Cochran's theorem
cochran's theorem definition, cochran's theorem paintingIn statistics, Cochran's theorem, devised by William G Cochran, is a theorem used to justify results relating to the probability distributions of statistics that are used in the analysis of variance
Contents
 1 Statement
 11 Proof
 2 Examples
 21 Sample mean and sample variance
 22 Distributions
 23 Estimation of variance
 3 Alternative formulation
 4 See also
 5 References
Statement
Suppose U1, , UN are iid standard normally distributed random variables, and there exist matrices B 1 , B 2 , … B k ,B^,\ldots B^} , with ∑ i = 1 k B i = I N ^B^=I_} Further suppose that r 1 + ⋯ + r k = N +\cdots +r_=N} , where ri is the rank of B i } If we write
Q i = ∑ j = 1 N ∑ ℓ = 1 N U j B j , ℓ i U ℓ =\sum _^\sum _^U_B_^U_}so that the Q i } are quadratic forms, then Cochran's theorem states that the Qi are independent, and each Qi has a chisquared distribution with ri degrees of freedom
Less formally, it is the number of linear combinations included in the sum of squares defining Qi, provided that these linear combinations are linearly independent
Proof
We first show that the matrices Bi can be simultaneously diagonalized and that their nonzero eigenvalues are all equal to +1 We then use the vector basis that diagonalize them to simplify their characteristic function and show their independence and distribution
Each of the matrices Bi has rank ri and thus ri nonzero eigenvalues For each i, the sum C i ≡ ∑ j ≠ i B j \equiv \sum _B^} has at most rank ∑ j ≠ i r j = N − r i r_=Nr_} Since B i + C i = I N × N +C^=I_} , it follows that Ci has exactly rank N − ri
Therefore Bi and Ci can be simultaneously diagonalized This can be shown by first diagonalizing Bi In this basis, it is of the form:
[ λ 1 0 0 ⋯ ⋯ 0 0 λ 2 0 ⋯ ⋯ 0 0 0 ⋱ ⋮ 0 ⋮ λ r i 0 ⋮ 0 0 ⋮ ⋱ 0 0 … 0 ] \lambda _&0&0&\cdots &\cdots &&0\\0&\lambda _&0&\cdots &\cdots &&0\\0&0&\ddots &&&&\vdots \\0&\vdots &&\lambda _}&&\\0&\vdots &&&0&\\0&\vdots &&&&\ddots \\0&0&\ldots &&&&0\end}}Thus the lower N − r i } rows are zero Since C i = I − B i =IB^} , it follows that these rows in Ci in this basis contain a right block which is a N − r i × N − r i \times Nr_} unit matrix, with zeros in the rest of these rows But since Ci has rank N − ri, it must be zero elsewhere Thus it is diagonal in this basis as well It follows that all the nonzero eigenvalues of both Bi and Ci are +1 Moreover, the above analysis can be repeated in the diagonal basis for C 1 = B 2 + ∑ j > 2 B j =B^+\sum _B^} In this basis C 1 } is the identity of an N − r 1 × N − r 1 \times Nr_} vector space, so it follows that both B2 and ∑ j > 2 B j B^} are simultaneously diagonalizable in this vector space and hence also together B1 By iteration it follows that all Bs are simultaneously diagonalizable
Thus there exists an orthogonal matrix S such that for all i, S T B i S }B^S} is diagonal with 1s between r i − 1 , ⋯ , r i ,\cdots ,r_}
