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Lecture 11 - Variance, Conditional Expectation

# Lecture 11 - Variance, Conditional Expectation - 1...

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1 CONDITIONAL EXPECTATION Lecture 11 ORIE3500/5500 Summer2011 Li 1 Conditional Expectation We have defined expectation of a random variable only in two cases: discrete and continuous. Expectation can be defined for all random variables but that is outside the scope of this class. We will define conditional expectation in these two cases only. 1. If ( X, Y ) is a discrete bivariate random vector with joint pmf p X,Y ( x i , y j ), then the conditional expectation of X given Y = y j is defined to be E ( X | Y = y j ) = X i x i p X | Y ( x i | y j ) . 2. If ( X, Y ) is a continuous bivariate random vector with joint pdf f X,Y ( x, y ), then the conditional expectation of X given Y = y is defined to be E ( X | Y = y ) = Z -∞ xf X | Y ( x | y ) dx. Let a function be g ( Y ) = E ( X | Y ), observe that g ( Y ) is also a random variable. We could define its expectation E ( g ( Y )) = E [ E ( X | Y = y )] j E [ X | Y = y j ] p Y ( y j ) Y is discrete R -∞ E [ X | Y = y ] f Y ( y ) dy Y is continuous Example Suppose that the joint density of X and Y is given by f ( x, y ) = e - x/y e - y y , 0 < x, y < . Compute E ( X | Y = y ). Solution: First compute the conditional density f ( x | y ) and then E ( X | Y = y ). Note that from Lecture 9 we know that f ( x | y ) = f ( x,y ) f Y ( y ) = y - 1 e - x/y . Therefore, E ( X | Y = y ) = R 0 xe - x/y /ydx = y R 0 te - t dt = y . 1

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1 CONDITIONAL EXPECTATION Law of Iterated Expectation: E ( X ) = E [ E ( X | Y )] Proof in the discrete case: Note that E ( E ( X | Y )) = X y E ( X | Y = y ) P ( Y = y ) = X y X x xP ( X = x | Y = y ) P ( Y = y ) = X y X x xP ( X = x, Y = y ) = X x X y xP ( X = x, Y = y ) = X x x X y P ( X = x, Y = y ) = X x xP ( X = x ) = EX. Intuition: To calculate EX , we may take a weighted average of the con- ditional expected value of X given Y = y , each of the terms E ( X | Y = y ) being weighted by the probability of the event on which it is conditioned. Example Here is a game: there are two different coins. One is fair and the other one has 0 . 9 chance to get tails when it is tossed. If you have 3 / 4 probability to get the fair one and you will win one dollar if you get a head and nothing otherwise. What is the expectation of the dollar you could win? Define the random variable X as the dollar you could win, and define a random variable Y , which equals to one if you pick the fair coin and zero otherwise. Then we have E [ X | Y = 1] = 1 · 1 2 + 0
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Lecture 11 - Variance, Conditional Expectation - 1...

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