Chapter 6 Distributions Derived from the Normal Distribution

6.2
2
, t, F Distribution (and gamma, beta)
Normal Distribution
Consider the integral
I =
_

e
y
2
/2
dy
To evaluate the intgral, note that I > 0 and
I
2
=
_

exp
_
_
_
y
2
+ z
2
2
_
_
_ dydz
This integral can be easily evaluated by changing to polar coordinates. y = rsin()
and z = rcos(). Then
1
I
2
=
_
2
0
_

0
e
r
2
/2
rdrd
=
_
2
0
_
e
r
2
/2
|

0
_
d
=
_
2
0
d = 2
This implies that I =

2 and
_

2
e
y
2
/2
dy = 1
2
If we introduce a new variable of integration
y =
x a
b
where b > 0, the integral becomes
_

1
b

2
exp
_

_
(x a)
2
2b
2
_

_ dx = 1
This implies that
f(x) =
1
b

2
exp
_

_
(x a)
2
2b
2
_

_
for x (, ) satises the conditions of being a pdf. A random variable of the
continuous type with a pdf of this form is said to have a normal distribution.
3
Lets nd the mgf of a normal distribution.
M(t) =
_

e
tx
1
b

2
exp
_

_
(x a)
2
2b
2
_

_ dx
=
_

1
b

2
exp
_
_
_
2b
2
tx + x
2
2ax + a
2
2b
2
_
_
_ dx
= exp
_

_
a
2
(a + b
2
t)
2
2b
2
_

_
_

1
b

2
exp
_

_
(x a b
2
t)
2
2b
2
_

_ dx
= exp
_
_
_at +
b
2
t
2
2
_
_
_
Note that the exponential form of the mgf allows for simple derivatives
M

(t) = M(t)(a + b
2
t)
4
and
M

(t) = M(t)(a + b
2
t)
2
+ b
2
M(t)
= M

(0) = a

2
= M

(0)
2
= a
2
+ b
2
a
2
= b
2
Using these facts, we write the pdf of the normal distribution in its usual form
f(x) =
1

2
exp
_

_
(x )
2
2
2
_

_
for x (, ). Also, we write the mgf as
M(t) = exp
_
_
_t +

2
t
2
2
_
_
_
5
Theorem If the random variable X is N(,
2
),
2
> 0, then the random variable
W = (X )/ is N(0, 1).
Proof:
F(w) = P[
X

w] = P[X w + ]
=
_
w+

2
exp
_

_
(x )
2
2
2
_

_ dx.
If we change variables letting y = (x )/ we have
F(w) =
_
w

2
e
y
2
/2
dy
Thus, the pdf f(w) = F

(w) is just
f(w) =
1

2
e
w
2
/2
for < w < , which shows that W is N(0, 1).
6
Recall, the gamma function is dened by
() =
_

0
y
1
e
y
dy
for > 0.
If = 1,
(1) =
_

0
e
y
dy = 1
If > 1, integration by parts can be used to show that
(a) = ( 1)
_

0
y
2
e
y
dy = ( 1)( 1)
By iterating this, we see that when is a positive integer () = ( 1)!.
7
In the integral dening () lets have a change of variables y = x/ for some > 0.
Then
() =
_

0
_
_
x

_
_
1
e
x/
_
_
1

_
_
dx
Then, we see that
1 =
_

0
1
()

x
1
e
x/
dx
When > 0, > 0 we have
f(x) =
1
()

x
1
e
x/
is a pdf for a continuous random variable with space (0, ). A random variable with
a pdf of this form is said to have a gamma distribution with parameters and
.
8
Recall, we can nd the mgf of a gamma distribution.
M(t) =
_

0
e
tx
()

x
1
e
x/
dx
Set y = x(1 t)/ for t < 1/. Then
M(t) =
_

0
/(1 t)
()