Let U i ′ ^} denote some specific linear combination of all U i } after transformation by S Note that ∑ i = 1 N U i ′ 2 = ∑ i = 1 N U i 2 ^U_^^=\sum _^U_^} due to the length preservation of the orthogonal matrix S
The characteristic function of Qi is:
φ i t = 2 π − N / 2 ∫ d u 1 ∫ d u 2 ⋯ ∫ d u N e i t Q i ⋅ e − u 1 2 2 ⋅ e − u 2 2 2 ⋯ e − u N 2 2 = 2 π − N / 2 ∏ j = 1 N ∫ d u j e i t Q i ⋅ e − ∑ j = 1 N u j 2 2 = 2 π − N / 2 ∏ j = 1 N ∫ d u j ′ e i t ⋅ ∑ m = r 1 + ⋯ + r i − 1 + 1 r 1 + ⋯ + r i u m ′ 2 ⋅ e − ∑ j = 1 N u j ′ 2 2 = 2 π − N / 2 ∫ e u 2 i t − 1 2 d u r i ∫ e − u 2 2 d u N − r i = 1 − 2 i t − r i / 2 \varphi _t=&2\pi ^\int du_\int du_\cdots \int du_e^}\cdot e^^}}}\cdot e^^}}}\cdots e^^}}}\\=&2\pi ^\left\prod _^\int du_\righte^}\cdot e^^^}}}\\=&2\pi ^\left\prod _^\int du_^\righte^+\cdots +r_+1}^+\cdots +r_}u_^^}\cdot e^^^}^}}}\\=&2\pi ^\left\int e^it}}du\right^}\left\int e^}}}du\right^}\\=&12it^/2}\end}}This is the Fourier transform of the chisquared distribution with ri degrees of freedom Therefore this is the distribution of Qi
Moreover, the characteristic function of the joint distribution of all the Qis is:
φ t 1 , t 2 , … , t k = 2 π − N / 2 ∏ j = 1 N ∫ d U j e i ∑ i = 1 k t i ⋅ Q i ⋅ e − ∑ j = 1 N U j 2 2 = 2 π − N / 2 ∏ j = 1 N ∫ d U j ′ e i ⋅ ∑ i = 1 k t i ∑ k = r 1 + ⋯ + r i − 1 + 1 r 1 + ⋯ + r i U k ′ 2 ⋅ e − ∑ j = 1 N U j ′ 2 2 = 2 π − N / 2 ∏ i = 1 k ∫ e u 2 i t i − 1 2 d u r i = ∏ i = 1 k 1 − 2 i t i − r i / 2 = ∏ i = 1 k φ i t i \varphi t_,t_,\ldots ,t_&=2\pi ^\left\prod _^\int dU_\righte^^t_\cdot Q_}\cdot e^^^}}}\\&=2\pi ^\left\prod _^\int dU_^\righte^^t_\sum _+\cdots +r_+1}^+\cdots +r_}U_^^}\cdot e^^^}^}}}\\&=2\pi ^\prod _^\left\int e^it_}}du\right^}\\&=\prod _^12it_^/2}=\prod _^\varphi _t_\end}}From this it follows that all the Qis are independent
Examples
Sample mean and sample variance
If X1, , Xn are independent normally distributed random variables with mean μ and standard deviation σ then
U i = X i − μ σ =\mu }}}is standard normal for each i It is possible to write
∑ i = 1 n U i 2 = ∑ i = 1 n X i − X ¯ σ 2 + n X ¯ − μ σ 2 ^U_^=\sum _^\left}}}\right^+n\left}\mu }}\right^}here X ¯ }} is the sample mean To see this identity, multiply throughout by σ 2 } and note that
∑ X i − μ 2 = ∑ X i − X ¯ + X ¯ − μ 2 \mu ^=\sum X_}+}\mu ^}and expand to give
∑ X i − μ 2 = ∑ X i − X ¯ 2 + ∑ X ¯ − μ 2 + 2 ∑ X i − X ¯ X ¯ − μ \mu ^=\sum X_}^+\sum }\mu ^+2\sum X_}}\mu }The third term is zero because it is equal to a constant times
∑ X ¯ − X i = 0 , }X_=0,}and the second term has just n identical terms added together Thus
∑ X i − μ 2 = ∑ X i − X ¯ 2 + n X ¯ − μ 2 , \mu ^=\sum X_}^+n}\mu ^,}and hence
∑ X i − μ σ 2 = ∑ X i − X ¯ σ 2 + n X ¯ − μ σ 2 = Q 1 + Q 2 \mu }}\right^=\sum \left}}}\right^+n\left}\mu }}\right^=Q_+Q_}Now the rank of Q2 is just 1 it is the square of just one linear combination of the standard normal variables The rank of Q1 can be shown to be n − 1, and thus the conditions for Cochran's theorem are met
