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In [[probability theory]] and [[statistics]], two real-valued [[random variable]]s, ''X'',''Y'', are said to be '''uncorrelated''' if their [[covariance]], E(''XY'') - E(''X'')E(''Y''), is zero. A set of two or more random variables is called uncorrelated if each pair of them are uncorrelated. If two variables are uncorrelated, there is no linear relationship between them.
 
Uncorrelated random variables have a [[Pearson correlation coefficient]] of zero, except in the trivial case when either variable has zero [[variance]] (is a constant).  In this case the correlation is undefined.
 
In general, uncorrelatedness is not the same as [[orthogonality]], except in the special case where either ''X'' or ''Y'' has an expected value of 0.  In this case, the [[covariance]] is the expectation of the product, and ''X'' and ''Y'' are uncorrelated [[if and only if]] E(''XY'') = 0.
 
If ''X'' and ''Y'' are [[statistical independence|independent]], then they are uncorrelated.  However, not all uncorrelated variables are independent. For example, if ''X'' is a continuous random variable [[uniform distribution (continuous)|uniformly distributed]] on [&minus;1, 1] and ''Y'' = ''X''<sup>2</sup>, then ''X'' and ''Y'' are uncorrelated even though ''X'' determines ''Y'' and a particular value of ''Y'' can be produced by only one or two values of ''X''.
 
==Example Of Dependence Without Correlation ==
;Uncorrelated random variables are not necessarily independent
* Let ''X'' be a random variable that takes the value 0 with probability 1/2, and takes the value 1 with probability 1/2.
* Let ''Z'' be a random variable that takes the value -1 with probability 1/2, and takes the value 1 with probability 1/2.
* Let ''U'' be a random variable constructed as ''U=XZ''.
The claim is that ''U'' and ''X'' have zero covariance (and thus are uncorrelated), but are not independent.
 
Proof:
 
First note:
* <math>E[U] =  0\times1/2 + 1\times1/4 + (-1)\times 1/4 = 0</math>
* <math>E[X] = 0\times1/2+1\times1/2 = 1/2</math>
 
Now, by definition <math>\mathrm{cov}(U,X) = E[(U-E[U])(X-E[X])] = E[ U (X-1/2)] = E[X^2Z - (1/2)XZ] = E[X^2Z] - (1/2)E[XZ]</math>
* <math>E[X^2Z] = 0\times1/2 + 1\times1/4 + (-1)\times 1/4 = 0</math>
* <math>E[XZ] = E[U] = 0</math>
Therefore <math>\mathrm{cov}(U,X)=0</math>
 
A necessary condition for showing that ''U'' and ''X'' are independent is showing that for any number ''a'' and ''b'', <math>Pr(U=a|X=b) = Pr(U=a)</math>.  We prove that this is not true.  PIck ''a=1'' and ''b=0''.
* <math>Pr(U=1|X=0) = Pr(XZ=1|X=0) = 0</math>
* <math>Pr(U=1) = Pr(XZ=1) = 1/4 </math>
Thus <math>Pr(U=1|X=0)\ne Pr(U=1)</math> so ''U'' and ''X'' are not independent.
 
Q.E.D.
 
==When uncorrelatedness implies independence==
There are cases in which uncorrelatedness does imply independence. One of these cases is the one in which both random variables are two-valued (so each can be linearly transformed to have a [[binomial distribution]] with ''n''=1).<ref>[http://www.math.uah.edu/stat/expect/Covariance.html Virtual Laboratories in Probability and Statistics: Covariance and Correlation], item 17.</ref>  Further, two jointly normally distributed random variables are independent if they are uncorrelated,<ref>{{cite book|chapter=Chapter 5.5 Conditional Expectation|pages=185–186|title=Introduction to Probability and Mathematical Statistics|year=1992|last1=Bain|first1=Lee|last2=Engelhardt|first2=Max|edition=2nd|isbn=0534929303}}</ref> although this does not hold for variables whose marginal distributions are normal and uncorrelated but whose joint distribution is not joint normal (see [[Normally distributed and uncorrelated does not imply independent]]).
 
==See also==
*[[Correlation and dependence]]
*[[Binomial distribution#Covariance between two binomials|Binomial distribution: Covariance between two binomials]]
 
==References==
{{reflist}}
 
==Further reading==
*''Probability for Statisticians'', Galen R. Shorack, Springer (c2000) ISBN 0-387-98953-6
 
[[Category:Covariance and correlation]]
[[Category:Statistical terminology]]
 
[[de:Korrelation]]

Revision as of 08:36, 4 March 2014

Greetings! I am Myrtle Shroyer. One of the very best issues in the world for me is to do aerobics and now I'm attempting to earn money with it. For a whilst I've been in South Dakota and my parents reside close by. She is a librarian but she's always wanted her personal business.

my webpage ... http://lewat.in/