According to the Chebyshev's inequality, if we take any distribution, we get >88.8889% of data in +-3 sigma interval.

For a normal distribution it is 99.97%.

How to calculate the interval for a Pareto or any other distribution?

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    $\begingroup$ You seem to be mixing up two different facts. Although 99.7% of the probability of a Normal random variable is within 3 sigmas of its mean, data realized by sampling this variable will not necessarily have that property. $\endgroup$ – whuber Dec 2 '20 at 16:14
  • $\begingroup$ If the CDF of $X$ is known and $\mu$ and $\sigma$ both exist, then you can use the CDF to find $P(\mu-3\sigma < X < \mu+3\sigma).$ For $X\sim\mathsf{Exp}(1)$ with $\mu=\sigma=1,$ you can find the probability in R (where pexp is a CDF) with code diff(pexp(c(-2,4),1)), which returns $0.9816844.$ $\endgroup$ – BruceET Dec 2 '20 at 17:24
  • $\begingroup$ For very large samples, you get nearly the theoretical proportion: For a sample of $10^6$ observations from $\mathsf{Exp}(1),$ R code set.seed(2020); x = rexp(10^6,1); mean(x < 4) returns $0.981702,$ which has a little better than the anticipated 2 or 3-place agreement with the value in my last Comment. $\endgroup$ – BruceET Dec 2 '20 at 17:48

Let $X$ be a random variable with mean $\mu$ and standard deviation $\sigma$. By simple probability rules, we have $$P\left(\mu-3\sigma \leq X \leq \mu+3\sigma\right) = P(X \leq \mu + 3\sigma) - P(X \leq \mu - 3\sigma).$$

The quantity $P(X \leq x) = \int_{-\infty}^xf(t)dt$ is called the cumulative distribution function (CDF), and is often written as $F(x)$. If the mean, standard deviation and CDF all exist in closed form, then this calculation is pretty straightforward.

For instance, if $X \sim \text{Pareto}(x_m=1, \alpha=3)$, then we have $\mu = 1.5$, $\sigma = \sqrt{0.75}$ and $$F(x) = \begin{cases} 1 - x^{-3}, & x > 1 \\ 0, & x \leq 1. \end{cases}$$

So the density contained within three standard deviations of the mean can be computed as \begin{align*} \approx P(-1.1 \leq X \leq 4.1) &= P(X \leq 4.1) - P(X \leq -1.1) \\ &= 1 - 4.1^{-3} - 0 = 0.985 \end{align*}

The normal distribution yields the same answer for all parameter values, but the Pareto (and most distributions) depends on the values of the parameters. For instance, if $\alpha = 4$, we get $0.9824$.

It is also worth considering the case where $\alpha = 1$. In this case, the mean and variance of the Pareto distribution is infinite, and the question no longer makes sense.


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