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Chapter 12: Problem 7

In Exercise 12.1.5, the influence function of the variance functional was derived directly. Assuming that the mean of \(Y\) is 0 , note that the variance functional, \(V\left(F_{Y}\right)\), also solves the equation $$ 0=\int_{-\infty}^{\infty}\left[t^{2}-V\left(F_{Y}\right)\right] f_{Y}(t) d t $$ (a) Determine the natural estimator of the variance by writing the defining equation at the empirical cdf \(F_{n}(t)\), for \(Y_{1}-\bar{Y}, \ldots Y_{n}-\bar{Y}\) iid with cdf \(F_{Y}(t)\), and solving for \(V\left(F_{n}\right)\)

Short Answer

Expert verified
The natural estimator for the variance when the mean of \(Y\) is 0 and the cumulative distribution function is the empirical CDF \(F_n\) is given by the integral \(V\left(F_{n}\right) = \int_{-\infty}^{\infty}(t - \bar{Y})^2 dF_n(t)\).

Step by step solution

01

Write the defining equation

The given integral equation can be rewritten as:\[ 0 = \int_{-\infty}^{\infty} t^2 f_{Y}(t) dt - V\left(F_{Y}\right)\int_{-\infty}^{\infty} f_{Y}(t) dt\]Given that \(\int_{-\infty}^{\infty}f_Y(t) dt = 1\), this simplifies to:\[0 = \int_{-\infty}^{\infty} t^2 f_{Y}(t) dt - V\left(F_{Y}\right)\]
02

Substitute the empirical CDF

Substitute the empirical CDF \(F_{n}(t)\) in place of \(F_Y(t)\) in the equation. This results in the equation:\[ 0 = \int_{-\infty}^{\infty}(t - \bar{Y})^2 dF_n(t) - V\left(F_{n}\right)\]where \(dF_n(t)\) is a term that represents the change in the empirical CDF with respect to \(t\), and \(\bar{Y}\) represents the mean of the samples \(Y_1, Y_2,...,Y_n\).
03

Find the natural estimator

The natural estimator of the variance is that for which the equation holds true. Solve the equation for \(V(F_n)\) to obtain the natural estimator:\[ V\left(F_{n}\right) = \int_{-\infty}^{\infty}(t - \bar{Y})^2 dF_n(t)\]This is the natural estimator of the variance when the mean of \(Y\) is 0 and the CDF is the empirical CDF \(F_n\).

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Key Concepts

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

Empirical Cumulative Distribution Function
The empirical cumulative distribution function (ECDF) is a fundamental statistical tool that provides a step-by-step representation of the data distribution in a given sample. It is constructed by ordering data points from smallest to largest and plotting the proportion of observations that are less than or equal to each data point.

For a sample of size n, with ordered values \(Y_1, Y_2, ..., Y_n\), the ECDF at a point t is given by:\[ F_n(t) = \frac{1}{n} \sum_{i=1}^{n} I(Y_i \leq t) \]where \(I\) is the indicator function, equal to 1 if \(Y_i \leq t\) and 0 otherwise.

Importance of the ECDF

  • It provides an intuitive visualization of the data distribution, highlighting where data points are concentrated and identifying outliers.
  • Unlike theoretical distribution functions, which may assume a particular distribution shape, the ECDF makes no such assumptions and is based purely on the observed data.
  • It serves as a non-parametric estimator of the cumulative distribution function (CDF), which can be particularly useful when the underlying distribution of the data is unknown.
Natural Estimator of Variance
The natural estimator of variance is a method for estimating the variability or spread of a random variable in a dataset. When the true mean of a population is known to be zero, which simplifies many statistical formulas, the natural estimator of variance is particularly straightforward.

Defining the Natural Estimator

If a sample comes from a population with mean zero, the variance can be estimated by the sample mean of the squared deviations from the sample mean. This is mathematically represented as:\[ V(F_n) = \frac{1}{n} \sum_{i=1}^{n} (Y_i - \bar{Y})^2 \]where \( \bar{Y} \) is the sample mean and \( F_n \) denotes the ECDF based on the sample. This formula effectively uses the ECDF to approximate the true variance of the population.

