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Topic 2: Special Continuous Distributions
When we think of a continuous probability distribution, we tend to think of the Normal distribution
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There are many things that are not well modeled by the normal distribution
...
g
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In this Topic we will consider variants of the normal distribution: distributions that are skewed
and bounded on one side or others bounded on both sides etc
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We
begin this section by defining a special function called the gamma function
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For α > 0, Equation (1) is called the gamma function and is denoted Γ (α) read ”gamma alpha”
hence:

Z∞
tα−1 e−t dt

Γ (α) =

(2)

0

Properties of the gamma function
1
...
e
...
Γ (4) = 3Γ (3)
2
...
g
...
Γ ( 21 ) =

√
π

Dividing both sides of Equation (2) by Γ (α) gives
Z ∞ α−1 −t
t
e
1=
dt
Γ (α)
0

1

(3)

The function in Equation (3) must be a probability density function
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The basic or standard gamma distribution
A random variable T is said to have a basic (or standard) gamma distribution if its probability




function is defined as
f(t) =

1
tα−1 e−t ,
Γ (α)


0,

t > 0, α > 0
(5)
elsewhere

We use the notation T ∼ Γ (α) - T has a gamma distribution with parameter α
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The full gamma distribution
Suppose T is a basic gamma random variable
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It is a stretch
parameter

From Equation (6),
x
β
1
dt = dx
β
t=

Writing Equation (3) in terms of x gives
Z∞
1=
0

x
x α−1 − β
e
β

Γ (α)

2

1
dx
β

(7)

which simplifies to

Z∞
1=
0

x
1
xα−1 e− β dx
α
Γ (α)β

(8)

A random variable X is said to have a full gamma distribution if its probability function is




defined as
f(x) =

x
1
xα−1 e− β ,
Γ (α)βα


0,

x > 0, α > 0, β > 0
(9)
elsewhere

We use the notation X ∼ Γ (α, β) - X has a gamma distribution with parameters α and β
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NB: The standard gamma distribution has β = 1
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Use the change of variable technique to evaluate the integral i
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β

Mean and Variance of the gamma distribution
If X ∼ Γ (α, β) then
µ = E(X)

σ2 = Var(X)

= αβ

= αβ2

NB: For the gamma distribution with β = 1, like the Poisson distribution, the mean and
variance are equal
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3

Derivation of Mean and Variance: The MGF technique
Mx (t) = E(etx )
Z∞
x
1
α−1 − β
=
etx
x
e
dx
Γ (α)βα
0
Z∞
x(1−βt)
1
α−1 −
β
x
e
dx
=
α
0 Γ (α)β
Let
1
x(1 − βt)
,t <
β
β
βy
x=
1 − βt
β
dx =
dy
1 − βt
y=

hence
Z∞

x(1−βt)
1
α−1 −
β
dx
x
e
α
0 Γ (α)β
α−1 


Z∞
1
βy
β
y
=
e
dy
α
1 − βt
1 − βt
0 Γ (α)β

α Z ∞
1
1 α−1 −y
=
y
e dy
1 − βt
0 Γ (α)
1
= (1 − βt)−α , t <
β

Mx (t) =

and
0

00

Mx (t) = −α × −β(1 − βt)−α−1

Mx (t) = αβ × −(α + 1) × β(1 − βt)−α−2

= αβ(1 − βt)−α−1

= α(α + 1)β2 (1 − βt)−α−2

therefore
0

E(X) = Mx (0)

00

0

Var(X) = Mx (0) − [Mx (0)]2
= α(α + 1)β2 − α2 β2

= αβ

= αβ2

4

What does the Gamma distribution Model?
Recall: Properties of Poisson happenings
a) Independence - the number of times an event occurs in a given time interval is
independent of the occurrences in a disjoint time interval
b) Lack of clustering - the probability of two or more occurrences happening simultaneously is zero
c) Rate - the average number of occurrences per unit time or space is constant and is
denoted by λ
In the approximate Poisson process with mean λ, the waiting time X until the first event occurs
follows an exponential distribution with mean θ = λ1
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X is a Poisson random variable
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The waiting time until the 1st customer arrives is a continuous random variable - W = w
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Therefore
f(w) = F 0 (w) = −e−λw (−λ) = λe−λw , 0 < w < ∞
Hence if λ is the mean number of events in an interval and θ is the mean waiting time
until the first customer arrives then θ =

1
λ

and λ = θ1
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Illustration 2
Suppose X is the number of customers arriving at a bank in an interval of length 1
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Let the mean number of events in the interval of length 1 be
λ
...
e
...

