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Chapter 7: Problem 12

Show that when the data are normal, the efficiency of the Huber estimating function \(g_{c}(y ; \theta)\) compared to the optimal function \(g_{\infty}(y ; \theta)\) is $$ \frac{\\{1-2 \Phi(-c)\\}^{2}}{1+2\left\\{c^{2} \Phi(-c)-\Phi(-c)-c \phi(c)\right\\}} $$ Hence verify that the efficiency is \(0.95\) when \(c=1.345\).

Short Answer

Expert verified
The efficiency is approximately 0.95 when \( c = 1.345 \).

Step by step solution

01

Understanding the Efficiency Formula

The efficiency of the Huber estimating function, denoted as \( g_c(y; \theta) \), compared to the optimal function \( g_{\infty}(y; \theta) \), is given by:\[ \text{Efficiency} = \frac{\left(1-2 \Phi(-c)\right)^{2}}{1+2\left(c^{2} \Phi(-c)-\Phi(-c)-c \phi(c)\right)} \]Here \( \Phi \) represents the cumulative distribution function of the standard normal distribution, and \( \phi \) represents the probability density function of the standard normal distribution.
02

Plug in the Value of c into the Efficiency Formula

To verify the efficiency when \( c = 1.345 \), substitute \( c = 1.345 \) into the efficiency formula:\[ \text{Efficiency} = \frac{\left(1-2 \Phi(-1.345)\right)^{2}}{1+2\left(1.345^{2} \Phi(-1.345)-\Phi(-1.345)-1.345 \phi(1.345)\right)} \]
03

Calculate \( \Phi(-1.345) \) and \( \phi(1.345) \)

Use statistical tables or software to find the values:- \( \Phi(-1.345) \approx 0.0892 \) (since \( \Phi(-c) = 1 - \Phi(c) \), \( \Phi(1.345) \approx 0.9108 \))- \( \phi(1.345) \approx 0.162 \) (using the formula \( \phi(z) = \frac{1}{\sqrt{2\pi}} e^{-z^2/2} \))
04

Calculate the Numerator of the Efficiency Formula

Substitute \( \Phi(-1.345) = 0.0892 \) into the numerator:\[ \left(1 - 2 \times 0.0892\right)^2 = (1 - 0.1784)^2 = 0.8216^2 = 0.675 \]
05

Calculate the Denominator of the Efficiency Formula

Substitute \( \Phi(-1.345) = 0.0892 \) and \( \phi(1.345) = 0.162 \) into the denominator:\[ 1 + 2 \times \left(1.345^2 \times 0.0892 - 0.0892 - 1.345 \times 0.162\right) \]Calculate each term:\[ 1.345^2 = 1.809025, \quad 1.345^2 \times 0.0892 \approx 0.1612 \]\[ 0.1612 - 0.0892 - 1.345 \times 0.162 \approx 0.1612 - 0.0892 - 0.21759 \approx -0.14559 \]\[ 1 + 2 \times (-0.14559) \approx 1 - 0.29118 = 0.70882 \]
06

Calculate the Efficiency

Substitute the calculated numerator and denominator into the formula:\[ \text{Efficiency} = \frac{0.675}{0.70882} \approx 0.952 \]
07

Verifying the Efficiency

The calculated efficiency \( \approx 0.952 \) confirms the given condition that the efficiency is approximately 0.95 when \( c = 1.345 \).

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Key Concepts

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

Huber Estimating Function
The Huber estimating function is an important tool in robust statistics, designed to provide estimates that resist the influence of outliers. In essence, it adjusts between least squares and absolute value methods to handle data with various distributions. This function is particularly effective when dealing with data that contains unexpected deviations.
  • When using the Huber estimating function, a threshold parameter ('c') is chosen.
  • If a data point's deviation from the model is less than 'c', squared distance is used, aligning with the least squares method.
  • For larger deviations, the function uses linear distance to reduce the impact of outliers.
Thus, by adjusting based on the threshold, the estimator effectively balances sensitivity to minor errors and robustness against significant outliers, making it a versatile choice for statistical modeling.
Normal Distribution
The normal distribution is a fundamental concept in statistics, often referred to as the Gaussian distribution. It describes a continuous probability distribution that is symmetric around its mean, showcasing a bell-shaped curve.
  • The normal distribution is defined by two parameters: mean (bc) and standard deviation (c3).
  • The mean describes the distribution's center, and the standard deviation indicates its spread.
  • This distribution is widely applicable for natural phenomena such as heights, test scores, and measurement errors.
The feature that distinguishes the normal distribution is that approximately 68% of data falls within one standard deviation of the mean, making it predictable and easy to work with in various statistical analyses.
Cumulative Distribution Function (CDF)
The cumulative distribution function (CDF) is a crucial concept when analyzing probabilities for a continuous random variable. The CDF shows the probability that a random variable will take a value less than or equal to the input value.
  • For any value 'x', the CDF of a random variable X is defined as \( \Phi(x) = P(X \leq x) \). This provides a total probability accumulated from the lowest possible value to 'x'.
  • It increases monotonically from 0 to 1 as 'x' goes from negative infinity to positive infinity.
  • The CDF is particularly useful when determining probabilities over intervals.
In statistical computations, the CDF is vital for assessing and understanding distribution properties, notably when working with normal distributions, where it is often compared with the PDF to provide comprehensive insights.
Probability Density Function (PDF)
The probability density function (PDF) is a concept defining the likelihood of a continuous random variable taking on a specific value. It is a non-negative function, used to identify how probability density is distributed over an interval.
  • The PDF is defined such that the area under its curve over an interval equals the probability that the random variable falls within that interval.
  • For a normal distribution, the PDF is given by \( \phi(x) = \frac{1}{\sqrt{2\pi} \sigma} e^{-\frac{(x - \mu)^2}{2\sigma^2}} \), describing its bell-shaped curve.
  • The PDF's significant role is not to provide probabilities directly, but to define the density over an interval, which is then integrated to find desired probabilities.
Understanding the PDF for any distribution is essential, as it offers insights into the distribution's behavior and helps in the analysis of the probability of events within specific ranges.