_
_
y
1 t
_
_
1
e
y
dy
=
_
_
1
1 t
_
_

_

0
1
()
y
1
e
y
dy
=
1
(1 t)

for t <
1

.
9
M

(t) = (1 t)
1
M

(t) = ( + 1)
2
(1 t)
2
So, we can nd the mean and variance by
= M

(0) =
and

2
= M

(0)
2
=
2
10
An important special case is when = r/2 where r is a positive integer, and = 2.
A random variable X with pdf
f(x) =
1
(r/2)2
r/2
x
r/21
e
x/2
for x > 0 is said to have a chi-square distribution with r degrees of freedom.
The mgf for this distribution is
M(t) = (1 2t)
r/2
for t < 1/2.
11
Example: Let X have the pdf
f(x) = 1
for 0 < x < 1. Let Y = 2ln(X). Then x = g
1
(y) = e
y/2
.
The space A is {x : 0 < x < 1}, which the one-to-one transformation y = 2ln(x)
maps onto B.
B= {y : 0 < y < }.
The Jacobian of the transformation is
J =
1
2
e
y/2
Accordingly, the pdf of Y is
12
f(y) = f(e
y/2
)|J| =
1
2
e
y/2
for 0 < y < .
Recall the pdf of a chi-square distribution with r degress of freedom.
f(x) =
1
(r/2)2
r/2
x
r/21
e
x/2
From this we see that f(x) = f(y) when r = 2.
Denition (Book) If Z is a standard normal random variable, the distribution of
U = Z
2
is called a chi-square distribution with 1 degree of freedom.
Theorem If the random variable X is N(,
2
), then the random variable V =
(X )
2
/
2
is
2
(1).
13
Beta Distribution
Let X
1
and X
2
be independent gamma variables with joint pdf
h(x
1
, x
2
) =
1
()()
x
1
1
x
1
2
e
x
1
x
2
for 0 < x
1
< and 0 < x
2
< , where > 0, > 0.
Let Y
1
= X
1
+ X
2
and Y
2
=
X
1
X
1
+X
2
.
y
1
= g
1
(x
1
, x
2
) = x
1
+ x
2
y
2
= g
2
(x
1
, x
2
) =
x
1
x
1
+ x
2
x
1
= h
1
(y
1
, y
2
) = y
1
y
2
x
2
= h
2
(y
1
, y
2
) = y
1
(1 y
2
)
14
J =

y
2
y
1
(1 y
2
) y
1

= y
1
The transformation is one-to-one and maps A, the rst quadrant of the x
1
x
2
plane
onto
B={(y
1
, y
2
) : 0 < y
1
< , 0 < y
2
< 1}.
The joint pdf of Y
1
, Y
2
is
f(y
1
, y
2
) =
y
1
()()
(y
1
y
2
)
1
[y
1
(1 y
2
)]
1
e
y
1
=
y
1
2
(1 y
2
)
1
()()
y
+1
1
e
y
1
for (y
1
, y
2
) B.
Because B is a rectangular region and because g(y
1
, y
2
) can be factored into a function
of y
1
and a function of y
2
, it follows that Y
1
and Y
2
are statistically independent.
15
The marginal pdf of Y
2
is
f
Y
2
(y
2
) =
y
1
2
(1 y
2
)
1
()()
_

0
y
+1
1
e
y
1
dy
1
=
( + )
()()
y
1
2
(1 y
2
)
1
for 0 < y
2
< 1.
This is the pdf of a beta distribution with parameters and .
Also, since f(y
1
, y
2
) = f
Y
1
(y
1
)f
Y
2
(y
2
) we see that
f
Y
1
(y
1
) =
1
( + )
y
+1
1
e
y
1
for 0 < y
1
< .
Thus, we see that Y
1
has a gamma distribution with parameter values + and 1.
16
To nd the mean and variance of the beta distribution, it is helpful to notice that
from the pdf, it is clear that for all > 0 and > 0,
_
1
0
y
1
(1 y)
1
dy =
()()
( + )
The expected value of a random variable with a beta distribution is
_
1
0
yg(y)dy =
( + )
()()
_
1
0
y

(1 y)
1
dy
=
( + 1)()
( + 1 + )

( + )
()()
=

+
This follows from applying the fact that
( + 1) = ()
17
To nd the variance, we apply the same idea to nd E[Y
2
] and use the fact that
var(Y ) = E[Y
2
]
2
.

