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Lecture 18 on BST 631: Statistical Theory I – Kui Zhang, 10/30/2007 Review for the previous lecture Definitions:
conditional pmf (pdf), conditional expected values, independence
Theorems:
how determine if two random variables are independence
Examples:
calculate conditional pmf (pdf)
Chapter 4 – Multiple Random Variables
Chapter 4.2 – Conditional Distributions and Independence Theorem 4.2.12:
Let
and
be independent random variables with moment generating functions
X
Y
X M
t^
and
Y M
t^
. Then the moment generating function of the random variable
Z
X
Y
is given by
Y
Z^
X
M
t^
M
t M
t
(^
)^
Xt
Yt
Xt
Yt
Xt
Yt
Z^
X^
Y^
t^
Ee
E e
e
Ee
Ee
M
t M
t
Proof:
M
Example 4.2.13:
Let
(^
X
n
μ
σ
and
Y
n
γ τ
and they are independent. Find the pdf of
Z
X
Y
. (This is
a very important result!!) Solution:
and
, thus
2
2
exp(
X M
t^
t^
t
μ
σ
2
2
exp(
Y M
t^
t^
t
γ
τ
2
2
2
exp((
)^
(^
)^
Z M
t^
t^
t
μ
γ
σ
τ
, thus
2
2
(^
Z
n
μ
γ σ
τ
1
Lecture 18 on BST 631: Statistical Theory I – Kui Zhang, 10/30/2007 Theorem 4.2.14:
Let
2
(^
X
n
μ
σ
and
2
Y
n
γ
τ
be independent random variables. Then the random variable
X
Y has an
2
2
(^
n
μ
γ σ
τ
distribution.
Z Technical Note:
If
f^
x y
is the joint pdf for the continuous random vector
where
(^
,^
X Y
)^
X^
Y^
f^
x y
f^
x f
y
on a set
A
such that
)^
A^
f^
x y dxdy
, then
and
are still called independent random
variables.
X
Y
Example:
{(
, ):
0,
0}
)^
x^
y
u v u
v
x y
e
I^
x y
−^
−
>^
>
and
{( , ):
0,
0,
}
)^
x^
y
u v u
v^
u^
v
f^
x y
e
I^
x y
−^
−
>^
>^
≠
f
Theorem:
Let
)^
be a bivariate random vector, then
(^
X Y
(^
(^
,^
)^
(^
,^
)^
)^
(^
(^
,^
,^
E
g X Y
h X Y
c
E g X Y
E h X Y
c
α
β
α
β
, specifically, we have
(^
)^
(^
)^
E
X
Y
E X
E Y
(2) If
and
are independent, then
X
Y
(^
(^
)^
(^
)^
)^
(^
(^
Var g X
h Y
c
Var g X
Var h Y
. Specifically, we have
(^
)^
(^
)^
(^
Var X
Y
Var X
Var Y
Chapter 4.3 – Bivariate Transformations Discrete Case: Let
be a discrete bivariate random vector with joint pmf
(^
,^
X Y
f^
x y
and support
)^
x
y
f^
x y
A
Define
x y
and
V
g
x y
with
2
u v
u
g
x
g
x y
U
g
1
) and y
v
B
for some
x
y
A
. For any
2
Lecture 18 on BST 631: Statistical Theory I – Kui Zhang, 10/30/2007 Continuous Case: Consider
a continuous bivariate random vector with joint pdf
(^
,^
X Y
f^
x y
and define
U
g
x y
and
V
g
. Let
2 ( , x
y
)^
x
y
f^
x y
and
2
)^
u v
u
g
x y
g
x y
A
1
and
v
B
for some
x
y
A
Assume that
U
and
are 1-1 transformation from
to
. Define
V
A
B
x
h u v
and
y
h
u v ( , ) 2
=
be the inverse
transformations. The Jacobian,
J
, of the transformation is defined as the determinant of a matrix of partial
derivatives:
x
x
x
y
y
x
u
v
J
y
y
u
v
u
v
u
v
Therefore, the joint pdf of
and
is given by
U
V
,^
,^
1
2
(^
U V
X Y
f^
u v
f^
h u v
h
u v
J
where
|^
is the absolute value of the Jacobian.
