Difference between revisions of "Chebyshev's Inequality"

Revision as of 19:47, 26 March 2020

Chebyshev's inequality, named after Pafnuty Chebyshev, states that if $a_1\geq a_2\geq ... \geq a_n$ and $b_1\geq b_2\geq ... \geq b_n$ then the following inequality holds:

$n \left(\sum_{i=1}^{n}a_ib_i\right)\geq\left(\sum_{i=1}^{n}a_i\right)\left(\sum_{i=1}^{n}b_i\right)$ .

On the other hand, if $a_1\geq a_2\geq ... \geq a_n$ and $b_n\geq b_{n-1}\geq ... \geq b_1$ then: $n \left(\sum_{i=1}^{n}a_ib_i\right)\leq\left(\sum_{i=1}^{n}a_i\right)\left(\sum_{i=1}^{n}b_i\right)$ .

Proof

Chebyshev's inequality is a consequence of the Rearrangement inequality, which gives us that the sum $S=a_1b_{i_1}+a_2b_{i_2}+...+a_nb_{i_n}$ is maximal when $i_k=k$ .

Now, by adding the inequalities:

$\sum_{i=1}^{n}a_ib_i\geq a_1b_1+a_2b_2+...+a_n b_{n}$

$\sum_{i=1}^{n}a_ib_i\geq a_1b_2+a_2b_3+...+a_nb_1$

$\cdots$

$\sum_{i=1}^{n}a_ib_i\geq a_1b_n+a_2b_1+...+a_nb_{n-1}$

we get the initial inequality.

@@ Line 16: / Line 16: @@
 <math>\sum_{i=1}^{n}a_ib_i\geq a_1b_2+a_2b_3+...+a_nb_1</math>
-...
+<math>\cdots</math>
 <math>\sum_{i=1}^{n}a_ib_i\geq a_1b_n+a_2b_1+...+a_nb_{n-1}</math>

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Difference between revisions of "Chebyshev's Inequality"

Revision as of 19:47, 26 March 2020

Proof