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Stat 5101 Lecture Notes - School of Statistics

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86 <strong>Stat</strong> <strong>5101</strong> (Geyer) Course <strong>Notes</strong>3.2 Conditional Probability DistributionsScalar VariablesThe conditional probability distribution <strong>of</strong> one random variable Y givenanother X is the probability model you are supposed to use in the situationwhen you have seen X and know its value but have not yet seen Y and don’tknow its value. The point is that X is no longer random. Once you know itsvalue x, it’s a constant not a random variable.We write the density <strong>of</strong> this probability model, the conditional distribution<strong>of</strong> Y given X as f(y | x). We write expectations with respect to this model asE(Y | x), and we write probabilities asP (Y ∈ A | x) =E{I A (Y)|x}(couldn’t resist an opportunity to reiterate the lesson <strong>of</strong> Section 2.6 that probabilityis a special case <strong>of</strong> expectation).We calculate probabilities or expectations from the density in the usual waywith integrals in the continuous case∫E{g(Y ) | x} = g(y)f(y | x) dy (3.3)∫P {Y ∈ A | x} = f(y | x) dy (3.4)and with the integrals replaced by sums in the discrete case.Note thatA conditional probability density is just an ordinary probability density,when considered as a function <strong>of</strong> the variable(s) in front <strong>of</strong> thebar alone with the variable(s) behind the bar considered fixed.This means that in calculating a conditional probability or expectation from aconditional densityalways integrate with respect to the variable(s) in front <strong>of</strong> the bar(with, <strong>of</strong> course, “integrate” replaced by “sum” in the discrete case).Example 3.2.1 (Exponential Distribution).Of course, one doesn’t always have to do an integral or sum, expecially when a“brand name” distribution is involved. Suppose the conditional distribution <strong>of</strong>Y given X is Exp(X), denotedAY | X ∼ Exp(X)for short. This means, <strong>of</strong> course, that the conditional density isf(y | x) =xe −xy , y > 0

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