Hilbert transform: Difference between revisions

From formulasearchengine
Jump to navigation Jump to search
en>Arlene47
References: Added two references
 
(One intermediate revision by one other user not shown)
Line 1: Line 1:
In [[statistics]], '''Cochran's theorem''', devised by [[William G. Cochran]],<ref name="Cochran">{{cite journal|last=Cochran|first=W. G.|authorlink=William Gemmell Cochran|title=The distribution of quadratic forms in a normal system, with applications to the analysis of covariance|journal=[[Mathematical Proceedings of the Cambridge Philosophical Society]]|date=April 1934|volume=30|issue=2|pages=178–191|doi=10.1017/S0305004100016595}}</ref> is a [[theorem]] used to justify results relating to the [[probability distribution]]s of statistics that are used in the [[analysis of variance]].<ref>{{cite book |author= Bapat, R. B.|title=Linear Algebra and Linear Models|edition=Second|publisher= Springer |year=2000|isbn=978-0-387-98871-9}}</ref>
Another day I woke up and realised - I have also been solitary for a while today and after much intimidation from friends I today locate myself signed-up for internet dating. They [http://En.Search.wordpress.com/?q=assured assured] me that there are plenty of standard, pleasant and fun people to meet up, therefore the pitch is gone by here!<br>My household and pals are   [http://www.hotelsedinburgh.org when is the luke bryan concert] awesome and spending time with them at tavern gigabytes or meals is constantly vital. As I discover you can do not own a decent dialogue with the sound I haven't ever been into night clubs. I likewise have 2 undoubtedly cheeky and quite cunning canines that are almost always excited to meet fresh individuals.<br>I attempt to keep as physically fit as potential being at the fitness center several-times per  Luke Bryan Discount Code ([http://www.banburycrossonline.com you could try this out]) week.   [http://www.senatorwonderling.com who is luke bryan on tour with] I appreciate my sports and try to play or see as several a possible. I will regularly at Hawthorn matches being winter. Note: I have observed the carnage of fumbling suits at stocktake revenue, In case that you really contemplated buying a  [http://lukebryantickets.flicense.com buy luke bryan tickets] hobby I don't mind.<br><br>Check out my blog post [http://lukebryantickets.lazintechnologies.com luke bryan tour dates]
 
== Statement ==
Suppose ''U''<sub>1</sub>, ..., ''U''<sub>''n''</sub> are [[statistical independence|independent]] standard [[normal distribution|normally distributed]] [[random variable]]s, and an identity of the form
 
:<math>
\sum_{i=1}^n U_i^2=Q_1+\cdots + Q_k
</math>
 
can be written, where each ''Q''<sub>''i''</sub> is a sum of squares of linear combinations of the ''U''s.  Further suppose that
 
:<math>
r_1+\cdots +r_k=n
</math>
 
where ''r''<sub>''i''</sub> is the [[rank (linear algebra)|rank]] of ''Q''<sub>''i''</sub>.  Cochran's theorem states that the ''Q''<sub>''i''</sub> are independent, and each ''Q''<sub>''i''</sub> has a  [[chi-squared distribution]] with ''r''<sub>''i''</sub> [[degrees of freedom (statistics)|degrees of freedom]].<ref name="Cochran"/> Here the rank of ''Q''<sub>''i''</sub> should be interpreted as meaning the rank of the matrix ''B''<sup>(''i'')</sup>, with elements ''B''<sub>''j,k''</sub><sup>(''i'')</sup>, in the representation of ''Q''<sub>''i''</sub> as a [[quadratic form]]:
 
:<math>Q_i=\sum_{j=1}^n\sum_{k=1}^n U_j B_{j,k}^{(i)} U_k .</math>
 
Less formally, it is the number of linear combinations included in the sum of squares defining ''Q''<sub>''i''</sub>, provided that these linear combinations are linearly independent.
<!--
Cochran's theorem is the converse of [[Fisher's theorem]]. -->
 
== Examples ==
 
=== Sample mean and sample variance ===
If ''X''<sub>1</sub>, ..., ''X''<sub>''n''</sub> are independent normally distributed random variables with mean μ and standard deviation σ
then
 
:<math>U_i = \frac{X_i-\mu}{\sigma}</math>
 
is [[standard normal]] for each ''i''. It is possible to write
 
:<math>
\sum_{i=1}^n U_i^2=\sum_{i=1}^n\left(\frac{X_i-\overline{X}}{\sigma}\right)^2
+ n\left(\frac{\overline{X}-\mu}{\sigma}\right)^2
</math>
 
(here <math>\overline{X}</math> is the [[Arithmetic mean|sample mean]]). To see this identity, multiply throughout by <math>\sigma^2</math> and note that
 
:<math>
\sum(X_i-\mu)^2=
\sum(X_i-\overline{X}+\overline{X}-\mu)^2
</math>
 
and expand to give
 
:<math>
\sum(X_i-\mu)^2=
\sum(X_i-\overline{X})^2+\sum(\overline{X}-\mu)^2+
2\sum(X_i-\overline{X})(\overline{X}-\mu).
</math>
 
The third term is zero because it is equal to a constant times
 
:<math>\sum(\overline{X}-X_i)=0,</math>
 
and the second term has just ''n'' identical terms added together. Thus
:<math>
\sum(X_i-\mu)^2=
\sum(X_i-\overline{X})^2+n(\overline{X}-\mu)^2 ,
</math>
 
and hence
 
:<math>
\sum\left(\frac{X_i-\mu}{\sigma}\right)^2=
\sum\left(\frac{X_i-\overline{X}}{\sigma}\right)^2
+n\left(\frac{\overline{X}-\mu}{\sigma}\right)^2
=Q_1+Q_2.
</math>
 
Now the rank of ''Q''<sub>2</sub> is just 1 (it is the square of just one linear combination of the standard normal variables). The rank of ''Q''<sub>1</sub> can be shown to be ''n'' &minus; 1, and thus the conditions for Cochran's theorem are met.
 
