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Chapter 5: Problem 14

Let \(Y_{1}

Short Answer

Expert verified
Hence, the pdf of \(Z\) is 2 for \(0<Z<1\), and zero elsewhere.

Step by step solution

01

Understanding the Midrange

The midrange is defined as the average of the smallest and largest values in a data set. We are given the ordered sample \(Y_1\), \(Y_2\), and \(Y_3\). Therefore, the midrange \(Z\) is \((Y_1 + Y_3) / 2\).
02

Expressing Z in Terms of Y

To find the pdf of \(Z\), we first express it in terms of \(Y\). We have \(Z = \frac{Y_1 + Y_3}{2}\), which equates to \(Z = \frac{1}{2}(Y_1 + Y_3)\). Now, we apply the Jacobian method for transformation of random variables. The Jacobian Transformation matrix, given by the absolute value of the determinant of the transformation, follows the following formula: \[J = \left| \frac{\partial(x,y)}{\partial(u,v)} \right| = \left| \begin{matrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{matrix} \right| \] Using \(Y_2\) as it is and substituting \(Z = \frac{1}{2}(Y_1 + Y_3)\), we can solve determinant.
03

Applying Jacobian Transformation

Applying the Jacobian transformation, we see that (u,v) = (Y_2, Z) and (x,y) = (Y_1, Y_3). Therefore, the Jacobian matrix becomes:\[ \left| \begin{matrix} \frac{\partial Y_1}{\partial Y_2} & \frac{\partial Y_1}{\partial Z} \ \frac{\partial Y_3}{\partial Y_2} & \frac{\partial Y_3}{\partial Z} \end{matrix} \right| \]After calculations, we get \(J = 2\).
04

Calculating PDF

With the Jacobian, we calculate the probability density function (pdf). The pdf of Z, \(f_Z(Z)\), is given by the absolute value of the Jacobian multiplied by the pdfs of \(Y_1\), \(Y_2\), and \(Y_3\). As \(Y_1\), \(Y_2\), and \(Y_3\) are from a uniform distribution on (0,1), the pdfs of \(Y_1\), \(Y_2\), and \(Y_3\) are all 1. Therefore, \(f_Z(Z)\) is equal to \(2*1*1*1 = 2\).
05

Finalizing the PDF

The final pdf for Z must be valid in the given domain. As \(Y_1\), \(Y_2\), and \(Y_3\) are all from a uniform distribution in the interval (0,1) and \(Z = \frac{Y_1 + Y_3}{2}\), the minimum value Z can take is 0, and the maximum is 1. Therefore, the pdf of \(Z\), \(f_Z(Z)\), is 2 for \(0<Z<1\) and zero elsewhere.

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Most popular questions from this chapter

Using the assumptions behind the confidence interval given in expression (5.4.17), show that $$ \sqrt{\frac{S_{1}^{2}}{n_{1}}+\frac{S_{2}^{2}}{n_{2}}} / \sqrt{\frac{\sigma_{1}^{2}}{n_{1}}+\frac{\sigma_{2}^{2}}{n_{2}}} \stackrel{P}{\rightarrow} 1 $$

Let the observed value of the mean \(\bar{X}\) of a random sample of size 20 from a distribution that is \(N(\mu, 80)\) be \(81.2 .\) Find a 95 percent confidence interval for \(\mu .\)

Proceeding similar to Example 5.8.6, use the Accept-Reject Algorithin to generate an observation from a \(t\) distribution with \(r>1\) degrees of freedom ussing the Cauchy distribution.

. To illustrate Exercise 5.4.22, let \(X_{1}, X_{2}, \ldots, X_{9}\) and \(Y_{1}, Y_{2}, \ldots, Y_{12}\) represent two independent random samples from the respective normal distributions \(N\left(\mu_{1}, \sigma_{1}^{2}\right)\) and \(N\left(\mu_{2}, \sigma_{2}^{2}\right) .\) It is given that \(\sigma_{1}^{2}=3 \sigma_{2}^{2}\), but \(\sigma_{2}^{2}\) is unknown. Define a random variable which has a \(t\) -distribution that can be used to find a 95 percent confidence interval for \(\mu_{1}-\mu_{2}\).

. Let \(X_{1}, X_{2}, \ldots, X_{n}\) be random sample from a distribution of either type. A measure of spread is Gini's mean difference $$ G=\sum_{j=2}^{n} \sum_{i=1}^{j-1}\left|X_{i}-X_{j}\right| /\left(\begin{array}{l} n \\ 2 \end{array}\right) $$ (a) If \(n=10\), find \(a_{1}, a_{2}, \ldots, a_{10}\) so that \(G=\sum_{i=1}^{10} a_{i} Y_{i}\), where \(Y_{1}, Y_{2}, \ldots, Y_{10}\) are the order statistics of the sample. (b) Show that \(E(G)=2 \sigma / \sqrt{\pi}\) if the sample arises from the normal distribution \(N\left(\mu, \sigma^{2}\right)\)
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