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A lecture note from a statistics course (bst 631) taught by kui zhang on october 18, 2007. The lecture covers the concepts of conditional probability mass functions (pmfs) and independence for bivariate random vectors. Definitions, theorems, examples, and solutions to problems related to these topics.
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Lecture 17 on BST 631: Statistical Theory I – Kui Zhang, 10/18/2007 Review for the previous lecture Definition:
n-dimensional random vector, joint pmf (pdf), marginal pmf (pdf)
Theorem:
How to calculate marginal pmf (pdf) given joint pmf (pdf)
Example:
How to calculate marginal pmf (pdf), joint pmf (pdf)
Example:
How to calculate probability based on marginal pmf (pdf), joint pmf (pdf)
Chapter 4 – Multiple Random Variables
Chapter 4.2 – Conditional Distributions and Independence Definition 4.2.1:
Let
)^ be a discrete bivariate random vector with joint pmf
( , f x y
and marginal pmfs
f^ xX
and
(^ ) f y^ Y
. For any such
x^
such that
X
x^
f^
x
=^
, the
conditional pmf
of^
Y^ given that
x = is
the function of
denoted by
y^
f^ y
x^
f^ y
x^
y^ X
x^
f^ x y
f
x
. For any such
y^
such that
y^
f =^
, the
conditional pmf
of^
given that
y
(^ ) y^ Y
is the function of
x^
denoted by
f^ x
y^
Y
f^ x
y^
= X x
y^
f^ x y
f^
y
Note:
(^ | f y^
x^
is a valid pmf since
f^ y
x^
, because
f^ x y
and
being joint and marginal pmfs,
and since
f^ xX
y^ x
f^ x y
f^
x
X
f^
x^
f^
x
y^
y
f
1
Lecture 17 on BST 631: Statistical Theory I – Kui Zhang, 10/18/2007 Example 4.2.
:^ Define the joint pmf of
by (^
f^
f =
f^
f =
f^
f^
, that is , y x^
(^ ) f yY
(^ ) f x^ X
(a) Obtain the conditional distribution of
given Y
x =.
(b) Find
Solution: Definition 4.2.3:
Let
be a continuous bivariate random vector with joint pdf (^
f^ x y
and marginal pdfs
( ) f x^ X
and
(^ ) f y^ Y
. For any such
x^
such that
0 , the
conditional pdf
of^
Y^ given that
x
f^ xX
is the function of
denoted by
y^
( f y^
x^ :
X
f^ y
x^
f^ x y
f^
x
. For any
y^
such that
, the
conditional pdf
of^
given
that
y^ is the function of
f^ y^ Y
x^ denoted by
f^ x
y^
f^ x
y^
f^ x y
f^
y
Example 4.2.4:
Let the continuous random vector
have joint pdf (^
,^
{(^ , ):
y
X Y^
u v^
u^ v^
}^
f^
x y
e^
x y
−
< < <∞
=^
(a) Find the marginal pdf of
(b) For any
x^
such that
, find ( )
f^ xX
f^ y
x^
2
Lecture 17 on BST 631: Statistical Theory I – Kui Zhang, 10/18/2007 Definition 4.2.5:
Let
be a bivariate random vector with joint pdf or pmf (^
f^ x y
and marginal pdfs or pmfs
f^ xX
and
(^ ) f y^ Y
. Then
and
are called
independent random variables
, if for every
x^ ∈
and
y^ ∈
Y
f^ x y
f^
x f
y
=^
.^ Consequently, if
and
are independent,
f^ y
x^
f^
y =^
and
f^ x
y^
f^
x =^
Technical Note:
If^
f^ x y
is the joint pdf for the continuous random vector
where (^
X^
Y
f^ x y
f^
x f
y
≠^
on a set
such that
f^ A x y dxdy
, then
and
are still called independent random
variables.
Example 4.2.6:
Consider the discrete bivariate random vector
, with joint pmf given by (^
f^
f^
f
f^
f =
, and
f^
Find the marginals of
and
. Are
and
independent?
Solution: Question:
Can we check for independence without knowing the marginals?
Lemma 4.2.7:
Let
)^ be a bivariate random vector with joint pdf or pmf
( , f x yy
. Then
and
are
independent random variables if and only if there exist functions
g x
and
h^
such that, for every
x^ ∈
and
y^ ∈
f^ x y
g x h y =
. In this case
f^ xX
cg x =
and
f^
y^
dh^
y =^
, where c and d are some constants that
would make
f^ xX
and
(^ ) f y^ Y
valid pdfs or pmfs.
4
Lecture 17 on BST 631: Statistical Theory I – Kui Zhang, 10/18/2007 Example 4.2.8:
Consider the joint pdf
2 4
(^ / 2
y^ x
f^ x y
x y e
−^ −
x^ >
and
. By Lemma 4.2.7,
and
are
independent random variables.
y^ >
Notes:
Consider the set {( ,
and y
x^ y
x
, where
X
A^
x^
f^
x
=^
and
Y
B^
y^
f^
x
=^
, then this set is
called a cross-product denoted by
If^
f^ x y
is a joint pdf or pmf such that the set
x^ y
f^ x y
is not a cross-product, then the random
variables
and
with joint pdf or pmf
f^ x y
are not independent.
If it is known that
and
Y^ are independent random variables with marginal pdfs (or pmfs)
( ) f x^ X
and
(^ ) f y^ Y
, then the joint pdf (or pmf) of
and
is given by
X^
Y
f^ x y
f^
x f
y
=^
. (See Example 4.2.9 for
discrete case.) Theorem 4.2.10:
Let
and
be independent random variables.
X^
a. For any
and
b. Let
)^ be a function only of ( g x
x^ and
)^ be a function only of
. Then
( h y^
y^
E^ g X h Y
E g X
E h Y
=^
Proof: Example 4.2.11:
Let
and
be independent exponential(1) random variables.
X^
a. Find the joint pdf of
and
b. Find
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