# Independent classes of random variables

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The concept of independence for classes of events is developed in terms of a product rule. Recall that for a real random variable X, the inverse image of each reasonable subset M on the real line (i.e., the set of all outcomes which are mapped into M by X) is an event. Similarly, the inverse image of N by random variable Y is an event. We extend the notion of independence to a pair of random variables by requiring independence of the events they determine in this fashion. This condition may be stated in terms of the product rule P(X in M, Y in N) = P(X in M)P(Y in N) for all Borel sets M, N.This product rule holds for the distribution functions FXY(t,u) = FX(t)FY(u) for all t, u. And similarly for density functions when they exist. This condition puts restrictions on the nature of the probability mass distribution on the plane. For a rectangle with sides M, Nthe probability mass in M x N is P(X in M)P(Y in N). Extension to general classes is simple and immediate.

## Introduction

The concept of independence for classes of events is developed in terms of a product rule. In this unit, we extend the concept to classes of random variables.

## Independent pairs

Recall that for a random variable X , the inverse image ${X}^{-1}\left(M\right)$ (i.e., the set of all outcomes $\omega \in \Omega$ which are mapped into M by X ) is an event for each reasonable subset M on the real line. Similarly, the inverse image ${Y}^{-1}\left(N\right)$ is an event determined by random variable Y for each reasonable set N . We extend the notion of independence to a pair of random variables by requiring independence of the events they determine. More precisely,

Definition

A pair $\left\{X,Y\right\}$ of random variables is (stochastically) independent iff each pair of events $\left\{{X}^{-1}\left(M\right),{Y}^{-1}\left(N\right)\right\}$ is independent.

This condition may be stated in terms of the product rule

$P\left(X\in M,\phantom{\rule{0.277778em}{0ex}}Y\in N\right)=P\left(X\in M\right)P\left(Y\in N\right)\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\text{for}\phantom{\rule{4.pt}{0ex}}\text{all}\phantom{\rule{4.pt}{0ex}}\text{(Borel)}\phantom{\rule{4.pt}{0ex}}\text{sets}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}M,\phantom{\rule{0.166667em}{0ex}}N$

Independence implies

${F}_{XY}\left(t,\phantom{\rule{0.166667em}{0ex}}u\right)=P\left(X\in \left(-\infty ,\phantom{\rule{0.166667em}{0ex}}t\right],\phantom{\rule{0.277778em}{0ex}}Y\in \left(-\infty ,\phantom{\rule{0.166667em}{0ex}}u\right]\right)=P\left(X\in \left(-\infty ,\phantom{\rule{0.166667em}{0ex}}t\right]\right)P\left(Y\in \left(-\infty ,\phantom{\rule{0.166667em}{0ex}}u\right]\right)=$
${F}_{X}\left(t\right){F}_{Y}\left(u\right)\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\forall \phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}t,\phantom{\rule{0.166667em}{0ex}}u$

Note that the product rule on the distribution function is equivalent to the condition the product rule holds for the inverse images of a special class of sets $\left\{M,N\right\}$ of the form $M=\left(-\infty ,t\right]$ and $N=\left(-\infty ,u\right]$ . An important theorem from measure theory ensures that if the product rule holds for this special classit holds for the general class of $\left\{M,N\right\}$ . Thus we may assert

The pair $\left\{X,\phantom{\rule{0.166667em}{0ex}}Y\right\}$ is independent iff the following product rule holds

${F}_{XY}\left(t,\phantom{\rule{0.166667em}{0ex}}u\right)={F}_{X}\left(t\right){F}_{Y}\left(u\right)\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\forall \phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}t,\phantom{\rule{0.166667em}{0ex}}u$

## An independent pair

Suppose ${F}_{XY}\left(t,u\right)=\left(1,-,{e}^{-\alpha t}\right)\left(1,-,{e}^{-\beta u}\right)\phantom{\rule{4pt}{0ex}}0\le t,0\le u$ . Taking limits shows

${F}_{X}\left(t\right)=\underset{u\to \infty }{lim}{F}_{XY}\left(t,\phantom{\rule{0.166667em}{0ex}}u\right)=1-{e}^{-\alpha t}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\text{and}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}{F}_{Y}\left(u\right)=\underset{t\to \infty }{lim}{F}_{XY}\left(t,\phantom{\rule{0.166667em}{0ex}}u\right)=1-{e}^{-\beta u}$

so that the product rule ${F}_{XY}\left(t,\phantom{\rule{0.166667em}{0ex}}u\right)={F}_{X}\left(t\right){F}_{Y}\left(u\right)$ holds. The pair $\left\{X,\phantom{\rule{0.166667em}{0ex}}Y\right\}$ is therefore independent.

If there is a joint density function, then the relationship to the joint distribution function makes it clear that the pair is independent iff the product rule holds for the density. That is, thepair is independent iff

${f}_{XY}\left(t,\phantom{\rule{0.166667em}{0ex}}u\right)={f}_{X}\left(t\right){f}_{Y}\left(u\right)\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\forall \phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}t,\phantom{\rule{0.166667em}{0ex}}u$

## Joint uniform distribution on a rectangle

Suppose the joint probability mass distributions induced by the pair $\left\{X,\phantom{\rule{0.166667em}{0ex}}Y\right\}$ is uniform on a rectangle with sides ${I}_{1}=\left[a,\phantom{\rule{0.166667em}{0ex}}b\right]$ and ${I}_{2}=\left[c,\phantom{\rule{0.166667em}{0ex}}d\right]$ . Since the area is $\left(b-a\right)\left(d-c\right)$ , the constant value of ${f}_{XY}$ is $1/\left(b-a\right)\left(d-c\right)$ . Simple integration gives

${f}_{X}\left(t\right)=\frac{1}{\left(b-a\right)\left(d-c\right)}{\int }_{c}^{d}du=\frac{1}{b-a}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}a\le t\le b\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\text{and}$
${f}_{Y}\left(u\right)=\frac{1}{\left(b-a\right)\left(d-c\right)}{\int }_{a}^{b}dt=\frac{1}{d-c}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}c\le u\le d$

Thus it follows that X is uniform on $\left[a,\phantom{\rule{0.166667em}{0ex}}b\right]$ , Y is uniform on $\left[c,\phantom{\rule{0.166667em}{0ex}}d\right]$ , and ${f}_{XY}\left(t,\phantom{\rule{0.166667em}{0ex}}u\right)={f}_{X}\left(t\right){f}_{Y}\left(u\right)$ for all $t,u$ , so that the pair $\left\{X,\phantom{\rule{0.166667em}{0ex}}Y\right\}$ is independent. The converse is also true: if the pair is independent with X uniform on $\left[a,\phantom{\rule{0.166667em}{0ex}}b\right]$ and Y is uniform on $\left[c,\phantom{\rule{0.166667em}{0ex}}d\right]$ , the the pair has uniform joint distribution on ${I}_{1}×{I}_{2}$ .

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