It is a continuous random variable
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or (α -1) events in [0, w])
=1−

α−1
X
k=0

=1−e

e−λw (λw)k
k!

−λw

−

α−1
X
k=1


1  −λw
e
(λw)k , w > 0
k!

NB: If the mean number of events in an interval of length 1 is λ then the mean number
of events in the interval of length w is λw
...
(−λe−λw ) + e−λw
...
(λw)k−1
...
+
−
2!
(α − 1)!
(α − 2)!


α−1
(λw)
= λe−λw + λe−λw −1 +
(α − 1)!
−λw
λe
(λw)α−1
= λe−λw − λe−λw +
(α − 1)!
−λw
α−1
λe
(λw)
=
(α − 1)!
= λe−λw −

Hence if λ is the mean number of events in an interval and θ is the mean waiting time
until the first customer arrives then θ =

1
λ

and λ = θ1
...
What is the probability that the shopkeeper will wait more than 5 minutes
before both of the first two customers arrive?

Let X be the waiting time until the 2nd customer arrives (a gamma random variable)
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= 2) [α = 2 means waiting time until the 2nd change (k = 2)]
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287

Incomplete Gamma Function
When X is a standard gamma random variable, the cdf of X which is
Zx
1 α−1 −y
F(x; α) =
y
e dy, x > 0
0 Γ (α)

(10)

is called the incomplete gamma function
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There are tables for F(x; α) available
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Determine:
a) p(3 6 X 6 5)
b) the probability that the reaction time is more than 4 seconds
8

Recall: For X continuous,
a) p(3 6 X 6 5) = F(5; 2) − F(3; 2) = 0
...
801 = 0
...
908 = 0
...
Let X have a gamma distribution with parameters α and β
...

β

Proof : Calculate the full incomplete gamma i
...

Zx
y
1
yα−1 e− β dy, x > 0
p(X 6 x) =
α
0 Γ (α)β
with the use of the substitution u =

y
β

(12)

y

...

β

means y = βu and dy = βdu
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a) What is the value of α and β?
b) What is the probability that a student uses the lab for at most 24 min?
c) What is the probability that a student spends between 20 and 40 min in the lab?
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Solution
a) α = 5, β = 4
b) p(X 6 24) = F(24; 5, 4) = F( 24
; 5) = F(6; 5) = 0
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4110
4
4

The Gamma curve
The gamma distribution represents a family of shapes
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The fundamental shapes are characterized by the following values of α:
1
...

2
...
e
...

When used to describe the sum of a series of exponentially distributed random variables, the
shape parameter represents the number of variables and the scale parameter is the mean of
the exponential distribution
...
α > 1 - the gamma distribution assumes a unimodal but skewed shape
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Additive property of the Gamma distribution
The sum of independent gamma variates is also a gamma variate
...
, Xn are independent gamma variates with parameters α1 , α2 ,
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Class Exercise 4
Show that the sum of independent gamma variates is also a gamma variate
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A random variable of this type has probability density function


 r1 r x 2r −1 e− x2 , x > 0
2
f(x) = Γ ( 2 )2

0,
elsewhere

(14)

and moment generating function
r

MX (t) = (1 − 2t)− 2 , t <

1
2

(15)

is said to have chi-square distribution and any probability density function as in Equation (14)
is called a chi-square probability density function
...
We say X has a
chi-square distribution with r degrees of freedom (df) denoted X ∼ χ2(r)
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Yet the chi-square distribution can also be considered as a derived distribution from the normal
distribution
...
This means that if X ∼ N(µ, σ2 ) then Z = X−µ
∼ N(0, 1) and Z2 = X−µ
∼ χ2(1)
...
, n) are n independent normal variates with mean µi and variance
σ2i then

2
n 
X
Xi − µ i
σi

i=1

∼ χ2(n)

(16)

NB: The term degrees of freedom (df) refers to the number of independent normal variates
which are added up to get the chi-square variate
...
∵ V = W 2 where W =

X−µ
σ






X−µ 2
σ

∼ χ2(1)

∼ N(0, 1)

G(v) = p(V 6 v) = p(W 2 6 v)
√
√
= p(− v 6 W 6 v)
√
= 2p(0 6 W 6 v) due to symmetry of the normal distribution
Z √v
1
w2
√ e− 2 dw since W ∼ N(0, 1)
=2
2π
0
Using the change of variable technique
...

