| > t) \le E|X| \iff P(|X| > t) \le \frac{E|X|}{t}$$

Observe that you can apply this to functions like $\phi(y) = y^n$ to get that

$$P(|X| > t) \le \frac{E|X|^n}{t^n}$$