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Chapter 10: Problem 1

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

Short Answer

Expert verified
Kendall's tau, being a measure of correlation, lies between -1 and 1. This is derived from its formula and the definition of concordant and discordant pairs. The value of \(\tau\) is 1 when all pairs are concordant and -1 when all are discordant, proving \(-1 \leq \tau \leq 1\).

Step by step solution

01

Understand Kendall's Tau

Kendall's Tau is a measure of correlation between two ranked variables. It's named after Maurice Kendall who introduced it in 1938. The formula for computing this correlation coefficient is: \[\tau = \frac{(\text{number of concordant pairs}) - (\text{number of discordant pairs})}{0.5 \cdot n \cdot (n-1)}\]where 'n' is the total number of observations. Concordant pairs are those pairs in which the ranking of both variables increase or decrease together. Meanwhile, discordant pairs are those pairs in which the ranking of one variable increases while the ranking of the other variable decreases.
02

Deriving the Range of Kendall's Tau

It can be observed from the formula that the maximum number of either concordant or discordant pairs can be \(0.5 \cdot n \cdot (n-1)\). Therefore, in the worst-case scenario, all the pairs are either concordant or discordant, making the value of \( \tau \) either -1 or 1, respectively.
03

Proof of Kendall's Tau Range

Therefore, as seen, in the best-case scenario, there are no discordant pairs (all pairs are concordant), which leads to Kendall's \( \tau = 1 \). Conversely, in the worst-case scenario, all pairs are discordant (no pairs are concordant), which leads to Kendall's \( \tau = -1 \). Given these two bounds, and all possible combinations of concordant and discordant pairs, we have demonstrated that Kendall's tau always lies within the range \(-1 \leq \tau \leq 1\). This completes the proof of the Kendall tau range.

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Key Concepts

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

Correlation Coefficient
The correlation coefficient is a statistical measure that calculates the strength, direction, and relationship between two variables. In the context of Kendall's Tau, it specifically addresses the correlation between two ranked variables. To grasp this concept, think of two lists of rankings, such as sports teams in a league or the popularity of products. If a high rank in one list tends to be associated with a high (or low) rank in the other list, then there is a strong correlation. The coefficient ranges from -1 to +1.

A value of +1 indicates a perfect positive correlation, meaning as one variable increases, so does the other. Conversely, a -1 indicates a perfect negative correlation, showing that as one variable increases, the other decreases. A value of 0 signifies no correlation; the variables do not affect each other. Kendall's Tau is unique as it's a non-parametric measure, meaning it does not assume a normal distribution of the variables, making it useful for ordinal data where the assumption of normality cannot be made.
Concordant and Discordant Pairs
In the evaluation of Kendall's Tau, the concepts of concordant and discordant pairs are paramount. Concordant pairs occur when the order of the rankings for two pairs is consistent. For instance, if one food item is both less expensive and more nutritious than another, these two items form a concordant pair. On the other hand, discordant pairs are exactly the opposite; if one item is less expensive but less nutritious when compared to another, they form a discordant pair.

For practical application, imagine you've ranked your top five movies. If a friend has a similar ranking, most pairs of movies you compare will be concordant. Kendall's Tau quantifies how often this concordance occurs across all pairs. The balance between concordant and discordant pairs鈥攚here one set does not greatly outnumber the other鈥攍eads to a Tau value close to 0, indicating a weak or nonexistent relationship between the variables under study.
Non-parametric Statistics
Non-parametric statistics refer to a branch of statistics that is not reliant on data belonging to any particular probability distribution. These methods are particularly useful when dealing with ordinal data or when the assumptions necessary for parametric tests (like normal distribution) are violated.

Kendall's Tau is an example of a non-parametric statistical measure because it does not require the assumption of normality of the data. It depends solely on the ranks of the data points, not their actual values. This attribute makes it highly versatile and robust for real-world data, which is often not perfectly normally distributed. Non-parametric methods like Kendall's Tau are valuable tools because they can be applied more broadly and with fewer assumptions about the data, which is especially important in fields like social sciences and market research where data often defy parametric norms.

