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(a) For \(n=3\), expand the mgf (10.3.6) to show that the distribution of the signed-rank Wilcoxon is given by \begin{tabular}{|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{tabular} (b) Obtain the distribution of the signed-rank Wilcoxon for \(n=4\).

Short Answer

Expert verified
The distribution of the Wilcoxon signed-rank for n = 3 is shown in the table with respective probabilities. For n = 4, an equivalent table can be derived by expanding the moment generating function and finding the probabilities of the \(j\) values.

Step by step solution

01

Understanding and Expanding The Moment Generating Function (mgf)

The moment generating function (mgf) \(M(T, t)\) of a random variable \(T\) is defined as: \(M(T, t) = E[e^{tT}]\), where \(E[.]\) is the expected value. First, recognize that the Moment Generating Function (mgf) for the Wilcoxon signed-ranks is given by \(M(T,t) = [(e^t+ e^{-t})/2]^n\). This function has to be expanded for different values of \(n\) as per the problem statement. Given \(n=3\), we can expand the Moment Generating Function (mgf) as follows, which gives us: \(M(T,t) = [(e^t + e^{-t})/2]^3\).
02

Extraction of Possibilities and Probabilities for n=3

The next step after expansion is to extract possible values of \(j\) and their respective probabilities \(P(T=j)\). This will involve identifying different powers of \(e^t\) and their coefficients in the expanded equation of \(M(T, t)\). Once we’ve expanded and simplified the equation, we will find the probabilities associated with each value of \(j\), which can be found by dividing the coefficient of \(e^{jt}\) by \(2^n\). The table given in the question already provides a list of possible \(j\) values and their respective probabilities for \(n=3\).
03

Determination of Distribution for n=4

The process to obtain the distribution for \(n=4\) is similar to the previous step. Here, the moment generating function (mgf) when \(n=4\) will be \(M(T,t) = [(e^t + e^{-t})/2]^4\). Expanding and simplifying this equation similar to what was done in the previous steps will render the probabilities of different \(j\) values by dividing each coefficient of \(e^{jt}\) by \(2^n\).

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Key Concepts

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

Moment Generating Function (mgf)
The Moment Generating Function (mgf) plays a critical role in probability and statistics. Imagine having a tool that could summarize all the information about the distribution of a random variable in one swooping function. That's precisely what the mgf does.

The mgf of a random variable, say, T, is denoted as M(T, t) and can be expressed as \(M(T, t) = E[e^{tT}]\), where \(E[.]\) stands for the expected value, and \(t\) is a real number. Put simply, the mgf is the expected value of \(e^{tT}\), the exponential function raised to the product of \(t\) and the random variable T.

When you compute the mgf for a specific value of \(n\), you can expand this function to reveal the weights of all possible values that T could take. It's like unboxing the essence of a probability distribution. For example, with the signed-rank Wilcoxon test for \(n=3\), the mgf \(M(T,t) = [(e^t + e^{-t})/2]^3\) can be expanded to uncover the probabilities for different ranks.

It's important to grasp that the mgf is more than just a formula; it's a framework to derive numerous properties of a distribution, such as mean and variance, and it plays a fundamental role in hypothesis testing, like in the Wilcoxon signed-rank test.
Probability Distribution
A probability distribution is the heartbeat of any statistical analysis. It describes how the values of a random variable are spread or distributed. This is crucial for understanding the behavior of random phenomena.

Visualize a probability distribution as a map that shows where you're likely to find the values of a random variable. The higher the peak on this map, the more likely you are to encounter that particular value. In the context of the signed-rank Wilcoxon test, the probability distribution tells us how likely it is to observe particular ranks. For instance, the table provided for \(n=3\) shows a uniform distribution of ranks, except for the rank 0, which has double the probability of occurring compared to the others.

To obtain a probability distribution, you identify all possible outcomes and assign a probability to each. As demonstrated in earlier steps, this can be achieved by expanding the mgf and interpreting the coefficients in terms of probabilities. The distribution can be discrete, as with the Wilcoxon test, or continuous. Each type provides a distinct way of understanding random variables and helps in predicting future observations.
Random Variable
A random variable is like a container that holds different outcomes of randomness with certain likelihoods attached to them. It's a fundamental concept in statistics that allows us to quantify random events.

