Introduction

We frequently have the type of problem where we have a random variable $X$ with known distribution and a function $g$ and wish to find the distribution of the random variable $Y=g(X)$. There are essentially 3 methods for finding the distribution of $Y$ and these are summarized briefly as follows.

1. Method of Distribution Functions

Let $F_Y(y)$ denote the cdf of $Y$. Then

$$\begin{eqnarray*} F_Y(y) & = & P(Y \leq y)\\ & = & P(g(X)\leq y)\\ & = & P(X\leq g^{-1}(y))\\ & = & F_X(g^{-1}(y)) \end{eqnarray*}$$

where $F_X$ is the cdf of random variable $X$.

2. Method of Transformations

In the case of a continuous random variable $X$ with pdf $f_X(x), \ x\in R_X$, and $g$ a strictly increasing or strictly decreasing function for $x\in R_X$, the random variable $Y$ has pdf given by

$$f_Y(y) = f_X(x)\left\vert \frac{dx}{dy}\right\vert$$ (3.1)

where the RHS is expressed as a function of $y$.

For example, if $f(x)=\alpha e^{-\alpha X}$ and $y=x^2$, write $f_X(x)=\alpha e^{-\alpha \sqrt{y}}$. The Jacobian keeps track of the scale change in going from $x$ to $y$.

A modification of the procedure enables us to deal with the situation where $g$ is piecewise monotone.

3. Method of Moment Generating Functions

This method is based on the uniqueness theorem, which states that if two mgf's are identical, the two random variables with those mgf's possess the same probability distribution. So we would need to find the mgf of $Y$ and compare it with the mgf's for the common distributions. If it is identical to some well-known mgf, the probability distribution of $Y$ will be identified.

The problem above was dealt with in a section called Change of Variable in the Statistics 260-1 course. The new work in this chapter concerns what may be called bivariate transformations. That is, we begin with the joint distribution of $2$ random variables, $X_1$ and $X_2$ say, and two functions, $g$ and $h$, and wish to find the joint distribution of the random variables $Y_1=g(X_1, X_2)$ and $Y_2=h(X_1,X_2)$. The marginal distribution of one or both of $Y_1$ and $Y_2$ can then be found. We may wish to do this if we changed coordinates from Cartesian $(X_1,X_2)$ to polar coordinates $(Y_1,Y_2)$.

This can, of course, be extended to multivariable transformations.

Before leaving this section, the following example should help you recall the technique.

Example 3..1
Suppose random variable $X$ is distributed $N(\mu , \sigma^2 )$, and random variable $Y$ is defined by $Y=X^2 /\sigma^2$, find the distribution of $Y$.

Method 1. Let $G_Y(y)$ be the cdf of $Y$. Then

$$\begin{eqnarray*} G_Y(y) & = & P(Y\leq y)\\ & = & P(\frac{X^2}{\sigma^2} \leq... ...\frac{\mu}{\sigma} ) - \Phi (-\sqrt{y} - \frac{\mu}{\sigma}). \end{eqnarray*}$$

The pdf of $Y, \ g_Y(y)$ is obtained by differentiating $G_Y(y)$ wrt $y$.

$$g_Y(y) = \phi (\sqrt{y} - \frac \mu\sigma ).\frac 12 y^{-\frac 12}+ \phi (-\sqrt{y} - \frac \mu\sigma )\frac 12 y^{-\frac 12}$$

$$= \frac 12 y^{-\frac 12} \left[ \frac{e^{-\frac 12 (\sqrt{y}- \frac... ...i}} + \frac{e^{-\frac 12 (\sqrt{y} + \frac \mu\sigma )^2}}{\sqrt{2\pi}} \right]$$

$$= \frac{e^{-\frac y2} y^{\frac 12 -1}}{2^{\frac 12} \Gamma (\frac 1... ...-\mu y^{\frac 12} / \sigma})}{2} \hspace{1cm} \mbox{where} \ \ y\in [0,\infty).$$

Note that the first part of the RHS is the pdf of a chi-square random variable with 1 df. In fact $Y$ is said to have a non-central $\chi^2$ distribution with 1 df and non-centrality parameter $\mu^2 /\sigma^2$. [This will be dealt with further in Chapter 5.]

Method 2. Noting that $y=x^2/\sigma^2$ is strictly decreasing for $x\in (-\infty ,0]$ and strictly increasing for $x\in (0,\infty )$, we use a modification of (3.1).

For $x\in (-\infty ,0]$ we have $x=-\sigma y^{\frac 12}$ and $\vert dx/dy\vert=\frac 12 \sigma y^{-\frac 12}$.

So

$$f^*_Y(y) = \frac{1}{\sqrt{2\pi }\sigma} e^{-\frac 12 (-\sigma... ...2 /\sigma ^2} . \textstyle{\frac 12} \sigma y^{-\frac 12} \ ,$$

replacing $x$ in the $N(\mu , \sigma^2 )$ pdf by $-\sigma y^{\frac 12}$.

For $x\in (0,\infty )$ we have $x= +\sigma y^{\frac 12}$ and $\vert dx/dy\vert=\frac 12 \sigma y^{-\frac 12}$.

So

$$f^{**}_Y(y) = \frac{1}{\sqrt{2\pi} \sigma} e^{-\frac 12 (\sig... ...} / \sigma^2} . \textstyle{\frac 12} \sigma y^{-\frac 12} \ .$$

The pdf of $Y$ is the sum of $f^*_Y(y)$ and $f^{**}_Y(y)$ which simplifies to (3.2).

Method 3.

$$\begin{eqnarray*} M_Y(t)&=& E(e^{tY}) \\ &=& \int_{-\infty}^{\infty} e^{tx^2/\... ...{ {{\mu^2}\over{\sigma^2}} {{t} \over {(1-2t)}} \right\} \ \ , \end{eqnarray*}$$

which is the M.G.F. of a non-central $\chi^2$ distribution (see Continuous Distributions by Johnson and Kotz, chapter 28.) The integral is a standard result obtained by completing the square in the exponent and using the result that $\int_{-\infty}^{\infty}e^{-u^2}du=\sqrt{\pi}$ giving

$$\int_\infty^\infty e^{-(ax^2 + bx + c)} dx = \sqrt{ {{\pi}\over{a}}} e^{(b^2-4ac)/4a} \ .$$