Download MTH/STA 561 GAMMA DISTRIBUTION, CHI SQUARE

MTH/STA 561
GAMMA DISTRIBUTION, CHI-SQUARE DISTRIBUTION,
AND EXPONENTIAL DISTRIBUTION
In this section, we introduce the gamma, chi-square, Erlang, and exponential distributions.
Gamma Distribution
According to Theorem 1 of the Appendix, it follows at once that
Z1
1
( )
1
y
e
y/
dy = 1
0
and thus we have the following de…nition:
De…nition 1. The continuous random variable Y is said to have a gamma distribution
with parameters and if and only if its density function of Y is de…ned by
f (y; ; ) =
1
y
( )
1
e
y/
if > 0, > 0, and y > 0
elsewhere
0
A continuous random variable Y that follows a gamma probability distribution with parameter and is referred to as a gamma random variable with parameters and .
The graphs of gamma distribution for
…gure below.
= 1; 2; 3; and 4 and
1
= 1 are sketched in the
Poisson Distribution As a Special Case.
If we let
be a positive integer
(that is, = 1; 2; 3: : : :) and = 1. Then gamma distribution has the form
1
y
( )
f (y; ) =
for y
0 and
1 = 0; 1; 2;
with parameter y 0.
Theorem 1.
1
y
e
=
1
y
(
e y
1)!
, which coincides with the form of a Poisson distribution
If Y has a gamma distribution with parameters
2
and
= E (Y ) =
and , then
2
= V ar (Y ) =
:
Proof Applying Theorem 1 of the Appendix in conjunction with the recursion formula
(Proposition 3 ) of the Appendix, it follows that
= E (Y ) =
Z1
1
y
( )
1
y
y/
e
Z1
1
dy =
( )
0
1
=
y/
y e
dy
0
+1
1
( + 1) =
( )
+1
( )=
( )
and that
E Y2
=
Z1
1
y2
( )
1
y
e
y/
1
dy =
( )
0
=
=
Z1
+1
y
e
y/
dy
0
1
+2
( )
( + 1)
2
1
( + 2) =
+2
( )
( + 1)
( )
:
Thus,
2
= E Y2
2 2
=
Theorem 2.
parameters and
+
=
2
( + 1)
2
2 2
=
2
2 2
:
The moment-generating function of a gamma random variable Y with
is given by
mY (t) =
Proof
2
1
(1
for t <
t)
1
By de…nition,
mY (t) = E etY =
Z1
ety
y
1
1
( )
y
1
0
=
1
( )
Z1
e
y/
dy =
1
( )
Z1
0
e
y/[ =(1
t)]
dy
0
2
y
1
e
y(1
t)/
dy
In view of Theorem 1 of the Appendix with substituting = (1
1
mY (t) =
( ) 1
1
=
(1
t)
t) for , it follows that
( )
t
for t < 1=
Now di¤erentiating mY (t) with respect to t gives
= E (Y ) =
and
E Y
2
d2
= 2 mY (t)
dt
d
mY (t)
dt
=
=
(1
1
t)
t=0
t=0
( + 1)
2
(1
2
t)
t=0
t=0
=
=
( + 1)
2
Hence,
2
=E Y2
2
=
( + 1)
2
2 2
=
2
:
Once the parameters and are speci…ed, the gamma distribution will be completely
determined as illustrated in the example below.
Example 1. In a certain city, the daily consumption of electric power, in millions of
kilowatt-hours, is a random variable having a gamma distribution with mean = 4 and
variance 2 = 8. What is the probability that on a given day the daily power consumption
will exceed 12 millions of kilowatt-hours?
2
Solution. By Theorem 2, it follows that =
= 4 and 2 =
= 8. This implies
that = 2 and = 2. Let Y stand for the daily power consumption. Then the desired
probability is given by
1
P (Y > 12) = 2
2 (2)
Z1
1
y 2 1 e y/2 dy =
4
12
Z1
ye
y/2
dy:
12
By the integration by parts with u = y and dv = e y/2 dy, we obtain du = dy and v =
2e y/2 so that
8
9
Z1
=
<
1
b
P (Y > 12) =
lim
2ye y/2 12 + 2 e y/2 dy
;
4 :b!1
12
1
2b
1
=
lim b/2 + 24e 6 + 2 2e y/2 12
b!1 e
4
1
1
=
0 + 24e 6 4 0 e 6 =
28e 6
4
4
= 7e 6 = (7) (0:002479) = 0:017353;
3
where we have
lim
b!1
2b
=0
e b/2
by L’Hôpital’s rule.
Erlang distribution
As seen in the preceding section, if we let Tr be the random variable representing the
waiting time to occurrence of the rth event in a Poisson process with parameter , then the
density function of Tr is
r
(r 1)!
f (t; r; ) =
tr 1 e
t
for t > 0
elsewhere
0
which is known as the Erlang distribution. It should be noted that the Erlang distribution
is a special case of the gamma distribution with = r and = 1= . Then it follows from
Theorems 2 and 3 that the mean and variance of the Erlang random variable are
=
r
2
and
=
r
2
and the moment-generating function is
mY (t) =
(1
1
t= )r
for t <
Chi-Square Distribution
Another special case of the gamma distribution is obtained by letting = =2 and = 2,
where is a positive integer. The probability density function so obtained is referred to as
the chi-square distribution with degrees of freedom.
