# If x and y are independent then it must be the case

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If X and Y are independent, then it must be the case that Cov( X, Y ) = 0. This is because the factorization f XY ( x, y ) = f X ( x ) f Y ( y ) implies that E ( XY ) = E ( X ) E ( Y ). On the other hand, it is possible for two random variables to have zero covariance, but not be independent. An important exception is when ( X, Y ) has the bivariate normal distribution; in this case, X and Y are independent if and only if they have zero covariance. But we will not dwell on this point. 7 Conditional expectation Given a discrete pair of random variables ( X, Y ) with joint pmf f XY , we define the conditional probability mass function for Y given X to be f Y | X ( y | x ) = f XY ( x, y ) f X ( x ) . 7

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When the pair ( X, Y ) is continuous, the same equation defines the conditional probability density function for Y given X . Note that if X and Y are independent, then the factorization f XY ( x, y ) = f X ( x ) f Y ( y ) implies that f Y | X ( y | x ) = f Y ( y ), so that the conditional probability mass or density function for Y given X is just the marginal probability mass or density function for Y . Conditional expectations may easily be defined using conditional probability density or mass functions. For any function g , we define E ( g ( Y ) | X = x ) = X y g ( y ) f Y | X ( y | x ) when ( X, Y ) is discrete, and E ( g ( Y ) | X = x ) = Z -∞ g ( y ) f Y | X ( y | x )d y when ( X, Y ) is continuous. Letting g ( y ) = y , we obtain expressions for the simple case E ( Y | X = x ). When X and Y are independent, f Y | X ( y | x ) = f Y ( y ), and so there is no difference between conditional and unconditional expectation – we simply have E ( g ( Y ) | X = x ) = E ( g ( Y )) for any function g . An important property of conditional expectations is the law of iterated expecta- tions . To state this law, we need to define a slightly different version of conditional expectation. For any function g , define E ( g ( Y ) | X ) = X y g ( y ) f Y | X ( y | X ) when ( X, Y ) is discrete, and E ( g ( Y ) | X ) = Z -∞ g ( y ) f Y | X ( y | X )d y when ( X, Y ) is continuous. This is almost the same as our definition of E ( g ( Y ) | X = x ) above, but we have left X as a random quantity, rather than specifying that we know it is equal to some fixed value x . This makes E ( g ( Y ) | X ) a random quantity. Think of E ( g ( Y ) | X ) as your prediction of g ( Y ) in terms of X . Different realizations of X will give you a different prediction of g ( Y ), so your prediction E ( g ( Y ) | X ) must itself be random, but in a way that depends only on X . Since E ( g ( Y ) | X ) is random, we can consider taking its expected value: E ( E ( g ( Y ) | X )). The law of iterated expectations states that E ( E ( g ( Y ) | X )) = E ( g ( Y )) . To see why the law of iterated expectations is true, suppose that the pair ( X, Y ) is continuous with joint pdf f XY . The conditional expectation E ( g ( Y ) | X ) 8
is a random quantity depending on X - in other words, it is a function of X . Therefore, its expected value is given by E ( E ( g ( Y ) | X )) = Z -∞ E ( g ( Y ) | x ) f X ( x )d x. Plugging in our definition for E ( g ( Y ) | X ), we obtain E ( E ( g ( Y ) | X )) = Z -∞ Z -∞ g ( y ) f Y | X ( y | x )d y f X ( x )d x = Z -∞ Z -∞ g ( y ) f Y | X ( y | x ) f X ( x )d x d y.

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