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Let \(Y_{1}

Short Answer

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The variables \(Z_{1}, Z_{2},..., Z_{n}\) defined as given are independent and each follows an exponential distribution. Any linear function of the \(Y_{i}\) variables can be expressed as a linear function of the independent \(Z_{j}\) variables.

Step by step solution

01

Show the Independence of Z Variables and Define their Distributions

Starting with the variable definitions, we note that each \(Z_{i}\) is a difference between two successive \(Y_{i}\) variables multiplied by a constant. The independence of \(Z_{i}\) variables can be shown by computing their joint distribution. Since each \(Z_{i}\) is dependent only on its corresponding and previous \(Y_{i}\) variable and since these joint distributions are known to be exponential, our job is to transform this joint exponential distribution into an exponential distribution of \(Z_{i}\) variables. After transforming the joint distribution and separating it in terms of each \(Z_{i}\) variable (using principles of transformations of random variables), the joint distribution will factorize into individual \(Z_{i}\) distributions, thus proving their independence. Further, each of these individual distributions will be exponential, proving the second part.
02

Express Linear Functions of Y Variables as Functions of Independent Z Variables

Given a linear function of \(Y_{i}\) variables, such as \(\sum_{1}^{n} a_{i} Y_{i}\), we can express \(Y_{i}\) in terms of \(Z_{j}\) variables as follows: \(Y_{i} = Z_{i} + Z_{i-1} + \cdots + Z_{1}\) for \(i = 2, 3, \cdots, n\), and \(Y_{1} = Z_{1}\). Replacing each \(Y_{i}\) in the linear function with the equivalent \(Z_{j}\) expression will result in a new linear function of \(Z_{j}\) variables. As shown in Step 1, \(Z_{j}\) variables are independent, so we have successfully expressed a linear function of dependent \(Y_{i}\) variables as a function of independent \(Z_{j}\) variables.
03

Simplify and Express the Final Result

Finally, simplify the expression obtained in Step 2 to have a more elegant and comprehensible linear function of \(Z_{j}\) variables. This is the final result.

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Most popular questions from this chapter

Let \(Y_{1}0\), provided that \(x \geq 0\), and \(f(x)=0\) elsewhere. Show that the independence of \(Z_{1}=Y_{1}\) and \(Z_{2}=Y_{2}-Y_{1}\) characterizes the gamma pdf \(f(x)\), which has parameters \(\alpha=1\) and \(\beta>0\) Hint: Use the change-of-variable technique to find the joint pdf of \(Z_{1}\) and \(Z_{2}\) from that of \(Y_{1}\) and \(Y_{2} .\) Accept the fact that the functional equation \(h(0) h(x+y) \equiv\) \(h(x) h(y)\) has the solution \(h(x)=c_{1} e^{c_{2} x}\), where \(c_{1}\) and \(c_{2}\) are constants.

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample from a \(\Gamma(1, \beta)\) distribution. (a) Show that the confidence interval \(\left(2 n \bar{X} /\left(\chi_{2 n}^{2}\right)^{(1-(\alpha / 2))}, 2 n \bar{X} /\left(\chi_{2 n}^{2}\right)^{(\alpha / 2)}\right)\) is an exact \((1-\alpha) 100 \%\) confidence interval for \(\beta\). (b) Show that value of a \(90 \%\) confidence interval for the data of the example is \((64.99,136.69)\)

. Let \(f(x)=\frac{1}{6}, x=1,2,3,4,5,6\), zero elsewhere, be the pmf of a distribution of the discrete type. Show that the pmf of the smallest observation of a random sample of size 5 from this distribution is $$ g_{1}\left(y_{1}\right)=\left(\frac{7-y_{1}}{6}\right)^{5}-\left(\frac{6-y_{1}}{6}\right)^{5}, \quad y_{1}=1,2, \ldots, 6 $$ zero elsewhere. Note that in this exercise the random sample is from a distribution of the discrete type. All formulas in the text were derived under the assumption that the random sample is from a distribution of the continuous type and are not applicable. Why?

A Weibull distribution with pdf $$ f(x)=\left\\{\begin{array}{ll} \frac{1}{\theta^{3}} 3 x^{2} e^{-x^{3} / \theta^{3}} & 0

Consider the sample of data: \(\begin{array}{rrrrrrrrrrr}13 & 5 & 202 & 15 & 99 & 4 & 67 & 83 & 36 & 11 & 301 \\ 23 & 213 & 40 & 66 & 106 & 78 & 69 & 166 & 84 & 64 & \end{array}\) (a) Obtain the five number summary of these data. (b) Determine if there are any outliers. (c) Boxplot the data. Comment on the plot. (d) Obtain a \(92 \%\) confidence interval for the median \(\xi_{1 / 2}\).

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