|
|
|
|
![[parent]](http://images.planetmath.org:8080/images/uparrow.png)
proof of Chebyshev's inequality
|
(Proof)
|
|
|
Let $x_1,x_2,\ldots,x_n$ and $y_1,y_2,\ldots,y_n$ be real numbers such that $x_1\le x_2\le\cdots\le x_n$ Write the product $(x_1+x_2+\cdots+x_n)(y_1+y_2+\cdots+y_n)$ as \begin{eqnarray} \nonumber &&(x_1y_1+x_2y_2+\cdots+x_ny_n)\\ \nonumber &+&(x_1y_2+x_2y_3+\cdots+x_{n-1}y_n+x_ny_1)\\ \nonumber &+&(x_1y_3+x_2y_4+\cdots+x_{n-2}y_n+x_{n-1}y_1+x_ny_2)\\ \nonumber &+&\cdots\\ &+&(x_1y_n+x_2y_1+x_3y_2+\cdots+x_ny_{n-1}). \label{expanded} \end{eqnarray}
- If $y_1\le y_2\le\cdots\le y_n$ each of the $n$ terms in parentheses is less than or equal to $x_1y_1+x_2y_2+\cdots+x_ny_n$ according to the rearrangement inequality. From this, it follows that $$ (x_1+x_2+\cdots+x_n)(y_1+y_2+\cdots+y_n)\le n(x_1y_1+x_2y_2+\cdots+x_ny_n) $$ or (dividing by $n^2$ $$ \left(\frac{x_1+x_2+\cdots+x_n}{n}\right) \left(\frac{y_1+y_2+\cdots+y_n}{n}\right)\le \frac{x_1y_1+x_2y_2+\cdots+x_ny_n}{n}. $$
- If $y_1\ge y_2\ge\cdots\ge y_n$ the same reasoning gives $$ \left(\frac{x_1+x_2+\cdots+x_n}{n}\right) \left(\frac{y_1+y_2+\cdots+y_n}{n}\right)\ge \frac{x_1y_1+x_2y_2+\cdots+x_ny_n}{n}. $$
It is clear that equality holds if $x_1=x_2=\cdots=x_n$ or $y_1=y_2=\cdots=y_n$ To see that this condition is also necessary, suppose that not all $y_i$ s are equal, so that $y_1\neq y_n$ Then the second term in parentheses of (![[*]](http://images.planetmath.org:8080/cache/objects/3582/js//usr/share/latex2html/icons/crossref.png "[*]") ) can only be equal to $x_1y_1+x_2y_2+\cdots+x_ny_n$ if $x_{n-1}=x_n$ the third term only if $x_{n-2}=x_{n-1}$ and so on, until the last term which can only be equal to $x_1y_1+x_2y_2+\cdots+x_ny_n$ if $x_1=x_2$ This implies that $x_1=x_2=\cdots=x_n$ Therefore, Chebyshev's inequality is an equality if and only if $x_1=x_2=\cdots=x_n$ or $y_1=y_2=\cdots=y_n$
|
"proof of Chebyshev's inequality" is owned by pbruin.
|
|
(view preamble | get metadata)
Cross-references: Chebyshev's inequality, implies, necessary, equality, clear, rearrangement inequality, terms, product, real numbers
This is version 1 of proof of Chebyshev's inequality, born on 2002-11-10.
Object id is 3582, canonical name is ProofOfChebyshevsInequality2.
Accessed 6832 times total.
Classification:
| AMS MSC: | 26D15 (Real functions :: Inequalities :: Inequalities for sums, series and integrals) |
|
|
|
|
|
|
Pending Errata and Addenda
|
|
|
|
|
|
|
|
|
|
|
