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Chapter 5: Problem 8

Let \(Y_{1}

Short Answer

Expert verified
\(\(Z_{1}\) and \( Z_{2}\) are independent as the joint pdf can be separated to the product of the marginal pdfs, which is the mathematical definition of independence.

Step by step solution

01

Calculate joint pdf of \(Y_{2}\) and \(Y_{4}\)

The joint pdf of the second and fourth order statistics from a sample of 5 from this exponential distribution is given by \( g(y_{2}, y_{4}) = 25e^{-y_{4}}(1-e^{-y_{2}+y_{1}})^{2}(e^{-y_{3}+y_{2}})(1-e^{-y_{5}+y_{4}}), 0 < y_{2} < y_{4} < \infty \).
02

Change of variables

We need to change our variables to \(Z_{1} = Y_{2}\) and \(Z_{2}=Y_{4}-Y_{2}\). Convert the limits of the joint pdf to respect to the new variables \(z_{1}\) and \(z_{2}\), yielding \( 0 < z_{1} < z_{1}+z_{2} < \infty \). The Jacobian of this transformation is 鈭鈭�=1.
03

Calculate joint pdf of \(Z_{1}\) and \(Z_{2}\)

We can now express the joint pdf in terms of \(z_{1}\) and \( z_{2}\), using the Jacobian determinant and the joint pdf of \(y_{2}\) and \(y_{4}\). The joint pdf of \(z_{1}\) and \( z_{2}\) becomes: \( h(z_{1}, z_{2}) = |J| * g(y_{2}, y_{4}) = 25e^{-(z_{1}+z_{2})}(1-e^{-z_1})^{2}(e^{-z_2})(1-e^{-z_2}) \).
04

Separate to marginal pdfs

The joint pdf of \(Z_{1}\) and \(Z_{2}\) can be reduced to the product of two marginal pdfs. The marginal pdf of \(Z_{1}\) is \( h_{Z1}(z_{1}) = \int_{z_1}^{\infty}h(z_{1}, z_{2}) dz_{2} = 5e^{-z_1}(1-e^{-z_1})^{2}, 0 < z_{1} < \infty \). The marginal pdf of \(Z_{2}\) is \( h_{Z2}(z_{2}) = \int_{0}^{z_2}h(z_{1}, z_{2}) dz_{1} = 5e^{-z_{2}}(1-e^{-z_{2}^2}), 0 < z_{2} < \infty \).
05

Prove independence

Now, observe that the joint pdf equals the product of the marginal pdfs. This is the mathematical definition of independence: \(h(z_{1}, z_{2}) = h_{Z1}(z_1) * h_{Z2}(z_2)\). Therefore, \(Z_{1}\) and \(Z_{2}\) are independent.

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