Lecture 4: Conditional expectation and independence
Lecture 4: Conditional expectation and independence
In elementry probability, conditional probability P(B|A) is defined as
P(B|A) = P(A B)/P(A) for events A and B with P(A) > 0.
For two random variables, X and Y , how do we define
P(X B|Y = y )?
Definition 1.6
Let X be an integrable random variable on (?, F , P).
(i) The conditional expectation of X given A (a sub- -field of F ),
denoted by E(X |A ), is the a.s.-unique random variable satisfying
the following two conditions:
(a) RE(X |A ) is measurable
from (?, A ) to (R, B);
R
(b) A E(X |A )dP = A XdP for any A A .
(ii) The conditional probability of B F given A is defined to be
P(B|A ) = E(IB |A ).
(iii) Let Y be measurable from (?, F , P) to (, G ).
The conditional expectation of X given Y is defined to be
E(X |Y ) = E[X | (Y )].
UW-Madison (Statistics)
Stat 709 Lecture 4
beamer-tu-logo
2018
1 / 15
Remarks
The existence of E(X |A ) follows from Theorem 1.4.
(Y ) contains the information in Y "
E(X |Y ) is the expectation of X given the information in Y
For a random vector X , E(X |A ) is defined as the vector of
conditional expectations of components of X .
Lemma 1.2
Let Y be measurable from (?, F ) to (, G ) and Z a function from
(?, F ) to R k .
Then Z is measurable from (?, (Y )) to (R k , B k ) iff there is a
measurable function h from (, G ) to (R k , B k ) such that Z = h ? Y .
By Lemma 1.2, there is a Borel function h on (, G ) such that
E(X |Y ) = h ? Y .
For y , we define E(X |Y = y ) = h(y ) to be the conditional
expectation of X given Y = y .
beamer-tu-logo
h(y ) is a function on , whereas h ? Y = E(X |Y ) is a function on ?.
UW-Madison (Statistics)
Stat 709 Lecture 4
2018
2 / 15
Example 1.21
Let X be an integrable random variable on (?, F , P), A1 , A2 , ... be
disjoint events on (?, F , P) such that Ai = ? and P(Ai ) > 0 for all i,
and let a1 , a2 , ... be distinct real numbers.
Define Y = a1 IA1 + a2 IA2 + . We now show that
E(X |Y ) =
i=1
R
Ai
XdP
P(Ai )
IAi .
We need to verify (a) and (b) in Definition 1.6 with A = (Y ).
Since (Y ) = ({A1 , A2 , ...}), it is clear that the function on the
right-hand side is measurable on (?, (Y )).
This verifies (a).
To verify (b), we need to show
#
" R
Z
Z
Ai XdP
XdP =
IAi dP.
Y ?1 (B)
Y ?1 (B) i=1 P(Ai )
beamer-tu-logo
for any B B,
UW-Madison (Statistics)
Stat 709 Lecture 4
2018
3 / 15
Example 1.21 (continued)
Using the fact that Y ?1 (B) = i:ai B Ai , we obtain
Z
Y ?1 (B)
Z
XdP =
XdP
i:ai B Ai
R
=
Ai
P Ai Y ?1 (B)
P(Ai )
" R
i=1
Z
=
XdP
Y ?1 (B)
i=1
Ai
XdP
P(Ai )
#
IAi dP,
where the last equality follows from Fubinis theorem.
This verifies (b) and thus the result.
Let h be a Borel function on R satisfying
h(ai ) =
Z
Ai
XdP/P(Ai ).
beamer-tu-logo
Then E(X |Y ) = h ? Y and E(X |Y = y ) = h(y ).
UW-Madison (Statistics)
Stat 709 Lecture 4
2018
4 / 15
Proposition 1.9
Let X be a random n-vector and Y a random m-vector.
Suppose that (X , Y ) has a joint p.d.f. f (x, y ) w.r.t. , where and
are -finite measures on (R n , B n ) and (R m , B m ), respectively.
Let g(x, y ) be a Borel function on R n+m for which E|g(X , Y )| < .
Then
R
g(x, Y )f (x, Y )d(x)
R
E[g(X , Y )|Y ] =
a.s.
f (x, Y )d(x)
Proof
Denote the right-hand side by h(Y ).
By Fubinis theorem, h is Borel.
Then, by Lemma 1.2, h(Y ) is Borel on (?, (Y )).
Also, by Fubinis theorem,
fY (y ) =
Z
f (x, y )d(x)
beamer-tu-logo
is the p.d.f. of Y w.r.t. .
UW-Madison (Statistics)
Stat 709 Lecture 4
2018
5 / 15
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