Editing Chebyshev's inequality (section)

==Integral Chebyshev inequality==

There is a second (less well known) inequality also named after Chebyshev<ref name=Fink1984>{{cite book |last1=Fink |first1=A. M. |last2=Jodeit |first2=Max Jr. |title= Inequalities in Statistics and Probability|isbn=978-0-940600-04-1 |mr=789242 |editor1-first=Y. L. |editor1-last=Tong |editor2-last=Gupta |editor2-first=Shanti S. |year=1984 |volume=5 |series=Institute of Mathematical Statistics Lecture Notes - Monograph Series |pages=115–120 |doi=10.1214/lnms/1215465637 |chapter-url=http://projecteuclid.org/euclid.lnms/1215465617 |access-date=7 October 2012|chapter=On Chebyshev's other inequality }}</ref>

If ''f'', ''g'' : [''a'', ''b''] → '''R''' are two [[monotonic]] [[function (mathematics)|function]]s of the same monotonicity, then

: <math> \frac{ 1 }{ b - a } \int_a^b \! f(x) g(x) \,dx \ge  \left[ \frac{ 1 }{ b - a } \int_a^b \! f(x) \,dx \right] \left[ \frac{ 1 }{ b - a } \int_a^b \! g(x) \,dx \right] .</math>

If ''f'' and ''g'' are of opposite monotonicity, then the above inequality works in the reverse way.

{{Math theorem|Let <math>f</math> and <math>g</math> be monotonic functions of the same monotonicity on <math>[a,b]</math>. Then for any <math>x, y \in [a,b]</math> we have
<math display="block">(f(x)-f(y))(g(x)-g(y)) \geq 0.</math>|name=Lemma}}

{{Math proof|Integrate this inequality with respect to <math>x</math> and <math>y</math> over <math>[a,b]</math>:
<math display="block">\int_a^b \int_a^b (f(x)-f(y))(g(x)-g(y)) \,dx\,dy \geq 0.</math>

Expanding the integrand gives:
<math display="block">\int_a^b \int_a^b \left[f(x)g(x) - f(x)g(y) - f(y)g(x) + f(y)g(y)\right] \,dx\,dy \geq 0.</math>

Separate the double integral into four parts:
<math display="block">\int_a^b \int_a^b f(x)g(x) \,dx\,dy - \int_a^b \int_a^b f(x)g(y) \,dx\,dy - \int_a^b \int_a^b f(y)g(x) \,dx\,dy + \int_a^b \int_a^b f(y)g(y) \,dx\,dy \geq 0.</math>

Since the integration variable in each inner integral is independent, we have:
* <math>\int_a^b \int_a^b f(x)g(x) \,dx\,dy = (b-a) \int_a^b f(x)g(x) \,dx,</math>
* <math>\int_a^b \int_a^b f(y)g(y) \,dx\,dy = (b-a) \int_a^b f(y)g(y) \,dy = (b-a) \int_a^b f(x)g(x) \,dx,</math>
* <math>\int_a^b \int_a^b f(x)g(y) \,dx\,dy = \left(\int_a^b f(x) \,dx\right)\left(\int_a^b g(y) \,dy\right),</math>
* <math>\int_a^b \int_a^b f(y)g(x) \,dx\,dy = \left(\int_a^b f(y) \,dy\right)\left(\int_a^b g(x) \,dx\right) = \left(\int_a^b f(x) \,dx\right)\left(\int_a^b g(x) \,dx\right).</math>

Let

<math display="block">I = \int_a^b f(x)g(x) \,dx, \quad F = \int_a^b f(x) \,dx, \quad G = \int_a^b g(x) \,dx.</math>

Substitute these into the inequality:

<math display="block">(b-a)I - FG - FG + (b-a)I \geq 0.</math>

Simplify:

<math display="block">2(b-a)I - 2FG \geq 0.</math>

Dividing by <math>2(b-a)</math> (noting that <math>b-a>0</math>):

<math display="block">I \geq \frac{FG}{(b-a)}.</math>

Divide both sides by <math>b-a</math> to obtain:

<math display="block">\frac{1}{b-a} \int_a^b f(x)g(x) \,dx \geq \left(\frac{1}{b-a}\int_a^b f(x) \,dx\right) \left(\frac{1}{b-a}\int_a^b g(x) \,dx\right).</math>

This completes the proof.}}

This inequality is related to [[Jensen's inequality]],<ref name=Niculescu2001>{{cite journal |last=Niculescu |first=Constantin P. |title=An extension of Chebyshev's inequality and its connection with Jensen's inequality |journal=Journal of Inequalities and Applications |year=2001 |volume=6 |issue=4 |pages=451–462 |doi=10.1155/S1025583401000273 |url=http://emis.matem.unam.mx/journals/HOA/JIA/Volume6_4/462.html |access-date=6 October 2012 |issn=1025-5834|citeseerx=10.1.1.612.7056 |doi-access=free }}</ref> [[Kantorovich's inequality]],<ref name=Niculescu2001a>{{cite journal |last1=Niculescu |first1=Constantin P. |last2=Pečarić |first2=Josip |author-link2=Josip Pečarić |title=The Equivalence of Chebyshev's Inequality to the Hermite–Hadamard Inequality |journal=Mathematical Reports |year=2010 |volume=12 |issue=62 |pages=145–156 |url=http://www.csm.ro/reviste/Mathematical_Reports/Pdfs/2010/2/Niculescu.pdf |access-date=6 October 2012 |issn=1582-3067}}</ref> the [[Hermite–Hadamard inequality]]<ref name="Niculescu2001a"/> and [[Walter's conjecture]].<ref name=Malamud2001>{{cite journal |last=Malamud |first=S. M. |title=Some complements to the Jensen and Chebyshev inequalities and a problem of W. Walter |journal=Proceedings of the American Mathematical Society |date=15 February 2001 |volume=129 |issue=9 |pages=2671–2678 |doi=10.1090/S0002-9939-01-05849-X |mr=1838791 |url=https://www.ams.org/journals/proc/2001-129-09/S0002-9939-01-05849-X/ |access-date=7 October 2012 |issn=0002-9939|doi-access=free }}</ref>

===Other inequalities===

There are also a number of other inequalities associated with Chebyshev:

*[[Chebyshev's sum inequality]]
*[[Chebyshev–Markov–Stieltjes inequalities]]