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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.

Short Answer

Expert verified
The mean of the empirical CDF \(\widehat{F}_{n}(x)\) is the sample mean \(\bar{X}\) and the median of the empirical CDF is the sample median, \(Q_{2}\).

Step by step solution

01

Understanding Empirical Cumulative Distribution Function and Sample Mean

The empirical cumulative distribution function (CDF) \(\widehat{F}_{n}(x)\) is a step function whose value at any specified value of the measured variable is the fraction of observations of the measured variable that are less than or equal to the specified value. The sample mean \(\bar{X}\) is the sum of the observed values divided by the count of observed values.
02

Calculation of Mean of Empirical CDF

Mean is a location parameter of the distribution and hence can also be found from the empirical CDF. Mean of empirical CDF is calculated by multiplying each observation with its probability (1/n) and summing these up. So, it will be \( \sum_{i=1}^{n} X_{i}/n = \bar{X}\)
03

Understanding Median

Median is also one of the location parameters and is defined as the value separating the higher half from the lower half of the data sample. For empirical distribution function, \(F^{{-1}}(0.5)\) gives the value of the median.
04

Calculation of Median of Empirical CDF

Empirical CDF \(F^{{-1}}(0.5)\) gives us the value of the variable which divides the area under the empirical CDF into two equal halves. This is nothing but the sample median, which is denoted by \(Q_{2}\). Hence, we have \(T(\widehat{F}_{n})=Q_{2}\)

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Key Concepts

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

Sample Mean
Understanding the sample mean is crucial to grasping the fundamental concepts of statistics. In essence, the sample mean is the average of all the data points in a given sample. This is calculated by adding all the observed values together and then dividing by the number of observations. The formula for the sample mean, often denoted as \( \bar{X} \), is:
\[ \bar{X} = \frac{1}{n} \sum_{i=1}^{n} X_{i} \]
where \( n \) is the number of observations, and \( X_i \) represents each value in the dataset. In the context of an empirical cumulative distribution function (CDF), each data point contributes equally to the average, hence the multiplication by the constant \( 1/n \), which reflects the probability of each value in the empirical CDF. It's crucial to comprehend that the sample mean is a measure of central tendency, providing a single value that summarizes the central position of a data distribution.
Median
The median is another measure of central tendency, which is particularly useful for understanding the distribution of a dataset. Unlike the mean, the median is less affected by outliers and skewed distributions. It is defined as the value that separates the higher half from the lower half of the data sample. To find the median, one must organize the sample in ascending order and then locate the middle value.

If there is an odd number of observations, the median is the middle number. If there's an even number of observations, the median is the average of the two middle numbers. In the realm of empirical CDFs, the median is denoted as \( F^{-1}(0.5) \). This representation corresponds to the value at which the area under the empirical CDF curve is evenly split, with half of the observations lying below and half lying above the median. This makes the median a special kind of location parameter that indicates the central position of a distribution in a different way than the mean.
Location Parameter
Location parameters are descriptive statistics that give some type of central value of a data distribution. Both the mean and median mentioned earlier are examples of location parameters.

These parameters describe the position or location of a distribution on a number line. While the sample mean takes all values into account and is influenced by outliers, the median identifies the middle value and is thus more robust to extreme values. In addition to the mean and median, there are other location parameters such as the mode (the most frequently occurring value) and quantiles (values that divide the data into equal-sized subsets).

Understanding location parameters is fundamental when analyzing data because they provide a quick snapshot of the data's central tendency, which can tell us a lot about the overall distribution. They are widely used in fields ranging from finance to social sciences to natural sciences to summarize and convey the key characteristics of data.

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Most popular questions from this chapter

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample that follows the location model (10.2.1). In this exercise we want to compare the sign tests and \(t\) -test of the hypotheses \((10.2 .2) ;\) so we assume the random errors \(\varepsilon_{i}\) are symmetrically distributed about \(0 .\) Let \(\sigma^{2}=\operatorname{Var}\left(\varepsilon_{i}\right) .\) Hence the mean and the median are the same for this location model. Assume, also, that \(\theta_{0}=0 .\) Consider the large sample version of the \(t\) -test, which rejects \(H_{0}\) in favor of \(H_{1}\) if \(\bar{X} /(\sigma / \sqrt{n})>z_{\alpha}\). (a) Obtain the power function, \(\gamma_{t}(\theta)\), of the large sample version of the \(t\) -test. (b) Show that \(\gamma_{t}(\theta)\) is nondecreasing in \(\theta\). (c) Show that \(\gamma_{t}\left(\theta_{n}\right) \rightarrow 1-\Phi\left(z_{\alpha}-\sigma \theta^{*}\right)\), under the sequence of local alternatives \((10.2 .13)\) (d) Based on part (c), obtain the sample size determination for the \(t\) -test to detect \(\theta^{*}\) with approximate power \(\gamma^{*}\). (e) Derive the \(\operatorname{ARE}(S, t)\) given in \((10.2 .27)\).

In Exercise \(10.9 .5\), the influence function of the variance functional was derived directly. Assuming that the mean of \(X\) is 0 , note that the variance functional, \(V\left(F_{X}\right)\), also solves the equation $$ 0=\int_{-\infty}^{\infty}\left[t^{2}-V\left(F_{X}\right)\right] f_{X}(t) d t $$ (a) Determine the natural estimator of the variance by writing the defining equation at the empirical cdf \(F_{n}(t)\), for \(X_{1}-\bar{X}, \ldots, X_{n}-\bar{X}\) iid with \(\operatorname{cdf} F_{X}(t)\) and solving for \(V\left(F_{n}\right)\). (b) As in Exercise \(10.9 .6\), write the defining equation for the variance functional at the contaminated \(\operatorname{cdf} F_{x, \epsilon}(t)\). (c) Then derive the influence function by implicit differentiation of the defining equation in part (b).

Let \(X\) be a continuous random variable with cdf \(F(x)\). Suppose \(Y=X+\Delta\), where \(\Delta>0\). Show that \(Y\) is stochastically larger than \(X\).

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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