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In [[probability theory]], two [[random variable]]s being [[Pearson product-moment correlation coefficient|uncorrelated]] does not imply their [[statistical independence|independence]].  In some contexts, uncorrelatedness implies at least [[pairwise independence]] (as when the random variables involved have [[Bernoulli distribution]]s).
== Straight burst into young woman came out of a belly came. ==


It is sometimes mistakenly thought that one context in which uncorrelatedness implies independence is when the random variables involved are [[normal distribution|normally distributed]]. However, this is incorrect if the variables are merely [[Marginal distribution|marginally]] normally distributed but not [[multivariate normal distribution|jointly normally distributed]].
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Suppose two random variables ''X'' and ''Y'' are ''jointly'' normally distributed.  That is the same as saying that the random vector (''X'',&nbsp;''Y'') has a [[multivariate normal distribution]]. It means that the [[joint probability distribution]] of ''X'' and ''Y'' is such that each linear combination of ''X'' and ''Y'' is normally distributed, i.e. for any two constant (i.e., non-random) scalars ''a'' and ''b'', the random variable ''aX''&nbsp;+&nbsp;''bY'' is normally distributed.  ''In that case'' if ''X'' and ''Y'' are uncorrelated, i.e., their [[covariance]] cov(''X'',&nbsp;''Y'') is zero, ''then'' they are independent.<ref>{{cite book |title=Probability and Statistical Inference |year=2001 |last1=Hogg |first1=Robert |authorlink1=Robert Hogg |last2=Tanis |first2=Elliot | authorlink2=Elliot Tanis |edition=6th |chapter=Chapter 5.4 The Bivariate Normal Distribution |pages=258–259 |isbn=0130272949}}</ref>  ''However'', it is possible for two random variables ''X'' and ''Y'' to be so distributed jointly that each one alone is marginally normally distributed, and they are uncorrelated, but they are not independent; examples are given below.
== am facing the mess left by the previous captain daze. ==


==Examples==
Qi nodded, but kate spade トートバッグ they all laughed, obviously cheap relative thing like this<br>One<br>rest II, III standard Columbia is the day before work, which the day before and the captain sat ケイトスペード バッグ sit, pour tea and a cigarette lighter in person, from the captain's Office came out, the two have been fraternizing.<br><br>mine was a brigade vice squad, sixty people, both will brigade is understaffed instructor volunteered after noon, and the captain together to clean up the backlog of cases, the transfer of residence of the suspects escorted to this day not been finished, according to a new instructor had to face Interpol names are written down, and to the evening shift home Interpol sooner arrived home, rear ケイトスペード 財布 店舗 instructor to come knock on the door, ouch, and captain together, festivals condolences Jiafu Li personally sent home.<br>within<br>day, new instructors bursting with popularity, ah .........<br><br>This day is the third day I took office sin, am facing the mess left by the previous captain daze.<br><br>invoices, meal kate spade ハンドバッグ tickets, gas vouchers of various open check, the shortest time is two months ago, the longest ケイトスペード 財布 ゴールド a year, Interpol investigators are going out all kinds of flowers
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=== A symmetric example ===
== can not let the brother ==


[[File:uncorrelated sym.png|thumb|alt=Two normally distributed, uncorrelated but dependent variables.|Joint range of ''X'' and ''Y''. Darker indicates higher value of the density funtion.]]
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Suppose ''X'' has a normal distribution with [[expected value]] 0 and variance 1. Let ''W'' have the [[Rademacher distribution]], so that ''W'' = 1 or &minus;1, each with probability 1/2, and assume ''W'' is independent of ''X''.  Let ''Y''&nbsp;=&nbsp;''WX''.  Then<ref>[http://www.math.uiuc.edu/~r-ash/Stat/StatLec21-25.pdf UIUC, Lecture 21. ''The Multivariate Normal Distribution''], 21.6:"Individually Gaussian Versus Jointly Gaussian".</ref>
== for example ==
* ''X'' and ''Y'' are uncorrelated;
* Both have the same normal distribution; and
* ''X'' and ''Y'' are not independent.
Note that the distribution of the simple linear combination ''X''&nbsp;+&nbsp;''Y'' concentrates positive probability at 0: Pr(''X''&nbsp;+&nbsp;''Y''&nbsp;=&nbsp;0)&nbsp; =&nbsp;1/2 and so is not normally distributed. By the definition above, ''X'' and ''Y'' are not jointly normally distributed.


