In the (approximate) Poisson process with mean
λ
λ
, we have seen that the waiting time until the first change has an exponential distribution. Let now W denote the waiting time until the
α
α
th change occurs and let find the distribution of W. The distribution function of W ,when
w≥0
w≥0
is given by
F(
w
)=P(
W≤w
)=1−P(
W>w
)=1−P(
fewer_than_α_changes_occur_in_[
0,w
]
)
=1−
∑
k=0
α−1
(
λw
)
k
e
−λw
k!
,
F(
w
)=P(
W≤w
)=1−P(
W>w
)=1−P(
fewer_than_α_changes_occur_in_[
0,w
]
)
=1−
∑
k=0
α−1
(
λw
)
k
e
−λw
k!
,
since the number of changes in the interval
[
0,w
]
[
0,w
]
has a Poisson distribution with mean
λw
λw
. Because W is a continuous-type random variable,
F'(
w
)
F'(
w
)
is equal to the p.d.f. of W whenever this derivative exists. We have, provided w>0, that
F'(
w
)=λ
e
−λw
−
e
−λw
∑
k=1
α−1
[
k
(
λw
)
k−1
λ
k!
−
(
λw
)
k
λ
k!
]
=λ
e
−λw
−
e
−λw
[
λ−
λ
(
λw
)
α−1
(
α−1
)!
]
=
λ
(
λw
)
α−1
(
α−1
)!
e
−λw
.
F'(
w
)=λ
e
−λw
−
e
−λw
∑
k=1
α−1
[
k
(
λw
)
k−1
λ
k!
−
(
λw
)
k
λ
k!
]
=λ
e
−λw
−
e
−λw
[
λ−
λ
(
λw
)
α−1
(
α−1
)!
]
=
λ
(
λw
)
α−1
(
α−1
)!
e
−λw
.
- Definition 1:
1.
If
w<0
w<0
, then
F(
w
)=0
F(
w
)=0
and
F'(
w
)=0
F'(
w
)=0
, a p.d.f. of this form is said to be one of the gamma type, and the random variable W is said to have the gamma distribution.
2.
The gamma function is defined by
Γ(
t
)=
∫
0
∞
y
t−1
e
−y
dy
,0<t.
Γ(
t
)=
∫
0
∞
y
t−1
e
−y
dy
,0<t.
This integral is positive for
0<t
0<t
, because the integrand id positive. Values of it are often given in a table of integrals. If
t>1
t>1
, integration of gamma fnction of t by parts yields
Γ(
t
)=
[
−
y
t−1
e
−y
]
0
∞
+
∫
0
∞
(
t−1
)
y
t−2
e
−y
dy
=(
t−1
)
∫
0
∞
y
t−2
e
−y
dy=
(
t−1
)Γ(
t−1
).
Γ(
t
)=
[
−
y
t−1
e
−y
]
0
∞
+
∫
0
∞
(
t−1
)
y
t−2
e
−y
dy
=(
t−1
)
∫
0
∞
y
t−2
e
−y
dy=
(
t−1
)Γ(
t−1
).
Let
Γ(
6
)=5Γ(
5
)
Γ(
6
)=5Γ(
5
)
and
Γ(
3
)=2Γ(
2
)=(
2
)(
1
)Γ(
1
)
Γ(
3
)=2Γ(
2
)=(
2
)(
1
)Γ(
1
)
. Whenever
t=n
t=n
, a positive integer, we have, be repeated application of
Γ(
t
)=(
t−1
)Γ(
t−1
)
Γ(
t
)=(
t−1
)Γ(
t−1
)
, that
Γ(
n
)=(
n−1
)Γ(
n−1
)=(
n−1
)(
n−2
)...(
2
)(
1
)Γ(
1
).
Γ(
n
)=(
n−1
)Γ(
n−1
)=(
n−1
)(
n−2
)...(
2
)(
1
)Γ(
1
).