Cochran's theorem then states that Q1 and Q2 are independent, with chisquared distributions with n − 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
Distributions
The result for the distributions is written symbolically as
∑ X i − X ¯ 2 ∼ σ 2 χ n − 1 2 }\right^\sim \sigma ^\chi _^} n X ¯ − μ 2 ∼ σ 2 χ 1 2 , }\mu ^\sim \sigma ^\chi _^,}Both these random variables are proportional to the true but unknown variance σ2 Thus their ratio does not depend on σ2 and, because they are statistically independent The distribution of their ratio is given by
n X ¯ − μ 2 1 n − 1 ∑ X i − X ¯ 2 ∼ χ 1 2 1 n − 1 χ n − 1 2 ∼ F 1 , n − 1 }\mu \right^}}\sum \leftX_}\right^}}\sim ^}}\chi _^}}\sim F_}where F1,n − 1 is the Fdistribution with 1 and n − 1 degrees of freedom see also Student's tdistribution The final step here is effectively the definition of a random variable having the Fdistribution
Estimation of variance
To estimate the variance σ2, one estimator that is sometimes used is the maximum likelihood estimator of the variance of a normal distribution
σ ^ 2 = 1 n ∑ X i − X ¯ 2 }^=}\sum \leftX_}\right^}Cochran's theorem shows that
n σ ^ 2 σ 2 ∼ χ n − 1 2 }^}}}\sim \chi _^}and the properties of the chisquared distribution show that
E n σ ^ 2 σ 2 = E χ n − 1 2 n σ 2 E σ ^ 2 = n − 1 E σ ^ 2 = σ 2 n − 1 n E\left}^}}}\right&=E\left\chi _^\right\\}}E\left}^\right&=n1\\E\left}^\right&=n1}}\end}}the expected value of σ ^ 2 }^} is σ2n − 1/n
Alternative formulation
The following version is often seen when considering linear regression Suppose that Y ∼ N n 0 , σ 2 I n 0,\sigma ^I_} is a standard multivariate normal random vector here I n } denotes the nbyn identity matrix, and if A 1 , … , A k ,\ldots ,A_} are all nbyn symmetric matrices with ∑ i = 1 k A i = I n ^A_=I_} Then, on defining r i = Rank A i =\operatorname A_} , any one of the following conditions implies the other two:
 ∑ i = 1 k r i = n , ^r_=n,}
 Y T A i Y ∼ σ 2 χ r i 2 A_Y\sim \sigma ^\chi _}^} thus the A i } are positive semidefinite
 Y T A i Y A_Y} is independent of Y T A j Y A_Y} for i ≠ j
See also
 Cramér's theorem, on decomposing normal distribution
 Infinite divisibility probability
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References
 ^ a b Cochran, W G April 1934 "The distribution of quadratic forms in a normal system, with applications to the analysis of covariance" Mathematical Proceedings of the Cambridge Philosophical Society 30 2: 178–191 doi:101017/S0305004100016595
 ^ Bapat, R B 2000 Linear Algebra and Linear Models Second ed Springer ISBN 9780387988719
 ^ Craig AT 1938 On The Independence of Certain Estimates of Variances Ann Math Statist 9, pp 48–55
 ^ Geary, RC 1936 "The Distribution of the "Student's" Ratio for the NonNormal Samples" Supplement to the Journal of the Royal Statistical Society 3 2: 178–184 doi:102307/2983669 JFM 63109003 JSTOR 2983669
 ^ "Cochran's Theorem A quick tutorial" PDF



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