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Chapter 2: Problem 9

The coefficient of variation of a random sample \(Y_{1}, \ldots, Y_{n}\) is \(C=S / \bar{Y}\), where \(\bar{Y}\) and \(S^{2}\) are the sample average and variance. It estimates the ratio \(\psi=\sigma / \mu\) of the standard deviation relative to the mean. Show that $$ \mathrm{E}(C) \doteq \psi, \quad \operatorname{var}(C) \doteq n^{-1}\left(\psi^{4}-\gamma_{3} \psi^{3}+\frac{1}{4} \gamma_{4} \psi^{2}\right)+\frac{\psi^{2}}{2(n-1)} $$

Short Answer

Expert verified
The expressions hold approximately for large samples due to asymptotic properties.

Step by step solution

01

Understand Definitions and Notations

First, let's clearly understand and define the quantities involved. The coefficient of variation (C) is a statistical measure of the relative variability, given by \( C = \frac{S}{\bar{Y}} \), where \( \bar{Y} \) is the sample mean and \( S^2 \) is the sample variance. The ratio \( \psi \) is defined as \( \psi = \frac{\sigma}{\mu} \) where \( \sigma \) and \( \mu \) are the population standard deviation and mean, respectively. Our objective is to find the expected value and variance of \( C \) and relate them approximately to \( \psi \).
02

Express Expected Value of C

To find the expected value of \( C \), we use the relation \( \mathrm{E}(C) \approx \psi \). Since \( C \) is approximating \( \psi \), we can expect that for large \( n \), \( \mathrm{E}(C) \to \psi \). This suggests that \( \mathrm{E}(C) \approx \psi \) under the assumption that the sample mean \( \bar{Y} \) and standard deviation \( S \) reasonably estimate their population counterparts \( \mu \) and \( \sigma \).
03

Find Variance of C

Next, we aim to show the variance relation: \( \operatorname{var}(C) \doteq n^{-1}\left(\psi^{4}-\gamma_{3} \psi^{3}+\frac{1}{4} \gamma_{4} \psi^{2}\right)+\frac{\psi^{2}}{2(n-1)} \). Here, \( \gamma_{3} \) and \( \gamma_{4} \) are, respectively, the third and fourth standardized moments (skewness and kurtosis minus 3 of the population distribution). The expression here suggests a first-order approximation where the variance of \( C \) includes contributions from higher moments of the distribution. This expression involves assumptions and derivations from the moments of \( Y_i \).
04

Justify the Approximation

The approximate equality in variance expression implies that the relationship holds asymptotically (as \( n \) becomes very large). Simplifications often arise from using large-sample theory, where the law of large numbers and central limit theorem make \( \bar{Y} \) and \( S^2 \) approach their population parameters, and their moments approximate normality or exhibit specific behavior captured by the moments \( \gamma_3 \) and \( \gamma_4 \). Therefore, assumptions about unbiasedness and efficiency of estimators are involved.

