Exponential Distribution
Definition
X is said to have an exponential distribution with parameter λ(λ > 0) if
the pdf of X is
(
λe −λx x ≥ 0
f (x; λ) =
0
otherwise
Remark:
1. Usually we use X ∼ EXP(λ) to denote that the random variable X has
an exponential distribution with parameter λ.
2. In some sources, the pdf of exponential distribution is given by
(
1 − θx
e
x ≥0
f (x; θ) = θ
0
otherwise
The difference is that λ → 1θ .
Liang Zhang (UofU)
Applied Statistics I
June 30, 2008
1 / 20
Exponential Distribution
Liang Zhang (UofU)
Applied Statistics I
June 30, 2008
2 / 20
Exponential Distribution
Proposition
If X ∼ EXP(λ), then
E (X ) =
1
λ
and
V (X ) =
1
λ2
And the cdf for X is
(
1 − e −λx
F (x; λ) =
0
Liang Zhang (UofU)
Applied Statistics I
x ≥0
x <0
June 30, 2008
3 / 20
Exponential Distribution
Proof:
Z
E (X ) =
=
=
=
=
=
∞
xλe −λx dx
0
Z
1 ∞
(λx)e −λx d(λx)
λ 0
Z
1 ∞ −y
ye dy
y = λx
λ 0
Z ∞
1
[−ye −y |∞
e −y dy ] integration by parts:u = y , v = −e −y
0 +
λ
0
1
−y ∞
[0 + (−e |0 )]
λ
1
λ
Liang Zhang (UofU)
Applied Statistics I
June 30, 2008
4 / 20
Exponential Distribution
Proof (continued):
Z ∞
2
E (X ) =
x 2 λe −λx dx
0
Z ∞
1
= 2
(λx)2 e −λx d(λx)
λ 0
Z ∞
1
= 2
y 2 e −y dy
λ 0
Z ∞
1
= 2 [−y 2 e −y |∞
+
2ye −y dy ]
0
λ
0
Z ∞
1
−y ∞
= 2 [0 + 2(−ye |0 +
e −y dy )]
λ
0
1
= 2 2[0 + (−ye −y |∞
0 )]
λ
2
= 2
λ
Liang Zhang (UofU)
Applied Statistics I
y = λx
integration by parts
integration by parts
June 30, 2008
5 / 20
Exponential Distribution
Proof (continued):
2
1
1
V (X ) = E (X 2 ) − [E (X )]2 = 2 − ( )2 = 2
λ
λ
λ
Z x
−λy
F (x) =
λe
dy
0
Z x
=
e −λy d(λy )
0
Z x
=
e −z dz
z = λy
0
= −e −z |x0
= 1 − e −x
Liang Zhang (UofU)
Applied Statistics I
June 30, 2008
6 / 20
Exponential Distribution
Example (Problem 108)
The article “Determination of the MTF of Positive Photoresists Using the
Monte Carlo method” (Photographic Sci. and Engr., 1983:
254-260) proposes the exponential distribution with parameter λ = 0.93
as a model for the distribution of a photon’s free path length (µm) under
certain circumstances. Suppose this is the correct model.
a. What is the expected path length, and what is the standard deviation
of path length?
b. What is the probability that path length exceeds 3.0?
c. What value is exceeded by only 10% of all path lengths?
Liang Zhang (UofU)
Applied Statistics I
June 30, 2008
7 / 20
Exponential Distribution
Proposition
Suppose that the number of events occurring in any time interval of length
t has a Poisson distribution with parameter αt (where α, the rate of the
event process, is the expected number of events occurring in 1 unit of
time) and that numbers of occurrences in nonoverlappong intervals are
independent of one another. Then the distribution of elapsed time
between the occurrence of two successive events is exponential with
parameter λ = α.
e.g.
the number of customers visiting Costco in each hour =⇒ Poisson
distribution;
the time between every two successive customers visiting Costco =⇒
Exponential distribution.
Liang Zhang (UofU)
Applied Statistics I
June 30, 2008
8 / 20
Exponential Distribution
Example (Example 4.22)
Suppose that calls are received at a 24-hour hotline according to a Poisson
process with rate α = 0.5 call per day.
Then the number of days X between successive calls has an exponential
distribution with parameter value 0.5.
The probability that more than 3 days elapse between calls is
P(X > 3) = 1 − P(X ≤ 3) = 1 − F (3; 0.5) = e −0.5·3 = 0.223.
The expected time between successive calls is 1/0.5 = 2 days.
Liang Zhang (UofU)
Applied Statistics I
June 30, 2008
9 / 20
Exponential Distribution
“Memoryless” Property
Let X = the time certain component lasts (in hours) and we
assume the component lifetime is exponentially distributed with parameter
λ. Then what is the probability that the component can last at least an
additional t hours after working for t0 hours, i.e. what is
P(X ≥ t + t0 | X ≥ t0 )?
