Home page for accesible maths 3 Probability Models 3.2 Uniform Distribution:

{\operatorname{\mathsf{Unif}}}(a,b)

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3.3 Exponential Distribution: $\operatorname{\mathsf{Exp}}(\beta)$

The pdf and cdf of a random variable, $X$ with an exponential distribution with rate $\beta>0$ are given by

\displaystyle f_{X}(x;\boldsymbol{\theta})=\left\{\begin{array}[]{ll}\beta\exp% (-\beta x)&\quad x\geq 0\\ 0&\quad\text{otherwise}\end{array}\right.

\displaystyle F_{X}(x)=\left\{\begin{array}[]{ll}1-\exp(-\beta x)&\quad x\geq 0% \\ 0&\quad\text{otherwise}\end{array}\right.

where $\boldsymbol{\theta}=\beta$ . The survivor function for $x>0$ is $S(x)=\exp(-\beta x)$ . We write $X\sim\operatorname{Exponential}(\beta)$ or sometimes $X\sim\operatorname{Exp}(\beta)$ . The pdf for two different values of $\beta$ is shown in Figure 3.2 (First Link, Second Link).

Figure 3.2: First Link, Second Link, Caption: The pdfs for two exponentially-distributed random variables

X

with

\beta=1

and

\beta=2

The exponential distribution arises in practice as the distribution of a waiting time when events follow a Poisson process (see Section 3.1); that is, events occur at random with a rate of $\beta$ per unit time.

Proof.

Consider some time $t$ . We know from Section 3.1 that the number of events, $N$ , in the interval $[0,t]$ has a Poisson distribution with mean $\beta t$ , $N\sim\operatorname{\mathsf{Poisson}}(\beta t)$ .

So the probability that there is at least one event in $[0,t]$ is

\displaystyle\operatorname{\mathsf{P}}\left({N\geq 1}\right)=1-\operatorname{% \mathsf{P}}\left({N=0}\right)=1-\frac{(\beta t)^{0}e^{-\beta t}}{0!}=1-e^{-% \beta t}.

Let $X$ be the time until the first event. We know that

•

If there is at least one event in in $[0,t]$ then the time until the first event must be less than $t$ .
•

If the time until the first event is less than $t$ then there must be at least one event in $[0,t]$ .

Or, more succinctly,

\displaystyle X\leq t\Leftrightarrow N\geq 1.

Hence

\displaystyle F_{X}(t)=\operatorname{\mathsf{P}}\left({X\leq t}\right)=% \operatorname{\mathsf{P}}\left({N\geq 1}\right)=1-e^{-\beta t}.

∎

The exponential distribution therefore arises either as the distribution of the time between events or as the distribution of the time to an event from a given start time. Examples include the amount of time from now until an earthquake occurs, or the time between receiving telephone calls, or the time until the next flood occurs.

Lack of memory

A key property of the exponential distribution is its lack of memory property, which arises due to the way the exponential distribution is obtained from the Poisson process. Exponential random variables are the only continuous random variables with this property. A random variable satisfies the memoryless property if

\displaystyle\operatorname{\mathsf{P}}\left({X>s+t\mid X>t}\right)=% \operatorname{\mathsf{P}}\left({X>s}\right)

for $s>0$ , $t>0$ , i.e. the conditional probability that a variable exceeds $s+t$ , given that it exceeds $t$ , is independent of $t$ so has no memory for how large it is already.

If we interpret $X$ as a waiting time to an event, this means that the probability that you have to wait a further time $s$ is independent of how long you have waited already.

To show this result holds for $X\sim\operatorname{Exponential}(\beta)$ recall that $\operatorname{\mathsf{P}}\left({X>x}\right)=\exp(-\beta x)$ for all $x>0$ . Hence, for $s>0$ , $t>0$

\displaystyle\operatorname{\mathsf{P}}\left({X>s+t\mid X>t}\right)={\color[rgb% ]{0.76,0.01,0}\frac{\operatorname{\mathsf{P}}\left({X>s+t}\right)}{% \operatorname{\mathsf{P}}\left({X>t}\right)}=\frac{\exp\{-\beta(s+t)\}}{\exp(-% \beta t)}=\exp(-\beta s)=\operatorname{\mathsf{P}}\left({X>s}\right).}

Also note that

\displaystyle\operatorname{\mathsf{P}}\left({X\in(t,t+\Delta t]|X>t}\right)=% \frac{\operatorname{\mathsf{P}}\left({X\in(t,t+\Delta t]}\right)}{% \operatorname{\mathsf{P}}\left({X>t}\right)}\approx\frac{\beta\exp(-\beta t)% \Delta t}{\exp(-\beta t)}=\beta\Delta t,

bringing us full circle back to the rate of the Poisson process.

Example 3.3.1.

Exam2016 Q2

(a)

A long-jump athlete, Jay, jumps a distance in metres which is modelled as $D=7.5+X$ , where $X\sim\operatorname{\mathsf{Exp}}(4.6)$ and $X$ is independent of the distance jumped on any previous attempt. The world long-jump record is $8.95$ m. Find the probability that, on any given jump, Jay exceeds the world record.
(b)

Just before his latest jump, Jay’s personal best was $b$ metres. Given that his latest jump exceeded his personal best, find the probability that it exceeded his personal best by at least $0.5$ m.
(c)

Comment on the appropriateness of the model for the length of each jump that was set out in Part (a).

Solution.

(a)

$\displaystyle\operatorname{\mathsf{P}}\left({D>8.95}\right)={\color[rgb]{% 0.76,0.01,0}\operatorname{\mathsf{P}}\left({X>1.45}\right)=e^{-4.6*1.45}% \approx 0.00127.}$
(b)

$\displaystyle\operatorname{\mathsf{P}}\left({D>b+0.5|D>b}\right)$ $\displaystyle={\color[rgb]{0.76,0.01,0}\operatorname{\mathsf{P}}\left({X>b-7.5% +0.5|X>b-7.5}\right)}$

$\displaystyle={\color[rgb]{0.76,0.01,0}\operatorname{\mathsf{P}}\left({X>0.5}% \right)=e^{-.5*4.6}\approx 0.100.}$
(c)

If there is a following wind (or Jay is tired) on one jump then there may be a following wind (or Jay is tired) on the next jump, so independence may not be entirely believable.

${\color[rgb]{0.76,0.01,0}\operatorname{\mathsf{P}}\left({\text{break world % record}}\right)\approx 1/1000}$ seems reasonable.

Less reasonable that when he breaks his own record there is a ${\color[rgb]{0.76,0.01,0}1/10}$ chance of breaking it by $50$ cm.

Will he really never jump less than 7.5m?

Moments

	$\displaystyle\operatorname{\mathsf{P}}\left({D>b+0.5\|D>b}\right)$	$\displaystyle={\color[rgb]{0.76,0.01,0}\operatorname{\mathsf{P}}\left({X>b-7.5% +0.5\|X>b-7.5}\right)}$
		$\displaystyle={\color[rgb]{0.76,0.01,0}\operatorname{\mathsf{P}}\left({X>0.5}% \right)=e^{-.5*4.6}\approx 0.100.}$

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3.3 Exponential Distribution: 𝖤𝗑𝗉⁡(β)

Proof.

Example 3.3.1.

3.3 Exponential Distribution: $\operatorname{\mathsf{Exp}}(\beta)$