91Ó°ÊÓ

Suppose we take a random sample of size \(n\) from a continuous distribution having median 0 so that the probability of any one observation being positive is .5. We now disregard the signs of the observations, rank them from smallest to largest in absolute value, and then let \(W=\) the sum of the ranks of the observations having positive signs. For example, if the observations are \(-.3,+.7\), \(+2.1\), and \(-2.5\), then the ranks of positive observations are 2 and 3, so \(W=5\). In Chapter \(14, W\) will be called Wilcoxon's signed-rank statistic. W can be represented as follows: $$ \begin{aligned} W &=1 \cdot Y_{1}+2 \cdot Y_{2}+3 \cdot Y_{3}+\cdots+n \cdot Y_{n} \\ &=\sum_{i=1}^{n} i \cdot Y_{i} \end{aligned} $$ where the \(Y_{i}^{\prime}\) s are independent Bernoulli rv's, each with \(p=.5\left(Y_{i}=1\right.\) corresponds to the observation with rank \(i\) being positive). Compute the following: a. \(E\left(Y_{i}\right)\) and then \(E(W)\) using the equation for \(W\) [Hint: The first \(n\) positive integers sum to \(n(n+1) / 2 .]\) b. \(V\left(Y_{i}\right)\) and then \(V(W)\) [Hint: The sum of the squares of the first \(n\) positive integers is \(n(n+1)(2 n+1) / 6 .]\)

Step by step solution

01

Compute E(Y_i)

Since each observation being positive is a Bernoulli random variable with probability \( p = 0.5 \), the expectation (mean) for each \( Y_i \) is given by the formula: \( E(Y_i) = p = 0.5 \).

02

Compute E(W)

Using the equation \( W = \sum_{i=1}^{n} i \cdot Y_{i} \), we apply the linearity of expectation to find \( E(W) \). This becomes \( E(W) = \sum_{i=1}^{n} i \cdot E(Y_i) = \sum_{i=1}^{n} i \cdot 0.5 \). The sum \( \sum_{i=1}^{n} i \) is the sum of the first \( n \) positive integers, which is \( n(n+1)/2 \). Thus, \( E(W) = 0.5 \times n(n+1)/2 = n(n+1)/4 \).

03

Compute V(Y_i)

For a Bernoulli random variable, the variance is given by \( V(Y_i) = p(1-p) = 0.5 \times 0.5 = 0.25 \).

04

Compute V(W)

For \( V(W) \), we use the formula for the variance of a weighted sum of independent random variables: \( V(W) = \sum_{i=1}^{n} i^2 \cdot V(Y_i) = \sum_{i=1}^{n} i^2 \cdot 0.25 \). The sum \( \sum_{i=1}^{n} i^2 \) is the sum of the squares of the first \( n \) positive integers, which is \( n(n+1)(2n+1)/6 \). Therefore, \( V(W) = 0.25 \times n(n+1)(2n+1)/6 = n(n+1)(2n+1)/24 \).

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Key Concepts

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

Bernoulli Random Variables

Bernoulli random variables are the simplest type of random variable in probability theory, only taking the values 0 and 1. They are named after Swiss mathematician Jacob Bernoulli. When we talk about a Bernoulli random variable, we often denote it as a variable that represents the outcome of a single trial of an experiment. This trial can only have two outcomes, commonly referred to as "success" and "failure."

• The probability of success is usually denoted by \( p \).
• The probability of failure is \( 1-p \).

When we consider them in the context of ranks, like in the Wilcoxon signed-rank test, each observation's rank is treated as a Bernoulli trial regarding whether that observation is positive or not. So, each \( Y_i \) is a Bernoulli random variable representing whether an observation has a positive sign, with \( p = 0.5 \) in our given exercise.
Understanding Bernoulli random variables is crucial when working with the Wilcoxon signed-rank test, as it helps in defining how we compute expectations and variances, given their trial-like nature.

Expectation and Variance

Expectation and variance are fundamental concepts in statistics and probability that help us understand the behavior of random variables. Let’s break these down further to see how they relate to our problem:

• The **expectation** (or expected value) of a random variable gives us a measure of the center of its distribution. For a Bernoulli random variable \( Y_i \), the expectation \( E(Y_i) \) is simply the probability \( p \). In our example, since the probability that an observation is positive is \( 0.5 \), it follows that \( E(Y_i) = 0.5 \).


• The **variance**, on the other hand, measures the spread of the distribution around its mean. For Bernoulli random variables, the variance \( V(Y_i) \) is given by \( p(1-p) \). Thus, for our Bernoulli variable where \( p = 0.5 \), the variance becomes \( 0.5 imes 0.5 = 0.25 \).

Now, moving to our sum of ranks, \( W \), it’s a combination of these Bernoulli variables weighed by their ranks. Therefore, the expectation \( E(W) \) is calculated through the formula \( E(W) = \sum_{i=1}^{n} i \cdot E(Y_i) \), ultimately resulting in \( n(n+1)/4 \). Similarly, the variance \( V(W) \) is calculated as \( V(W) = \sum_{i=1}^{n} i^2 \cdot V(Y_i) \), yielding \( n(n+1)(2n+1)/24 \).
These calculations are vital for understanding how likely different outcomes of \( W \) are, especially in hypothesis testing scenarios.

Rank Statistics

Rank statistics play an important role in non-parametric statistics, which particularly shines in the situation where data do not necessarily follow a normal distribution. Ranking is one approach to handling such data efficiently. Instead of considering the actual measured values, rank statistics focuses on their ordinal positions when sorted.
In the context of the Wilcoxon Signed-Rank Test, each observation is ranked based on its absolute magnitude, regardless of its sign. For example, if you have a set of data points, both negative and positive, you ignore their signs, rank them by their size, and then restore the signs to their ranks when reconstructing \( W \).

• When we compute \( W \), we sum up the ranks corresponding to the observations that are positive, effectively incorporating both the magnitude and sign.
• Rank sum statistics like \( W \) are robust against outliers since ranks do not get disproportionately influenced by extreme values.

Understanding rank statistics is crucial because it allows us to use data effectively even when assumptions concerning their distribution are violated. This makes the Wilcoxon Signed-Rank Test particularly powerful as it uses rank statistics to assess differences in median values among paired measurements.