Characteristics of the Natural Estimator

  • It is an unbiased estimator of the population variance when the population mean is known to be zero.
  • The estimator incorporates all the sample data points, making it sensitive to outliers which can affect the variance significantly.
  • In practice, when the population mean is unknown and estimated from the data, a correction factor of \((n-1)/n\) is usually applied, resulting in the sample variance, which is an unbiased estimator of the population variance under more general conditions.
Integral Equation for Variance
The integral equation for variance arises from the definition of variance in probability theory and is a foundational component in the field of functional estimation. This equation offers a continuous analogue to the discrete sum used in the natural estimator.

Understanding the Integral Equation

The variance functional for a random variable \(Y\) with probability density function \(f_Y(t)\) and cumulative distribution function \(F_Y(t)\) is derived from:\[ 0 = \int_{-\infty}^{\infty} (t^2 - V(F_Y)) f_Y(t) dt \]This expresses the balance between the mean squared distance of the variable from zero and the variance functional \(V(F_Y)\).

Role in Estimating Variance

  • The equation serves as the basis for determining a theoretical value of variance for a given distribution, which can then be compared to empirical estimates from sample data.
  • When applied to the empirical CDF, the integral equation adapts to a sum over the observed data points, leading to the natural estimator of variance.
  • This integral equation underpins many statistical techniques and intuitive methods for estimating and understanding variability within datasets.

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Most popular questions from this chapter

Establish the identity $$ \|\mathbf{v}\|_{W}=\frac{\sqrt{3}}{2(n+1)} \sum_{i=1}^{n} \sum_{j=1}^{n}\left|v_{i}-v_{j}\right| $$ for all \(\mathbf{v} \in R^{n}\). Thus we have shown that $$ \widehat{\beta}_{W}=\operatorname{Argmin} \sum_{i=1}^{n} \sum_{j=1}^{n}\left|\left(y_{i}-y_{j}\right)-\beta\left(x_{c i}-x_{c j}\right)\right| . $$ Note that the formulation of \(\widehat{\beta}_{W}\) given in expression \((12.2 .29)\) allows an easy way to compute the Wilcoxon estimate of slope by using an \(L_{1}\), (least absolute deviations), routine. This was used in the cited article by Terpstra, et al. for their \(\mathrm{R}\) or S-PLUS functions which compute the Wilcoxon fit.

. The following data set is taken from Shirley (1981) discussed from a robust point-of-view in McKean and Vidmar (1994). The response is the time it takes a rat to enter a chamber after receiving a treatment designed to delay the time of entry. There were 30 rats in the experiment and they were divided evenly into three groups. The rats in Groups 2 and 3 received an antidote to the treatment. The covariate is the time taken by the rat to enter the chamber prior to its treatment. The data are: $$ \begin{array}{|rr|rr|rr|} \hline \multicolumn{2}{|c|} {\text { Group 1 }} & \multicolumn{2}{|c|} {\text { Group 2 }} & \multicolumn{2}{|c|} {\text { Group 3 }} \\ \begin{array}{c} \text { Initial } \\ \text { time } \end{array} & \begin{array}{c} \text { Final } \\ \text { time } \end{array} & \begin{array}{c} \text { Initial } \\ \text { time } \end{array} & \begin{array}{r} \text { Final } \\ \text { time } \end{array} & \begin{array}{c} \text { Initial } \\ \text { time } \end{array} & \begin{array}{c} \text { Final } \\ \text { time } \end{array} \\ \hline 1.8 & 79.1 & 1.6 & 10.2 & 1.3 & 14.8 \\ 1.3 & 47.6 & 0.9 & 3.4 & 2.3 & 30.7 \\ 1.8 & 64.4 & 1.5 & 9.9 & 0.9 & 7.7 \\ 1.1 & 68.7 & 1.6 & 3.7 & 1.9 & 63.9 \\ 2.5 & 180.0 & 2.6 & 39.3 & 1.2 & 3.5 \\ 1.0 & 27.3 & 1.4 & 34.0 & 1.3 & 10.0 \\ 1.1 & 56.4 & 2.0 & 40.7 & 1.2 & 6.9 \\ 2.3 & 163.3 & 0.9 & 10.5 & 2.4 & 22.5 \\ 2.4 & 180.0 & 1.6 & 0.8 & 1.4 & 11.4 \\ 2.8 & 132.4 & 1.2 & 4.9 & 0.8 & 3.3 \\ \hline \end{array} $$(a) Obtain scatterplots of the data by group. (b) Obtain the LS and Wilcoxon fits of the full model \(y_{i j}=\mu_{i}+\beta_{i} x_{i j}+e_{i j}, i=\) \(1,2,3, j=1, \ldots, 10\), where \(y_{i j}\) denotes the response for the \(j\) th rat in Group \(i\) and \(x_{i j}\) denotes the corresponding covariate. (c) Overlay the LS and Wilcoxon fits on your scatterplots in (a). Comment on the effect of outliers in groups 2 and 3 . (d) Form the hypothesis matrix to test the homogeneity of the slopes, i.e., \(H_{0}\) : \(\beta_{1}=\beta_{2}=\beta_{3}\) (e) Carry out the LS and Wilcoxon tests of the hypothesis in Part (d).

