# Proof of Chebyshev's inequality

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In English: "The probability that the outcome of an experiment with the random variable will fall more than standard deviations beyond the mean of , , is less than ."

Or: "The proportion of the total area under the probability distribution function of outside of standard deviations from the mean is at most ."

## Proof

Let be the sample space for a random variable, , and let stand for the pdf of . Let partition , such that for every sample point

Then

Clearly

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

For any sample point

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

And for any sample point

So, for any sample point or , it can be said that , and so

Dividing each side of the inequality by results in

Or, in other terms

And thus the original claim is proven.

QED