2
=

( + + 1)( + )
2
18
t distribution
Let W and V be independent random variables for which W is N(0, 1) and V is

2
(r).
f(w, v) =
1

2
e
w
2
/2
1
(r/2)2
r/2
v
r/21
e
r/2
for < w < , 0 < v < .
Dene a new random variable T by
T =
W
_
V/r
To nd the pdf f
T
(t) we use the change of variables technique with transformations
t =
w

v/r
and u = v.
19
These dene a one-to-one transformation that maps
A={(w, v) : < w < , 0 < v < } to
B={(t, u) : < t < , 0 < u < }.
The inverse transformations are
w =
t

r
and v = u.
Thus, it is easy to see that
|J| =

u/

r
20
By applying the change of variable technique, we see that the joint pdf of T and U
is
f
TU
(t, u) = f
WV
(
t

r
, u)|J|
=
u
r/21

2(r/2)2
r/2
exp
_

u
2
(1 + t
2
/r)
_

r
for < t < , 0 < u < .
To nd the marginal pdf of T we compute
f
T
(t) =
_

0
f(t, u)du
=
_

0
u
(r+1)/21

2r(r/2)2
r/2
exp
_

u
2
(1 + t
2
/r)
_
du
This simplies with a change of variables z = u[1 + (t
2
/r)]/2.
21
f
T
(t) =
_

0
1

2r(r/2)2
r/2
_
_
_
2z
1 + t
2
/r
_
_
_
(r+1)/21
e
z
_
_
_
2
1 + t
2
/r
_
_
_ dz
=
[(r + 1)/2]

r(r/2)(1 + t
2
/2)
(r+1)/2
for < t < .
A random variable with this pdf is said to have a t distribution with r degrees
of freedom.
22
F Distribution
Let U and V be independent chi-square random variables with r
1
and r
2
degrees of
freedom, respectively.
f(u, v) =
u
r
1
/21
v
r
2
/21
e
(u+v)/2
(r
1
/2)(r
2
/2)2
(r
1
+r
2
)/2
Dene a new random variable
W =
U/r
1
V/r
2
To nd f
W
(w) we consider the transformation
w =
u/r
1
v/r
2
and z = v.
This maps
A={(u, v) : 0 < u < , 0 < v < } to
B={(w, z) : 0 < w < , 0 < z < }.
23
The inverse transformations are
u = (r
1
/r
2
)zw and v = z.
This results in
|J| = (r
1
/r
2
)z
The joint pdf of W and Z by the change of variables technique is
f(w, z) =
_
r
1
zw
r
2
_
r
1
/21
z
r
2
/21
(r
1
/2)(r
2
/2)2
(r
1
+r
2
)/2
exp
_
_

z
2
_
_
r
1
w
r
2
+ 1
_
_
_
_
r
1
z
r
2
for (w, z) B.
The marginal pdf of W is
f
W
(w) =
_

0
f(w, z)dz
24
=
_

0
(r
1
/r
2
)
r
1
/2
(w)
r
1
/21
z
r
1
+r
2
/21
(r
1
/2)(r
2
/2)2
(r
1
+r
2
)/2
exp
_
_

z
2
_
_
r
1
w
r
2
+ 1
_
_
_
_
dz
We simplify this by changing the variable of integration to
y =
z
2
_
_
r
1
w
r
2
+ 1
_
_
Then the pdf f
W
(w) is
_

0
(r
1
/r
2
)
r
1
/2
(w)
r
1
/21
(r
1
/2)(r
2
/2)2
(r
1
+r
2
)/2
_
_
_
2y
r
1
w/r
2
+ 1
_
_
_
(r
1
+r
2
)/21
e
y
_
_
_
2
r
1
w/r
2
+ 1
_
_
_ dy
=
[(r
1
+ r
2
)/2](r
1
/r
2
)
r
1
/2
(w)
r
1
/21
(r
1
/2)(r
2
/2)(1 + r
1
w/r
2
)
(r
1
+r
2
)/2
for 0 < w < .
A random variable with a pdf of this form is said to have an F-distribution with
numerator degrees of freedom r
1
and denominator degrees of freedom
r
2
.
25