/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 8 If \(X_{1}, X_{2}, \ldots, X_{n}... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 8

If \(X_{1}, X_{2}, \ldots, X_{n}\) is a random sample from a distribution that has a pdf which is a regular case of the exponential class, show that the pdf of \(Y_{1}=\sum_{1}^{n} K\left(X_{i}\right)\) is of the form \(f_{Y_{1}}\left(y_{1} ; \theta\right)=R\left(y_{1}\right) \exp \left[p(\theta) y_{1}+n q(\theta)\right]\). Hint: Let \(Y_{2}=X_{2}, \ldots, Y_{n}=X_{n}\) be \(n-1\) auxiliary random variables. Find the joint pdf of \(Y_{1}, Y_{2}, \ldots, Y_{n}\) and then the marginal pdf of \(Y_{1}\).

Short Answer

Expert verified
By transforming the random variables \(X_{i}\) into \(Y_{1}, Y_{2}, \ldots, Y_{n}\) and finding the joint and marginal pdfs, you can derive that the pdf of \(Y_{1}=\sum_{1}^{n} K\left(X_{i}\right)\) follows the given form \[f_{Y_{1}}\left(y_{1} ; \theta\right)=R\left(y_{1}\right) \exp \left[p(\theta) y_{1}+ n q(\theta)\right]\]

Step by step solution

01

Identify the pdf of \(X_{i}\)

The pdf of \(X_{i}\) following the exponential distribution is, \(f_{X_{i}}\left(x_{i} ; \theta\right)=h(x_{i}) \exp \left[p(\theta) K(x_{i})+ q(\theta)\right]\) for \(i = 1, 2, \ldots, n\).
02

Find the Joint pdf of \(Y_{1}, Y_{2}, \ldots, Y_{n}\)

As per the definition, \(Y_{1}=\sum_{1}^{n} K\left(X_{i}\right)\) and \(Y_{i}=X_{i}\) for \(i = 2, \ldots, n\), we can write the joint pdf \(f_{Y_{1}, Y_{2}, \ldots, Y_{n}}\) as:\[f_{Y_{1}, Y_{2}, \ldots, Y_{n}}\left(y_{1}, y_{2}, \ldots, y_{n} ; \theta\right)=c(\theta^n) \exp \left[p(\theta) y_{1}+ (n-1) q(\theta)\right]\]where \(c(\theta^n)\) is a normalizing constant.
03

Find the marginal pdf of \(Y_{1}\)

The marginal pdf of \(Y_{1}\) is obtained by integrating the joint pdf over all values of \(Y_{2}, \ldots, Y_{n}\). This yields:\[f_{Y_{1}}\left(y_{1} ; \theta\right)=R\left(y_{1}\right) \exp \left[p(\theta) y_{1}+ n q(\theta)\right]\]where \(R\left(y_{1}\right)\) is a function only depending on \(y_{1}\), being the result of the integrations.

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Most popular questions from this chapter

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

Let \(X_{1}, X_{2}, \ldots, X_{n}\) denote a random sample from a distribution that is \(b(1, \theta), 0 \leq \theta \leq 1 .\) Let \(Y=\sum_{1}^{n} X_{i}\) and let \(\mathcal{L}[\theta, \delta(y)]=[\theta-\delta(y)]^{2} .\) Consider decision functions of the form \(\delta(y)=b y\), where \(b\) does not depend upon \(y .\) Prove that \(R(\theta, \delta)=b^{2} n \theta(1-\theta)+(b n-1)^{2} \theta^{2}\). Show that $$\max _{\theta} R(\theta, \delta)=\frac{b^{4} n^{2}}{4\left[b^{2} n-(b n-- 1)^{2}\right]}$$ provided that the value \(b\) is such that \(b^{2} n \geq 2(b n-1)^{2} .\) Prove that \(b=1 / n\) does not maximize \(\max _{\theta} R(\theta, \delta)\).

As in Example 7.6.2, let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample of size \(n>1\) from a distribution that is \(N(\theta, 1) .\) Show that the joint distribution of \(X_{1}\) and \(\bar{X}\) is bivariate normal with mean vector \((\theta, \theta)\), variances \(\sigma_{1}^{2}=1\) and \(\sigma_{2}^{2}=1 / n\), and correlation coefficient \(\rho=1 / \sqrt{n}\)

Let \(Y_{1}

. In a personal communication, LeRoy Folks noted that the inverse Gaussian pdf $$f\left(x ; \theta_{1}, \theta_{2}\right)=\left(\frac{\theta_{2}}{2 \pi x^{3}}\right)^{1 / 2} \exp \left[\frac{-\theta_{2}\left(x-\theta_{1}\right)^{2}}{2 \theta_{1}^{2} x}\right], \quad 00\) and \(\theta_{2}>0\) is often used to model lifetimes. Find the complete sufficient statistics for \(\left(\theta_{1}, \theta_{2}\right)\), if \(X_{1}, X_{2}, \ldots, X_{n}\) is a random sample from the distribution having this pdf.
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