Math 300 , Spring 2008, Day37, M, April 28 Hit reload .After class.

HW:  Conditional probability distributions and Expectation: Ash 8.1 covers conditional for continuous only.  Does a good job.
Finish HW on Day 36. Try all, bring questions.
Also: Go through the writeups below for f(x,y) = x+y, 0<x<1, 0<y<1.,  f(x,y) = 2e^-x-y , y>x, x>0, checking my work.  Slice the DPGraph pictures and check on the plausibility of the conditional formulas.
f(x,y) = 2e^-x-y , y>x, x>0, I'll go over in class, the rest. See if you can do it ahead of me, so mine will be reinforcement.

= = = = = = = = = = = = = = = = = = = = = = = = = = = =
Conditional distributions: E(Y|x_o) :  The "Average" of the y-column  at x_o .
Continuous density:  same formulas as discrete, integration instead of sums.
      Think of slicing through the joint density at x_o--look at f(x_o,y)--the slice.
      The conditional density f_Y|x(y|x_o) has the shape of the slice, but the slice may not have area 1.
          If we divide by  f_X(x_o), it will. So   f_Y|x(y|x_o) = f(x_o, y)/f_X(x_o) 
                    Watch out for regions of positive density: y-region may depend on x_o.
Details, Day 36.
New DPGraph models, for the following: (in "Conditionals" folder, in "For 300 class" folder.  Copy whole"For 300 class" folder; added/improved)
  See DPGraph pictures--slicing x  will give shape of conditional density given x.  Must  divide by f_X(x) to get area = 1.

f(x,y) = x(1+y)/2, 0<x<1, 0<y<2.  f_X(x) = 2x, 0<x<1.   f_Y(y) = (1+y)/4, 0<y<2.  Independent, so conditionals = marginals.
f(x,y) = x+y, 0<x<1, 0<y<1.  f_X(x) = x + 1/2, 0<x<1  Symmetric in x and y so only need to do one conditional.
    f_Y|x(y|x_o) = (x_o+y)/(x_o + 1/2), 0<y<1, 0<x_o<1.  Note it's linear in y, with slope 1/(x_o + 1/2)
   E(Y|x) =  (3x+2)/(6x+3)  Note it's not linear in x! 
     Look at DPGraph; also Excel graph of E(Y|x)

f(x,y) = 2e^-x-y , y>x, x>0.     f_X(x) = 2e^-2x ,  x>0.     f_Y(y) =  2e^-y(1-e^-y), y>0
     Caution: if support of f(x,y) isn't rectangular, you have to keep track of possible values of x and y in conditionals.
f_X|y(x|y) = e^-x/(1-e^-y) , 0<x <y, for y >0  Make it clearer (?) by making all the y's into y_o's
Pick up here Wed.
f_Y|x(y|x) = e^x-y , x<y<oo, for each x_o >0, an "exponential" (lambda=1) with starting point x_o.
                  (x_o-y) = -(y-x_o). y-x_o is time after starting point x_o, where "real" times are x_o and y.)
E(Y|x) = _xS^ooye^x-ydy  =e^x_xS^ooye^-ydy {integrate by parts or use ash p. 95, a = -1} =  e^x[e^-y(-y -1) _y=x]^oo
     =  e^x[0 - (e^-x(-x -1))] = 1+x.  Note this is linear in x!
            (E(Y|x) is Consistent with our interpretation of  f_Y|x  as exponential (lambda = 1) with starting point x.)
Astonishing(?) fact:  IF E(Y|x) is linear in x, it will coincide with the (abstraction of) the  familiar least squares regression line formula. (formula 2, Ash p. 236)
    2nd page bottom, Discrete conditional handout:  (M&M 5th ed. page number 137)