Tchebysheff's theorem
By John Thompson
Let k≥1. Show that, for any set of n measurements, the fraction included in the interval $\bar{y} − ks$ to $\bar{y} + ks$ is at least $(1−1/k^2)$. This result is known as Tchebysheff's theorem.
Hint: $s^2=(1/(n−1))\displaystyle\sum\limits_{i=0}^n (y_i - \bar{y})^2$. In this expression, replace all deviations for which $|y_i − \bar{y}| \geq ks$ with ks. Simplify.
Can someone tell me where I am suppose to start with this problem? We haven't covered probabilities yet so no proofs with them.
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$\begingroup$The "Hint" in your question is really the key to answer the question.
Assume the contrary. If the fraction of $|y_i - \bar{y}| \ge ks$ is greater than or equal to $\frac{1}{k^2}$, their part of contribution in $s^2$: $$\frac{1}{n-1}\sum_{\stackrel{i=1}{|y_i - \bar{y}| \ge ks}}^n (y_i - \bar{y})^2 \ge \frac{1}{n-1}\frac{n}{k^2} (ks)^2 = \frac{n}{n-1} s^2 > s^2 $$ is bigger than $s^2$ itself. This is absurd!
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