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8.1 One-to-one Bivariate Transformations

Suppose that there are two random variables $X$ and $Y$ which have joint pdf $f_{XY}$ . We are interested in the joint distribution of two new random variables, $S=g_{1}(X,Y)$ and $T=g_{2}(X,Y)$ , which are functions of $(X,Y)$ . We assume that the transformation from $(X,Y)\rightarrow(S,T)$ is a one-to-one bivariate transformation, so that there exist functions $X=h_{1}(S,T)$ and $Y=h_{2}(S,T)$ . Then the joint pdf of $(S,T)$ is (see Appendix B)

\displaystyle f_{ST}(s,t)=f_{XY}(x,y)\mid\det J\mid_{x=h_{1}(s,t),y=h_{2}(s,t)}

where $\mid\det J\mid$ is the absolute value of the determinant of $J$ where $J$ is the Jacobian of the transformation

\displaystyle J={\frac{\partial(x,y)}{\partial(s,t)}}=\begin{bmatrix}\frac{% \partial x}{\partial s}&\frac{\partial x}{\partial t}\\ \frac{\partial y}{\partial s}&\frac{\partial y}{\partial t}\end{bmatrix}

and both $f(x,y)$ and $J$ are written as functions of $s$ and $t$ .

So the transformation procedure is:

1.

Check one-to-one bivariate transformation. (Given $x$ and $y$ can we find $s$ and $t$ uniquely, and given $s$ and $t$ can we find $x$ and $y$ uniquely?)
2.

Invert the transformation – find $s$ and $t$ as functions of $x$ and $y$ . (Again this might be an easy way of checking whether it is a one-to-one transformation).
3.

Find the Jacobian (as a function of $s$ and $t$ ).
4.

Use the formula, replacing $x$ and $y$ in $f(x,y)$ by the appropriate functions of $s$ and $t$ .
5.

Summarise, taking care with the ranges of $S$ and $T$ .

As in the univariate case it is sometimes easier to calculate the inverse $\mid\det J\mid^{-1}$ using

\displaystyle\mid\det J\mid^{-1}=\left|\det\begin{bmatrix}{\partial x\over% \partial s}&{\partial x\over\partial t}\\ &\\ {\partial y\over\partial s}&{\partial y\over\partial t}\end{bmatrix}\right|^{-% 1}=\left|\det\begin{bmatrix}{\partial s\over\partial x}&{\partial s\over% \partial y}\\ &\\ {\partial t\over\partial x}&{\partial t\over\partial y}\end{bmatrix}\right|.

Example 8.1.1.

Suppose $X$ and $Y$ are independent $\operatorname{\mathsf{N}}(0,1)$ random variables. Find the joint and marginal pdfs of $S=X+Y$ and $T=X-Y$ .

Solution. Since $X$ and $Y$ are independent their joint pdf is the product of the marginal pdfs

\displaystyle f_{XY}(x,y)=f_{X}(x)f_{Y}(y)=\frac{1}{2\pi}\exp\left(-\frac{x^{2% }}{2}-\frac{y^{2}}{2}\right).

Rearranging the transformation we have

$X=h_{1}(S,T)={\color[rgb]{0.76,0.01,0}\frac{S+T}{2},}$
$Y=h_{2}(S,T)={\color[rgb]{0.76,0.01,0}\frac{S-T}{2},}$

\displaystyle\mid\det J\mid={\color[rgb]{0.76,0.01,0}\left|{\det}\begin{% bmatrix}{\color[rgb]{0.76,0.01,0}1/2}&{\color[rgb]{0.76,0.01,0}1/2}\\ {\color[rgb]{0.76,0.01,0}1/2}&{\color[rgb]{0.76,0.01,0}-1/2}\end{bmatrix}% \right|=|-1/4-1/4|=|-1/2|=1/2.}

Thus

	$\displaystyle f_{ST}(s,t)$	$\displaystyle=\frac{1}{2\pi}{\color[rgb]{0.76,0.01,0}\exp\left(-\frac{(s+t)^{2% }}{8}-\frac{(s-t)^{2}}{8}\right)\times\frac{1}{2}}$
		$\displaystyle={\color[rgb]{0.76,0.01,0}\frac{1}{\sqrt{2\pi}\sqrt{2}}\exp(-s^{2% }/4)\frac{1}{\sqrt{2\pi}\sqrt{2}}\exp(-t^{2}/4)}.$

The marginal ranges of $s$ and $t$ are ${\color[rgb]{0.76,0.01,0}~{}-\infty<s<\infty}~{}$ and $~{}{\color[rgb]{0.76,0.01,0}-\infty<t<\infty}~{}$ . Further, for any given value of $s$ the range of $t$ is always $-\infty<t<\infty$ so ${\color[rgb]{0.76,0.01,0}S}$ and ${\color[rgb]{0.76,0.01,0}T}$ are iid ${\color[rgb]{0.76,0.01,0}\operatorname{\mathsf{N}}(0,2)}$ variables.

Example 8.1.2.

Suppose $X$ and $Y$ are independent, $X$ with an $\operatorname{\mathsf{Exp}}(1)$ distribution and $Y$ with a ${\operatorname{\mathsf{Unif}}}(0,2\pi)$ distribution. Find the joint and marginal pdfs of

\displaystyle(S,T)=(\sqrt{2X}\cos(Y),\sqrt{2X}\sin(Y)),

i.e. if $(\sqrt{2X},Y)$ are the polar coordinates of a point in the plane then $(S,T)$ are the corresponding Cartesian coordinates.