√
dy = 2wdw hence dy = 2 ydw and dw =

dy
√
2 y

Change of limits: since y = w2 , when w = 0, y = 0 and when w =

√
v, y = v
...
e V ∼ χ2(1)
...
, n) are n independent normal variates with mean µi and variance σ2i , we want


Pn  Xi −µi 2 Pn
Xi −µi
2
to find the distribution of Y = i=1
= i=1 Ui where Ui =

...
Hence
MY (t) =MPni=1 U2i (t)
=

n
Y

MU2i (t)

i=1

h
in
= MU2i (t)
since the Ui0 s identically distributed N(0, 1) rvs
Now
2

MU2i (t) =E(eUi t )
Z∞
2
=
eUi t f(xi )dxi
Z−∞

2
∞
x −µ
1
− 21 iσ i
U2i t
i
√ e
=
e
dxi
σi 2π
−∞
Z∞
1
1 2
2
=
eUi t √ e− 2 Ui σi dUi
σ 2π
−∞
Z ∞ i (1−2t)U2
1
i
=√
e− 2 dUi
2π −∞
√
π
1
=√ ×

1 using Equation (17)
1−2t 2
2π
2

=(1 − 2t)

− 21

Hence
h
in
MY (t) = MU2i (t)
h
in
− 12
= (1 − 2t)
n

=(1 − 2t)− 2

13

which is the moment generating function of a gamma variate with parameters 2 and

n

...




fY (y) =

1
n
2
Γ(n
2 )2

n

y

y 2 −1 e− 2 , y > 0
(18)


0,

elsewhere

which is the required probability function of a chi-square distribution with n df
...

As n −→ ∞ the chi-square distribution tends to the normal distribution
...
Scientists and statisticians have hence developed other families of distributions which are
appropriate in practice
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A random variable X is said to have a Weibull distribution with parameters α and β (α > 0,
β > 0) if its pdf is given by



f(x; α, β) =

x α
)
α α−1 −( β
,
x
e
βα


0,

x>0

(19)

x<0

In some cases there is a theoretical justification for the Weibull distribution but it most cases
it provides a good fit to the observed data for various values of α and β
...

β

So the

exponential distribution is a special case of both the gamma and Weibull distributions
...
There are gamma distributions which are not Weibull distributions and vice versa, so one
family is not a subset of the other
...
Both α and β can be varied to obtain different distribution shapes
...
β is a scale parameter so different values stretch or compress the curve in the x - direction
Integrating to obtain E[X] and E[X2 ] gives
 


  
2 
1
2
1
µ = βΓ 1 +
and σ2 = β2 Γ 1 +
− Γ 1+
α
α
α
The cdf of the Weibull distribution having parameters α and β is


0,
x<0
F(x; α, β) =

1 − e−( βx )α , x > 0

(20)

(21)

Class Exercise 5
1) The lifetime X (in hundreds of hours) of a type of vacuum tube has a Weibull distribution
with parameters α = 2 and β = 3
...
5 6 X 6 6)

Introduction: Beta Function
Consider the beta function:
Z1
xα−1 (1 − x)β−1 dx

B(α, β) =

(22)

0

Again B(α, β) is a notation for the integral
...
The distribution of a random variable X with probability
function as in Equation (24) is called the Standard Beta distribution with parameters α and
β
...


Definition: Beta Distribution
The beta distribution provides positive density only for X in an interval of finite length
...

The mean and variance of X are
µ = E[x] = A + (B − A)
...
For this reason, the Beta distribution is used
extensively in PERT (Programme Evaluation and Review Technique) - used to plan and control
projects when there are restricted resources
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Class Exercise 5
Look up the Cauchy Distribution
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S C Gupta & V K Kapoor (2000) Fundamentals of Mathematical Statistics (A Modern
Approach); 10th Edition, Sultan Chand & Sons
2
...
William Mendenhall, III, Robert J Beaver & Barbara M Beaver (2013) Introduction to
Probability and Statistics 14th Edition, Brooks/Cole Cengage Learning

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