In formal terms, a random variable is a function that assigns real numbers to outcomes of a random process. It's 'random' because we can't predict the exact value it will take, and it’s 'variable' because it can take on a range of values. Consider our Wilcoxon test's ranks as a random variable. You know which ranks are possible (from -6 to 6 for \(n=3\)), but you don't know which rank will show up after an experiment.

There are two main types of random variables: discrete and continuous. In the case of the signed-rank Wilcoxon test, we're dealing with a discrete random variable because the ranks are countable outcomes. By understanding the behavior of this random variable through its distribution, one can gain insights into the nature of the underlying statistical test - the Wilcoxon signed-rank test in this case - and hence make well-informed decisions or predictions in statistical inference.

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Most popular questions from this chapter

Let \(\widehat{F}_{n}(x)\) denote the empirical cdf of the sample \(X_{1}, X_{2}, \ldots, X_{n} .\) The distribution of \(\hat{F}_{n}(x)\) puts mass \(1 / n\) at each sample item \(X_{i} .\) Show that its mean is \(\bar{X}\). If \(T(F)=F^{-1}(1 / 2)\) is the median, show that \(T\left(\widehat{F}_{n}\right)=Q_{2}\), the sample median.

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 $$ \text { (i) } T\left(F_{a X}\right)=a T\left(F_{X}\right), \text { for } a>0 $$ (ii) \(T\left(F_{X+b}\right)=T\left(F_{X}\right), \quad\) for all \(b\) $$ \text { (iii) } T\left(F_{-X}\right)=T\left(F_{X}\right) \text { . } $$ 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 $$ X_{i}=\theta+e_{i}, \quad(10.5 .39) $$ 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 $$ \varphi_{f}(1-u)=-\varphi_{f}(u), \quad 00 $$ Our decision rule for the statistic \(W_{\varphi^{+}}\) is to reject \(H_{0}\) in favor of \(H_{1}\) if \(W_{\varphi^{+}} \geq\) \(k\), for some \(k\). Write \(W_{\varphi^{+}}\) in terms of the anti-ranks, \((10.3 .5) .\) Show that \(W_{\varphi^{+}}\) is distribution-free under \(H_{0}\). (f) Determine the mean and variance of \(W_{\varphi^{+}}\) under \(H_{0}\). (g) Assuming that, when properly standardized, the null distribution is asymptotically normal, determine the asymptotic test.

Consider the location model as defined in expression (10.9.1). Let $$ \widehat{\theta}=\operatorname{Argmin}_{\theta}\|\mathbf{X}-\theta \mathbf{1}\|_{\mathrm{LS}}^{2} $$ where \(\|\cdot\|_{\mathrm{LS}}^{2}\) is the square of the Euclidean norm. Show that \(\widehat{\theta}=\bar{x}\).

Consider the general score rank correlation coefficient \(r_{a}\) defined in Exercise 10.8.5. Consider the null hypothesis \(H_{0}: X\) and \(Y\) are independent. (a) Show that \(E_{H_{0}}\left(r_{a}\right)=0\). (b) Based on part (a) and \(H_{0}\), as a first step in obtaining the variance under \(H_{0}\), show that the following expression is true: $$ \operatorname{Var}_{H_{0}}\left(r_{a}\right)=\frac{1}{s_{a}^{4}} \sum_{i=1}^{n} \sum_{j=1}^{n} E_{H_{0}}\left[a\left(R\left(X_{i}\right)\right) a\left(R\left(X_{j}\right)\right)\right] E_{H_{0}}\left[a\left(R\left(Y_{i}\right)\right) a\left(R\left(Y_{j}\right)\right)\right] $$ (c) To determine the expectation in the last expression, consider the two cases \(i=j\) and \(i \neq j\). Then using uniformity of the distribution of the ranks, show that $$ \operatorname{Var}_{H_{0}}\left(r_{a}\right)=\frac{1}{s_{a}^{4}} \frac{1}{n-1} s_{a}^{4}=\frac{1}{n-1} $$

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