9
$\begingroup$

I am interested in constructing random variables for which Markov or Chebyshev inequalities are tight.

A trivial example is the following random variable.

$P(X=1)=P(X=-1) = 0.5$. Its mean is zero, variance is 1 and $P(|X| \ge 1) = 1$. For this random variable chebyshev is tight (holds with equality).

$$P(|X|\ge 1) \le \frac{\text{Var}(X)}{1^2} = 1$$

Are there more interesting(non-uniform) random variables for which Markov and Chebyshev are tight ? Some examples would be great.

$\endgroup$
5
$\begingroup$

The class of distributions for which the limiting case of the Chebyshev bound holds is well known (and not that hard to simply guess). Normalized for location and scale it is

$$Z=\begin{cases}-k,&{\text{with probability }}{\:\frac {1}{2k^2}}\\\:0,&{\text{with probability }}1-\frac {1}{k^2}\\\:k,&{\text{with probability }}{\:\frac {1}{2k^2}}\end{cases}$$

This is (up to scale) the solution given at the Wikipedia page for the Chebyshev inequality.

[You can write a sequence of distributions (by placing $\epsilon>0$ more probability at the center with the same removed evenly from the endpoints) that strictly satisfy the inequality and approach that limiting case as closely as desired.]

Any other solution can be obtained by location and scale shifts of this: Let $X=\mu+\sigma Z$.

For the Markov inequality, let $Y=|Z|$ so you have probability $1-1/k^2$ at 0 and $1/k^2$ at $k$. (One can introduce a scale parameter here but not a location parameter)

Chebyshev and Markov limiting cases

Moment inequalities - and indeed many other similar inequalities - tend to have discrete distributions as their limiting cases.

| cite | improve this answer | |
$\endgroup$
2
$\begingroup$

I believe that getting a continuous distribution over the whole real axis that follows Chebyshev's bound exactly may be impossible.

Assume that a continuous distribution's mean and standard deviation are 0 and 1, or make it so via rescaling. Then require $P(\mid X\mid >x)=1/x^2$. For simplicity consider $x>0$; the negative values will be defined symmetrically. Then the CDF of the distribution is $1-1/x^2$. And so the pdf, the derivative of the cdf, is $2/x^3$. Obviously this must be defined only for $x>0$ because of the discontinuity. In fact, this can't even be true everywhere, or the integral of the pdf is not finite. Instead, if discontinuities are to be avoided (e.g. the pdf cat just be 0 for $\mid x \mid <\alpha$) the pdf must be piecewise, equal to $\mid x \mid^3$ for $\mid x\mid \geq \alpha$.

However, this distribution fails the hypothesis - it does not have finite variance. To get a continuous distribution over the real axis with a finite variance, the expected values of $x$ and $x^2$ must be finite. Examining inverse polynomials, tails that go like $x^{-3}$ lead to a finite $E[x]$, but an undefined $E[x^2]$ because this involves an integral with asymptotically logarithm behavior.

So, Chebychev's bound can't be satisfied exactly. You can require $P(\mid X \mid > x ) = x^{-(2+\epsilon)}$ for arbitrarily small $\epsilon$, however. The tail of the pdf goes like $x^{-(3+\epsilon)}$ and has a defined variance on the order of $1/\epsilon$.

If you're willing to let the distribution live on only part of the real line, but still be continuous, then defining $pdf(x) = 2/\mid x \mid^3$ for $\epsilon < \mid x \mid < \Lambda$ works for $$ \epsilon = \sqrt{2 \left( 1 - \frac{1}{\sqrt{e}} \right) } $$ and $$ \Lambda = \epsilon = \sqrt{2 \left( \sqrt{e} - 1 \right) } $$ or any linear scaling thereof -- but this is basically $0.887 < |x| < 1.39$, which isn't much of a range. And its doubtful whether this restriction is still in line with the original motivation.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I don't think it's hard to prove no infinite-support continuous variable can achieve the lower bound $\endgroup$ – MichaelChirico Sep 22 '16 at 3:09
  • $\begingroup$ @MichaelChirico I don't think so either; I just didn't want to go through the effort. $\endgroup$ – jwimberley Sep 23 '16 at 13:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.