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4.4 One-to-one Transformations

Returning to more general transformation of random variables we now restrict attention to general one-to-one transformations $Y=g(X)$ , so $X=g^{-1}(Y)$ exists.

Theorem 4.4.1.

If $X$ has pdf $f_{X}(x)$ and $Y=g(X)$ defines a one-to-one transformation, then $Y$ has pdf

\displaystyle f_{Y}(y)=f_{X}(x)\left|{dx\over dy}\right|

evaluated at $x=g^{-1}(y)$ .

Proof.

If $g$ is increasing then

\displaystyle F_{Y}(y)=\operatorname{\mathsf{P}}\left({g(X)\leq y}\right)=% \operatorname{\mathsf{P}}\left({X\leq g^{-1}(y)}\right)=F_{X}(g^{-1}(y))

so by differentiating wrt. $y$

\displaystyle f_{Y}(y)=f_{X}\{g^{-1}(y)\}\frac{dg^{-1}(y)}{dy}=f_{X}\{g^{-1}(y% )\}\frac{dx}{dy}.

If $g$ is decreasing then

\displaystyle F_{Y}(y)=\operatorname{\mathsf{P}}\left({g(X)\leq y}\right)=% \operatorname{\mathsf{P}}\left({X\geq g^{-1}(y)}\right)=1-\operatorname{% \mathsf{P}}\left({X\leq g^{-1}(y)}\right)=1-F_{X}(g^{-1}(y))

\displaystyle f_{Y}(y)=-f_{X}\{g^{-1}(y)\}\frac{dg^{-1}(y)}{dy}=f_{X}\{g^{-1}(% y)\}\left(-\frac{dx}{dy}\right).

Combining these results gives the stated result. ∎

Notation and hints:

•

We call $|dx/dy|$ the Jacobian of the transformation.
•

Sometimes it is most easy to evaluate $|dx/dy|$ by $|dy/dx|^{-1}$ .
•
It is often hard to remember which way up the Jacobian term should be. It is helpful to think in terms of probabilities of small sets, i.e. for suitable $x$ and $y$
1. $\operatorname{\mathsf{P}}\left({y<Y\leq y+\delta y}\right)=\operatorname{% \mathsf{P}}\left({x<X\leq x+\delta x}\right)$ ,
2. $f_{Y}(y)\delta y\approx f_{X}(x)\delta x$ ,
3. $f_{Y}(y)=f_{X}(x)|dx/dy|$ .

In practice the procedure for finding the pdf of $Y$ from this type of transformation is:

1.

Check you have a one-to-one transformation $g$ over the range of $X$ .
2.

Invert it – find $x$ as a function of $y$ . (This gives a way of checking it is a one-to-one transformation: can it be inverted?)
3.

Find $dx/dy$ (as a function of $y$ ).
4.

Use the theorem, replacing $x$ in $f_{X}(x)$ with $g^{-1}(y)$ .
5.

Summarise, remembering to state the range of $Y$ .

Example 4.4.1.

The Rayleigh distribution is used in modelling wave heights. Its pdf is

\displaystyle f_{X}(x)=\left\{\begin{array}[]{ll}2\lambda x\exp(-\lambda x^{2}% )&\quad x>0\\ 0&\quad\text{otherwise}\end{array}\right.

Find the pdf of the variable $Y=X^{2}$ which relates to wave power required to assess the force of the waves.

Solution. The range of $X$ is $(0,\infty)$ and so the range of $Y=X^{2}$ is the same.

The transform $y=x^{2}$ is one-to-one on $x>0$ so

$x=y^{1/2}=g^{-1}(y)$ ,
$\frac{dx}{dy}={\color[rgb]{0.76,0.01,0}\frac{1}{2}y^{-1/2},}$

	$\displaystyle f_{Y}(y)$	$\displaystyle={\color[rgb]{0.76,0.01,0}2\lambda y^{1/2}\exp(-\lambda y)\frac{1% }{2}y^{-1/2}}$
		$\displaystyle={\color[rgb]{0.76,0.01,0}\lambda\exp(-\lambda y)}$

for ${\color[rgb]{0.76,0.01,0}y>0.}$

Example 4.4.2.

If $X$ has pdf $f_{X}(x)$ then the linear transformation $Y=a+bX$ , $b\neq 0$ , has pdf

\displaystyle f_{Y}(y)=\frac{1}{|b|}f_{X}\{(y-a)/b\}.

This follows from the theorem with $g(x)=a+bx$ since $g^{-1}(y)=(y-a)/b$ and $|dx/dy|=1/|b|$ .

