Math 300 , Spring 2004, Day38, W, May 5Hit reload ...After class

  Conditional Expectation E(Y|x_o):The mean y value, when x is a particular fixed value x_o.
      Treat x_o as a constant and find the Integral  of  y f_Y|x(y|x_o) dy.
    Remember "Regression problem" in statistics:  For a particular x_o, predict the "best" y-value.
           ("best" in some sense or other--average, typical...)
       In probability (the abstraction from data), E(Y|x_o)  can play that role.
             Find it for "all" x_o's, and graph it on the x-y plane.
Handout: Multivariate distributions of the Continuous Type: Example 3.7-2 ff.   Uniform (flat) so intuition can check asnwers.
--Find the conditional densities and the conditional expectations for the two functions above:
   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.
  (didn't look at.  X,Y indep. so conditionals = marginals.)
f(x,y) = x+y,  0<x<1, 0<y<1.  f_X(x) = x + 1/2, 0<x<1  (did)
Another? f(x,y) = 2e^-x-y , y>x, x>0. (looked at DPGraph. )
  FYI, f_X(x) = 2e^-2x, 0<x.   f_Y(y) =2e^-y (1-e^-y), 0<y
   f_X|y(x|y) = e^-x/(1-e^-y)= , 0<x <y, for y >0 
  f_Y|x(y|x) =e^x-y , x<y<oo, for x >0, an "exponential" (lambda=1) with starting point x.
                  (x-y = -(y-x). y-x is time after starting point x, where "real" times are x 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.)
Caution: if support of f(x,y) isn't rectangular, you have to keep track of possible values of x and y.

Astonishing(?) fact:  IF E(Y|x) is linear in x, it will coincide with the (abstraction of) the  familiar least squares regression line formula.
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HW:  Ash 8.1 covers conditional for continuous only.  Does a good job.   Read it and handout. Use DPGraph to visualize examples.
Multivariate ...Continuous handout:
  problem 3.7-6 .  Flat surface so you can check your answers with your intuition.
  B. (this is checking the equation in the middle of the second page of the handout.)
  C. (I  did  this in class.  The cardboard model.  DPGraph " x+y f(y given x)" shows the density and f_X. Slicing on x will show the shape of the conditional on x, and the number (f_X(x)) to divide by to get area 1.  DPGraph "x+y E(y given x)"  is the same, with an added surface which draws the curve E(Y|x) in the x-y plane.  Slicing on x will show that it does look like  the mean of the y's at each x. Write it up. ALSO graph E(Y|x) for 0<x<1, plotting at least 5 points.)
Added problem:  Go through the writeup above for  f(x,y) = 2e^-x-y , y>x, x>0, checking my work.  Slice the DPGraph picture and check on the plausibility of the conditional formulas.
Ash, p. 249, #2, #5a, #3.

I figured out "slicing by x".  If it loads with slicing by x chosen in the Scrollbar menu it won't slice by x.  Choose Off, close the scrollbar menu, open it again and choose slice by x.  Then it should work, on any of the pictures. (Making cutting with the parameter a redundant, where i put that in.)  Use whatever helps you with the visualization.