Lecture 21
Agenda
1. Exponential Distribution
There are several random variables X that are observed in real life experiments, which have the property that “the chance that X takes values close
to x, decreases exponentially with x”. Often lifetimes of various objects have
been found to satisfy this property. How do we convert this intuitive notion
into a mathematical density function fX ?
First we notice that this means, for small h > 0,
P (x −
i.e.
h
h
< X < x + ) ∝ exp (−λx)
2
2
h
h
< X < x + ) = c(h) exp (−λx)
2
2
P (x −
for some c(h) > 0.
We also know,
lim
P (x −
h
2
h→0
< X < x + h2 )
= fX (x)
h
Thus,
lim
h→0
c(h)
h
exp (−λx) = fX (x)
i.e.
fX (x) = k exp (−λx)
where k = limh→0 c(h)
.
h
Since lifetimes are always non-negative, we have proved until this point that
fX (x) = k exp (−λx)
= 0
if x ≥ 0
otherwise
Note that k is unknown, but fX is a density, so
Z ∞
Z ∞
k
1=
fX (x)dx =
k exp (−λx)dx =
λ
−∞
0
and thus
k=λ
1
. Hence,
fX (x) = λ exp (−λx)
= 0
if x ≥ 0
otherwise
For reasons which will be clear soon we will replace λ by θ = λ1 .
Definition 1. A random variable X is said to have an Exponential distribution with parameter θ > 0 if
x
1
exp −
fX (x) =
θ
θ
= 0
if x ≥ 0
otherwise
Distribution Function
For x ≥ 0,
x
1
t
FX (x) =
exp − dt
θ
θ
0
x
t
= − exp −
θ 0
h
x i
= 1 − exp −
θ
Z
and for x < 0, FX (x) = 0.
Expectation and Variance
Before hopping into the calculation, let’s first learn something which will help
us now and also later.
In mathematics for α > 0, the Gamma Function is defined as follows,
Z ∞
Γ(α) =
xα−1 exp (−x)dx
0
R∞
Notice that Γ(1) = 0 exp (−x)dx = 1. Let’s derive another result. You
don’t have to memorize this proof, just go through it once.
2
Lemma 1. For α > 0,
Γ(α + 1) = αΓ(α)
Proof. For any differentiable function g(x),
d
{exp (−x)g(x)} = exp (−x){g 0 (x) − g(x)}
dx
Consider for x > 0, g(x) = xα , hence
d
{exp (−x)xα } = exp (−x){αxα−1 − xα }
dx
Integrating the above equation from 0 to ∞,
Z ∞
α ∞
[exp (−x)x ]0 =
exp (−x){αxα−1 − xα }dx
0
Thus
0 = αΓ(α) − Γ(α + 1)
and we are done.
We have seen Γ(1) = 1, and we use the above result repeatedly to get,
Γ(2)
Γ(3)
Γ(4)
Γ(5)
=
=
=
=
1 × Γ(1) = 1
2 × Γ(2) = 2
3 × Γ(3) = 3 × 2 = 3!
4 × Γ(4) = 4 × 3! = 4!
and continuing like this we get for any positive integer n, Γ(n) = (n − 1)!.
Now if X ∼ Exponential(θ),
Z ∞
x
1
E(X) =
x exp − dx
θ
θ
0
Z ∞
x
= θ
z exp (−z)dz by putting z =
θ
0
= θ × Γ(2)
= θ×1
= θ
3
2
Z
∞
x
x exp − dx
θ
θ
Z
∞
E(X ) =
0
= θ
2
21
z 2 exp (−z)dz by putting z =
0
x
θ
2
= θ × Γ(3)
= 2θ2
Hence,
V (X) = E(X 2 ) − [E(X)]2 = 2θ2 − θ2 = θ2
Memoryless Property of Geometric Distribution
For a, b > 0
P (X > a + b|X > a) =
=
=
=
=
P (X > a + b)
P (X > a)
1 − 1 − exp − a+b
θ 1 − 1 − exp − aθ
exp − a+b
θ
exp − aθ
b
exp −
θ
P (X > b)
If we remember the Geometric distribution, we understand why the above
property is called the memoryless property.
Example
The magnitude of earthquakes recorded in a region of North America can
be modelled by an exponential distribution with mean 2.4, as measured on
Richter scale.
(a) Find the probability that the next earthquake will be no more than 2.5
on the Richter scale.
Let X denote the magnitude of the earthquake.
X ∼ Exponential(2.4)
4
Thus,
P (X ≤ 2.5) = 1 − exp (−
2.5
) = 0.6471
2.4
(b) Given that the next earthquake will be more than 2 on Richter scale
what’s the probability that it will be more than 3 ?
1
= 0.6592
P (X > 3|X > 2) = P (X > 1) = exp −
2.4
Homework ::
No Homework
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