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Chapter 2: Problem 3

Let \(X_{1}\) and \(X_{2}\) have the joint pdf \(h\left(x_{1}, x_{2}\right)=2 e^{-x_{1}-x_{2}}, 0

Short Answer

Expert verified
The joint probability density function of \(Y_{1}\) and \(Y_{2}\) is \(f(y_{1}, y_{2}) = e^{-2y_{1}-2y_{2}}\) for \(0 < y_{2} < y_{1} < \infty, 0 < y_{2}\) and zero elsewhere.

Step by step solution

01

Identify the given information and new functions

The problem is defined as follows. We have two random variables \(X_{1}\) and \(X_{2}\) with the joint pdf \(h(x_{1}, x_{2}) = 2e^{-x_{1}-x_{2}}\) for \(0<x_{1}<x_{2}<\infty\), zero elsewhere. Then we are given two functions of these variables: \(Y_{1}=2 X_{1}\) and \(Y_{2}=X_{2}-X_{1}\). Our aim is to find the joint pdf of \(Y_{1}\) and \(Y_{2}\).
02

Write down the expressions to derive the Jacobian

To find the joint pdf of \(Y_{1}\) and \(Y_{2}\), we have to compute the Jacobian of the transformation, which is the determinant of matrix of the partial derivatives of the original variables (\(X_{1}\) and \(X_{2}\)) with respect to the transformed variables (\(Y_{1}\) and \(Y_{2}\)). So, write down the expressions, \(X_{1} = Y_{1}/2\) and \(X_{2} = Y_{1}/2 + Y_{2}\), to find the Jacobian.
03

Compute the Jacobian

The Jacobian \(J\) is given by the determinant \(\partial(x_{1}, x_{2})/\partial(y_{1}, y_{2})\). In our case, it will be \(J = |\partial(x_{1}, x_{2})/\partial(y_{1}, y_{2})| = |1/2, -1/2; 0, 1| = 1/2.\)
04

Compute the joint pdf

Substitute \(x_{1} = y_{1}/2\) and \(x_{2} = y_{1}/2 + y_{2}\) into original pdf and multiply by absolute Jacobian (which is 1/2 here), you get the joint pdf of \(Y_{1}\) and \(Y_{2}\) as \(f(y_{1}, y_{2}) = |J| h(x_{1}, x_{2}) = |1/2| * 2e^{-y_{1}-y_{2}-y_{1}/2} = e^{-2y_{1}-2y_{2}},\) where \(0 < y_{2} < y_{1} < \infty, 0 < y_{2}\) and zero elsewhere.

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Most popular questions from this chapter

Let \(f\left(x_{1}, x_{2}, x_{3}\right)=\exp \left[-\left(x_{1}+x_{2}+x_{3}\right)\right], 0

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