Statistical Theory I: Conditional Probability Mass Functions and Independence, Study notes of Biostatistics

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 )(,XY (,
f
xy and marginal pmfs
)(
X
f
x and ( )
Y
f
y. For any such
x
such that ( ) ( ) 0
X
PX x f x
=
=>, the conditional pmf of Y given that Xx
=
is
the function of denoted by (
y| )
f
yx: )(| (,)/ (
X
) ( | )
f
yx PY yX x fxy f x
===
XY
. For any such y such that
() 0
PY y f== >, the conditional pmf of given that y
()
Yy
=
is the function of
x
denoted by ( | )
f
xy:
) (,)/ ()(|) |( Y
f
xy P==X xY y fxy f y== .
Note: )(|
f
yx is a valid pmf since ( | ) 0
f
yx, because ( , ) 0
f
xy
)1=
and being joint and marginal pmfs,
and since .
∑∑
( ) 0
X
fx
(|) (,) ()/ (
X
yx fxy fx==/ ()
XX
fx fx
yy
f
1
pf3
pf4
pf5

Partial preview of the text

Download Statistical Theory I: Conditional Probability Mass Functions and Independence and more Study notes Biostatistics in PDF only on Docsity!

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

(^

, X Y^

( , f x y

and marginal pmfs

f^ xX

and

(^ ) f y^ Y

. For any such

x^

such that

)^

X

P X

x^

f^

x

=^

=^

>^

, the

conditional pmf

of^

Y^ given that

X^

x = is

the function of

denoted by

y^

|^ )

f^ y

x^

:^

(^ |

( X

)^

(^

|^

f^ y

x^

P Y

y^ X

x^

f^ x y

f

x

=^

=^

=^

X^

Y

. For any such

y^

such that

(^

)^

P Y

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^

P =

= X x

Y^

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

(^ |

)^

)^

( X

y^ x

f^ x y

f^

x

=^

/^

( ) 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 (^

,^ ) X Y^

f^

f =

=^

,^ (1,10)

f^

f =

=^

f^

=^

,^

f^

=^

, that is , y x^

(^ ) f yY

(^ ) f x^ X

(a) Obtain the conditional distribution of

given Y

X^

x =.

(b) Find

(^

P Y

X

>^

Solution: Definition 4.2.3:

Let

be a continuous bivariate random vector with joint pdf (^

,^ ) X Y^

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^

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^

X^

given

that

Y^

y^ is the function of

f^ y^ Y

=^

x^ denoted by

|^

f^ x

y^

:^

(^ |

( ,^

)^

(^ ) Y

f^ x

y^

f^ x y

f^

y

=^

Example 4.2.4:

Let the continuous random vector

have joint pdf (^

,^ ) X Y

,^

{(^ , ):

( ,^

)^

y

X Y^

u v^

u^ v^

}^

f^

x y

e^

I^

x y

< < <∞

=^

(a) Find the marginal pdf of

. X

(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 (^

,^ ) X Y^

f^ x y

and marginal pdfs or pmfs

f^ xX

and

(^ ) f y^ Y

. Then

and

Y^

are called

independent random variables

, if for every

X^

x^ ∈

R^

and

y^ ∈

R^ ,

( ,^

)^

(^ )

( ) X

Y

f^ x y

f^

x f

y

=^

.^ Consequently, if

X^

and

Y^

are independent,

(^

|^ )

( Y

f^ y

x^

f^

y =^

and

(^ |

)^

( X

f^ x

y^

f^

x =^

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

)^

f^ A x y dxdy

∫∫^

, then

and

are still called independent random

variables.

X^

Y

Example 4.2.6:

Consider the discrete bivariate random vector

, with joint pmf given by (^

,^ ) X Y

f^

f^

f

=^

=^

=^

,^ (10,2)

f^

f =

=^

, and

f^

=^

Find the marginals of

and

. Are

and

independent?

X^

Y^

X^

Y

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

(^

, X Y^

( , 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^

Y

(^

(^

x^ ∈

R^

and

y^ ∈

R^ ,

( ,^

)^

f^ x y

g x h y =

. In this case

f^ xX

cg x =

and

(^ ) Y

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^ >

X^

Y

Notes:

1.^

Consider the set {( ,

and y

x^ y

x

A

B

∈^

∈^

, where

{^ :

X

A^

x^

f^

x

=^

>^

and

{^

:^

Y

B^

y^

f^

x

=^

>^

, then this set is

called a cross-product denoted by

A^

B ×.

2.^

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

X^

and

Y^

with joint pdf or pmf

f^ x y

are not independent.

3.^

If it is known that

X^

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

X^

and

Y^

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^

Y

a. For any

A^

⊂^ R

and

B^

⊂^ R

,^

(^

,^

)^

(^

)^ (

P X

A Y

B^

P X

A P Y

B

∈^

∈^

=^

∈^

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^

Y

a. Find the joint pdf of

and

X^

Y

b. Find

(^

4,^

P X

Y

≥^

5