|^
J
Example 4.3.3 (Distribution of the product of two independent beta variables)
:. Let
(^
,^
X
beta
α
β
and
(^
,^
Y
beta
α
β γ
, then the joint pdf of (
,^
X Y
is
1
1
1
1
,
(^
)^
(^
)^
)^
)^
(^
)^
(^
)^
(^
)^
X Y f^
x y
x
x
x
x
x
y
α
β
α^
β
γ
α
β
α
β
γ
α
β
α
β
γ
−^
−^
+^
−^
−
Consider
Y
and
, then this an one to one transformation from
U
X
V
X
x
y
x
y
to
u v
. In addition,
V
u
v
,^
X
V Y
U
,^
v
|^
|^
J
, therefore
1
1
1
1 γ
,^
,
(^
)^
(^
)^
/^
)^
)^
(^
/^
)^
/^
(^
)^
(^
)^
(^
)^
U V
X Y
f^
u v
f^
v u v
v
v
u v
u v
v
α
β
α^
β
α
β
α
β
γ
α
β
α
β
γ
−^
−^
+^
−^
−
4
Lecture 18 on BST 631: Statistical Theory I – Kui Zhang, 10/30/2007 and
1
,
1 1
1
1
1
1
(^
)^
)^
(^
(^
)^
(^
)^
(^
)^
)^
(^
)^
(^
U
U V u
u
f^
u
f^
u v dv
u
v
v
v
u
dv
u
u
u
α^
β
β^
γ
β
γ
α
β^
γ
α
β
γ
α
β
γ
α
β
γ
α
β
γ
+^
−^
−^
−^
−^
−
−^
+^
−
Example 4.3.4 (Sum and difference of two independent normal random variables):
Let
(^
X
n
α
σ
and
2
(^
Y
n
β
σ
and they are independent. Consider
Y
U
X
and
Y
V
X
, then
X
U
V
,^
Y
U
V
and
Therefore,
|^
|^
J
,^
,
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
)^
exp(
)^
)^
exp(
(^
(^
(^
(^
exp(
)^
exp(
U V
X Y
f^
u v
f^
u
v
u
v
u
v
u
v
u
u
v
v
u
v
α
β
πσ
σ
α
β
α
β
α
β
πσ
σ
α
β
α
β
σ
σ
π
σ
π
σ
Theorem 4.3.5:
Let
and
be independent random variables. Let
be a function only of
X
Y
g x
x
and
be a
function only of y. Then the random variables
U
g x
(^
h y
and
(^
V
h
Y
are independent.
5
Lecture 18 on BST 631: Statistical Theory I – Kui Zhang, 10/30/
7
2
2
2
2
0
2
2
2
2
0
2
2
2
2
2
2
0 1
exp(
)^
exp(
)^
exp(
)^
exp(
exp(
)^
U
x
x
x
x
x
x
f^
u
dx
dx
u
u
u
u
u x
x
x
dx
u
u
u
∞
−∞
∞
∫^
2
2
2
2
0
2
2
0
(^
(^
exp(
)^
exp(
,^
)^
U
v
u
v
u
v
f^
u
dv
u
u
u
π
π
π
∞^
∞
Example 4.3.6:
Show that the distribution of the ratio of two independent normal variables is a Cauchy random
variable. That is, if
X
n
,^
Y
n
, then
/^
U
X
Y
Cauchy
2
2
,^
,^
,
(^
(^
|^
(^
,^
|^
exp(
U V
X Y
X Y
v
u
v
f^
u v
f^
uv v
v
f^
uv
v
v
u
v
π
Similarly, you can get
f^ U
u
for
. You also need to verify the conditions such that the integration and
differentiation can be exchanged.
u
Solution 1.
Let
|^ , then |
V
Y
1
A
x y
y
,^
2
A
x y
y
,^
and
0
A
x y
y
2
2
exp(
0
/^
0
0
0
/
(^
)^
(^
/^
(^
U
x u
X^
y^
X^
y
x u
F
u
P U
u
P X
Y
u
f^
x f
y dydx
f^
x f
y dydx
∞
−∞
Take the derivative respect to
, we have u
,
)^
X Y
x
y
f^
x y
π
. Therefore,
Solution 2.
For
, we have 0
u
