Mathematics 9709 · Probability & Statistics 2

Uniform & exponential models

Two PDFs come up so often they have ready-made formulas — learn these and you skip the integration. The uniform (rectangular) is a flat box: every value equally likely. The exponential is a decay curve: the waiting time between rare (Poisson) events.

Uniform & exponential: know the results

Left: a uniform (rectangular) density is a flat box. Right: an exponential density decays from a peak at t = 0.

Uniform (rectangular)

$f(x)=\dfrac{1}{b-a}$ on $[a,b]$ (flat, so total area $=\tfrac{1}{b-a}\cdot(b-a)=1$ ). By symmetry the centre is the mean and the median; the box has no mode (no single peak).

$\text{mean}=\text{median}=\tfrac{a+b}{2},\qquad \operatorname{Var}(X)=\dfrac{(b-a)^2}{12}$

Uniform on [a, b]: mean = median = (a+b)/2, variance = (b−a)²/12, no mode. No integration needed.

Worked example

Easiest first: a uniform distribution

Arrival time is uniform on $[0,8]$ , so $f(x)=\tfrac18$ . Read the summaries straight off.

$\text{mean}=\text{median}=\tfrac{0+8}{2}=4,\qquad \operatorname{Var}(X)=\tfrac{8^2}{12}=\tfrac{16}{3}\approx5.33$

Answer

\text{mean}=\text{median}=4,\ \operatorname{Var}(X)=\tfrac{16}{3}\approx5.33

Exponential

$f(t)=\lambda e^{-\lambda t}$ for $t\ge0$ . The tail has a shortcut: $P(T>a)=e^{-\lambda a}$ (so $P(T<a)=1-e^{-\lambda a}$ ). The mean and variance are also fixed: $E(T)=\tfrac1\lambda$ , $\operatorname{Var}(T)=\tfrac{1}{\lambda^2}$ .

Exponential f = λe^(−λt): P(T > a) = e^(−λa), E(T) = 1/λ, Var(T) = 1/λ². It models the wait between Poisson events.

Worked example

An exponential distribution

A battery's lifetime is exponential with $f(t)=0.2e^{-0.2t}$ , $t\ge0$ (so $\lambda=0.2$ ).

Tail: $P(T>1.5)=e^{-0.2\times1.5}=e^{-0.3}=0.741$ (no integration needed).
Mean: $E(T)=\tfrac{1}{0.2}=5$ years.
Variance: $\operatorname{Var}(T)=\tfrac{1}{0.2^2}=25$ .

Answer

P(T>1.5)\approx0.741,\ E(T)=5,\ \operatorname{Var}(T)=25

	Uniform on [a,b]	Exponential λe^(−λt)
f	1/(b−a)	λe^(−λt), t ≥ 0
P(X>c)	(b−c)/(b−a)	e^(−λc)
mean	(a+b)/2	1/λ
median	(a+b)/2	(ln 2)/λ
variance	(b−a)²/12	1/λ²

Why exponential = Poisson wait-time

If events arrive as $\text{Po}(\lambda)$ per unit time, then the wait $T$ until the next one is exponential with that same $\lambda$ . “No event before time $a$ ” is $P(\text{Po}=0)=e^{-\lambda a}$ — exactly the exponential tail $P(T>a)$ . The two distributions are the same fact seen from two angles.

Examiner note

Quote the formula and state which model you are using; for an exponential tail use

e^{-\lambda a}

directly rather than re-integrating. (9709/62/O/N/24 Q2)

Common mistake

Mixing up

E(T)=\tfrac1\lambda

with

\lambda

itself; forgetting the box has no mode; and a sign slip on the

\infty

limit (

e^{-\lambda t}\to0

, so

P(T>a)=e^{-\lambda a}

, not

1-e^{-\lambda a}

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Uniform & exponential models

Uniform & exponential: know the results

Left: a uniform (rectangular) density is a flat box. Right: an exponential density decays from a peak at t = 0.

Uniform (rectangular)

$f(x)=\dfrac{1}{b-a}$ on $[a,b]$ (flat, so total area $=\tfrac{1}{b-a}\cdot(b-a)=1$ ). By symmetry the centre is the mean and the median; the box has no mode (no single peak).

$\text{mean}=\text{median}=\tfrac{a+b}{2},\qquad \operatorname{Var}(X)=\dfrac{(b-a)^2}{12}$

Uniform on [a, b]: mean = median = (a+b)/2, variance = (b−a)²/12, no mode. No integration needed.

Worked example

Easiest first: a uniform distribution

Arrival time is uniform on $[0,8]$ , so $f(x)=\tfrac18$ . Read the summaries straight off.

$\text{mean}=\text{median}=\tfrac{0+8}{2}=4,\qquad \operatorname{Var}(X)=\tfrac{8^2}{12}=\tfrac{16}{3}\approx5.33$

Answer

\text{mean}=\text{median}=4,\ \operatorname{Var}(X)=\tfrac{16}{3}\approx5.33

Exponential

Exponential f = λe^(−λt): P(T > a) = e^(−λa), E(T) = 1/λ, Var(T) = 1/λ². It models the wait between Poisson events.

Worked example

An exponential distribution

A battery's lifetime is exponential with $f(t)=0.2e^{-0.2t}$ , $t\ge0$ (so $\lambda=0.2$ ).

Tail: $P(T>1.5)=e^{-0.2\times1.5}=e^{-0.3}=0.741$ (no integration needed).
Mean: $E(T)=\tfrac{1}{0.2}=5$ years.
Variance: $\operatorname{Var}(T)=\tfrac{1}{0.2^2}=25$ .

Answer

P(T>1.5)\approx0.741,\ E(T)=5,\ \operatorname{Var}(T)=25

	Uniform on [a,b]	Exponential λe^(−λt)
f	1/(b−a)	λe^(−λt), t ≥ 0
P(X>c)	(b−c)/(b−a)	e^(−λc)
mean	(a+b)/2	1/λ
median	(a+b)/2	(ln 2)/λ
variance	(b−a)²/12	1/λ²

Why exponential = Poisson wait-time

Examiner note

Quote the formula and state which model you are using; for an exponential tail use

e^{-\lambda a}

directly rather than re-integrating. (9709/62/O/N/24 Q2)

Common mistake

Mixing up

E(T)=\tfrac1\lambda

with

\lambda

itself; forgetting the box has no mode; and a sign slip on the

\infty

limit (

e^{-\lambda t}\to0

, so

P(T>a)=e^{-\lambda a}

, not

1-e^{-\lambda a}