lecture31.pdf

Lecture 31

Zhihua (Sophia) Su

University of Florida

Apr 1, 2015

STA 4321/5325 Introduction to Probability 1

Agenda

Conditional Expectation of Random VariablesProperties Involving Conditional Expectations

Reading assignment: Chapter 5: 5.11


Conditional Expectation of Random Variables

Let us recollect that for two random variables X and Y , theconditional distribution of X given Y = y, describes theprobability behavior of the random variables X given theinformation that Y = y. In the same spirit, we define the notionof the conditional expectation of X given Y = y.



DefinitionIf X and Y are two discrete random variables with jointprobability mass function pX,Y , then the conditionalexpectation of X given Y = y is defined as:

E(X | Y = y) =∑x∈X

xpX|Y=y(x),

where pX|Y=y(x) is the conditional probability mass of X givenY = y.



DefinitionIf X and Y are two continuous random variables with jointprobability density function fX,Y , then the conditionalexpectation of X given Y = y is defined as:

E(X | Y = y) =

ˆ ∞−∞

xfX|Y=y(x)dx,

where fX|Y=y(x) is the conditional probability density of Xgiven Y = y.



Example: A soft-drink machine has a random supply Y at thebeginning of a given day and dispenses a random amount Xduring the day (with measurements in gallons). It has beenobserved that X and Y have joint density

fX,Y (x, y) =

{12 if 0 ≤ x ≤ y, 0 ≤ y ≤ 2,

0 otherwise.

If it is known that on a particular day 1 gallon was supplied atthe beginning of the day, find the expected value of the amountof soft-drink consumed in that day.


Properties Involving Conditional Expectations

Results:Let X and Y be two random variables. Then

E(X) = E[E(X | Y )],

where on the right-hand side, the inside expectation is statedwith respect to the conditional distribution of X given Y , andthe outside expectation is stated with respect to the distributionof Y .


Properties Involving Conditional ExpectationsSometimes, the way the distribution of the random variable X isspecified, it is not possible to directly evaluate its expectation,and the previous identity comes is very handy.

Here is an alternative way of evaluating E(X) based on theprevious identity.

1 For every y, find E(X | Y = y).2 Define the function h as h(y) = E(X | Y = y).3 E(X) = E(h(Y )).

Hence, the way to interpret the previous relationship is asfollows.

E(X) = E(h(Y )),where h(y) = E(X | Y = y).



Example: A quality control plan for an assembly line involvessampling n finished items per day and counting X, the numberof defective items. If p denotes the probability that an item isdefective, then given p, X has a binomial distribution withparameters n and p. However, it is observed that p is a randomquantity, and has a uniform distribution on the interval [0, 14 ].(a) Find E(X).(b) Find SD(X).



Results:Let X and Y be two random variables. Then

V (X) = E(V (X | Y )) + V (E(X | Y )).



Here is an algorithm based on this identity to evaluate V (X).1 For every y, find E(X | Y = y) and V (X | Y = y).2 Define the function h as h(y) = E(X | Y = y).3 Define the function h̃ as h̃(y) = V (X | Y = y).4 V (X) = V (h(Y )) + E(h̃(Y )).

Hence, the way to integrate the previous relationship is asfollows:

V (X) = V (h(Y )) + E(h̃(Y )),

where h(y) = E(X | Y = y) and h̃(y) = V (X | Y = y).



Note that for any function g,

E(g(X) | Y = y) =

{´∞−∞ g(x)fX|Y=y(x)dx if X is continuous,∑x∈X g(x)pX|Y=y(x) if X is discrete.



Example (cont’d)


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