How do I show Tchebysheffs's theorem,

Question

Let a k ≥ 1. show that for any set of n measurements, the fraction included in the interval y¯ − ks to y¯ + ks is at least (1 - 1/k²).

 

hint: 

s² = 1/(n-1)[ ∑(y - y¯)²]

In this expression , replace all deviation for which |y - y¯| ≥ ks with ks. Simplify. (I don't understand the hint or ow to go about the problem)

Answer 1

Let n_k be the number of measurements y that fall in the interval y¯ − ks to y¯ + ks, or |y - y¯| < ks. We are trying to show that n_k/n ≥1-1/k². Before we can use their hint, we need to split the sum  ∑(y - y¯)² that runs from 1 to n into two sums, the first from 1 to n_k and the second from n_k+1 to n. The first sum includes all those deviations within the interval, |y - y¯| < ks. The second sum includes all the other deviations, i.e., those outside the interval: |y - y¯| ≥ ks. Now the full sum is obviously greater than just the second sum:

∑_i=1ⁿ(y - y¯)² ≥ ∑_i=nk+1ⁿ(y - y¯)²

Now we use the hint and replace the deviations in the second sum with ks:

∑_i=1ⁿ(y - y¯)² ≥ ∑_i=nk+1ⁿ(ks)²

The second sum can now be written as just (n-n_k)(ks)², so that

∑_i=1ⁿ(y - y¯)² ≥(n-n_k)(ks)²

Now use the definition of standard deviation:

s² =(1/(n-1)) ∑_i=1ⁿ(y - y¯)²

With the inequality we just derived this becomes

s² ≥(1/(n-1)) (n-n_k)(ks)²

Cancel s² and simplify:

1 ≥(1/(n-1)) (n-n_k) k²

1/k²≥(n-n_k)/(n-1)

(n_k-1)/(n-1)≈n_k/n ≥1-1/k²

which proves the Tchebysheff inequality!