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Chapter 7: Problem 3

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

Expert verified
The nth order statistic of a random sample from the given uniform distribution is a sufficient statistic for \( \theta \). This statement is also valid for any pdf having the form \( f(x; \theta) = Q(\theta) M(x) \) for \( 0 < x < \theta \).

Step by step solution

01

Formulate the joint density function

Given the uniform distribution, \( f(x; \theta) = 1/ \theta \) for \( 0 < x < \theta \) and zero everywhere else. We can write down the joint density function for a sample \(X_1, X_2, ..., X_n\) as: \( f(x_1, x_2, ..., x_n; \theta) = \prod_{i=1}^n f(x_i; \theta) = 1/ \theta^n \) for \( 0 < x(i) < \theta \) and zero elsewhere, where \( x(i) \) is the ith order statistic of \(X_1, X_2, ..., X_n\).
02

Derive the density of the nth order statistic

The probability density function of the nth order statistic \(X_n\) can be derived from the joint density function. \(X_n\) is the maximum of \(X_1, X_2, ..., X_n\). So, we integrate the joint density over all possible values of \(x_1, x_2, ..., x_{n-1}\) that are less than \(x_n\). The result is \( g(x_n; \theta) = n/ \theta^n * x_n^{n-1} \) for \( 0 < x_n < \theta \) and zero elsewhere.
03

Apply the factorization theorem

The factorization theorem states that a statistic T(X) is sufficient for \( \theta \) if and only if the joint density \( f(x_1, x_2, ..., x_n; \theta) \) can be factored into a product of two functions one of which depends on the sample through \(T(x)\) only, and the other of which does not depend on \( \theta \). Here, the joint density can be expressed as \( f(x_1, x_2, ..., x_n; \theta) = \{ n/ \theta^{n} \} \{x_n^{n-1}\} \), which factors into a function of \(x_n\) only and a function of \( \theta \) only. Therefore, according to the factorization theorem, \(x_n\), the nth order statistic, is a sufficient statistic for \( \theta \).
04

Generalize for any pdf

The above result can be generalized to any pdf in the form \( f(x; \theta) = Q(\theta) M(x) \), for \( 0 < x < \theta \) and zero everywhere else. In such a case, by following an analogous procedure to the one above, the nth order statistic remains a sufficient statistic for \( \theta \). This is because regardless of the forms of M(x) and Q(\( \theta \)), the joint density function synthesizes to involve the maximum value of x only, which is the nth order statistic.

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Most popular questions from this chapter

Let a random sample of size \(n\) be taken from a distribution of the discrete type with pmf \(f(x ; \theta)=1 / \theta, x=1,2, \ldots, \theta\), zero elsewhere, where \(\theta\) is an unknown positive integer. (a) Show that the largest observation, say \(Y\), of the sample is a complete sufficient statistic for \(\theta\). (b) Prove that $$ \left[Y^{n+1}-(Y-1)^{n+1}\right] /\left[Y^{n}-(Y-1)^{n}\right] $$ is the unique MVUE of \(\theta\).

Let \(X_{1}, X_{2}, \ldots, X_{n}, n>2\), be a random sample from the binomial distribution \(b(1, \theta)\). (a) Show that \(Y_{1}=X_{1}+X_{2}+\cdots+X_{n}\) is a complete sufficient statistic for \(\theta\). (b) Find the function \(\varphi\left(Y_{1}\right)\) which is the MVUE of \(\theta\). (c) Let \(Y_{2}=\left(X_{1}+X_{2}\right) / 2\) and compute \(E\left(Y_{2}\right)\). (d) Determine \(E\left(Y_{2} \mid Y_{1}=y_{1}\right)\).

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be iid with the distribution \(N\left(\theta, \sigma^{2}\right),-\infty<\theta<\infty\). Prove that a necessary and sufficient condition that the statistics \(Z=\sum_{1}^{n} a_{i} X_{i}\) and \(Y=\sum_{1}^{n} X_{i}\), a complete sufficient statistic for \(\theta\), are independent is that \(\sum_{1}^{n} a_{i}=0\)

Let \(f(x, y)=\left(2 / \theta^{2}\right) e^{-(x+y) / \theta}, 0

Let \(X_{1}, X_{2}, \ldots, X_{n}\) denote a random sample from a distribution that is \(N(\theta, 1),-\infty<\theta<\infty\). Find the MVUE of \(\theta^{2}\). Hint: \(\quad\) First determine \(E\left(\bar{X}^{2}\right)\).
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