FANDOM


$ P(|X-\mu|>k\sigma)<\frac{1}{k^2} $

In English: "The probability that the outcome of an experiment with the random variable $ X $ will fall more than $ k $ standard deviations beyond the mean of $ X $ , $ \mu $ , is less than $ \frac1{k^2} $ ."

Or: "The proportion of the total area under the probability distribution function of $ X $ outside of $ k $ standard deviations from the mean $ \mu $ is at most $ \frac1{k^2} $ ."

Proof

Let $ S $ be the sample space for a random variable, $ X $ , and let $ f_X(x) $ stand for the pdf of $ X $ . Let $ R_1,R_2,R_3 $ partition $ S $ , such that for every sample point $ x\in S $

$ x\in \begin{cases} R_1&x<\mu-k\sigma\\ R_2&|x-\mu|\le k\sigma\\ R_3&x>\mu+k\sigma \end{cases} $

Then

$ \sigma^2=\sum_{R_1}(x-\mu)^2f_X(x)+\sum_{R_2}(x-\mu)^2f_X(x)+\sum_{R_3}(x-\mu)^2f_X(x) $

Clearly

$ \sigma^2\ge\sum_{R_1}(x-\mu)^2f_X(x)+\sum_{R_3}(x-\mu)^2f_X(x) $

since the term that evaluates to the variance in $ R_2 $ has been subtracted on the right-hand side.

For any sample point $ x\in R_1 $

$ x<\mu-k\sigma $
$ =x-\mu<-k\sigma $
$ =(x-\mu)^2>k^2\sigma^2 $

Notice that the direction of the inequality changes since squaring causes the right-hand expression to become positive.

And for any sample point $ x\in R_3 $

$ x>\mu+k\sigma $
$ =x-\mu>k\sigma $
$ =(x-\mu)^2>k^2\sigma^2 $

So, for any sample point $ x\in R_1 $ or $ x\in R_3 $ , it can be said that $ (x-\mu)^2>k^2\sigma^2 $ , and so

$ \sigma^2>\sum_{R_1}(k\sigma)^2f_X(x)+\sum_{R_3}(k\sigma)^2f_X(x) $

Dividing each side of the inequality by $ (k\sigma)^2 $ results in

$ \frac1{k^2}>\sum_{R_1}f_X(x)+\sum_{R_3}f_X(x) $

Or, in other terms

$ \frac1{k^2}>P(|x-\mu|>k\sigma) $

And thus the original claim is proven.

QED