To see that ''X'' and ''Y'' are uncorrelated, consider
Most of the time, and society plays a robber, for example, inflation is printing money, you're a big tycoons also turn you into a pauper; such as social change, it is likely that you will become bandits from tyrannical overnight money ...... to measure a person's wealth, it would be too shallow a. '<br><br>than sin laugh, whenever you meet different people can always find some rare in bizarre idea, Wei Jincheng is even worse, and perhaps the reason a super-rich, more than sin against him extra attention [http://www.dmwai.com/webalizer/kate-spade-9.html ケイトスペードバッグセール] a few, wondering authentic: 'What do you What is considered a rich feel? '<br><br>'you feel rich to the rich, according to the survey, living [http://www.dmwai.com/webalizer/kate-spade-4.html kate spade バッグ] in the United [http://www.dmwai.com/webalizer/kate-spade-4.html ケイトスペード バッグ 新作] States [http://www.dmwai.com/webalizer/kate-spade-14.html ケイトスペード ハンドバッグ] of paradise, and residents of small African country warlord, the difference is not large compared to the happiness index.' Wei Jincheng laughed.<br><br>'you ignore a problem.' over sin, and he was eating explaining: 'You talk about the spirit of the rich, I am talking about is material wealth, the material is the basis of the spirit of the ah, you can talk, but I did not so no room money is not the girl without a [http://www.dmwai.com/webalizer/kate-spade-6.html ケイトスペード マザーズバッグ] home, talk about the rich that not a joke? '<br><br>'No, no, the mentality of the rich
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: <math> \begin{align}
== demobilization Secret Service task in a few years ==
\operatorname{cov}(X,Y) &{} = E(XY) - E(X)E(Y) = E(XY) - 0 = E(E(XY\mid W)) \\
& {} = E(X^2)\Pr(W=1) + E(-X^2)\Pr(W=-1) \\
& {} = 1\cdot\frac12 + (-1)\cdot\frac12 = 0.
\end{align}
</math>


To see that ''Y'' has the same normal distribution as&nbsp;''X'', consider
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: <math>\begin{align}
<ul>
\Pr(Y \le x) & {} = E(\Pr(Y \le x\mid W)) \\
 
& {} = \Pr(X \le x)\Pr(W = 1) + \Pr(-X\le x)\Pr(W = -1) \\
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& {} = \Phi(x) \cdot\frac12 + \Phi(x)\cdot\frac12
 
\end{align}</math>
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(since ''X'' and &minus;''X'' both have the same normal distribution), where <math>\Phi(x)</math> is the [[cumulative distribution function]] of the normal distribution..
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To see that ''X'' and ''Y'' are not independent, observe that |''Y''|&nbsp;=&nbsp;|''X''| or that Pr(''Y''&nbsp;>&nbsp;1&nbsp;|&nbsp;''X''&nbsp;=&nbsp;1/2)&nbsp;=&nbsp;0.
</ul>
 
=== An asymmetric example ===
 
[[File:Uncorrelated asym.png|thumb|The joint density of X and Y. Darker indicates a higher value of the density.]]
 
