# 2.1 The uniform and exponential distributions

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This course is a short series of lectures on Introductory Statistics. Topics covered are listed in the Table of Contents. The notes were prepared by EwaPaszek and Marek Kimmel. The development of this course has been supported by NSF 0203396 grant.

## The uniform distribution

Let the random variable X denote the outcome when a point is selected at random from the interval $\left[a,b\right]$ , $-\infty . If the experiment is performed in a fair manner, it is reasonable to assume that the probability that the point is selected from the interval $\left[a,x\right]$ , $a\le x is $\left(x-a\right)\left(b-a\right)$ . That is, the probability is proportional to the length of the interval so that the distribution function of X is

$F\left(x\right)=\left(\begin{array}{l}0,x

Because X is a continuous-type random variable, $F\text{'}\left(x\right)$ is equal to the p.d.f. of X whenever $F\text{'}\left(x\right)$ exists; thus when $a , we have $f\left(x\right)=F\text{'}\left(x\right)=1/\left(b-a\right).$

DEFINITION OF UNIFORM DISTRIBUTION
The random variable X has a uniform distribution if its p.d.f. is equal to a constant on its support. In particular, if the support is the interval $\left[a,b\right]$ , then
$f\left(x\right)=\frac{1}{b=a},a\le x\le b.$

Moreover, one shall say that X is $U\left(a,b\right)$ . This distribution is referred to as rectangular because the graph of $f\left(x\right)$ suggest that name. See Figure1. for the graph of $f\left(x\right)$ and the distribution function F(x).

We could have taken $f\left(a\right)=0$ or $f\left(b\right)=0$ without alerting the probabilities, since this is a continuous type distribution, and it can be done in some cases.

The mean and variance of X are as follows:

$\mu =\frac{a+b}{2}$ and ${\sigma }^{2}=\frac{{\left(b-a\right)}^{2}}{12}.$

An important uniform distribution is that for which a =0 and b =1, namely $U\left(0,1\right)$ . If X is $U\left(0,1\right)$ , approximate values of X can be simulated on most computers using a random number generator. In fact, it should be called a pseudo-random number generator (see the pseudo-numbers generation ) because the programs that produce the random numbers are usually such that if the starting number is known, all subsequent numbers in the sequence may be determined by simple arithmetical operations.

## An exponential distribution

Let turn to the continuous distribution that is related to the Poisson distribution . When previously observing a process of the approximate Poisson type, we counted the number of changes occurring in a given interval. This number was a discrete-type random variable with a Poisson distribution. But not only is the number of changes a random variable; the waiting times between successive changes are also random variables. However, the latter are of the continuous type, since each of then can assume any positive value.

Let W denote the waiting time until the first change occurs when observing the Poisson process in which the mean number of changes in the unit interval is $\lambda$ . Then W is a continuous-type random variable, and let proceed to find its distribution function.

Because this waiting time is nonnegative, the distribution function $F\left(w\right)=0$ , $w<0$ . For $w\ge 0$ ,

$F\left(w\right)=P\left(W\le w\right)=1-P\left(W>w\right)=1-P\left(no_changes_in_\left[0,w\right]\right)=1-{e}^{-\lambda w},$

since that was previously discovered that ${e}^{-\lambda w}$ equals the probability of no changes in an interval of length w is proportional to w , namely, $\lambda w$ . Thus when w >0, the p.d.f. of W is given by $F\text{'}\left(w\right)=\lambda {e}^{-\lambda w}=f\left(w\right).$

DEFINITION OF EXPONENTIAL DISTRIBUTION
Let $\lambda =1/\theta$ , then the random variable X has an exponential distribution and its p.d.f. id defined by
$f\left(x\right)=\frac{1}{\theta }{e}^{-x/\theta },0\le x<\infty ,$
where the parameter $\theta >0$ .

Accordingly, the waiting time W until the first change in a Poisson process has an exponential distribution with $\theta =1/\lambda$ . The mean and variance for the exponential distribution are as follows: $\mu =\theta$ and ${\sigma }^{2}={\theta }^{2}$ .

So if $\lambda$ is the mean number of changes in the unit interval, then $\theta =1/\lambda$ is the mean waiting for the first change. Suppose that $\lambda$ =7 is the mean number of changes per minute; then that mean waiting time for the first change is 1/7 of a minute.

Let X have an exponential distribution with a mean of 40. The p.d.f. of X is

$f\left(x\right)=\frac{1}{40}{e}^{-x/40},0\le x<\infty .$

The probability that X is less than 36 is

$P\left(X<36\right)=\underset{0}{\overset{36}{\int }}\frac{1}{40}{e}^{-x/40}dx=1-{e}^{-36/40}=0.593.$

Let X have an exponential distribution with mean $\mu =\theta$ . Then the distribution function of X is

$F\left(x\right)=\left\{\begin{array}{l}0,-\infty

The p.d.f. and distribution function are graphed in the Figure 3 for $\theta$ =5.

For an exponential random variable X , we have that $P\left(X>x\right)=1-F\left(x\right)=1-\left(1-{e}^{-x/\theta }\right)={e}^{-x/\theta }.$

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