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How do I show Tchebysheffs's theorem,

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)
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1 Answer

Let nk 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 nk/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 nk and the second from nk+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-nk)(ks)², so that
i=1n(y - y¯)² ≥(n-nk)(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-nk)(ks)²
Cancel s² and simplify:
1 ≥(1/(n-1)) (n-nk) k²
1/k²≥(n-nk)/(n-1)
(nk-1)/(n-1)≈nk/n ≥1-1/k²
which proves the Tchebysheff inequality!

Comments

There are n measurements. nk of these measurements lie between y¯ - ks and y¯ + ks.
I showed that the fraction nk/n is at least 1-1/k².

Comment