/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 1 Show that Kendall's \(\tau\) sat... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Problem 1

Show that Kendall's \(\tau\) satisfies the inequality \(-1 \leq \tau \leq 1\).

Short Answer

Expert verified
Kendall's \(\tau\) is a measure of rank correlation which estimates the number of pairs in agreement minus those in disagreement, divided by the total number of pairs. Given this, when all pairs are either concordant or discordant, \(\tau\) becomes 1 and -1, respectively. All other conditions (a mix of concordant and discordant pairs) yield a \(\tau\) value that falls between -1 and 1, hence proving that \(-1 \leq \tau \leq 1\).

Step by step solution

01

Understanding Kendall's Tau

Kendall's \(\tau\) is a measure of rank correlation. It evaluates relationships between variables and gives a value between -1 and 1. The \(\tau\) value estimates the number of pairs which are in agreement minus the number in disagreement, all divided by the total number of pairs. More formally, for two observed variables \(x\) and \(y\) and their corresponding pairs \((x_i,x_j)\) and \((y_i,y_j)\), we have: \(\tau = \frac{\text{Number of concordant pairs - Number of discordant pairs}}{\text{Total number of pairs}}\)
02

Explaining the Ranges of Tau

The numerator of the definition contains the difference between the number of concordant and discordant pairs. When all pairs are concordant (i.e., agree on their relative rankings), the numerator and denominator become equal, and hence \(\tau = 1\). When all pairs are discordant (i.e., disagree on their relative rankings), the numerator is equal to \(-\)denominator and hence \(\tau = -1\). In most cases, we will have a mixture of concordant and discordant pairs that yield a value between -1 and 1.
03

Concluding the proof

By understanding the nature of Kendall's \(\tau\) as descibred in the previous steps, it can be seen that the maximum value is gained when all pairs are concordant (\(\tau = 1\)), whereas the minimum value obtained is when all pairs are discordant (\(\tau = -1\)). Consequently, all other values of \(\tau\) fall between -1 and 1, including these limits. This completes the proof that \(-1 \leq \tau \leq 1\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

In this section, as discussed above expression \((10.5 .2)\), the scores \(a_{\varphi}(i)\) are generated by the standardized score function \(\varphi(u) ;\) that is, \(\int_{0}^{1} \varphi(u) d u=0\) and \(\int_{0}^{1} \varphi^{2}(u) d u=1 .\) Suppose that \(\psi(u)\) is a square-integrable function defined on the interval \((0,1)\). Consider the score function defined by $$ \varphi(u)=\frac{\psi(u)-\bar{\psi}}{\int_{0}^{1}[\psi(v)-\bar{\psi}]^{2} d v} $$ where \(\bar{\psi}=\int_{0}^{1} \psi(v) d v\). Show that \(\varphi(u)\) is a standardized score function.

(a) For \(n=3\), expand the mgf (10.3.6) to show that the distribution of the signed-rank Wilcoxon is given by $$ \begin{array}{|l|ccccccc|} \hline j & -6 & -4 & -2 & 0 & 2 & 4 & 6 \\ \hline P(T=j) & \frac{1}{8} & \frac{1}{8} & \frac{1}{8} & \frac{2}{8} & \frac{1}{8} & \frac{1}{8} & \frac{1}{8} \\ \hline \end{array} $$$$ \text { (b) Obtain the distribution of the signed-rank Wilcoxon for } n=4 \text { . } $$

Let \(X\) be a random variable with cdf \(F(x)\) and let \(T(F)\) be a functional. We say that \(T(F)\) is a scale functional if it satisfies the three properties $$ T\left(F_{a X}\right)=a T\left(F_{X}\right), \quad \text { for } a>0 $$ (ii) \(T\left(F_{X+b}\right)=T\left(F_{X}\right), \quad\) for all \(b\) $$ \text { (iii) } T\left(F_{-X}\right)=T\left(F_{X}\right) $$ Show that the following functionals are scale functionals. (a) The standard deviation, \(T\left(F_{X}\right)=(\operatorname{Var}(X))^{1 / 2}\). (b) The interquartile range, \(T\left(F_{X}\right)=F_{X}^{-1}(3 / 4)-F_{X}^{-1}(1 / 4)\).

Optimal signed-rank based methods also exist for the one-sample problem. In this exercise, we briefly discuss these methods. Let \(X_{1}, X_{2}, \ldots, X_{n}\) follow the location model $$ X_{i}=\theta+e_{i} $$ where \(e_{1}, e_{2}, \ldots, e_{n}\) are iid with pdf \(f(x)\), which is symmetric about 0 ; i.e., \(f(-x)=\) \(f(x)\) (a) Show that under symmetry the optimal two-sample score function \((10.5 .26)\) satisfies $$ \varphi_{f}(1-u)=-\varphi_{f}(u), \quad 0

Suppose \(X\) is a random variable with mean 0 and variance \(\sigma^{2} .\) Recall that the function \(F_{x, \epsilon}(t)\) is the cdf of the random variable \(U=I_{1-\epsilon} X+\left[1-I_{1-\epsilon}\right] W\), where \(X, I_{1-\epsilon}\), and \(W\) are independent random variables, \(X\) has \(\operatorname{cdf} F_{X}(t), W\) has cdf \(\Delta_{x}(t)\), and \(I_{1-\epsilon}\) has a binomial \((1,1-\epsilon)\) distribution. Define the functional \(\operatorname{Var}\left(F_{X}\right)=\operatorname{Var}(X)=\sigma^{2} .\) Note that the functional at the contaminated \(\operatorname{cdf} F_{x, \epsilon}(t)\) has the variance of the random variable \(U=I_{1-\epsilon} X+\left[1-I_{1-\epsilon}\right] W .\) To derive the influence function of the variance, perform the following steps: (a) Show that \(E(U)=\epsilon x\). (b) Show that \(\operatorname{Var}(U)=(1-\epsilon) \sigma^{2}+\epsilon x^{2}-\epsilon^{2} x^{2}\). (c) Obtain the partial derivative of the right side of this last equation with respect to \(\epsilon\). This is the influence function.
See all solutions

Recommended explanations on Math Textbooks

Discrete Mathematics

Read Explanation

Applied Mathematics

Read Explanation

Logic and Functions

Read Explanation

Probability and Statistics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Pure Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.