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Key Concepts

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

Sample Mean
The sample mean is a fundamental concept in statistics, representing an average value of a sample data set. It is denoted by \( \bar{Y} \) and serves as an estimator for the population mean \( \mu \). Calculating the sample mean involves summing up all the data points and dividing by the number of observations, \( n \).
\[ \bar{Y} = \frac{1}{n} \sum_{i=1}^{n} Y_i \]
The sample mean provides a central value, which is used in various statistical computations.
  • It helps in understanding the distribution of data points around this central value.
  • It is essential in estimating population parameters.
  • It plays a vital role in constructing other statistical measures like the coefficient of variation, where it appears in the denominator.
Understanding the sample mean is crucial in grasping more advanced statistical concepts. Its simplicity and intuitive nature make it an accessible starting point for statistical analysis.
Sample Variance
Sample variance provides a measure of the data's spread or dispersion around the sample mean, \( \bar{Y} \). Denoted as \( S^2 \), it is calculated by averaging the squared differences between each data point and the sample mean.
\[ S^2 = \frac{1}{n-1} \sum_{i=1}^{n} (Y_i - \bar{Y})^2 \]
This formula uses \( n-1 \) instead of \( n \), making it an unbiased estimator for the population variance. The sample variance:
  • Indicates how spread out the data points are within the sample.
  • Forms the basis for calculating standard deviation, which is the square root of the variance.
  • Is essential in understanding variability, especially when analyzing the coefficient of variation as it appears in the numerator.
Mastering the concept of sample variance aids in comprehending the variability of data and serves as a foundation for understanding other variability measures.
Expected Value
The expected value, also known as the mean of a random variable, is a key concept in probability and statistics, representing the average or predicted value of a random variable over an infinite number of experiments or samples. For a discrete random variable, it is computed as:
\[ \mathrm{E}(X) = \sum (x_i \cdot P(x_i)) \]
For continuous variables, integrals are used instead of summations.
  • The expected value is a measure of central tendency, similar to the concept of mean in a set of observations.
  • It provides predictions about the potential outcomes of a random variable.
  • In the context of the coefficient of variation, it helps estimate the ratio of the standard deviation to the mean.
The expected value plays a vital role in statistical inference, as it underpins many statistical estimates and models.
Variance
Variance measures the degree of spread in a probability distribution. It indicates how much the data points are likely to deviate from the expected value. Defined for a random variable \( X \) as:
\[ \operatorname{Var}(X) = \mathrm{E}[(X - \mathrm{E}(X))^2] \]
Variance is foundational in analysis:
  • It tells us how much variability to expect in the data.
  • High variance indicates data points are widely dispersed.
  • Low variance suggests data points are closely clustered around the mean.
  • In the context of the coefficient of variation, it contributes significantly to understanding overall variability.
Being squared, variance amplifies the effect of outliers, making it a robust measure of spread in data analysis.
Standardized Moments
Standardized moments are used to describe the shape of a distribution. Third and fourth standardized moments, skewness, and kurtosis, respectively, often appear in analyzing distribution characteristics.
Skewness (c3):
- Measures the degree of asymmetry of a distribution around its mean.
- Positive skewness indicates a distribution with an elongated tail on the right.
- Negative skewness suggests a tail on the left. Kurtosis (c4):
- Indicates the "tailedness" of the distribution.
- High kurtosis implies more of the variance is due to infrequent, extreme deviations. Standardized moments help in understanding:
  • How data deviates from a normal distribution.
  • The level of asymmetry (from skewness).
  • The extremity of deviations (from kurtosis).
They are critical in advanced statistical analysis as they reveal underlying data tendencies, impacting estimates like the coefficient of variation.

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Most popular questions from this chapter

(a) Let \(X\) be the number of trials up to and including the first success in a a sequence of independent Bernoulli trials having success probability \(\pi\). Show that \(\operatorname{Pr}(X=k)=\) \(\pi(1-\pi)^{k-1}, k=1,2, \ldots\), and deduce that \(X\) has moment-generating function \(\pi e^{t} /\\{1-\) \(\left.(1-\pi) e^{t}\right\\}\); hence find its mean and variance. \(X\) has the geometric distribution. (b) Now let \(Y_{n}\) be the number of trials up to and including the \(n\)th success in such a sequence of trials. Show that $$ \operatorname{Pr}\left(Y_{n}=k\right)=\left(\begin{array}{l} k-1 \\ n-1 \end{array}\right) \pi^{n}(1-\pi)^{k-n}, \quad k=n, n+1, \ldots $$ this is the negative binomial distribution. Find the mean and variance of \(Y_{n}\), and show that as \(n \rightarrow \infty\) the sequence \(\left\\{Y_{n}\right\\}\) satisfies the conditions of the Central Limit Theorem. Deduce that $$ \lim _{n \rightarrow \infty} 2^{1-n} \sum_{k=0}^{n}\left(\begin{array}{c} k+n-1 \\ n-1 \end{array}\right) \frac{1}{2^{k}}=1 $$ (c) Find the limiting cumulant-generating function of \(\pi Y_{n} /(1-\pi)\) as \(\pi \rightarrow 0\), and hence show that the limiting distribution is gamma.