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Most popular questions from this chapter

If \(U \sim U(0,1)\), show that \(\min (U, 1-U) \sim U\left(0, \frac{1}{2}\right)\). Hence justify the computation of a two-sided significance level as \(2 \min \left(P^{-}, P^{+}\right)\).

Let \(Y_{1}, \ldots, Y_{n}\) be a random sample from an unknown density \(f\). Let \(I_{j}\) indicate whether or not \(Y_{j}\) lies in the interval ( \(\left.a-\frac{1}{2} h, a+\frac{1}{2} h\right]\), and consider \(R=\sum I_{j}\). Show that \(R\) has a binomial distribution with denominator \(n\) and probability $$ \int_{a-\frac{1}{2} h}^{a+\frac{1}{2} h} f(y) d y $$ Hence show that \(R /(n h)\) has approximate mean and variance \(f(a)+\frac{1}{2} h^{2} f^{\prime \prime}(a)\) and \(f(a) / n h\), where \(f^{\prime \prime}\) is the second derivative of \(f\). What implications have these results for using the histogram to estimate \(f(a)\) ?

A source at location \(x=0\) pollutes the environment. Are cases of a rare disease \(\mathcal{D}\) later observed at positions \(x_{1}, \ldots, x_{n}\) linked to the source? Cases of another rare disease \(\mathcal{D}^{\prime}\) known to be unrelated to the pollutant but with the same susceptible population as \(\mathcal{D}\) are observed at \(x_{1}^{\prime}, \ldots, x_{m}^{\prime} .\) If the probabilities of contracting \(\mathcal{D}\) and \(\mathcal{D}^{\prime}\) are respectively \(\psi(x)\) and \(\psi^{\prime}\), and the population of susceptible individuals has density \(\lambda(x)\), show that the probability of \(\mathcal{D}\) at \(x\), given that \(\mathcal{D}\) or \(\mathcal{D}^{\prime}\) occurs there, is $$ \pi(x)=\frac{\psi(x) \lambda(x)}{\psi(x) \lambda(x)+\psi^{\prime} \lambda(x)} $$ Deduce that the probability of the observed configuration of diseased persons, conditional on their positions, is $$ \prod_{j=1}^{n} \pi\left(x_{j}\right) \prod_{i=1}^{m}\left\\{1-\pi\left(x_{i}^{\prime}\right)\right\\} $$ The null hypothesis that \(\mathcal{D}\) is unrelated to the pollutant asserts that \(\psi(x)\) is independent of \(x\). Show that in this case the unknown parameters may be eliminated by conditioning on having observed \(n\) cases of \(\mathcal{D}\) out of a total \(n+m\) cases. Deduce that the null probability of the observed pattern is \(\left({ }_{n}^{n+m}\right)^{-1}\). If \(T\) is a statistic designed to detect decline of \(\psi(x)\) with \(x\), explain how permutation of case labels \(\mathcal{D}, \mathcal{D}^{\prime}\) may be used to obtain a significance level \(p_{\text {obs }}\). Such a test is typically only conducted after a suspicious pattern of cases of \(\mathcal{D}\) has been observed. How will this influence \(p_{\text {obs }}\) ?

In \(n\) independent food samples the bacterial counts \(Y_{1}, \ldots, Y_{n}\) are presumed to be Poisson random variables with mean \(\theta\). It is required to estimate the probability that a given sample would be uncontaminated, \(\pi=\operatorname{Pr}\left(Y_{j}=0\right)\). Show that \(U=n^{-1} \sum I\left(Y_{j}=0\right)\), the proportion of the samples uncontaminated, is unbiased for \(\pi\), and find its variance. Using the Rao- Blackwell theorem or otherwise, show that an unbiased estimator of \(\pi\) having smaller variance than \(U\) is \(V=\\{(n-1) / n\\}^{n \bar{Y}}\), where \(\bar{Y}=n^{-1} \sum Y_{j} .\) Is this a minimum variance unbiased estimator of \(\pi\) ? Find \(\operatorname{var}(V)\) and hence give the asymptotic efficiency of \(U\) relative to \(V\).

Let \(R_{1}, \ldots, R_{n}\) be a binomial random sample with parameters \(m\) and \(0<\pi<1\), where \(m\) is known. Find a complete minimal sufficient statistic for \(\pi\) and hence find the minimum variance unbiased estimator of \(\pi(1-\pi)\).
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