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 ≥11/k^{2}. 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=1}^{n}(y  y¯)² ≥ ∑_{i=n}_{k+1}^{n}(y  y¯)²
Now we use the hint and replace the deviations in the second sum with ks:
∑_{i=1}^{n}(y  y¯)² ≥ ∑_{i=nk+1}^{n}(ks)²
The second sum can now be written as just (nn_{k})(ks)², so that
∑_{i=1}^{n}(y  y¯)² ≥(nn_{k})(ks)²
Now use the definition of standard deviation:
s^{2 }=(1/(n1)) ∑_{i=1}^{n}(y  y¯)²
With the inequality we just derived this becomes
s^{2} ≥(1/(n1)) (nn_{k})(ks)²
Cancel s² and simplify:
1 ≥(1/(n1)) (nn_{k}) k²
1/k²≥(nn_{k})/(n1)
(n_{k}1)/(n1)≈n_{k}/n ≥11/k²
which proves the Tchebysheff inequality!
9/3/2013

Andre W.
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