Statistics and Tchebysheff's theorum

[major_maths]

Homework Statement

Let k\geq1. 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(y_i-\overline{y})²]. In this expression, replace all deviations for which the absolute value of (y_i-\overline{y})\geqks with ks. Simplify.] This result is known as Tchebysheff's theorem.

2. Homework Equations are the above.

The Attempt at a Solution

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

 

[Stephen Tashi]

Homework Statement

Let k\geq1. 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(y_i-\overline{y})²]. In this expression, replace all deviations for which the absolute value of (y_i-\overline{y})\geqks with ks. Simplify.] This result is known as Tchebysheff's theorem.

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}.

 

[hassman]

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

To be really pedantic, should not the last equation have > sign instead of >=?

 

[BigBrain]

When we take a look at the definition of theorem number two, we see that the theorem refers to the standard deviation of the possible sample means computed from all possible random samples. Theorem number one is similar in that it says for any population, the average value of all possible sample means computed from all possible random samples of a given size from the population equal the population mean. What does that mean? Does that mean that the mean of my sample will automatically be equal to the population mean?