Cochran's theorem then states that ''Q''<sub>1</sub> and ''Q''<sub>2</sub> are independent, with chi-squared distributions with ''n'' &minus; 1 and 1 degree of freedom respectively. This shows that the sample mean and [[sample variance]] are independent. This can also be shown by [[Basu's theorem]], and in fact this property ''characterizes'' the normal distribution – for no other distribution are the sample mean and sample variance independent.<ref>{{cite journal
|doi=10.2307/2983669
|first=R.C. |last=Geary |authorlink=Roy C. Geary
|year=1936
|title=The Distribution of the "Student's" Ratio for the Non-Normal Samples
|journal=Supplement to the Journal of the Royal Statistical Society
|volume=3 |issue=2 |pages=178–184
|jfm=63.1090.03
|jstor=2983669
}}</ref>
 
===Distributions===
 
The result for the distributions is written symbolically as
:<math>
n(\overline{X}-\mu)^2\sim \sigma^2 \chi^2_1,
</math>
:<math>
\sum\left(X_i-\overline{X}\right)^2  \sim \sigma^2 \chi^2_{n-1}.
</math>
 
Both these random variables are proportional to the true but unknown variance σ<sup>2</sup>. Thus their ratio is does not depend on σ<sup>2</sup> and, because they are statistically independent, the distribution of their ratio is given by
 
:<math>
\frac{n\left(\overline{X}-\mu\right)^2}
{\frac{1}{n-1}\sum\left(X_i-\overline{X}\right)^2}\sim \frac{\chi^2_1}{\frac{1}{n-1}\chi^2_{n-1}}
  \sim F_{1,n-1}
</math>
 
where ''F''<sub>1,''n''&nbsp;&minus;&nbsp;1</sub> is the [[F-distribution]] with 1 and ''n''&nbsp;&minus;&nbsp;1 degrees of freedom (see also [[Student's t-distribution]]). The final step here is effectively the definition of a random variable having the F-distribution.
 
=== Estimation of variance ===
To estimate the variance σ<sup>2</sup>, one estimator that is sometimes used is the [[maximum likelihood]] estimator of the variance of a normal distribution
 
:<math>
\widehat{\sigma}^2=
\frac{1}{n}\sum\left(
X_i-\overline{X}\right)^2. </math>
 
Cochran's theorem shows that
 
:<math>
\frac{n\widehat{\sigma}^2}{\sigma^2}\sim\chi^2_{n-1}
</math>
 
and the properties of the chi-squared distribution show that the expected value of <math>\widehat{\sigma}^2</math> is σ<sup>2</sup>(''n'' &minus; 1)/''n''.
 
==Alternative formulation==
The following version is often seen when considering linear regression.{{Citation needed|date=July 2011}} Suppose that <math>Y\sim N_n(0,\sigma^2I_n)</math> is a standard [[Multivariate normal distribution|multivariate normal]] [[random vector]] (here <math>I_n</math> denotes the n-by-n [[identity matrix]]), and if <math>A_1,\ldots,A_k</math> are all n-by-n [[symmetric matrices]] with <math>\sum_{i=1}^kA_i=I_n</math>. Then, on defining <math>r_i=Rank(A_i)</math>, any one of the following conditions implies the other two:
 
* <math>\sum_{i=1}^kr_i=n ,</math>
* <math>Y^TA_iY\sim\sigma^2\chi^2_{r_i}</math>  (thus the <math>A_i</math> are [[positive semidefinite]])
* <math>Y^TA_iY</math> is independent of <math>Y^TA_jY</math> for <math>i\neq j .</math>
 
== See also ==
* [[Cramér's theorem]], on decomposing normal distribution
* [[Infinite divisibility (probability)]]
 
{{refimprove|date=July 2011}}
 
==References==
<references/>
 
{{Experimental design|state=expanded}}
 
{{DEFAULTSORT:Cochran's Theorem}}
[[Category:Statistical theorems]]
[[Category:Characterization of probability distributions]]

Latest revision as of 03:17, 9 January 2015

Another day I woke up and realised - I have also been solitary for a while today and after much intimidation from friends I today locate myself signed-up for internet dating. They assured me that there are plenty of standard, pleasant and fun people to meet up, therefore the pitch is gone by here!
My household and pals are when is the luke bryan concert awesome and spending time with them at tavern gigabytes or meals is constantly vital. As I discover you can do not own a decent dialogue with the sound I haven't ever been into night clubs. I likewise have 2 undoubtedly cheeky and quite cunning canines that are almost always excited to meet fresh individuals.
I attempt to keep as physically fit as potential being at the fitness center several-times per Luke Bryan Discount Code (you could try this out) week. who is luke bryan on tour with I appreciate my sports and try to play or see as several a possible. I will regularly at Hawthorn matches being winter. Note: I have observed the carnage of fumbling suits at stocktake revenue, In case that you really contemplated buying a buy luke bryan tickets hobby I don't mind.

Check out my blog post luke bryan tour dates