Use: if $Z\sim N(0,1)$ and therefore has pdf

\displaystyle f_{Z}(z)=\frac{1}{\sqrt{2\pi}}e^{-z^{2}/2},z\in(-\infty,\infty)

then $Y=\mu+\sigma Z$ has pdf

\displaystyle f_{y}(y)=\frac{1}{\sigma\sqrt{2\pi}}e^{-(y-\mu)^{2}/(2\sigma^{2}% )},y\in(-\infty,\infty).

i.e. $Y\sim N(\mu,\sigma^{2})$ (proving Theorem 3.7.1)

Example 4.4.3.

Show that if $U\sim U(-\frac{\pi}{2},\frac{\pi}{2})$ then $Y=\tan(U)$ has a Cauchy distribution.

Solution. First note that the range of $Y$ is $(-\infty,\infty)$ as required for a Cauchy random variable.

It is easier to differentiate $g(u)=\tan(u)$ than its inverse

	$\displaystyle\frac{dy}{du}$	$\displaystyle=\frac{dg}{du}$
		$\displaystyle={\color[rgb]{0.76,0.01,0}1+\tan^{2}(u)}$
		$\displaystyle={\color[rgb]{0.76,0.01,0}1+y^{2},}$

	$\displaystyle f_{Y}(y)$	$\displaystyle=f_{U}(g^{-1}(y))\left\|\frac{du}{dy}\right\|$
		$\displaystyle={\color[rgb]{0.76,0.01,0}f_{U}(g^{-1}(y))\left\|\frac{dy}{du}% \right\|^{-1}}$
		$\displaystyle={\color[rgb]{0.76,0.01,0}\frac{1}{\pi(1+y^{2})},}$

which is the density of a Cauchy random variable.

Example 4.4.4.

Show that if $X\sim$ Cauchy then also $Y=1/X\sim$ Cauchy.

Solution. The range of $X$ is $(-\infty,\infty)$ and so the range of $Y=1/X$ is the same. Here $y=g(x)=1/x$ so that $x=1/y=g^{-1}(y)$ and

\displaystyle\frac{dy}{dx}=-\frac{1}{x^{2}}={\color[rgb]{0.76,0.01,0}-y^{2}},

	$\displaystyle f_{Y}(y)$	$\displaystyle=f_{X}(g^{-1}(y))\left\|\frac{dx}{dy}\right\|$
		$\displaystyle={\color[rgb]{0.76,0.01,0}\frac{1}{\pi(1+1/y^{2})}y^{-2}}$
		$\displaystyle={\color[rgb]{0.76,0.01,0}\frac{1}{\pi(1+y^{2})}}$

for ${\color[rgb]{0.76,0.01,0}-\infty<y<\infty.}$

Example 4.4.5.

Example 4.2.4 revisited. $X\sim N(0,1)$ . Show that $Y=X^{2}$ has a $\operatorname{\mathsf{Gam}}(1/2,1/2)$ distribution. Note: this is also a $\chi^{2}_{1}$ distribution – see Section 11.1.

Solution. Note that $g(X)=X^{2}$ is not a one-to-one map. However we can create a new random variable $X_{*}=|X|$ which, by symmetry, has density

\displaystyle f_{X_{*}}(x_{*})=\left\{\begin{array}[]{ll}{\color[rgb]{% 0.76,0.01,0}\frac{2}{\sqrt{2\pi}}}e^{-x_{*}^{2}/2}&\quad x_{*}\geq 0\\ 0&\quad x_{*}<0\end{array}\right.

Now $Y=g(X_{*})=X_{*}^{2}$ and $g$ is a one-to-one map on the region where $X_{*}$ can be, $[0,\infty)$ . Since $x_{*}=y^{1/2}$ ,

\displaystyle\left|\frac{dx_{*}}{dy}\right|={\color[rgb]{0.76,0.01,0}\frac{1}{% 2}y^{-1/2}}

and we can use the transformation method to see that for $y\geq 0$ ,

	$\displaystyle f_{Y}(y)$	$\displaystyle={\color[rgb]{0.76,0.01,0}\frac{2}{\sqrt{2\pi}}e^{-\left((x_{*}(y% )\right)^{2}/2}\frac{1}{2}y^{-1/2}}$
		$\displaystyle={\color[rgb]{0.76,0.01,0}\frac{1}{\sqrt{2\pi}}y^{-1/2}e^{-\frac{% 1}{2}y},}$

which is the density of a $\operatorname{\mathsf{Gam}}(1/2,1/2)$ random variable.

Which method?

If the transformation from $X$ to $Y$ is not 1-1 and there is no simple trick to make an equivalent random variable using a 1-1 transformation, such as in Example 4.4.5 then the cdf (distribution function) method is the only option.

If either option is available then both will give the same answer, but one may require much less work. If you have $F_{X}(x)$ and it is in a simple form then the cdf (distribution function) method may be preferable. Alternatively, if you have $f_{X}(x)$ and the Jacobian has a simple form then the pdf (density function) method may be preferable.