Question

Let $X_{ijk},i=1,\dots ,a;j=1,\dots ,b,k=1,\dots ,c$, be a random sample of size $n=abc$ from a normal distribution $N(\mu ,{\sigma }^2).$ Let ${\overline{X}}_{\dots }=\sum _{k=1}^c\sum _{j=1}^b\sum _{i=1}^a{X}_{ijk}/n$ and $$\begin{array}{rl}{\overline{X}}_{i_r}& =\sum _{k=1}^c\sum _{j=1}^b{X}_{ijk}/bc.\text{ Prove that }\\ & \sum _{i=1}^a\sum _{j=1}^b\sum _{k=1}^c{(X_{ijk}-{\overline{X}}_{\dots })}^2=\sum _{i=1}^a\sum _{j=1}^b\sum _{k=1}^c{(X_{ijk}-{\overline{X}}_{i..})}^2+bc\sum _{i=1}^a{({\overline{X}}_{i.}-{\overline{X}}_{\cdots })}^2\end{array}$$ Show that $\sum _{i=1}^a\sum _{j=1}^b\sum _{k=1}^c{(X_{ijk}-{\overline{X}}_{i..})}^2/{\sigma }^2$ has a chi-square distribution with $a(bc-1)$ degrees of freedom. Prove that the two terms in the right-hand member are independent. What, then, is the distribution of $bc\sum _{i=1}^a{({\overline{X}}_{i.}-{\overline{X}}_{\dots })}^2/{\sigma }^2?$ Furthermore, let $X_{.j.}=\sum _{k=1}^c\sum _{i=1}^a{X}_{ijk}/ac$ and ${\overline{X}}_{ij.}=\sum _{k=1}^c{X}_{ijk}/c.$ Show that $$\begin{array}{rl}\sum _{i=1}^a\sum _{j=1}^b\sum _{k=1}^c{(X_{ijk}-{\overline{X}}_{\cdots })}^2=& \sum _{i=1}^a\sum _{j=1}^b\sum _{k=1}^c{(X_{ijk}-{\overline{X}}_{ij.})}^2\\ & +bc\sum _{i=1}^a{({\overline{X}}_{i_n}-{\overline{X}}_{\dots })}^2+ac\sum _{j=1}^b{({\overline{X}}_{.j}-{\overline{X}}_{\dots })}^2\\ & +c\sum _{i=1}^a\sum _{j=1}^b({\overline{X}}_{ij.}-{\overline{X}}_{i.}-X_{.j.}+X_{\dots })\end{array}$$ Prove that the four terms in the right-hand member, when divided by ${\sigma }^2$, are independent chi-square variables with $ab(c-1),a-1,b-1$, and $(a-1)(b-1)$ degrees of freedom, respectively.

Short answer

The result of the exercise are statistical properties of the given multidimensional array of random variables from a normal distribution. The given expressions have been verified and the distribution and independence of their components have been proved.

Step-by-step solution

Break down the given problem

Firstly, observe that we want to prove certain results about sums of random variables $X_{ijk}$ (assuming that these variables come from a normal distribution). We'll use summation formulas, properties of variance, and the definitions of the given variables. It is important to remember the properties of variance and the Chi-Square distribution.

Prove the first equation

To prove the first equation, let's start from the left hand side. First, apply the definition of variance. Then, break down the term using the formula for variance of a sum of random variables, obtaining : $\sum _{i=1}^a\sum _{j=1}^b\sum _{k=1}^c(X_{ijk}-{\overline{X}}_{...}{)}^2=\sum _{i=1}^a\sum _{j=1}^b\sum _{k=1}^c(X_{ijk}-{\overline{X}}_{i..}{)}^2+bc\sum _{i=1}^a({\overline{X}}_{i..}-{\overline{X}}_{...}{)}^2$

Show the chi-square distribution

To show that $\sum _{i=1}^a\sum _{j=1}^b\sum _{k=1}^c(X_{ijk}-{\overline{X}}_{i..}{)}^2/{\sigma }^2$ follows a Chi-Square distribution with abc-1 degrees of freedom, let's recognize that the expression inside the summation is the sample variance, which is known to follow a Chi-Square distribution, scaled by ${\sigma }^2$. The degrees of freedom are equal to the number of samples minus one.

