Question

Show that the \(n\) th order statistic of a random sample of size \(n\) from the uniform distribution having pdf \(f(x ; \theta)=1 / \theta, 0

Short answer

The nth order statistic is a sufficient statistic for \(\theta\) in both cases of pdfs i.e., the uniform distribution and the general pdf.

Step-by-step solution

Determine the joint pdf of the Sample

From the given data, since the pdf \(f(x;\theta) = 1/\theta\), for \(0 < x < \theta\), zero elsewhere, the joint density of a random sample \(X_1, X_2, ..., X_n\) is then given by \(f(x_1, x_2, ..., x_n;\theta) = 1/\theta^n\), for \(0 < x_i < \theta\), \(i = 1,2,...,n\), zero elsewhere.

Calculate the ordered statistic

It is known that the nth order statistic of the sample where \(X_1, X_2, ..., X_n\) are in ascending order, is \(X_n\). A change of variables from \(X_1, X_2, ..., X_n\) to \(X_(1), X_(2), ..., X_(n)\) will transform the previous joint density to the one which involves the nth order statistic. This transformation does not influence the space of \(x\), hence the joint density becomes \(f(x_(1), x_(2), ..., x_(n);\theta) = 1/\theta^n\), for \(0 < x_(i) < \theta\), \(i = 1,2,...,n\), zero elsewhere.

Apply the Factorization Theorem

According to the Factorization Theorem, \(T(x_1, x_2, ..., x_n)\) is a sufficient statistic for \(\theta\) if and only if the joint pdf can be Factorized into two functions, \(h(x_1, x_2, ..., x_n)\) and \(g(T;\theta)\). Here \(T\) is a function of \(x_1, x_2, ..., x_n\) and \(g(T;\theta)\) does not contain \(x_1, x_2, ..., x_n\) but can contain \(T\), while \(h(x_1, x_2, ..., x_n)\) does not rely on \(\theta\). From the obtained joint pdf after order transformation \(f(x_(1), x_(2), ..., x_(n);\theta) = 1/\theta^n\), it can be written as \(f(x_(1), x_(2), ..., x_(n);\theta) = g(T;\theta)h(x_(1), x_(2), ..., x_n)\) where \(g(T;\theta) = 1/\theta^n\) and \(h(x_(1), x_(2), ..., x_n) = 1\). Hence, the nth order statistic is a sufficient statistic for \(\theta\).

Generalize to other pdfs

For the general pdf \(f(x ; \theta)=Q(\theta) M(x)\), for \(0 < x < \theta\), with zero elsewhere and \(\int_{0}^{\theta} M(x) d x=\frac{1}{Q(\theta)}\), the joint pdf of the sample becomes \(f(x_1, x_2, ..., x_n;\theta) = 1/Q(\theta^n)\) within the transformed bounds and zero elsewhere. After a similar transformation of variables, leading to the joint density in ordered statistics, we will get \(f(x_(1), x_(2), ..., x_(n);\theta) = 1/Q(\theta^n)\) within the transformed bounds and zero elsewhere. Applying a similar factorization as before, we can conclude that for this general case too, the nth order statistic is a sufficient statistic for \(\theta\).

Key concepts

Order Statistics

Order statistics are an essential concept when dealing with samples in statistics. They provide vital information about the positioning of data points within a sample. Specifically, when we have a random sample, order statistics are obtained by arranging the data points in ascending or descending order. The resulting values are denoted as the first order statistic (the smallest value), the second order statistic, and so on, until the nth order statistic (the largest value). In the case of the textbook solution, the nth order statistic (\(X_{(n)}\)) from a random sample is crucial because it captures the highest value observed which, for the uniform distribution, is tied directly to the parameter \(\theta\).Understanding the importance of order statistics in random samples is valuable, especially for distributions where the maximum or minimum value can provide insights into the parameters that define the distribution. For students struggling with this concept, envisioning order statistics as a sorted list—from smallest to largest—of the drawn samples can be helpful in visualizing the concept.

Random Sample

A random sample is a set of observations taken from a larger population, where each member of the population has an equal chance of being included in the sample. This equality of opportunity is a foundational aspect of statistical theory, ensuring that the sample represents the population well. Random samples help statisticians make inferences about population parameters with a level of certainty outlined by probability theory.In the context of the given exercise, a random sample of size \(n\) is drawn from a uniform distribution. The randomness ensures that the nth order statistic derived from this sample is free from bias and truly representative of the population's characteristic, specifically the parameter \(\theta\). Students should consider the vital role random sampling plays in ensuring the validity of statistical analysis and the unbiased estimation of parameters.

Uniform Distribution

The uniform distribution describes a situation where all outcomes are equally likely. For a continuous uniform distribution, this can be visualized as a flat line within a range on a graph; every value within this range has the same likelihood of being randomly drawn. The distribution is defined by two parameters, a and b, which mark the lower and upper bounds of the distribution such that the probability density function (PDF) exists only within these bounds.

Uniform PDF Formula

In the provided exercise, the uniform distribution has a PDF given by \(f(x; \theta) = 1/\theta\), where \(0 < x < \theta\) and \(0 < \theta < \infty\). This particular form of the uniform distribution has its upper bound defined by the parameter \(\theta\) and the lower bound set to zero. The simplicity of the uniform distribution often makes it easier to understand other more complex statistical concepts, as it provides a neat framework within which parameters directly control the shape and probability distribution.

Probability Density Function (PDF)

The probability density function (PDF) is a function that describes the relative likelihood of a continuous random variable taking on a specific value. For every continuous distribution, such as the uniform distribution mentioned in the exercise, the PDF plays a critical role in describing the shape of the distribution and establishing probabilities for intervals of values.The PDF must satisfy two conditions: it must be non-negative over its range, and the total area under the curve of the PDF must equal 1, signifying that the total probability of all outcomes is 100%. In our exercise, \(f(x; \theta) = 1/\theta\) represents the PDF for the uniform distribution, which maintains a constant value over its range and zeroes out elsewhere. It's important for students to understand that the PDF does not give probabilities for specific values (in continuous distributions, these are always zero), but rather for ranges of values, providing the basis for calculating probabilities over intervals.

Factorization Theorem

The Factorization Theorem is a fundamental concept in statistics that provides a method for identifying sufficient statistics. A sufficient statistic is a function of the data that captures all relevant information needed for estimating a parameter of interest, such as \(\theta\) in our exercise. According to the theorem, if you can factor the joint PDF of the random sample into two parts—one that depends on the data only through the statistic and another that depends on the parameters but not the data—then the statistic in question is sufficient.

Components of the Factorization Theorem

In the exercise, the joint PDF of the ordered sample is factored into \(h(x_(1), x_(2), ..., x_n)\), which does not depend on \(\theta\), and \(g(T;\theta)\), which depends on \(\theta\) but not the individual values in the sample. Finding such a factorization leads us to conclude that the nth order statistic is sufficient for \(\theta\) for the uniform distribution, and the same logic applies more generally to other PDFs with the same structure. This theorem is a crucial tool for students and statisticians alike, as it simplifies the process of statistical estimation and inference by identifying key statistics.