23.1 - Change-of-Variables Technique

Recall, that for the univariate (one random variable) situation: Given \(X\) with pdf \(f(x)\) and the transformation \(Y=u(X)\) with the single-valued inverse \(X=v(Y)\), then the pdf of \(Y\) is given by

\(\begin{align*} g(y) = |v^\prime(y)| f\left[ v(y) \right]. \end{align*}\)

Now, suppose \((X_1, X_2)\) has joint density \(f(x_1, x_2)\). and support \(S_X\).

Let \((Y_1, Y_2)\) be some function of \((X_1, X_2)\) defined by \(Y_1 = u_1(X_1, X_2)\) and \(Y_2 = u_2(X_1, X_2)\) with the single-valued inverse given by \(X_1 = v_1(Y_1, Y_2)\) and \(X_2 = v_2(Y_1, Y_2)\). Let \(S_Y\) be the support of \(Y_1, Y_2\).

Then, we usually find \(S_Y\) by considering the image of \(S_X\) under the transformation \((Y_1, Y_2)\). Say, given \(x_1, x_2 \in S_X\), we can find \((y_1, y_2) \in S_Y\) by

\(\begin{align*} x_1 = v_1(y_1, y_2), \hspace{1cm} x_2 = v_2(y_1, y_2) \end{align*}\)

The joint pdf \(Y_1\) and \(Y_2\) is

\(\begin{align*} g(y_1, y_2) = |J| f\left[ v_1(y_1, y_2), v_2(y_1, y_2) \right] \end{align*}\)

In the above expression, \(|J|\) refers to the absolute value of the Jacobian, \(J\). The Jacobian, \(J\), is given by

\(\begin{align*} \left| \begin{array}{cc} \frac{\partial v_1(y_1, y_2)}{\partial y_1} & \frac{\partial v_1(y_1, y_2)}{\partial y_2} \\ \frac{\partial v_2(y_1, y_2)}{\partial y_1} & \frac{\partial v_2(y_1, y_2)}{\partial y_2} \end{array} \right| \end{align*}\)

i.e. it is the determinant of the matrix

\(\begin{align*} \left( \begin{array}{cc} \frac{\partial v_1(y_1, y_2)}{\partial y_1} & \frac{\partial v_1(y_1, y_2)}{\partial y_2} \\ \frac{\partial v_2(y_1, y_2)}{\partial y_1} & \frac{\partial v_2(y_1, y_2)}{\partial y_2} \end{array} \right) \end{align*}\)

Example 23-1 Section

Suppose \(X_1\) and \(X_2\) are independent exponential random variables with parameter \(\lambda = 1\) so that

\(\begin{align*} &f_{X_1}(x_1) = e^{-x_1} \hspace{1.5 cm} 0< x_1 < \infty \\&f_{X_2}(x_2) = e^{-x_2} \hspace{1.5 cm} 0< x_2 < \infty \end{align*}\)

The joint pdf is given by

\(\begin{align*} f(x_1, x_2) = f_{X_1}(x_1)f_{X_2}(x_2) = e^{-x_1-x_2} \hspace{1.5 cm} 0< x_1 < \infty, 0< x_2 < \infty \end{align*}\)

Consider the transformation: \(Y_1 = X_1-X_2, Y_2 = X_1+X_2\). We wish to find the joint distribution of \(Y_1\) and \(Y_2\).

We have

\(\begin{align*} x_1 = \frac{y_1+y_2}{2}, x_2=\frac{y_2-y_1}{2} \end{align*}\)

OR

\(\begin{align*} v_1(y_1, y_2) = \frac{y_1+y_2}{2}, v_2(y_1, y_2)=\frac{y_2-y_1}{2} \end{align*}\)

The Jacobian, \(J\) is

\(\begin{align*} \left| \begin{array}{cc} \frac{\partial \left( \frac{y_1+y_2}{2} \right) }{\partial y_1} & \frac{\partial \left( \frac{y_1+y_2}{2} \right)}{\partial y_2} \\ \frac{\partial \left( \frac{y_2-y_1}{2} \right)}{\partial y_1} & \frac{\partial \left( \frac{y_2-y_1}{2} \right)}{\partial y_2} \end{array} \right| \end{align*}\)

\(\begin{align*} =\left| \begin{array}{cc} \frac{1}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{1}{2} \end{array} \right| = \frac{1}{2} \end{align*}\)

So,

\(\begin{align*} g(y_1, y_2) & = e^{-v_1(y_1, y_2) - v_2(y_1, y_2) }|\frac{1}{2}| \\ & = e^{- \left[\frac{y_1+y_2}{2}\right] - \left[\frac{y_2-y_1}{2}\right] }|\frac{1}{2}| \\ & = \frac{e^{-y_2}}{2} \end{align*}\)

Now, we determine the support of \((Y_1, Y_2)\). Since \(0< x_1 < \infty, 0< x_2 < \infty\), we have \(0< \frac{y_1+y_2}{2} < \infty, 0< \frac{y_2-y_1}{2} < \infty\) or \(0< y_1+y_2 < \infty, 0< y_2-y_1 < \infty\). This may be rewritten as \(-y_2< y_1 < y_2, 0< y_2 < \infty\).

Using the joint pdf, we may find the marginal pdf of \(Y_2\) as

\(\begin{align*} g(y_2) & = \int_{-\infty}^{\infty} g(y_1, y_2) dy_1 \\& = \int_{-y_2}^{y_2}\frac{1}{2}e^{-y_2} dy_1 \\& = \left. \frac{1}{2} \left[ e^{-y_2} y_1 \right|_{y_1=-y_2}^{y_1=y_2} \right] \\& = \frac{1}{2} e^{-y_2} (y_2 + y_2) \\& = y_2 e^{-y_2}, \hspace{1cm} 0< y_2 < \infty \end{align*}\)

Similarly, we may find the marginal pdf of \(Y_1\) as

\(\begin{align*} g(y_1)=\begin{cases} \int_{-y_1}^{\infty} \frac{1}{2}e^{-y_2} dy_2 = \frac{1}{2} e^{y_1} & -\infty < y_1 < 0 \\ \int_{y_1}^{\infty} \frac{1}{2}e^{-y_2} dy_2 = \frac{1}{2} e^{-y_1} & 0 < y_1 < \infty \\ \end{cases} \end{align*}\)

Equivalently,

\(\begin{align*} g(y_1) = \frac{1}{2} e^{-|y_1|} & 0 < y_1 < \infty \end{align*}\)

This pdf is known as the double exponential or Laplace pdf.