De…nition 2. Let be a positive integer. The continuous random variable Y is said
to have a chi-square distribution with degrees of freedom if and only if Y is a gammadistributed random variable with parameters
= =2 and
= 2; that is, the density
function of Y is de…ned by
(
1
y ( =2) 1 e y/2
if > 0 and y 0
2 =2 ( 2 )
f (y; ) =
0
elsewhere
Applying Theorems 1 and 2 with
Theorem 3.
= =2 and
= 2 yields the following results.
If Y has a chi-square distribution with
= E (Y ) =
2
and
4
degrees of freedom, then
= V ar (Y ) = 2 :
The moment-generating function of a chi-square random variable Y with degrees of freedom
is given by
1
1
mY (t) =
for t < :
=2
2
(1 2t)
Exponential Distribution
The gamma distribution with parameters
1
f (y; ) =
= 1 and
y/
e
has the form
if > 0 and y
elsewhere,
0
0
which is easily seen to be an exponential distribution with parameters
= 1= .
By virtue of Theorems 1 and 2 with = 1, it follows that the mean, variance of an
exponential random variable Y with parameter = 1= are, respectively, given by
= E (Y ) =
2
and
= 1=
= V ar (Y ) =
2
= 1= 2 :
and that the moment-generating function is given by
mY (t) =
or
mY (t) =
1
1
for t <
t
1
=
1 t=
1
for t < :
t
Appendix
In order to de…ne the gamma probability distribution, we need to study a special function,
called the gamma function. It is proved in books on advanced calculus that the integral
Z1
1
t
e t dt
0
exists for > 0 and that the value of the integral is a positive real number. The integral is
called the gamma function of and we formally de…ne the function as follows.
De…nition 1. The gamma function of
( )=
is de…ned by
Z1
0
for
> 0.
5
t
1
e t dt
In particular, the value of the gamma function at
(1) = 1.
Proposition 1.
Proof
= 1 is equal to one as shown below.
By de…nition,
(1) =
Z1
e t dt = lim
e
b!1
t b
0
= 1:
0
The gamma function may be written in an alternative form as follows.
For any
Proposition 2.
> 0,
( )=2
Z1
t2
1
e
t2
dt:
0
Proof
Let u = t2 . Then du = 2tdt so that
( )=
Z1
u
1
e
u
du =
0
Z1
1
t2
e
t2
(2t) dt = 2
0
Z1
t2
1
e
t2
dt:
0
The following proposition presents a recursion formula for evaluating the gamma function.
Proposition 3 (Recursion Formula).
(
If
> 1, then
( )=(
1
Proof If > 1, an integration by parts with u = t
1) t 2 dt and v = e t so that
( ) =
lim
1
t
b!1
e
t b
0
+(
1)
Z1
t
=
lim
b!1
1
b
eb
+(
1)
t
0
Repeated application of L’Hôpital’s rule shows that
lim
b!1
and by de…nition,
Z1
t
2
1
b
eb
e t dt =
0
Hence,
( )=(
1) (
1).
6
=0
(
1) :
2
1).
and dv = e t dt yields du =
2
e t dt
0
Z1
1) (
e t dt:
When = n is a positive integer, repeated application of the recursion formula leads to
the following result.
Proposition 4.
Proof
For any positive integer n,
(n) = (n
1)!.
Repeated application of the recursion formula (Proposition 3 ) gives
(n) =
=
=
=
=
=
=
=
(n
(n
(n
1) (n 1)
1) (n 2) (n 2)
1) (n 2) (n 3) (n
(n
(n
(n
(n
1) (n
1) (n
1) (n
1)!:
2) (n
2) (n
2) (n
3)
3)
3)
3)
2 (2)
2 1 (1)
2 1
The gamma function has been proved handy in evaluating certain improper integrals as
demonstrated below.
Proposition 5.
For any
> 0,
Z1
e
t
dt =
1
1
:
0
Proof
Let u = t . Then t = u1= and dt = (1= ) u(1=
Z1
Z1
1
1
t
e dy =
u(1= ) 1 e u du =
0
) 1
du so that
1
:
0
The value of the gamma function at = 12 can be shown to equal to
useful in evaluating certain improper integrals.
Proposition 6.
1
2
=
p
p
, which is also
.
By the de…nition of the standard normal distribution, it follows that
Z1
Z1
p
1
2
z 2 /2
p e
dz = 1 or
e z /2 dz = 2 :
2
1
1
p
p
p
2. Then dt = dz
2 or 2dt = dz so that
Let t = z
Proof
p
2 =
Z1
1
e
z 2 /2
1
p Z
dz = 2
e
1
7
t2
dt
which implies that
p
=
Z1
e
t2
dy = 2
1
Z1
e
t2
dy:
0
because the integrand is an even function. However, it follows from Proposition 2 that
1
2
Z1
e
p
:
=2
t2
dt:
0
Consequently,
1
2
=
Another interested result concerning gamma function is presented as follows.
