Bivariate Transformations

Both univariate and bivariate transformations of the discrete type are covered in HC 4.2, whereas transformations for continuous variables are covered in 4.3. The main result here, which is the two-dimensinal extension of (3.1), can be stated as follows.

For $(X,Y)$ continuous with joint pdf $f_{X,Y}(x,y), \ (x,y) \in \cal A$, and defining $U=g(X,Y),\\ V = h(X,Y)$, the joint pdf of $U$ and $V$, $f_{U,V} (u,v)$ is given by

$$f_{U,V}(u,v) = f_{X,Y} (x,y).\mbox{abs}\vert J\vert$$ (3.2)

providing the inverse transformation

$$\left. \begin{array}{l} x=G(u,v)\\ y=H(u,v) \end{array} \right\}$$

is one-to-one. Here abs$\vert J\vert$ refers to the absolute value of the Jacobian,

$$\left\vert \begin{array}{ll} \partial x / \partial u & \par... ...rtial u & \partial y / \partial v \end{array} \right\vert \ .$$

The RHS of (2.3) has to be expressed in terms of $u$ and $v$, and could be written more precisely as $f_{X,Y} (G(u,v), \ H(u,v))\mbox{abs}\vert J\vert.$

The diagonal elements of J account for scale change and the off-diagonal elements account for rotations.

Comments

1. In examples, it is essential to draw diagrams showing

   (i)

   the range space of $(X,Y)$, and

   (ii)

   the region this maps into under the transformation.

2. The distribution of the new random variables $U$ and $V$ is not complete unless the range space is specified.
3. Frequently we use this technique to find the distribution of some function of random variables, e.g., $X/Y$. That is, we are not mainly interested in the joint distribution of $U=g(X,Y)$ and $V=h(X,Y)$, but in the marginal distribution of one of them.

These points will be illustrated by the following examples.

Example 3..2
[This is HC Example 3 in 4.3.]

Given independent random variables $X$ and $Y$, each with uniform distributions on $(0,1)$, find the joint pdf of $U$ and $V$ defined by $U=X+Y, \ V=X-Y$, and the marginal pdf of $U$.

The joint pdf of $X$ and $Y$ is

$$f_{X,Y} (x,y)=1, \ \ 0\leq x\leq 1, \ \ 0\leq y \leq 1 \ .$$

The inverse transformation, written in terms of observed values is

$$x = (u+v)/2 \ \mbox{ and } y = (u-v)/2.$$

and is clearly one-to-one. The Jacobian is

$$J = \frac{\partial (x,y)}{\partial (u,v)} = \left\vert \begin... ...ht\vert = -\frac 12 \ , \mbox{ so abs}\vert J\vert=\frac{1}{2}.$$

Following the notation of HC, we will use $\cal A$ to denote the range space of $(X,Y)$, and $\cal B$ to denote that of $(U,V)$, and these are shown in the diagrams below. Firstly, note that there are 4 inequalities specifying ranges of $x$ and $y$, and these give 4 inequalities concerning $u$ and $v$, from which $\cal B$ can be determined. That is,

$$\begin{array}{llll} x\geq 0 & \Rightarrow & u+v\geq 0, & \mb... ...ow & u-v\leq 2, & \mbox{ that is, } \ \ v\geq u-2 \end{array}$$

Drawing the four lines

$$v = -u, \ v = 2-u, \ v = u, \ v = u-2$$

on the graph, enables us to see the region specified by the 4 inequalities.

Now, using (3.2) we have

$$f_{U,V}(u,v) = 1.\frac 12 \ , \ \ \left\{ \begin{array}{l} -u... ...\ u-2\leq v\leq 2-u, \ \ 1\leq u \leq 2 \end{array} \right.$$

The importance of having the range space correct is seen when we find the marginal pdf of $U$.

$$\begin{eqnarray*} f_{U}(u) & = & \int^\infty_{-\infty} f_{U,V} (u,v)\,dv\\ [0.2... ...u) + (2-u)I_{(1,2]}(u), \ \ \mbox{using indicator functions.} \end{eqnarray*}$$

Example 3..3
[HC Example 6, 4.3]
Given $X$ and $Y$ are independent random variables each with pdf
$f_X(x)=\frac 12\,e^{-x/2}, \ x\in [0,\infty )$, find the distribution of $(X-Y)/2$.

We note that the joint pdf of $X$ and $Y$ is

$$f_{X,Y}(x,y) = {\textstyle\frac 14}\,e^{-(x+y)/2}, \ \ 0 \leq x <\infty , \ \ 0\leq y <\infty .$$

Define $U=(X-Y)/2$. Now we need to introduce a second random variable $V$ which is a function of $X$ and $Y$. We wish to do this in such a way that the resulting bivariate transformation is one-to-one and our actual task of finding the pdf of $U$ is as easy as possible. Our choice for $V$ is of course, not unique. Let us define $V=Y$. Then the inverse transformation is, (using $u,v,x,y$, since we are really dealing with the range spaces here).

$$\begin{eqnarray*} x & = & 2u+v\\ y & = & v \end{eqnarray*}$$

from which we find the Jacobian,

$$J = \left\vert \begin{array}{cc} 2 & 1\\ 0 & 1 \end{array} \right\vert = 2 \ .$$

To determine $\cal B$, the range space of $U$ and $V$, we note that

$$\begin{eqnarray*} x \geq 0 & \Rightarrow & 2u +v \geq 0, \ \ \mbox{ that is, },... ...Rightarrow & v \geq 0\\ y < \infty & \Rightarrow & v < \infty \end{eqnarray*}$$

So $\cal B$ is as indicated in the diagram below

Now using (3.2) we have

$$\begin{eqnarray*} f_{U,V}(u,v)& = & {\textstyle \frac 14}\,e^{-(2u+v+v)/2}.2\\ & = & {\textstyle\frac 12}\, e^{-(u+v)} \ , \ \ (u,v) \in \cal B. \end{eqnarray*}$$

The marginal pdf of $U$ is obtained by integrating $f_{U,V} (u,v)$ with respect to $v$, giving

$$\begin{eqnarray*} f_U(u) & = & \left\{ \begin{array}{ll} \int^\infty_{-2u} {\t... ...style\frac 12}\,e^{-\vert u\vert} \ , \ \ -\infty < u < \infty \end{eqnarray*}$$

[This is sometimes called the folded (or double) exponential distribution.]

Example 3..4
Given $Z$ is distributed $N(0,1)$ and $Y$ is distributed as $\chi^2_\nu$, and $Z$ and $Y$ are independent, find the pdf of a random variable $T$ defined by

$$T = \frac{Z}{(Y/\nu )^{\frac 12}} \ .$$

Now the joint pdf of $Z$ and $Y$ is

$$f_{Z,Y}(z,y)=\frac{e^{-z^2/2}}{(2\pi )^{\frac 12}}.\frac{e^{-... ...u /2} \Gamma (\nu /2)} \ , \ \ y>0, \ \ -\infty < z < \infty.$$

Let $V=Y$, and we will find the joint pdf of $T$ and $V$ and then the marginal pdf of $T$. The inverse transformation is

$$\begin{eqnarray*} z & = & tv^{\frac 12}/\nu^{\frac 12}\\ y & = & v \end{eqnarray*}$$

from which $\vert J\vert=(v/\nu )^{\frac 12}$.

It is easy to check that ${\cal B}=\{ (t,v): \ -\infty < t < \infty, \ 0 <v<\infty\}$. So the joint pdf of $T$ and $V$ is

$$f_{T,V}(t,v)=\frac{e^{-t^2v/2\nu}}{(2\pi )^{\frac 12}}.\frac{... ... /2}\Gamma (\nu /2)}\,\frac{v^{\frac 12}}{\nu^{\frac 12}} \ ,$$

for $(t,v)\in \cal B$. The marginal pdf of $T$ is found by integrating $f_{T,V}(t,v)$ with respect to $v$, the limits on the integral being $0$ and $\infty$. Carry out this integration, substituting $x$ (say) for $\frac v2 (1+\frac{t^2}{\nu})$, and reducing the integral to a gamma function. The answer should be

$$f_T(t)=\frac{\Gamma (\frac{\nu +1}{2} )}{(\nu \pi )^{\frac 12... ...}{\nu} \right)^{-(\nu +1)/2} \ , \ \ -\infty < t < \infty \ ,$$

which you will recognise as the pdf of a random variable with a $t$-distribution with $\nu$ degrees of freedom. ($X$ is the sample mean and $Y$ is the sample variance)

Exercise. [See HC 4.4.]
Given random variables $X$ and $Y$ are independently distributed as chi-square with $\nu_1 ,\nu_2$ degrees of freedom, respectively, find the pdf of the random variable $F$ defined by $F=\nu_2 X / \nu_1 Y$.

Let $V=Y$ and find the joint pdf of $F$ and $V$, noting that the range space ${\cal B}=\{ (f,v):f >0, \ v>0\}$. You should find that $\vert J\vert=\nu_1 v / \nu_2$. Find the marginal pdf of $F$, which you should recognize as that for an $F_{\nu_1, \nu_2}$ distribution. You should try the following substitution to simplify the integration. Let $s=\frac v2 (1+\frac{\nu_1 f}{\nu_2})$.