Cryoscopic constant - Revision history

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2014-10-22T02:11:35Z

removed Category:Thermodynamics; added Category:Thermodynamic properties using HotCat

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en>Rotideypoc41352: Tagged {[refimprove}}, formatting, conciseness edits

2014-02-21T08:02:29Z

Tagged {[refimprove}}, formatting, conciseness edits

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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)~~.		I woke up yesterday and realised - At the moment I've been single for some time and following much bullying from buddies I now locate [http://www.ffpjp24.org justin bieber meet and greet] myself signed up for internet dating. They guaranteed me that there are a lot of normal, pleasant and [http://www.ehow.com/search.html?s=enjoyable+individuals enjoyable individuals] to meet, so the pitch is gone by here!<br>I try to stay as toned as possible staying at the gym several times a week. I appreciate my sports and strive to play or see since many a potential. Being wintertime I am going to frequently at Hawthorn suits. Notice: I've observed the carnage of wrestling fits at stocktake sales, Supposing that you [http://lukebryantickets.citizenswebcasting.com tickets online] would considered purchasing an athletics I do not brain.<br>My buddies and family [http://www.netpaw.org luke bryan tickets go on sale] are awesome and hanging out together at tavern gigabytes or meals is definitely critical. I is the luke bryan concert sold out [[http://www.cinemaudiosociety.org www.cinemaudiosociety.org]] have never been in to clubs as I find that one may not get a good conversation against the noise. Additionally, I got two really cunning and definitely cheeky dogs that are invariably keen to meet up fresh individuals.<br><br>

	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]].		Review my web site; [http://okkyunglee.com where to buy luke bryan tickets]

	Suppose two random variables ''X'' and ''Y'' are ''jointly'' normally distributed. That is the same as saying that the random vector (''X'', ''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'' + ''bY'' is normally distributed. ''In that case'' if ''X'' and ''Y'' are uncorrelated, i.e., their [[covariance]] cov(''X'', ''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.

	~~==Examples==~~

	~~=== A symmetric example ===~~

	~~[[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.]]~~

	Suppose ''X'' has a normal distribution with [[expected value]] 0 and variance 1. Let ''W'' have the [[Rademacher distribution]], so that ''W'' = 1 or −1, each with probability 1/2, and assume ''W'' is independent of ''X''. Let ''Y'' = ''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>~~
	* ''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'' + ''Y'' concentrates positive probability at 0: Pr(''X'' + ''Y'' = 0)  = 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~~

	~~: <math> \begin{align}~~
	~~\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 ''X'', consider~~

	~~: <math>\begin{align}~~
	~~\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) \\~~
	~~& {} = \Phi(x) \cdot\frac12 + \Phi(x)\cdot\frac12~~
	~~\end{align}</math>~~

	~~(since ''X'' and −''X'' both have the same normal distribution), where <math>\Phi(x)</math> is the [[cumulative distribution function]] of the normal distribution..~~

	~~To see that ''X'' and ''Y'' are not independent, observe that \|''Y''\| = \|''X''\| or that Pr(''Y'' > 1 \| ''X'' = 1/2) = 0.~~

	~~=== 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'', ''Y'') is near −1; if ''c'' is very large, then corr(''X'', ''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 0. That value is approximately 1.54. In that case, ''X'' and ''Y'' are uncorrelated, but they are clearly not independent, since ''X'' completely determines ''Y''.

	~~To see that ''Y'' is normally distributed—indeed, that its distribution is the same as that of ''X''—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''\| ≤ ''c''.~~

	Observe that the difference ''X'' − ''Y'' is nowhere near being normally distributed, since it has a substantial probability (about 0.88) of it being equal to 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 372–373</ref>

	~~== References ==~~
	~~{{reflist}}~~

	~~[[Category:Theory of probability distributions]]~~
	~~[[Category:Statistical dependence]~~]

112.210.102.79 at 05:40, 7 January 2014

2014-01-07T05:40:19Z

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en>ZéroBot: r2.7.1) (Robot: Adding ca:Constant crioscòpica

2012-07-27T19:23:17Z

r2.7.1) (Robot: Adding ca:Constant crioscòpica

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