# Statistics and Tchebysheff's theorum

1. Sep 6, 2011

### major_maths

1. The problem statement, all variables and given/known data

Let k$\geq$1. Show that, for any set of n measurements, the fraction included in the interval $\overline{y}$-ks to $\overline{y}$+ks is at least (1-1/k2).

[Hint: s2 = 1/(n-1)[$\sum$(yi-$\overline{y}$)2]. In this expression, replace all deviations for which the absolute value of (yi-$\overline{y}$)$\geq$ks with ks. Simplify.] This result is known as Tchebysheff's theorem.

2. Relevant equations are the above.

3. The attempt at a solution

I've got no clue what the problem wants, much less how to start a solution.

2. Sep 7, 2011

### Stephen Tashi

Let there be $M$ measurements where $| y_i - \overline{y}| \geq ks$
If in the sum $\sum(y_i -\overline{y})^2$ we replace those $M$ measurements by $ks$ and leave out the other $N-M$ measurements, we get a smaller sum. The smaller sum is $M (ks)^2$

Hence

$$s^2 = \frac{1}{n-1} \sum(y_i - \overline{y})^2 \geq \frac{1}{n-1} M (ks)^2$$

Since $\frac{1}{n-1} > \frac{1}{n}$

$$s^2 \geq \frac{1}{n-1}M(ks)^2 > \frac{1}{n}M(ks)^2$$
$$s^2 \geq \frac{1}{n}M(ks)^2$$

The "fraction of measurements" that $M$ constitutes is $\frac{M}{n}$ and the above inequality can be used to bound it.

The original problem concerns the fraction of measurements other than those M measurements, so that fraction is $1.0 - \frac{M}{n}$.
That needs to be bounded by using the bound for $\frac{M}{n}$.

3. Sep 7, 2011

### hassman

Thank you Stephen. That was part of my homework I was struggling with. I wonder which school OP goes :-).