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The product of the conditional density f y x y x and

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The product of the conditional density f Y | X ( y | x ) and the marginal density f X ( x ) is the joint density f XY ( x, y ), so we now have E ( E ( g ( Y ) | X )) = Z -∞ Z -∞ g ( y ) f XY ( x, y )d x d y = Z -∞ g ( y ) Z -∞ f XY ( x, y )d x d y = Z -∞ g ( y ) f Y ( y )d y = E ( g ( Y )) . A similar argument can be used to justify the law of iterated expectations when the pair ( X, Y ) is discrete. 8 Convergence in probability Suppose we have an infinite sequence of random variables Z 1 , Z 2 , Z 3 , . . . . Imagine that, as we move along this sequence, the random variables Z n start to settle down to some constant value. For instance, it might be the case that the n th variable in our sequence, Z n , has the uniform distribution on (1 - 1 /n, 1 + 1 /n ). In this case, as n increases, the distribution of Z n becomes more and more tightly concentrated around one. In some sense, Z n converges to one as n → ∞ . But since each Z n is random, we need to find a suitable way to define this convergence that involves a statement about the random behavior of the different Z n ’s. A sequence of random variables Z 1 , Z 2 , Z 3 , . . . is said to converge in probability to some constant c if, for every strictly positive number ε > 0, we have lim n →∞ P ( | Z n - c | > ε ) = 0 . 9
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In words, this means that for any positive ε we can think of – no matter how small – the probability that Z n differs from c by more than ε will eventually decrease to zero as n increases toward infinity. If Z 1 , Z 2 , Z 3 , . . . converges in probability to c , we say that Z n p c as n → ∞ . Returning to our example where Z n U (1 - 1 /n, 1 + 1 /n ), let us verify that in fact Z n p 1. Choose any number ε > 0. We need to show that lim n →∞ P ( | Z n - 1 | > ε ) = 0. Since Z n U (1 - 1 /n, 1+1 /n ), it must lie in the interval [1 - 1 /n, 1+ 1 /n ] with probability one. If n > 1 , then the interval [1 - 1 /n, 1 + 1 /n ] lies strictly inside the interval [1 - ε, 1+ ε ]. Therefore, we will have P ( | Z n - 1 | > ε ) = 0 for all n > 1 . No matter how small we chose ε , as n → ∞ we will eventually have n > 1 . This shows that lim n →∞ P ( | Z n - 1 | > ε ) = 0. Let us consider another example that is slightly less trivial. Suppose this time that the n th random variable in our sequence, Z n , is continuous with pdf f n given by f n ( x ) = (2 n - 1) x - 2 n for x 1 0 for x < 1 . The pdf f n is highest when x = 1, where it is equal to 2 n - 1. As we move rightward along the x -axis from one, f n ( x ) decays smoothly to zero at the rate x - 2 n . We can see that when n increases, f n spikes higher and sharper at one, and decays more rapidly to zero as we move along the x -axis. In a sense, the sequence of pdfs f n are gathering into an infinitely tall spike at x = 1 as n → ∞ . This is what we would expect if Z n p 1, which is in fact the case, as we will now show. Choose any number ε > 0. Since f n ( x ) = 0 for x < 1, it must be true that P ( | Z n - 1 | > ε ) = P ( Z n > 1 + ε ). Therefore, P ( | Z n - 1 | > ε ) = Z 1+ ε (2 n - 1) x - 2 n d x = - x 1 - 2 n 1+ ε = (1 + ε ) 1 - 2 n .
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