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Chapter 7: Problem 10

Let \(Y_{1}

Short Answer

Expert verified
The cdf of Z is given by \((\frac{\theta - \frac{z}{n}}{\theta})^n\). Its limiting distribution as n approaches infinity is an exponential distribution with parameter \(1/\theta\)

Step by step solution

01

Find the distribution function of Z

Firstly, we transform the random variable Yn into Z as given by the problem, \(Z = n (\theta - Y_n)\). We are given the pdf of Yn, to find the cumulative distribution function (cdf) of Z, we calculate the integral of the pdf from 0 to z. In this case: \[H_{n}(z;\theta) = P(Z\theta - \frac{z}{n}) = 1 - P(Y_{n}<\theta - \frac{z}{n})\] Now, substituting \[\int^{Y_{n}<\theta - z/n}_{0} g(y_{n}; \theta) dy_{n} \] in place of \(P(Y_{n}<\theta - z/n)\} we get \[H_{n}(z;\theta) = 1 - \int^{Y_{n}<\theta - z/n}_{0} \frac{n y_{n}^{n-1}}{\theta^{n}} dy_{n}\]
02

Calculate the integral to find Hn(z;theta)

Evaluating the above integral over the interval from 0 to \(\theta - z/n\), we have \[H_{n}(z;\theta) = 1 - \int^{Y_{n}<\theta - z/n}_{0} \frac{n y_{n}^{n-1}}{\theta^{n}} dy_{n} = 1 - \left(1 - \left(\frac{\theta-\frac{z}{n}}{\theta}\right)^n\right) = \left(\frac{\theta - \frac{z}{n}}{\theta}\right)^n\]
03

Find the limit of Hn(z;theta)

Next, we find the limit of the distribution function as n approaches infinity. It's \[ \lim_{n \rightarrow \infty} H_{n}(z;\theta) = \lim_{n \rightarrow \infty} \left( \frac{\theta - \frac{z}{n}}{\theta}\right)^n \] = \[ e^{-\frac{z}{\theta}} \] Thus, the limiting distribution of Z is an exponential distribution with parameter \(1/\theta\).

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Most popular questions from this chapter

Let \(Y_{1}

Let \(\bar{X}\) denote the mean of the random sample \(X_{1}, X_{2}, \ldots, X_{n}\) from a gammatype distribution with parameters \(\alpha>0\) and \(\beta=\theta \geq 0 .\) Compute \(E\left[X_{1} \mid \bar{x}\right]\) Hint: \(\quad\) Can you find directly a function \(\psi(X)\) of \(X\) such that \(E[\psi(X)]=\theta ?\) Is \(E\left(X_{1} \mid \bar{x}\right)=\psi(\bar{x}) ?\) Why?

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample from \(N\left(\theta_{1}, \theta_{2}\right) .\) (a) If the constant \(b\) is defined by the equation \(P(X \leq b)=0.90\), find the mle and the MVUE of \(b\). (b) If \(c\) is a given constant, find the mle and the MVUE of \(P(X \leq c)\).

Let \(X_{1}, X_{2}, \ldots, X_{5}\) be iid with pdf \(f(x)=e^{-x}, 0

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 \(b^{2} n>(b n-1)^{2} .\) Prove that \(b=1 / n\) does not \(\operatorname{minimize} \max _{\theta} R(\theta, \delta)\)
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