The binomial distribution models the number of 'successes' among independent variables with two outcomes such as success/failure or white/black. The multinomial distribution extends this to \(p\) possible outcomes, for example total failure/failure/success or white/black/red/blue/.... That is, each of the discrete variables \(X_{1}, \ldots, X_{m}\) takes values \(1, \ldots, p\), independently with probability \(\operatorname{Pr}\left(X_{j}=r\right)=\pi_{r}\), where \(\sum \pi_{r}=1, \pi_{r} \geq 0\). Let \(Y_{r}=\sum_{j} I\left(X_{j}=r\right)\) be the number of \(X_{j}\) that fall into category \(r\), for \(r=1, \ldots, p\), and consider the distribution of \(\left(Y_{1}, \ldots, Y_{p}\right)\). (a) Show that the marginal distribution of \(Y_{r}\) is binomial with probability \(\pi_{r}\), and that \(\operatorname{cov}\left(Y_{r}, Y_{s}\right)=-m \pi_{r} \pi_{s}\), for \(r \neq s\). Is it surprising that the covariance is negative? (b) Hence give consistent estimators of positive probabilities \(\pi_{r}\). What happens if some \(\pi_{r}=0 ?\) (d) Suppose that \(p=4\) with \(\pi_{1}=(2+\theta) / 4, \pi_{2}=(1-\theta) 4, \pi_{3}=(1-\theta) / 4\) and \(\pi_{4}=\) \(\theta / 4\). Show that \(T=m^{-1}\left(Y_{1}+Y_{4}-Y_{2}-Y_{3}\right)\) is such that \(\mathrm{E}(T)=\theta\) and \(\operatorname{var}(T)=a / m\) for some \(a>0\). Hence deduce that \(T\) is consistent for \(\theta\) as \(m \rightarrow \infty\). Give the value of \(T\) and its estimated variance when \(\left(y_{1}, y_{2}, y_{3}, y_{4}\right)\) equals \((125,18,20,34)\)

If \(Y_{1}, \ldots, Y_{n} \stackrel{\text { iidd }}{\sim} N\left(\mu, \sigma^{2}\right)\), show that \(n^{1 / 2}(\bar{Y}-\mu)^{2} \stackrel{P}{\longrightarrow} 0\) as \(n \rightarrow \infty\). Given that \(\operatorname{var}\left\\{\left(Y_{j}\right.\right.\) \(\left.\mu)^{2}\right\\}=2 \sigma^{4}\), deduce that \(\left(S^{2}-\sigma^{2}\right) /\left(2 \sigma^{4} / n\right)^{1 / 2} \stackrel{D}{\longrightarrow} Z\), where \(Z \sim N(0,1)\). When is this true for non-normal data?

Let \(Y=\exp (X)\), where \(X \sim N\left(\mu, \sigma^{2}\right) ; Y\) has the log-normal distribution. Use the moment-generating function of \(X\) to show that \(\mathrm{E}\left(Y^{r}\right)=\exp \left(r \mu+r^{2} \sigma^{2} / 2\right)\), and hence find \(\mathrm{E}(Y)\) and \(\operatorname{var}(Y)\). If \(Y_{1}, \ldots, Y_{n}\) is a log-normal random sample, show that both \(T_{1}=\bar{Y}\) and \(T_{2}=\exp (\bar{X}+\) \(\left.S^{2} / 2\right)\) are consistent estimators of \(\mathrm{E}(Y)\), where \(X_{j}=\log Y_{j}\) and \(S^{2}\) is the sample variance of the \(X_{j} .\) Give the corresponding estimators of \(\operatorname{var}(Y)\). Are the estimators based on the \(Y_{j}\) or on the \(X_{j}\) preferable? Why?

Let \(M\) and IQR be the median and interquartile range of a random sample \(Y_{1}, \ldots, Y_{n}\) from a density of form \(\tau^{-1} g\\{(y-\eta) / \tau\\}\), where \(g(u)\) is symmetric about \(u=0\) and \(g(0)>0\). Show that as \(n \rightarrow \infty\), $$n^{1 / 2} \frac{M-\eta}{\mathrm{IQR}} \stackrel{D}{\longrightarrow} N(0, c)$$ for some \(c>0\), and give \(c\) in terms of \(g\) and its integral \(G\). Give \(c\) when \(g(u)\) equals \(\frac{1}{2} \exp (-|u|)\) and \(\exp (u) /\\{1+\exp (u)\\}^{2}\).
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