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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 $p(\lambda)$?

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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

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Thank you, Henry! – Ilya Sep 15 '11 at 11:31

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