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Most popular questions from this chapter

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample that follows the location model (10.2.1). In this exercise we want to compare the sign tests and \(t\) -test of the hypotheses \((10.2 .2) ;\) so we assume the random errors \(\varepsilon_{i}\) are symmetrically distributed about \(0 .\) Let \(\sigma^{2}=\operatorname{Var}\left(\varepsilon_{i}\right) .\) Hence the mean and the median are the same for this location model. Assume, also, that \(\theta_{0}=0 .\) Consider the large sample version of the \(t\) -test, which rejects \(H_{0}\) in favor of \(H_{1}\) if \(\bar{X} /(\sigma / \sqrt{n})>z_{\alpha}\). (a) Obtain the power function, \(\gamma_{t}(\theta)\), of the large sample version of the \(t\) -test. (b) Show that \(\gamma_{t}(\theta)\) is nondecreasing in \(\theta\). (c) Show that \(\gamma_{t}\left(\theta_{n}\right) \rightarrow 1-\Phi\left(z_{\alpha}-\sigma \theta^{*}\right)\), under the sequence of local alternatives \((10.2 .13)\) (d) Based on part (c), obtain the sample size determination for the \(t\) -test to detect \(\theta^{*}\) with approximate power \(\gamma^{*}\). (e) Derive the \(\operatorname{ARE}(S, t)\) given in \((10.2 .27)\).

_{j}\left\\{R\left(Y_{j}\right)>\frac鈥� # Consider the sign scores test procedure discussed in Example \(10.5 .4\). (a) Show that \(W_{S}=2 W_{S}^{*}-n_{2}\), where \(W_{S}^{*}=\\#_{j}\left\\{R\left(Y_{j}\right)>\frac{n+1}{2}\right\\} .\) Hence \(W_{S}^{*}\) is an equivalent test statistic. Find the null mean and variance of \(W_{S}\). (b) Show that \(W_{S}^{*}=\\#_{j}\left\\{Y_{j}>\theta^{*}\right\\}\), where \(\theta^{*}\) is the combined sample median. (c) Suppose \(n\) is even. Letting \(W_{X S}^{*}=\\#_{i}\left\\{X_{i}>\theta^{*}\right\\}\), show that we can table \(W_{S}^{*}\) in the following \(2 \times 2\) contingency table with all margins fixed: $$ \begin{array}{|c|c|c|c|} \hline & Y & X & \\ \hline \text { No. items }>\theta^{*} & W_{S}^{*} & W_{X S}^{*} & \frac{n}{2} \\\ \hline \text { No. items }<\theta^{*} & n_{2}-W_{S}^{*} & n_{1}-W_{X S}^{*} & \frac{n}{2} \\ \hline & n_{2} & n_{1} & n \\ \hline \end{array} $$ Show that the usual \(\chi^{2}\) goodness-of-fit is the same as \(Z_{S}^{2}\), where \(Z_{S}\) is the standardized \(z\) -test based on \(W_{S}\). This is often called Mood's median test; see Example \(10.5 .4\).

Suppose the random variable \(e\) has cdf \(F(t)\). Let \(\varphi(u)=\sqrt{12}[u-(1 / 2)]\), \(0

By considering the asymptotic power lemma, Theorem \(10.4 .2\), show that the equal sample size situation \(n_{1}=n_{2}\) is the most powerful design among designs with \(n_{1}+n_{2}=n, n\) fixed, when level and alternatives are also fixed. Hint: Show that this problem is equivalent to maximizing the function $$ g\left(n_{1}\right)=\frac{n_{1}\left(n-n_{1}\right)}{n^{2}} $$ and then obtain the result.

Let \(x_{1}, x_{2}, \ldots, x_{n}\) be a realization of a random sample. Consider the Hodges-Lehmann estimate of location given in expression (10.9.4). Show that the breakdown point of this estimate is \(0.29 .\) Hint: Suppose we corrupt \(m\) data points. We need to determine the value of \(m\) that results in corruption of one-half of the Walsh averages. Show that the corruption of \(m\) data points leads to $$ p(m)=m+\left(\begin{array}{c} m \\ 2 \end{array}\right)+m(n-m) $$ corrupted Walsh averages. Hence the finite sample breakdown point is the "correct" solution of the quadratic equation \(p(m)=n(n+1) / 4\).
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