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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

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