Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

For the normally distributed r.v. $\xi$ there is a rule of $3\sigma$ which says that $$ \mathsf P\{\xi\in (\mu-3\sigma,\mu+3\sigma)\}\geq 0.99. $$

Clearly, this rule not necessary holds for other distributions. I wonder if there are lower bounds for $$ p(\lambda) = P\{\xi\in (\mu-\lambda\sigma,\mu+\lambda\sigma)\} $$ regardless of the distribution of real-valued random variable $\xi$. If we are focused only on absolute continuous distributions, a naive approach is to consider the variational problem $$ \int\limits_{\int\limits xf(x)\,dx - \lambda\sqrt{\int\limits x^2f(x)\,dx-(\int\limits xf(x)\,dx)^2}}^{\int\limits xf(x)\,dx + \lambda\sqrt{\int\limits x^2f(x)\,dx-(\int\limits xf(x)\,dx)^2}} f(x)\,dx \to\inf\limits_f $$ which may be too naive. The other problem is that dsitributions can be not necessary absolutely continuous.

So my question is if there are known lower bounds for $p(\lambda)$?

share|improve this question
add comment

1 Answer

up vote 8 down vote accepted

In general this is Chebyshev's inequality

$$\Pr(|X-\mu|\geq k\sigma) \leq \frac{1}{k^2}.$$

Equality is achieved by the discrete distribution $\Pr(X=\mu)=1-\frac{1}{k^2}$, $\Pr(X=\mu-k\sigma)=\frac{1}{2k^2}$, $\Pr(X=\mu+k\sigma)=\frac{1}{2k^2}$. This can be approached arbitrarily closely by an absolutely continuous distribution.

Letting $k=3$, this gives

$$\Pr(|X-\mu|\geq 3\sigma) \leq \frac{1}{9} \approx 0.11;$$

while letting $k=10$, this gives

$$\Pr(|X-\mu|\geq 10\sigma) \leq \frac{1}{100} =0.01.$$

so these bounds are relatively loose for a normal distribution. This diagram (from my page here) compares the bounds. Red is Chebyshev's inequality; blue is a one-tailed version of Chebyshev's inequality; green is a normal distribution; and pink is a one-tailed normal distribution.

Henry's Chebyshev's inequality

share|improve this answer
    
Thank you, Henry! –  Ilya Sep 15 '11 at 11:31
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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