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Chapter 4: Problem 22

Let \(Y_{1}

Short Answer

Expert verified
All the \(Z_i\)'s have exponential distribution and they are independent. Therefore, for any given \(a_i\), we can express \(a_1Y_1 + a_2Y_2 + ... + a_nY_n\) as a linear function of the \(Z_i\)'s, which are independent random variables.

Step by step solution

01

Define given Variables

Start by defining \(Z_{i}\) variables for \(i= 1,...,n\) as mentioned in the problem. Lets Analyze \(Z_1 = nY_1\). The distribution function will be \(F_{Z_{1}}(z) = P(Z_1 \leq z) = P(nY_1 \leq z) = P(Y_1 \leq z/n)\). Since Y1 has an exponential distribution, We know that for a standard exponential random variable X, its distribution function is \[ F_X(x) = \[1 - e^{-x}], for x \geq 0.\] We can then use this with our \(Y_1\) to solve our problem.
02

Calculate distribution for \(Z_1\)

As noted in the problem, \(Y_1\)'s distribution is exponentially based. Therefore, we can substitute and calculate the required distribution: \(F_{Z_{1}}(z) = P(Y_1 \leq z/n) = 1 - e^{-z/n)\). This implies that \(Z_1\) also has an exponential distribution. Similar steps can be followed to show for other \(Z_i\)'s.
03

Prove independence of \(Z_i\)'s

The independence of \(Z_i\)'s can be shown by demonstrating that the joint probability density function (pdf) of \(Z_1, Z_2, ..., Z_n\) factorizes into the product of the pdfs of the individual \(Z_i\)'s. This requires integrating the pdf of \(Z_i\)'s over the ranges of all \(Y_i\)'s in such a manner that the resulting expression turns into the product of individual densities.
04

Demonstrating linear functions

In order to demonstrate the second part of the exercise, express the given linear function in terms of \(Z_i\)'s. That is, write \(a_1Y_1 + a_2Y_2 + ... + a_nY_n\) in terms of \(Z_1, Z_2,..., Z_n\), then use the result from step 3 to show that the function can be expressed as a linear function of \(Z_i\), which are independent random variables.

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Most popular questions from this chapter

Let the result of a random experiment be classified as one of the mutually exclusive and exhaustive ways \(A_{1}, A_{2}, A_{3}\) and also as one of the mutually exhaustive ways \(B_{1}, B_{2}, B_{3}, B_{4}\). Say that 180 independent trials of the experiment result in the following frequencies: \begin{tabular}{|c|c|c|c|c|} \hline & \(B_{1}\) & \(B_{2}\) & \(B_{3}\) & \(B_{4}\) \\ \hline\(A_{1}\) & \(15-3 k\) & \(15-k\) & \(15+k\) & \(15+3 k\) \\ \hline\(A_{2}\) & 15 & 15 & 15 & 15 \\ \hline\(A_{3}\) & \(15+3 k\) & \(15+k\) & \(15-k\) & \(15-3 k\) \\ \hline \end{tabular} where \(k\) is one of the integers \(0,1,2,3,4,5\). What is the smallest value of \(k\) that leads to the rejection of the independence of the \(A\) attribute and the \(B\) attribute at the \(\alpha=0.05\) significance level?

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