# 23.1 - Change-of-Variables Technique

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

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.

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