Often influence functions are derived by differentiating implicitly the defining equation for the functional at the contaminated cdf \(F_{y, \epsilon}(t),(12.1 .18)\). Consider the mean functional with the defining equation (12.1.15). Using the linearity of the differential, first show that the defining equation at the \(\operatorname{cdf} F_{y, \epsilon}(t)\) can be expressed as $$ \begin{aligned} 0=\int_{-\infty}^{\infty}\left[t-T\left(F_{y, \epsilon}\right)\right] d F_{y, \epsilon}(t)=&(1-\epsilon) \int_{-\infty}^{\infty}\left[t-T\left(F_{y, \epsilon}\right)\right] f_{Y}(t) d t \\ &+\epsilon \int_{-\infty}^{\infty}\left[t-T\left(F_{y, \epsilon}\right)\right] d_{\Delta}(t) \end{aligned} $$ Recall that we want \(\partial T\left(F_{y, \epsilon}\right) / \partial \epsilon\). Obtain this by differentiating implicitly the above equation with respect to \(\epsilon\).

. Suppose \(Y\) is a random variable with mean 0 and variance \(\sigma^{2} .\) Recall that the function \(F_{y, \epsilon}(t)\) is the cdf of the random variable \(U=I_{1-\epsilon} Y+\left[1-I_{1-\epsilon}\right] W\) where \(Y, I_{1-\epsilon}\), and \(W\) are independent random variables, \(Y\) has cdf \(F_{Y}(t), W\) has \(\operatorname{cdf} \Delta_{y}(t)\), and \(I_{1-\epsilon}\) is \(b(1,1-\epsilon) .\) Define the functional \(V\left(F_{Y}\right)=\operatorname{Var}(Y)=\sigma^{2} .\) Note that the functional at the contaminated cdf \(F_{y, \epsilon}(t)\) is the variance of the random variable \(U=I_{1-\epsilon} Y+\left[1-I_{1-\epsilon}\right] W .\) To derive the influence function of the variance perform the following steps: (a) Show that \(E(U)=\epsilon y\). (b) Show that \(\operatorname{Var}(U)=(1-\epsilon) \sigma^{2}+\epsilon y^{2}-\epsilon^{2} y^{2}\) (c) Obtain the partial derivative of the right side of this last equation with respect to \(\epsilon\). This is the influence function. Hint: Because \(I_{1-\epsilon}\) is a Bernoulli random variable, \(I_{1-\epsilon}^{2}=I_{1-\epsilon} .\) Why?

The data below are generated from the model \(Y_{i}=0+5 i+i^{2}+\varepsilon_{i}\), for \(i=1, \ldots, 10\), and \(\varepsilon_{i}\) iid \(N\left(0,4^{2}\right)\) $$ \begin{array}{|l|r|r|r|r|r|r|r|r|r|r|} \hline i & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ \hline Y_{i} & 3.1 & 20.1 & 20.4 & 31.6 & 57.0 & 61.7 & 86.9 & 107.5 & 125.7 & 148.0 \\ \hline \end{array} $$(a) Fit the misspecified model \(Y_{i}=\alpha+\beta_{1} i+\varepsilon_{i}\) by LS and obtain the residual plot. Comment on the plot (Is it random? If not, does it suggest another model to try?). (b) Same as Part (a) for the fit of the model \(Y_{i}=\alpha+\beta_{1} i+\beta_{2} i^{2}+\varepsilon_{i}\) by LS.
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