P({X ≥ t + t0 } ∩ {X ≥ t0 })
P(X ≥ t0 )
P(X ≥ t + t0 )
=
P(X ≥ t0 )
1 − F (t + t0 ; λ)
=
F (t0 ; λ)
P(X ≥ t + t0 | X ≥ t0 ) =
= e −λt
Liang Zhang (UofU)
Applied Statistics I
June 30, 2008
10 / 20
Exponential Distribution
“Memoryless” Property
However, we have
P(X ≥ t) = 1 − F (t; λ) = e −λt
Therefore, we have
P(X ≥ t) = P(X ≥ t + t0 | X ≥ t0 )
for any positive t and t0 .
In words, the distribution of additional lifetime is exactly the same as the
original distribution of lifetime, so at each point in time the component
shows no effect of wear. In other words, the distribution of remaining
lifetime is independent of current age.
Liang Zhang (UofU)
Applied Statistics I
June 30, 2008
11 / 20
Gamma Distribution
Definition
For α > 0, the gamma function Γ(α) is defined by
Z ∞
x α−1 e −x dx
Γ(α) =
0
Properties for gamma function:
1. For any α > 1, Γ(α) = (α − 1) · Γ(α − 1) [via integration by parts];
2. For any positive integer, n, Γ(n) = (n − 1)!;
√
3. Γ( 21 ) = π.
√
e.g. Γ(4) = (4 − 1)! = 6 and Γ( 52 ) = 23 · Γ( 32 ) = 23 [ 12 · Γ( 12 )] = 34 π
Liang Zhang (UofU)
Applied Statistics I
June 30, 2008
12 / 20
Gamma Distribution
Definition
A continuous random variable X is said to have a gamma distribution if
the pdf of X is
(
1
x α−1 e −x/β x ≥ 0
α
f (x; α, β) = β Γ(α)
0
otherwise
where the parameters α and β satisfy α > 0, β > 0. The standard
gamma distribution has β = 1, so the pdf of a standard gamma rv is
(
1
x α−1 e −x x ≥ 0
f (x; α) = Γ(α)
0
otherwise
Liang Zhang (UofU)
Applied Statistics I
June 30, 2008
13 / 20
Gamma Distribution
Remark:
1. We use X ∼ GAM(α, β) to denote that the rv X has a gamma
distribution with parameter α and β.
2. If we let α = 1 and β = 1/λ, then we get the exponential distribution:
1
f (x; 1, ) =
λ
(
1
1
Γ(1)
λ
1
x 1−1 e −x/ λ = λe −λx
0
x ≥0
otherwise
3. When X is a standard gamma rv (β = 1), the cdf of X ,
Z
F (x; α) =
0
x
y α−1 e −y
dy
Γ(α)
is called the incomplete gamma function.
There are extensive tables of F (x; α) available (Appendix Table A.4).
Liang Zhang (UofU)
Applied Statistics I
June 30, 2008
14 / 20
Gamma Distribution
Liang Zhang (UofU)
Applied Statistics I
June 30, 2008
15 / 20
Gamma Distribution
Proposition
If X ∼ GAM(α, β), then
E (X ) = αβ
and
V (X ) = αβ 2
Furthermore, for any x > 0, the cdf of X is given by
x
;α
P(X ≤ x) = F (x; α, β) = F
β
where F (•; α) is the incomplete gamma function.
Liang Zhang (UofU)
Applied Statistics I
June 30, 2008
16 / 20
Gamma Distribution
Example:
The survival time (in days) of a white rat that was subjected to a certain
level of X-ray radiation is a random variable X ∼ GAM(5, 4). Then what is
a. the probability that the survival time is at most 16 days;
b. the probability that the survival time is between 16 days and 20 days
(not inclusive);
c. the expected survival time.
Liang Zhang (UofU)
Applied Statistics I
June 30, 2008
17 / 20
Chi-Squared Distribution
Definition
Let ν be a positive integer. Then a random variable X is said to have a
chi-squared distribution with parameter ν if the pdf of X is the gamma
density with α = ν/2 and β = 2. The pdf of a chi-squared rv is thus
(
1
x (ν/2)−1 e −x/2 x ≥ 0
ν/2
f (x; ν) = 2 Γ(ν/2)
0
x <0
The parameter ν is called the number of degrees of freedom (df) of X .
The symbol χ2 is often used in place of “chi-squared”.
Liang Zhang (UofU)
Applied Statistics I
June 30, 2008
18 / 20
Chi-Squared Distribution
Remark:
1. Usually, we use X ∼ χ2 (ν) to denote that X is a chi-squared rv with
parameter ν;
2. If X1 , X2 , . . . , Xn is n independent standard normal rv’s, then
X12 + X22 + · · · + Xn2 has the same distribution as χ2 (n).
Liang Zhang (UofU)
Applied Statistics I
June 30, 2008
19 / 20
Chi-Squared Distribution
Liang Zhang (UofU)
Applied Statistics I
June 30, 2008
20 / 20