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                            52                  Conditional Distributions and Expectation

                            Hence, the conditional distribution of X given Y = y is absolutely continuous with density
                            function
                                                          1 − x − y
                                              p X|Y (x|y) = 2     ,  0 < x < 1 − y
                                                           (1 − y) 2
                            where 0 < y < 1.

                            Example 2.12 (Trinomial distribution). Let (X 1 , X 2 , X 3 )bea random vector with a
                            multinomial distribution with parameters n and (θ 1 ,θ 2 ,θ 3 ); see Example 2.2. This is some-
                            times called a trinomial distribution. It was shown in Example 2.2 that the marginal distri-
                            bution of X 1 is binomial with parameters n and θ 1 .
                              Hence, (X 1 , X 2 , X 3 ) has frequency function

                                                                n
                                                                       x 1 x 2 x 3
                                                  p(x 1 , x 2 ) =     θ θ θ
                                                                       1  2  3
                                                             x 1 , x 2 , x 3
                            and the marginal frequency function of X 1 is given by
                                                             n

                                                                 x 1
                                                      (x 1 ) =  θ (1 − θ 1 ) n−x 1 .
                                                                 1
                                                   p X 1
                                                             x 1
                            It follows that the conditional distribution of (X 2 , X 3 )given X 1 = x 1 is discrete with fre-
                            quency function
                                                           n     x 1 x 2 x 3        x 2 x 3 )
                                                            θ θ θ                  θ θ
                                                        x 1 ,x 2  1  2  3  n − x 1  2  3
                                                                      =                    ,
                                                       n  x 1
                                    p X 2 ,X 3 |X 1  (x 2 , x 3 |x 1 ) =
                                                         θ (1 − θ 1 ) n−x 1  x 2 , x 3  (θ 2 + θ 3 ) n−x 1
                                                       x 1  1
                            where x 2 , x 3 = 0,..., n − x 1 with x 2 + x 3 = n − x 1 , for x 1 = 0,..., n; recall that θ 1 +
                            θ 2 + θ 3 = 1.
                              That is, the conditional distribution of (X 2 , X 3 )given X 1 = x 1 is multinomial
                            with parameters n − x 1 and (θ 2 /(θ 2 + θ 3 ),θ 3 /(θ 2 + θ 3 )). Alternatively, we can say
                            X 2 has a binomial distribution with parameters n − x 1 and θ 2 /(θ 2 + θ 3 ) with X 3 =
                            n − x 1 − X 2 .
                            Example 2.13 (A mixed distribution). Let (X, Y) denote a two-dimensional random vec-
                            tor with the distribution described in Example 2.3. Recall that this distribution is neither
                            absolutely continuous nor discrete.
                              First consider the conditional distribution of X given Y = y. Recall that for A ⊂ R and
                                                                                               +
                            y = 1, 2,
                                                                1
                                              Pr(X ∈ A, Y = y) =     y exp(−yx) dx.
                                                                2  A
                            Note that
                                                          1
                                        Pr(X ≤ x, Y = y) =  [1 − exp(−yx)],  x > 0, y = 1, 2
                                                          2
                            so that, for y = 1, 2,

                                              Pr(X ≤ x|Y = y) = 1 − exp(−yx),  x > 0.
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