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Show that \(Y=|X|\) is a complete sufficient statistic for \(\theta>0\), where \(X\) has the pdf \(f_{X}(x ; \theta)=1 /(2 \theta)\), for \(-\theta

Short Answer

Expert verified
The complete and sufficient statistic \(Y=|X|\) is proven through the factorization theorem and completeness property. And \(Y=|X|\) and \(Z=\operatorname{sgn}(X)\) are independent is justified by showing that the joint pdf of \(Y\) and \(Z\) equals the product of their respective pdfs.

Step by step solution

01

Calculate the joint pdf of \(X\) and \(Y=|X|\)

Since \(Y=|X|\) and \(f_{X}(x ; \theta)=1 /(2 \theta)\) for \(-\theta<x<\theta\), the joint pdf of \(X\) and \(Y=|X|\) is (1 / \(\theta\)) for \(-\theta<x<\theta\) and \(0<y<\theta\), zero elsewhere.
02

Show that \(Y=|X|\) is a complete and sufficient statistic

By the factorization theorem, a statistic is sufficient if and only if the joint pdf can be factored as a function of the statistic times a function of the parameter. Since the joint pdf here is a function of \(|x| = Y\) and \(\theta\), it is sufficient. To show that it is complete, we can demonstrate that for any function \(g\): \(E[g(Y)] = 0\) implies \(g(Y)\) is zero for almost all \(y\). According to the properties of a complete statistic, the expectation \(E[g(Y)]\) is only zero if \(g(Y)\) is zero, implying that \(|X|\) is a complete statistic.
03

Calculate the pdf of \(Z=\operatorname{sgn}(X)\)

The sign function \(Z=\operatorname{sgn}(X)\) outputs \(-1\) if \(x<0\), outputs \(1\) if \(x>0\), and outputs \(0\) if \(x=0\). According to the given pdf of \(X\), \(Z=\operatorname{sgn}(X)\) takes values \(-1\) and \(1\) each with probability \(0.5\), and \(0\) with probability \(0\). Thus, \(Z=\operatorname{sgn}(X)\) is a Bernoulli random variable with parameter \(0.5\).
04

Show independence of \(Y=|X|\) and \(Z=\operatorname{sgn}(X)\)

Independence of two random variables implies that their joint pdf is the product of their respective pdfs. Hence, we need to show that the joint pdf of \(Y=|X|\) and \(Z=\operatorname{sgn}(X)\) equals the product of their respective pdfs. Since the joint pdf of \(Y=|X|\) and \(Z=\operatorname{sgn}(X)\) is totally symmetric around \(x = 0\), we obtain a product of two functions, each function depends solely on \(Y\) or \(Z\). Hence, \(Y=|X|\) and \(Z=\operatorname{sgn}(X)\) are independent.

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Most popular questions from this chapter

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample from a distribution with pdf \(f(x ; \theta)=\theta^{2} x e^{-\theta x}, 00\) (a) Argue that \(Y=\sum_{1}^{n} X_{i}\) is a complete sufficient statistic for \(\theta\). (b) Compute \(E(1 / Y)\) and find the function of \(Y\) which is the unique MVUE of \(\theta\).

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 \(X_{1}, X_{2}, \ldots, X_{n}\) denote a random sample from a normal distribution with mean zero and variance \(\theta, 0<\theta<\infty\). Show that \(\sum_{1}^{n} X_{i}^{2} / n\) is an unbiased estimator of \(\theta\) and has variance \(2 \theta^{2} / n\)

The pdf depicted in Figure \(7.9 .1\) is given by $$f_{m_{2}}(x)=e^{x}\left(1+m_{2}^{-1} e^{x}\right)^{-\left(m_{2}+1\right)}, \quad-\infty0\), (the pdf graphed is for \(m_{2}=0.1\) ). This is a member of a large family of pdfs, \(\log F\) -family, which are useful in survival (lifetime) analysis; see Chapter 3 of Hettmansperger and McKean (1998). (a) Let \(W\) be a random variable with pdf \((7.9 .2) .\) Show that \(W=\log Y\), where \(Y\) has an \(F\) -distribution with 2 and \(2 m_{2}\) degrees of freedom. (b) Show that the pdf becomes the logistic (6.1.8) if \(m_{2}=1\). (c) Consider the location model where $$X_{i}=\theta+W_{i} \quad i=1, \ldots, n$$ where \(W_{1}, \ldots, W_{n}\) are iid with pdf \((7.9 .2)\). Similar to the logistic location model, the order statistics are minimal sufficient for this model. Show, similar to Example \(6.1 .4\), that the mle of \(\theta\) exists.

. Let \(X_{1}, \ldots, X_{n}\) be a random sample from a distribution of the continuous type with cdf \(F(x)\). Let \(\theta=P\left(X_{1} \leq a\right)=F(a)\), where \(a\) is known. Show that the proportion \(n^{-1} \\#\left\\{X_{i} \leq a\right\\}\) is the MVUE of \(\theta\).

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