Independent classes of random variables  (Page 3/6)

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The joint normal distribution

A pair $\left\{X,\phantom{\rule{0.166667em}{0ex}}Y\right\}$ has the joint normal distribution iff the joint density is

${f}_{XY}\left(t,\phantom{\rule{0.166667em}{0ex}}u\right)=\frac{1}{2\pi {\sigma }_{X}{\sigma }_{Y}{\left(1-{\rho }^{2}\right)}^{1/2}}{e}^{-Q\left(t,u\right)/2}$

where

$Q\left(t,\phantom{\rule{0.166667em}{0ex}}u\right)=\frac{1}{1-{\rho }^{2}}\left[{\left(\frac{t-{\mu }_{X}}{{\sigma }_{X}}\right)}^{2},-,2,\rho ,\left(\frac{t-{\mu }_{X}}{{\sigma }_{X}}\right),\left(\frac{u-{\mu }_{Y}}{{\sigma }_{Y}}\right),+,{\left(\frac{u-{\mu }_{Y}}{{\sigma }_{Y}}\right)}^{2}\right]$

The marginal densities are obtained with the aid of some algebraic tricks to integrate the joint density. The result is that $X\sim N\left({\mu }_{X},\phantom{\rule{0.166667em}{0ex}}{\sigma }_{X}^{2}\right)$ and $Y\sim N\left({\mu }_{Y},{\sigma }_{Y}^{2}\right)$ . If the parameter ρ is set to zero, the result is

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

so that the pair is independent iff $\rho =0$ . The details are left as an exercise for the interested reader.

Remark . While it is true that every independent pair of normally distributed random variables is joint normal, not every pair of normally distributed random variables has thejoint normal distribution.

A normal pair not joint normally distributed

We start with the distribution for a joint normal pair and derive a joint distribution for a normal pair which is not joint normal. The function

$\phi \left(t,u\right)=\phantom{\rule{4pt}{0ex}}\frac{1}{2\pi }exp\left(-,\frac{{t}^{2}}{2},-,\frac{{u}^{2}}{2}\right)$

is the joint normal density for an independent pair ( $\rho =0$ ) of standardized normal random variables. Now define the joint density for a pair $\left\{X,\phantom{\rule{0.166667em}{0ex}}Y\right\}$ by

${f}_{XY}\left(t,\phantom{\rule{0.166667em}{0ex}}u\right)=2\phi \left(t,\phantom{\rule{0.166667em}{0ex}}u\right)\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\text{in}\phantom{\rule{4.pt}{0ex}}\text{the}\phantom{\rule{4.pt}{0ex}}\text{first}\phantom{\rule{4.pt}{0ex}}\text{and}\phantom{\rule{4.pt}{0ex}}\text{third}\phantom{\rule{4.pt}{0ex}}\text{quadrants,}\phantom{\rule{4.pt}{0ex}}\text{and}\phantom{\rule{4.pt}{0ex}}\text{zero}\phantom{\rule{4.pt}{0ex}}\text{elsewhere}$

Both $X\sim N\left(0,\phantom{\rule{0.166667em}{0ex}}1\right)$ and $Y\sim N\left(0,\phantom{\rule{0.166667em}{0ex}}1\right)$ . However, they cannot be joint normal, since the joint normal distribution is positive for all $\left(t,\phantom{\rule{0.166667em}{0ex}}u\right)$ .

Independent classes

Since independence of random variables is independence of the events determined by the random variables, extension to general classes is simple and immediate.

Definition

A class $\left\{{X}_{i}:i\in J\right\}$ of random variables is (stochastically) independent iff the product rule holds for every finite subclass of two or more.

Remark . The index set J in the definition may be finite or infinite.

For a finite class $\left\{{X}_{i}:1\le i\le n\right\}$ , independence is equivalent to the product rule

${F}_{{X}_{1}{X}_{2}\cdots {X}_{n}}\left({t}_{1},\phantom{\rule{0.166667em}{0ex}}{t}_{2},\phantom{\rule{0.166667em}{0ex}}\cdots ,\phantom{\rule{0.166667em}{0ex}}{t}_{n}\right)=\prod _{i=1}^{n}{F}_{{X}_{i}}\left({t}_{i}\right)\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\text{for}\phantom{\rule{4.pt}{0ex}}\text{all}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\left({t}_{1},\phantom{\rule{0.166667em}{0ex}}{t}_{2},\phantom{\rule{0.166667em}{0ex}}\cdots ,\phantom{\rule{0.166667em}{0ex}}{t}_{n}\right)$

Since we may obtain the joint distribution function for any finite subclass by letting the arguments for the others be (i.e., by taking the limits as the appropriate t i increase without bound), the single product rule suffices to account for all finite subclasses.

Absolutely continuous random variables

If a class $\left\{{X}_{i}:i\in J\right\}$ is independent and the individual variables are absolutely continuous (i.e., have densities), then any finite subclass is jointly absolutelycontinuous and the product rule holds for the densities of such subclasses

${f}_{{X}_{i1}{X}_{i2}\cdots {X}_{im}}\left({t}_{i1},\phantom{\rule{0.166667em}{0ex}}{t}_{i2},\phantom{\rule{0.166667em}{0ex}}\cdots ,\phantom{\rule{0.166667em}{0ex}}{t}_{im}\right)=\prod _{k=1}^{m}{f}_{{X}_{ik}}\left({t}_{ik}\right)\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\text{for}\phantom{\rule{4.pt}{0ex}}\text{all}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\left({t}_{1},\phantom{\rule{0.166667em}{0ex}}{t}_{2},\phantom{\rule{0.166667em}{0ex}}\cdots ,\phantom{\rule{0.166667em}{0ex}}{t}_{n}\right)$

Similarly, if each finite subclass is jointly absolutely continuous, then each individual variable is absolutely continuous and the product rule holds for the densities.Frequently we deal with independent classes in which each random variable has the same marginal distribution. Such classes are referred to as iid classes (an acronym for i ndependent, i dentically d istributed). Examples are simple random samplesfrom a given population, or the results of repetitive trials with the same distribution on the outcome of each component trial. A Bernoulli sequence is a simple example.

Simple random variables

Consider a pair $\left\{X,\phantom{\rule{0.166667em}{0ex}}Y\right\}$ of simple random variables in canonical form

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