However,
Γ(
1
)=
∫
0
∞
e
−y
dy=1
.
Γ(
1
)=
∫
0
∞
e
−y
dy=1
.
Thus when n is a positive integer, we have that
Γ(
n
)=(
n−1
)!
Γ(
n
)=(
n−1
)!
; and, for this reason, the gamma is called the generalized factorial.
Incidentally,
Γ(
1
)
Γ(
1
)
corresponds to 0!, and we have noted that
Γ(
1
)=1
Γ(
1
)=1
, which is consistent with earlier discussions.
The random variable x has a gamma distribution if its p.d.f. is defined by
f(
x
)=
1
Γ(
α
)
θ
α
x
α−1
e
−x/θ
,0≤x<∞.
f(
x
)=
1
Γ(
α
)
θ
α
x
α−1
e
−x/θ
,0≤x<∞.
(1)
Hence, w, the waiting time until the
α
α
th change in a Poisson process, has a gamma distribution with parameters
α
α
and
θ=1/λ
θ=1/λ
.
Function
f(
x
)
f(
x
)
actually has the properties of a p.d.f., because
f(
x
)≥0
f(
x
)≥0
and
∫
−∞
∞
f(
x
)dx=
∫
0
∞
x
α−1
e
−x/θ
Γ(
α
)
θ
α
dx,
∫
−∞
∞
f(
x
)dx=
∫
0
∞
x
α−1
e
−x/θ
Γ(
α
)
θ
α
dx,
which, by the change of variables
y=x/θ
y=x/θ
equals
∫
0
∞
(
θy
)
α−1
e
−y
Γ(
α
)
θ
α
θdy=
1
Γ(
α
)
∫
0
∞
y
α−1
e
−y
dy
=
Γ(
α
)
Γ(
α
)
=1.
∫
0
∞
(
θy
)
α−1
e
−y
Γ(
α
)
θ
α
θdy=
1
Γ(
α
)
∫
0
∞
y
α−1
e
−y
dy
=
Γ(
α
)
Γ(
α
)
=1.
The mean and variance are:
μ=αθ
μ=αθ
and
σ
2
=α
θ
2
σ
2
=α
θ
2
.
Suppose that an average of 30 customers per hour arrive at a shop in accordance with Poisson process. That is, if a minute is our unit, then
λ=1/2
λ=1/2
. What is the probability that the shopkeeper will wait more than 5 minutes before both of the first two customers arrive? If X denotes the waiting time in minutes until the second customer arrives, then X has a gamma distribution with
α=2,θ=1/λ=2.
α=2,θ=1/λ=2.
Hence,
p(
X>5
)=
∫
5
∞
x
2−1
e
−x/2
Γ(
2
)
2
2
dx=
∫
5
∞
x
e
−x/2
4
dx=
1
4
[
(
−2
)x
e
−x/2
−4
e
−x/2
]
5
∞
=
7
2
e
−5/2
=0.287.
p(
X>5
)=
∫
5
∞
x
2−1
e
−x/2
Γ(
2
)
2
2
dx=
∫
5
∞
x
e
−x/2
4
dx=
1
4
[
(
−2
)x
e
−x/2
−4
e
−x/2
]
5
∞
=
7
2
e
−5/2
=0.287.
We could also have used equation with
λ=1/θ
λ=1/θ
, because
α
α
is an integer
P(
X>x
)=
∑
k=0
α−1
(
x/θ
)
k
e
−x/θ
k!
.
P(
X>x
)=
∑
k=0
α−1
(
x/θ
)
k
e
−x/θ
k!
.
Thus, with x=5,
α
α
=2, and
θ=2
θ=2
, this is equal to
P(
X>x
)=
∑
k=0
2−1
(
5/2
)
k
e
−5/2
k!
=
e
−5/2
(
1+
5
2
)=(
7
2
)
e
−5/2
.
P(
X>x
)=
∑
k=0
2−1
(
5/2
)
k
e
−5/2
k!