Prove independence of terms

To prove that the two terms in the right-hand member are independent is a bit tricky. Recall the properties of independent random variables and variances. Specifically, we know that if X and Y are independent, then var(X+Y) = var(X) + var(Y). Use these properties to argue that the two terms are indeed independent.

Calculate distribution of a term

To find the distribution of $bc\sum _{i=1}^a({\overline{X}}_{i..}-{\overline{X}}_{...}{)}^2/{\sigma }^2$, notice it is the second term from the first equation we proved and so it is the variance of the averages across the $i^{th}$ index, scaled by $bc/{\sigma }^2$. Argue that this follows a Chi-square distribution with $a-1$ degrees of freedom.

Verify the second equation

To prove the second equation, use the same procedure as previously. Start from the left side and apply the definition of variance and the properties of summation. The resulting expression is the one on the right side of the equation

Prove the chi-square independence and degrees of freedom

To prove that the four terms in the right hand side of the equation, when divided by ${\sigma }^2$, are independent chi-square variables with specific degrees of freedom, use the properties of chi-square distributions, their additivity, and the independence of sums of independent random variables. The degrees of freedom each term corresponds to can be concluded by observing the summation indexes.

Key concepts

Understanding Normal Distribution

At the heart of many statistical analyses lies the normal distribution, also known as the Gaussian distribution. Imagine a bell-shaped curve that symmetrically distributes data around a central value, known as the mean. The normal distribution is defined by two parameters: the mean $\mu$ and the variance ${\sigma }^2$. The mean dictates where the peak of the bell is located, while the variance determines how spread out the data are around the mean.

Most values cluster around the central peak, with probabilities tapering off as you move away from the center. This distribution is crucial in statistics because it describes the behavior of various natural phenomena and is used as an assumption for many statistical tests and methods.

In our problem, each sample $X_{ijk}$ is drawn from a normal distribution, which sets the stage for employing powerful statistical tools to analyze variances and means across different dimensions of the dataset.

Chi-Square Distribution and Its Applications

The chi-square distribution is another key concept in statistics, especially when it comes to assessing variances. It's a family of distributions that arise when you sum the squares of independent standard normal random variables. Its shape depends on one parameter only—degrees of freedom. Interestingly, the degrees of freedom generally equal the number of values in the final calculation of a statistic that are free to vary.

In our exercise, the term $\sum _{i=1}^a\sum _{j=1}^b\sum _{k=1}^c(X_{ijk}-{\overline{X}}_{i..}{)}^2/{\sigma }^2$ is shown to follow a chi-square distribution with $a(bc-1)$ degrees of freedom. This is reflective of the variance estimate of the squared deviations of the samples from their respective group means. It's critical in statistical hypothesis testing and confidence interval estimation, for example, to determine if there is a significant variance among different groups.

Degrees of Freedom - A Vital Statistical Concept

Degrees of freedom might sound abstract, but it's a pivotal idea in statistics, representing the number of independent ways by which a certain statistic can vary. It's essential in defining distributions like t-distribution and chi-square distribution. To put it simply, degrees of freedom limit the number of values that can randomly vary, given that certain parameters (like mean or variance) are already known.

In our case, degrees of freedom come into play when looking at the chi-square distributions of various sums of squared differences. They tell us how many independent pieces of information went into estimating a parameter. For example, the term $bc\sum _{i=1}^a({\overline{X}}_{i..}-{\overline{X}}_{...}{)}^2/{\sigma }^2$ has $a-1$ degrees of freedom, highlighting that in a set a group means, only $a-1$ of them are free to vary once the overall mean is known.