Suppose ''X'' has a normal distribution with [[expected value]] 0 and variance 1. Let
 
: <math>Y=\left\{\begin{matrix} X & \text{if } \left|X\right| \leq c \\
-X & \text{if } \left|X\right|>c \end{matrix}\right.</math>
 
where ''c'' is a positive number to be specified below.  If ''c'' is very small, then the [[correlation]] corr(''X'',&nbsp;''Y'') is near &minus;1; if ''c'' is very large, then corr(''X'',&nbsp;''Y'') is near 1. Since the correlation is a [[continuous function]] of ''c'', the [[intermediate value theorem]] implies there is some particular value of ''c'' that makes the correlation&nbsp;0. That value is approximately&nbsp;1.54. In that case, ''X'' and ''Y'' are uncorrelated, but they are clearly not independent, since ''X'' completely determines&nbsp;''Y''.
 
To see that ''Y'' is normally distributed&mdash;indeed, that its distribution is the same as that of ''X''&mdash;let us find its [[cumulative distribution function]]:
 
: <math>\begin{align}\Pr(Y \leq x) &= \Pr(\{|X| \leq c\text{ and }X \leq x\}\text{ or }\{|X|>c\text{ and }-X \leq x\})\\
&= \Pr(|X| \leq c\text{ and }X \leq x) + \Pr(|X|>c\text{ and }-X \leq x)\\
&= \Pr(|X| \leq c\text{ and }X \leq x) + \Pr(|X|>c\text{ and }X \leq x) \\
&= \Pr(X \leq x). \end{align}\,</math>
 
where the next-to-last equality follows from the symmetry of the distribution of ''X'' and the symmetry of the condition that |''X''|&nbsp;≤&nbsp;''c''.
 
Observe that the difference ''X''&nbsp;−&nbsp;''Y'' is nowhere near being normally distributed, since it has a substantial probability (about&nbsp;0.88) of it being equal to&nbsp;0, whereas the normal distribution, being a continuous distribution, has no discrete part, i.e., does not concentrate more than zero probability at any single point. Consequently ''X'' and ''Y'' are not ''jointly'' normally distributed, even though they are separately normally distributed.<ref>Edward L. Melnick and Aaron Tenenbein, "Misspecifications of the Normal Distribution", ''[[The American Statistician]]'', volume 36, number 4 November 1982, pages 372&ndash;373</ref>
 
== References ==
{{reflist}}
 
[[Category:Theory of probability distributions]]
[[Category:Statistical dependence]]

Latest revision as of 04:11, 22 October 2014

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Straight burst into young woman ケイトスペード バッグ 激安 came ケイトスペード かごバッグ out of a belly came.

'small Liao ...... fast access case to come, are coming!' Mouse invited the people, the police report to the pull of into the house, ケイトスペード 財布 新作 sat down and began weeping much more than that if you No strokes fell great oaks patience, I am afraid not heard that the nose a tear, and speak out the merits of what it is.

mouse can not this patient, it is estimated is a simple case, but it will bring out much complicated story, I am afraid it will take time to cracked.

Fortunately, this is not a police officer surnamed Yan deceived, as officers of the ケイトスペードのバッグ surname Yan.

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'talent na, na ...... me how wonderful encounter no idiot girl, Qugelaopo Nima wages of all confiscated.'

mouse secretly 相关的主题文章：

am facing the mess left by the previous captain daze.

Qi nodded, but kate spade トートバッグ they all laughed, obviously cheap relative thing like this
One
rest II, III standard Columbia is the day before work, which the day before and the captain sat ケイトスペード バッグ sit, pour tea and a cigarette lighter in person, from the captain's Office came out, the two have been fraternizing.

mine was a brigade vice squad, sixty people, both will brigade is understaffed instructor volunteered after noon, and the captain together to clean up the backlog of cases, the transfer of residence of the suspects escorted to this day not been finished, according to a new instructor had to face Interpol names are written down, and to the evening shift home Interpol sooner arrived home, rear ケイトスペード 財布 店舗 instructor to come knock on the door, ouch, and captain together, festivals condolences Jiafu Li personally sent home.
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'instructors, we ケイトスペード時計人気 have no way ah, poor team like this, field reimbursement list put for a year, let the old captain difficulties at home, the team perks support him, we have no opinion on the face ...... can not we matter asked funds to implement or not, can not let the brother 相关的主题文章：

for example

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