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Chapter 7: Problem 6

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

Short Answer

Expert verified
By using the transformation theorem to find the joint pdf and subsequently the marginal pdfs of Z and Y2, it is confirmed that the random variables \( Z = Y_1/Y_2 \) and \( Y_2 \) are independent.

Step by step solution

01

Applying the Transformation Method

Define new random variables \( Y_1 = X_1 + X_2 \) and \( Y_2 = X_1+X_2+X_3+X_4+X_5 \). Then, we can express this in terms of \( Y_1 \) and \( Y_2 \) i.e., \( Z = Y_1 / Y_2 \) and \( Y_2 \) and find the Jacobian matrix of the transformation from \( (X_1, X_2, X_3, X_4, X_5) \) to \( (Z, Y_2) \).
02

Find the Joint PDF of Z and Y2

Once the Jacobian matrix has been obtained, the joint pdf of \( Z \) and \( Y_2 \) is determined by applying the transformation theorem which involves multiplication of the absolute value of determinant of the Jacobian matrix to the pdf of \( X_1, X_2, X_3, X_4, X_5 \). Remember that since \( X_1, X_2, X_3, X_4, X_5 \) are iid, their joint pdf is simply the product of individual pdfs.
03

Find the Marginal PDFs

The marginal pdf of \( Z \) and \( Y_2 \) are determined by integrating the joint pdf of \( Z \) and \( Y_2 \) over the respective other variable.
04

Check for Independence

Finally, the independence is confirmed by verifying whether the joint pdf equals the product of the two marginal pdfs. If it holds true, then the two random variables \( Z \) and \( Y_2 \) are independent.

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