proof of Chebyshev’s inequality
Let and be real numbers such that . Write the product as
(1) | |||||
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If , each of the terms in parentheses is less than or equal to , according to the rearrangement inequality. From this, it follows that
or (dividing by )
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If , the same reasoning gives
It is clear that equality holds if or
. To see that this condition is also necessary,
suppose that not all ’s are equal, so that .
Then the second term in parentheses of (1) can only be
equal to if , the third
term only if , and so on, until the last term which
can only be equal to if . This
implies that . Therefore, Chebyshev’s inequality
is an equality if and only if or
.
Title | proof of Chebyshev’s inequality |
---|---|
Canonical name | ProofOfChebyshevsInequality |
Date of creation | 2013-03-22 13:08:38 |
Last modified on | 2013-03-22 13:08:38 |
Owner | pbruin (1001) |
Last modified by | pbruin (1001) |
Numerical id | 4 |
Author | pbruin (1001) |
Entry type | Proof |
Classification | msc 26D15 |