Unnumbered Figure: Link

Solution. The joint pdf of $(X,Y)$ is

\displaystyle f_{XY}(x,y)=\frac{1}{2\pi}\exp(-x)

for $0<x<\infty$ , $0<y<2\pi$ . In this case it is easier to find the inverse $\mid\det J\mid^{-1}$

\displaystyle\mid\det J\mid^{-1}=\left|\det\begin{bmatrix}{\partial s\over% \partial x}&{\partial s\over\partial y}\\ {\partial t\over\partial x}&{\partial t\over\partial y}\end{bmatrix}\right|=% \left|\det\begin{bmatrix}\frac{1}{\sqrt{2x}}\cos(y)&-\sqrt{2x}\sin(y)\\ \frac{1}{\sqrt{2x}}\sin(y)&\sqrt{2x}\cos(y)\end{bmatrix}\right|=1.

Since $X=(S^{2}+T^{2})/2$ we get

	$\displaystyle f_{ST}(s,t)$	$\displaystyle=\frac{1}{2\pi}\exp\left(-\frac{s^{2}+t^{2}}{2}\right)$
		$\displaystyle=\frac{1}{\sqrt{2\pi}}\exp(-s^{2}/2)\frac{1}{\sqrt{2\pi}}\exp(-t^% {2}/2).$

The marginal ranges of $s$ and $t$ are $-\infty<s<\infty$ and $-\infty<t<\infty$ . Further, for any given value of $s$ the range of $t$ is always $-\infty<t<\infty$ so $S$ and $T$ are independent and identically distributed $\operatorname{\mathsf{N}}(0,1)$ random variables.

The transformation in Example 8.1.2 is called the Box-Muller transformation, which is useful for simulating Normal random variables. Figure 8.1 (First Link, Second Link) illustrates the transformation.

Remember that we can generate an Exponential $(1)$ random variable from a uniform by the transformation $X=-\log(U)$ , thus we can generate two independent $N(0,1)$ random variables $N_{1}$ and $N_{2}$ from two independent Uniform $(0,1)$ random variables $U_{1}$ and $U_{2}$ by

\displaystyle(N_{1},N_{2})=(\sqrt{-2\log(U_{1})}\cos(2\pi U_{2}),\sqrt{-2\log(% U_{1})}\sin(2\pi U_{2})).

Figure 8.1: First Link, Second Link, Caption: The Box-Muller transformation in Example 8.1.2 for generating standard Normal random variables.

Example 8.1.3.

Suppose $X\sim\operatorname{\mathsf{Gam}}(a,1)$ and $Y\sim\operatorname{\mathsf{Gam}}(b,1)$ are independent random variables. Find the joint and marginal pdfs of $S=X+Y$ and $T=X/(X+Y)$ . Are $S$ and $T$ independent? Give your reasoning.

The marginal ranges of $S$ and $T$ are $~{}{\color[rgb]{0.76,0.01,0}s>0}~{}$ , $~{}{\color[rgb]{0.76,0.01,0}0\leq t\leq 1}~{}$ . Further, the range of $t$ does not depend on $s$ , so $s$ and $t$ are variationally independent.

Since $X$ and $Y$ are independent their joint pdf is the product of the marginal pdfs

	$\displaystyle f_{XY}(x,y)$	$\displaystyle=f_{X}(x)f_{Y}(y)$
		$\displaystyle=\frac{1}{\Gamma(a)}x^{a-1}\exp(-x)\frac{1}{\Gamma(b)}y^{b-1}\exp% (-y),$

for $0<x<\infty$ and $0<y<\infty$ .

Inverting the transformation, $X=~{}{\color[rgb]{0.76,0.01,0}ST}~{}$ and $Y=~{}{\color[rgb]{0.76,0.01,0}S(1-T)~{}.}$ The Jacobian matrix of partial derivatives is

\displaystyle J=\frac{\partial(x,y)}{\partial(s,t)}={\color[rgb]{0.76,0.01,0}% \begin{bmatrix}{\color[rgb]{0.76,0.01,0}t}&{\color[rgb]{0.76,0.01,0}s}\\ {\color[rgb]{0.76,0.01,0}1-t}&{\color[rgb]{0.76,0.01,0}-s}\end{bmatrix}}

Its determinant is $~{}{\color[rgb]{0.76,0.01,0}-s}~{}$ , so $|\det(J)|=~{}{\color[rgb]{0.76,0.01,0}s.}$

Thus the joint pdf is

	$\displaystyle f_{ST}(s,t)$	$\displaystyle=f_{XY}(x,y)\mid\det(J)\mid$
		$\displaystyle=\frac{1}{\Gamma(a)\Gamma(b)}x^{a-1}y^{b-1}\exp(-x-y)s$
		$\displaystyle=\frac{1}{\Gamma(a)\Gamma(b)}{\color[rgb]{0.76,0.01,0}(st)^{a-1}(% s[1-t])^{b-1}\exp(-s)s}$
		$\displaystyle=\frac{1}{\Gamma(a)\Gamma(b)}{\color[rgb]{0.76,0.01,0}s^{a+b-1}% \exp(-s)t^{a-1}(1-t)^{b-1}}$
		$\displaystyle={\color[rgb]{0.76,0.01,0}\frac{1}{\Gamma(a+b)}s^{a+b-1}\exp(-s)% \frac{\Gamma(a+b)}{\Gamma(a)\Gamma(b)}t^{a-1}(1-t)^{b-1},}$

with the range given above.

Joint pdf factorises and variationally independent so $~{}{\color[rgb]{0.76,0.01,0}S}~{}$ and $~{}{\color[rgb]{0.76,0.01,0}T}~{}$ are independent.
The pdfs can be recognised as $~{}{\color[rgb]{0.76,0.01,0}S\sim\operatorname{\mathsf{Gam}}(a+b,1),~{}T\sim% \operatorname{\mathsf{Beta}}(a,b)}.$