Solution for Homework for Section 6.3, page 426
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5.  page 426

This problem is just about identical to Example 1 on page 414 of the
text.  You can use any of the three methods explained there to solve this
problem,  although on my scratch sheet I used second and third
methods. 
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7.  page 426

First we have to find the value of k.  In order to do so we must first
identify the region where random variable are defined,  i.e.,  the
domain of (Y, Z).  Sketch the unit square in the yz-plane and draw its
main axis,  the line  y=z.  Then, the region of interest is the region
above the main diagonal but inside the unit square.  Thus,

1  =  int_{0}^{1} int_{y}^{1}  k (z-y)  dz  dy

=>   k = 6.

a.   f_{Y} (y)  =  int  f(y, z)  dz
  
                        =  6  int_{y}^{1}  (z - y)  dz

                        =  3 (1 - y)^{2}

b.  By definition,

P( Z < 2/3  |  Y = 1/2 )  =  int_{1/2}^{2/3}  f_{Z} (z | y=1/2)  dz

Hence, we need to find  the conditional density function
f_{Z} (z | y=1/2).  Again,  by definition,

f_{Z} (z | y=1/2)  =  f(z, 1/2)  /  f_{Y} (1/2)

                                =  6 (z - 1/2)  /  3 (1 - 1/2)^{2}

                                =  8 z - 4
Thus,

P( Z < 2/3  |  Y = 1/2 )  =  int_{1/2}^{2/3}  f_{Z} (z | y=1/2)  dz

                                         =  int_{1/2}^{2/3}  (8 z - 4)  dz

                                         =  1/9
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