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4.1 Introduction

Sometimes we are interested in a function of a random variable $X$ , say $Y=g(X)$ . For example, we have already discussed interest in the linear transformation

\displaystyle Y=\frac{X-\mu_{X}}{\sigma_{X}}.

Other applied examples include

$X$	$Y$
Diameter of imperfection in material	Area of imperfection
Speed of vehicle	Time to complete journey
Level of liquid in a vessel	Volume of liquid
Length of phone call	Cost of phone call

It is easy to show that in general the expectation of a function is not equal to the function of the expectation, i.e.

\displaystyle\operatorname{\mathsf{E}}\left[{Y}\right]=\operatorname{\mathsf{E% }}\left[{g(X)}\right]\neq g(\operatorname{\mathsf{E}}\left[{X}\right]).

For example $\operatorname{\mathsf{E}}\left[{X^{2}}\right]>\operatorname{\mathsf{E}}\left[{% X}\right]^{2}$ unless $X$ is constant, i.e. unless ${\operatorname{\mathsf{Var}}}\left[{X}\right]=0$ . Therefore what can we say about the new random variable $Y=g(X)$ ?

First consider the case of a discrete random variable $X$ and let $Y=g(X)$ for some real-valued function, $g$ . Then

	$\displaystyle p_{Y}(y)$	$\displaystyle=\sum_{\omega\in\Omega\colon Y(\omega)=y}\operatorname{\mathsf{P}% }\left({\omega}\right)$
		$\displaystyle={\color[rgb]{0.76,0.01,0}\sum_{\omega\in\Omega\colon g(X(\omega)% )=y}\operatorname{\mathsf{P}}\left({\omega}\right)}$
		$\displaystyle={\color[rgb]{0.76,0.01,0}\sum_{x\colon g(x)=y}\sum_{\omega\in% \Omega\colon X(\omega)=x}\operatorname{\mathsf{P}}\left({\Omega}\right)}$
		$\displaystyle={\color[rgb]{0.76,0.01,0}\sum_{x\colon g(x)=y}p_{X}(x).}$

In the discrete case finding the distribution of the transformed random variable $Y=g(X)$ is a simple matter of adding up the corresponding probabilities for $X$ .

Example 4.1.1.

Let $X$ have the following pmf

Value of $X$	0	1	2
$p_{X}(x)$	1/6	1/3	1/2

What are the pmfs of $Y=X^{2}$ and $V=(X-1)^{2}$ ?

Solution.

Value of $X$	0	1	2
$p_{X}(x)$	1/6	1/3	1/2
Value of $Y$	0	1	4
Value of $V$	1	0	1

Hence the mass functions are:

Value of $Y$	0	1	4
$p_{Y}(y)$	1/6	1/3	1/2

Value of $V$	0	1
$p_{V}(v)$	1/3	2/3

For continuous random variables, we have $\operatorname{\mathsf{P}}\left({X=x}\right)=0$ for all $x$ and we cannot find the density of $T=g(X)$ at $T=t$ by adding up the densities of all values of $x$ for which $g(x)=t$ .

This is illustrated in the figure. The left panel shows the histogram of a $1000$ independent samples from a random variable $U$ with a ${\operatorname{\mathsf{Unif}}}(0,2)$ distribution. The right histogram is of $Y=U^{2}$ . Just as in the first part of Example 4.1.1, the transformation is 1-1, so for each $U$ there is exactly one corresponding $Y$ ; hence if we could map densities in the same way that we map mass functions the second histogram would also be approximately flat.

⬇

par(mfrow=c(1,2))

U<-runif(10000,0,2)

Y<-U^2

hist(U,br=20,prob=TRUE)

hist(Y,br=20,prob=TRUE)

Unnumbered Figure: First link, Second Link

In the following we will see different methods of obtaining the cdf $F_{Y}(y)$ and pdf $f_{Y}(y)$ of the transformed random variable $Y=g(X)$ when $X$ is a continuous random variable with cdf $F_{X}(x)$ and pdf $f_{X}(x)$ .