For any positive integer n,
Proposition 7.
n
2
n+1
2
=
p
(n
2n
1)!
1
:
Proof Since either n or n+1 is a positive even integer, we may assume loss of generality
that n is a positive even integer. Then n=2 is an positive integer. Repeated application of
the recursion formula (Proposition 3 ) gives
n
2
n
2
=
n
=
n
2
1 !=
2 n
2
1
4
2
n
2
4 2
:
2 2
2
2 1
and
n+1
2
n
2
=
=
n
1
2
1 n
2
n
2
3
3
2
3 1
2 2
3 1
1
2 2
2
2
1
2
:
Combining the preceding two equalities in conjunction with Proposition 6 gives
n
2
n+1
2
=
n
1 n
2
2
(n 1)!
=
2n 1
2 n
3 n
2
1
2
=
p
Now let us review the famous Taylor’s Theorem.
8
2
(n
2n
4
4 3 2 1
2 2 2 2
1)!
1
:
1
2
Lemma 1 (Taylor’s Theorem).
If f has continuous derivatives of order up to and
including m + 1 in some neighborhood of x = a, then
f (x) =
m
X
f (k) (a)
k!
k=0
where
Rm (x) =
Zx
a)k + Rm (x)
(x
f (m+1) (u)
(x
m!
u)m du
a
is the remainder after the mth power of x
Proof
De…ne
a.
m
X
f (k) (u)
g (u) = f (x)
k!
k=0
Then
u)k :
(x
(4.2)
f (0) (x)
= f (x) f (x) = 0:
0!
When g 0 (u) is computed using the product rule, the sum telescopes to just one term
g (x) = f (x)
0
g (u) =
=
m
X
f (k+1) (u)
k=0
m
X
k=0
k!
m
X
f (k) (u)
u) +
(x
(k 1)!
k=1
k
(x
f (k+1) (u)
(x
k!
k
u) +
m
X1
u)k
1
f (k+1) (u)
(x
k!
u)k
f (m+1) (u)
(x
m!
u)m du:
k=0
f (m+1) (u)
=
(x u)m :
m!
Now with the fundamental theorem for calculus, we have
g (a) =
[g (x)
Zx
g (a)] =
g 0 (u) du =
a
Zx
a
By (4:2), we also see that
g (a) = f (x)
m
X
f (k) (a)
k!
k=0
Hence,
f (x) =
m
X
f (k) (a)
k=0
The function
k!
(x
a)k +
Zx
(x
f (m+1) (u)
(x
m!
a
( ) de…ned by the integral of the form
Z1
tn 1 e t dt
9
a)k :
u)m du:
is referred to as the incomplete gamma function. The relationship between the incomplete
gamma function and the Poisson probability distribution can be found in the following result.
Proposition 8.
For any positive integer n,
Z1
1
(n
1)!
n 1
t
t
e dt =
n 1
X
k
e
:
k!
k=0
Proof Let f (x) = ex , then f (k) (x) = ex for all k. Applying Lemma 1 to f (x) with
x = 0, a =
and m = n 1, we have
n 1
X
e
1 = f (0) =
k!
k=0
=
n 1
X
e
+
k!
k=0
=
k
n 1
X
k
+
1
(n
1)!
k
e
k!
k=0
Z0
+
eu
(n
Z0
1
(n
1)!
1)!
( u)n
Z
( u)n
du
1 u
e du
tn 1 e t dt
0
where the last equality follows from the substitution of t =
1
(n
1
1)!
Z
n 1
t
n 1
X
t
e dt = 1
u. Thus,
k
e
k!
k=0
0
(4.3)
Note now that
(n
1)! =
(n) =
Z1
n 1
t
t
e dt =
0
Z
n 1
t
Z1
t
e dt +
Z
tn 1 e t dt
0
tn 1 e t dt
which implies that
Z1
so that
1
(n
1)!
n 1
t
t
e dt = (n
1)!
0
Z1
n 1
t
1
t
e dt = 1
(n
1)!
Z
tn 1 e t dt:
0
Therefore, it follows from (4:3) and (4:4) that
1
(n
1)!
Z1
tn 1 e t dt =
n 1
X
k=0
10
k
e
:
k!
(4.4)
By making the change of variable y = t, we obtain the following useful result.
Theorem 1.
For
> 0 and
Z1
> 0,
1
y
e
y/
dy =
( ):
0
Proof Let y = t (or t = y= ). Then dy = dt (or dt = dy= ). Also, y ! 1 as t ! 1
and y ! 0 as t ! 0. Thus,
( )=
Z1
t
1
t
e dt =
0
Hence,
Z1
y
1
e
y=
1
dy =
0
Z1
y
1
1
Z1
0
e
y/
dy =
0
11
( ):
y
1
e
y=
dy: