Question

Let \(Y_{1}

Short answer

\(\(Z_{1}\) 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

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 \( g(y_{2}, y_{4}) = 25e^{-y_{4}}(1-e^{-y_{2}+y_{1}})^{2}(e^{-y_{3}+y_{2}})(1-e^{-y_{5}+y_{4}}), 0 < y_{2} < y_{4} < \infty \).

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 ∣J∣=1.

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: \( h(z_{1}, z_{2}) = |J| * g(y_{2}, y_{4}) = 25e^{-(z_{1}+z_{2})}(1-e^{-z_1})^{2}(e^{-z_2})(1-e^{-z_2}) \).

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 \( h_{Z2}(z_{2}) = \int_{0}^{z_2}h(z_{1}, z_{2}) dz_{1} = 5e^{-z_{2}}(1-e^{-z_{2}^2}), 0 < z_{2} < \infty \).

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.