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Chapter 5: Problem 25

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be independent, \(N(0,1)\)-distributed random variables, and set \(\bar{X}_{k}=\frac{1}{k-1} \sum_{i=1}^{k-1} X_{i}, 2 \leq k \leq n\). Show that $$ Q=\sum_{k=2}^{n} \frac{k-1}{k}\left(X_{k}-\bar{X}_{k}\right)^{2} $$ is \(\chi^{2}\)-distributed. What is the number of degrees of freedom?

Short Answer

Expert verified
Q is \(\chi^2\)-distributed with \(n-1\) degrees of freedom.

Step by step solution

01

Understanding the Problem

We need to show that the expression \( Q = \sum_{k=2}^{n} \frac{k-1}{k}\left(X_{k}-\bar{X}_{k}\right)^{2} \) is \( \chi^2 \)-distributed. The independent random variables \(X_{1}, X_{2}, \ldots, X_{n}\) are each \(N(0,1)\)-distributed, and \(\bar{X}_{k}\) is the average of these variables up to \(k-1\). We need to find the degrees of freedom as well.
02

Simplifying \(\bar{X}_{k}\)

The expression for \(\bar{X}_{k}\) is given by \(\bar{X}_{k} = \frac{1}{k-1} \sum_{i=1}^{k-1} X_{i}\). This is the average of \(X_1, X_2, \ldots, X_{k-1}\), which are all independent and \(N(0,1)\) distributions.
03

Calculating \(X_k - \bar{X}_k\)

Since \(X_k\) is \(N(0,1)\) and \(\bar{X}_k\) is a sum of \((k-1)\) independent variables divided by \(k-1\), \(X_k - \bar{X}_k\) is the difference of two Gaussian random variables. The distribution of the difference between a \(N(0,1)\) random variable and a \(N(0, \frac{1}{k-1})\) random variable is \(N(0, 1 + \frac{1}{k-1})\).
04

Simplifying \(Q\)

Rewriting \(Q\) with the distributed terms: \[ Q = \sum_{k=2}^{n} \frac{k-1}{k} \left(X_{k} - \bar{X}_{k}\right)^{2} = \sum_{k=2}^{n} \frac{k-1}{k} \cdot N(0, 1 + \frac{1}{k-1})^2. \] This simplifies because each term is a squared standard normal component scaled by \(\frac{k-1}{k}\).
05

Identifying \(\chi^2\) Distribution

Since \(\frac{k-1}{k}(X_k - \bar{X}_k)^2\) is a transformation of normally distributed variables, it follows a chi-squared distribution according to the properties of normal distributions and chi-squared distributions. It is summed over \(k=2\) to \(n\), resulting in \(n-1\) independent \(\chi^2\) components.
06

Conclusion: Degrees of Freedom

The expression \(Q\) follows a \(\chi^2\) distribution with \(n-1\) degrees of freedom as there are \(n-1\) terms in the summation, each contributing one degree of freedom.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Degrees of Freedom
In statistics, the concept of "Degrees of Freedom" refers to the number of independent values or quantities that can vary in the calculation of a statistic. When dealing with the chi-squared distribution, each component of the sum contributes one degree of freedom. This is because each is derived from an independent normally distributed variable. In the current problem, the sum is over expressions that are transformations of independent normal variables.
  • Each subtracted mean, \(\bar{X}_k\), takes away one degree of freedom due to it being based on \(k-1\) terms.
  • The total is \(n-1\), which reflects the number of degrees we can choose freely.
Understanding how to calculate degrees of freedom is essential for analyzing data with any statistical model.
Normal Distribution
A normal distribution, often called a Gaussian distribution, is a probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean.The standard normal distribution has a mean of 0 and a standard deviation of 1. This is denoted as \(N(0, 1)\).
  • It forms a bell-shaped curve where the majority of outcomes are near the mean.
  • In our problem, each \(X_i\) is sampled from this \(N(0, 1)\) distribution.
The normal distribution is a crucial concept because many statistical tests assume data follows this pattern. It also underpins other distributions such as the chi-squared distribution, which is used here in the transformation and summation of these variables.
Independent Random Variables
Independent random variables are those where the occurrence of one event does not affect the probability of another. This is critical because it ensures that the information about one variable provides no information about the other.In our exercise, each \(X_i\) is independent, meaning:
  • The value of one \(X_i\) does not influence another \(X_j\).
  • This independence is crucial for the properties of the linear combinations and transformations applied to them.
The independence makes it possible to apply statistical rules for transformations. This allows us to derive new distributions like the chi-squared distribution through the summation of independent squared terms.
Statistical Transformations
Statistical transformations involve changing a random variable into another form. These transformations are defined so that a variable retains some meaningful properties, such as converting standard normal variables into chi-squared variables.In the given solution, the transformation of \(X_k - \bar{X}_k\) involves:
  • Finding a variance-adjusted difference between each \(X_k\) and its mean \(\bar{X}_k\).
  • Summing the scaled squared results according to the formula \(\sum_{k=2}^{n} \frac{k-1}{k}\left(X_{k}-\bar{X}_{k}\right)^{2}\).
These transformations follow theoretical statistical distributions. Thus, the expression becomes chi-squared due to the aggregate effect of these transformations.

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Most popular questions from this chapter

The moment generating function of \((X, Y, Z)^{\prime}\) is $$ \psi(s, t, u)=\exp \left\\{\frac{s^{2}}{2}+t^{2}+2 u^{2}-\frac{s t}{2}+\frac{3 s u}{2}-\frac{t u}{2}\right\\} . $$ Determine the conditional distribution of \(X\) given that \(X+Z=0\) and \(Y+Z=1\).

Let \(X\) and \(Y\) be jointly normal with means 0 , variances 1 , and correlation coefficient \(\rho\). Compute the moment generating function of \(X \cdot Y\) for (a) \(\rho=0\), and (b) general \(\rho\).

Suppose \(X\) and \(Y\) have a two-dimensional normal distribution with means 0, variances 1, and correlation coefficient \(\rho,|\rho|<1\). Let \((R, \Theta)\) be the polar coordinates. Determine the distribution of \(\Theta\).

Let \(X_{1}, X_{2}\), and \(X_{3}\) be independent, \(N(1,1)\)-distributed random variables. Set \(U=X_{1}+X_{2}+X_{3}\) and \(V=X_{1}+2 X_{2}+3 X_{3}\). Determine the constants \(a\) and \(b\) so that \(E(U-a-b V)^{2}\) is minimized.

Let \(\mathbf{X} \in N(\mathbf{0}, \mathbf{\Lambda})\), where $$ \boldsymbol{\Lambda}=\left(\begin{array}{rrr} 1 & -\frac{1}{2} & \frac{3}{2} \\ -\frac{1}{2} & 2 & -1 \\ \frac{3}{2} & -1 & 4 \end{array}\right) $$ Determine the conditional distribution of \(\left(X_{1}, X_{1}+X_{2}\right)^{\prime}\) given that \(X_{1}+X_{2}+X_{3}=0\).
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