Statistics Assignment: Chi-square distribution, analysis of data

This homework assignment focuses on the analysis of Chi-square distribution. The solution provided delves into the properties of the Chi-square distribution, specifically examining its relationship with idempotent matrices and non-centrality parameters. The assignment explores the decomposition of matrices (A’ΛA) and the derivation of the Chi-square distribution from a set of independent and normally distributed variables. The solution provides a step-by-step breakdown of the process, including the calculation of the χ2 statistic, the use of eigenvalues, and the application of the formula to calculate the final result, χ2(r, µ’ Aµ/ 2δ2). The assignment likely requires the student to demonstrate an understanding of statistical concepts, including hypothesis testing, and data analysis and the ability to apply them to solve a specific problem involving the Chi-square distribution. The assignment is available on Desklib.

SOLUTION
For A’YA/ δ 2 to be χ2(r, μ A μ/2 δ2) then y~ NK (μ, δ2 ∑), must be idempotent of
rank r.
Since y1, y2, y3…yk are independently distributed with common mean μ and
common variance δ2, then it follows that 1
δ2 ∑
i=1
k
yi
2 with χ2 with K degrees of
freedom and non-centrality parameter λ= 1
δ2 ∑
i=1
k
μi
2
A∑ is Idempotent of rank r and has rank(A) = r,
We first need to assume that:
A∑ IS Idempotent of rank r
we then proceed to compute the eigen decomposition as follows
A’ΛA where we define Λ as orthogonal while A is n by n matrix whose
diagonal r equals to rank(A)
Let λj be 1 where j is values from 1 to r and λj=0 j=1,2---n and χ is the Chi-
square distribution with Ni degrees of freedom and no centrality parameter ’ A 
Then χ is A’Y
We proceed as follows
A’YA/ δ2 = (A’YA Λ A’YA/2δ2) where the A is (n x n) matrix
=χ’ Λ χ
=∑
j=1
r χj 2
δ 2 We therefore note that this summation will go until when j equals
to r
Since χj
δ ~ N (A’, μ/ δ,1) then χj2
δ 2 ~ χ12 (A’j μ/ δ )2
We proceed next as follows

∑
j=1
r xj 2
δ 2 is approximate χ2r (λ)2
and λ2 = {(μ’A Λ A’ μ)/ 2δ2}
Take note that f χ2(x) =
1
Г ( k
2 )2
k / 2 x
k
2 −1
exp(-x/2) 0<x<∞
Therefore,
= χ2r (μ’ Aμ/ 2δ2)
which can be written as χ2 (r, μ’ Aμ/ 2δ2)