=
e
−5/2
(
1+
5
2
)=(
7
2
)
e
−5/2
.
Let now consider the special case of the gamma distribution that plays an important role in statistics.
- Definition 2:
Let
X have a gamma distribution with
θ=2
θ=2
and
α=r/2
α=r/2
, where
r is a positive integer. If the p.d.f. of
X is
f(
x
)=
1
Γ(
r/2
)
2
r/2
x
r/2−1
e
−x/2
,0≤x<∞.
f(
x
)=
1
Γ(
r/2
)
2
r/2
x
r/2−1
e
−x/2
,0≤x<∞.
(2)
We say that
X has
chi-square distribution with
r degrees of freedom, which we abbreviate by saying is
χ
2
(
r
)
χ
2
(
r
)
.
The mean and the variance of this chi-square distributions are
μ=αθ=(
r
2
)2=r
μ=αθ=(
r
2
)2=r
and
σ
2
=α
θ
2
=(
r
2
)
2
2
=2r.
σ
2
=α
θ
2
=(
r
2
)
2
2
=2r.
That is, the mean equals the number of degrees of freedom and the variance equals twice the number of degrees of freedom.
In the fugure 2 the graphs of chi-square p.d.f. for r=2,3,5, and 8 are given.
the relationship between the mean
μ=r
μ=r
, and the point at which the p.d.f. obtains its maximum.
Because the chi-square distribution is so important in applications, tables have been prepared giving the values of the distribution function for selected value of r and x,
F(
x
)=
∫
0
x
1
Γ(
r/2
)
2
r/2
w
r/2−1
e
−w/2
dw
.
F(
x
)=
∫
0
x
1
Γ(
r/2
)
2
r/2
w
r/2−1
e
−w/2
dw
.
(3)
Let X have a chi-square distribution with r =5 degrees of freedom. Then, using tabularized values,
P(
1.145≤X≤12.83
)=F(
12.83
)−F(
1.145
)=0.975−0.050=0.925
P(
1.145≤X≤12.83
)=F(
12.83
)−F(
1.145
)=0.975−0.050=0.925
and
P(
X>15.09
)=1−F(
15.09
)=1−0.99=0.01.
P(
X>15.09
)=1−F(
15.09
)=1−0.99=0.01.
If X is
χ
2
(
7
)
χ
2
(
7
)
, two constants, a and b, such that
P(
a<X<b
)=0.95
P(
a<X<b
)=0.95
, are a=1.690 and b=16.01.
Other constants a and b can be found, this above are only restricted in choices by the limited table.
Probabilities like that in Example 4 are so important in statistical applications that one uses special symbols for a and b. Let
α
α
be a positive probability (that is usually less than 0.5) and let X have a chi-square distribution with r degrees of freedom. Then
χ
α
2
(
r
)
χ
α
2
(
r
)
is a number such that
P[
X≥
χ
α
2
(
r
)
]=α
P[
X≥
χ
α
2
(
r
)
]=α
That is,
χ
α
2
(
r
)
χ
α
2
(
r
)
is the 100(1-
α
α
) percentile (or upper 100a percent point) of the chi-square distribution with r degrees of freedom. Then the 100
α
α
percentile is the number
χ
1−α
2
(
r
)
χ
1−α
2
(
r
)
such that
P[
X≤
χ
1−α
2
(
r
)
]=α
P[
X≤
χ
1−α
2
(
r
)
]=α
.
This is, the probability to the right of
χ
1−α
2
(
r
)
χ
1−α
2
(
r
)
is 1-
α
α
.
SEE fugure 3.
Let X have a chi-square distribution with seven degrees of freedom. Then, using tabularized values,
χ
0.05
2
(
7
)=14.07
χ
0.05
2
(
7
)=14.07
and
χ
0.95
2
(
7
)=2.167.
χ
0.95
2
(
7
)=2.167.
These are the points that are indicated on Figure 3.
