Math 300 , Spring 2008, Day36, F, April 25 Hit reload .After class..

HW: A.  Here's a density that's not independent: f(x,y) = x+y, 0<x<1, 0<y<1.  It's a plane slicing through the origin. (See it in DPGraph)
       a) Find f_X(x). Argue by symmetry that  f_Y(y) has the same form, and write it down.
       b) Multiply f_X(x) f_Y(y) .  Are X and Y independent?  why not?
       c) Find E(X).  Find E(Y).   Find E(XY).
       d) cov(X,Y) = E(XY) - E(X)E(Y).  Find cov(X,Y).  Is it positive or negative?

Ash 8.1 covers conditional for continuous only.  Does a good job.
Read  Conditional (discrete) handout.
Recall Regression line y(hat) = a + bx, where b = r (s_x/s_y) and a = ybar - b xbar.
           So y(hat) = ybar + b(x-xbar).   Compare to E(Y|x) formula, bottom of 2nd page of handout. 
HW: Conditional, discrete:
B. For the chips below,  x red/y blue   1/4  2/2  2/4  3/4 
     Find f_Y|x(y|x_o) when x_o = 3. Find f_X|y(x|y_o) when y_o = 4
On Conditional  (discrete) handout:   Problem 2.10-1  Can do all put part e after Friday
  "A"(also on handout)  (Show that X, Y independent -->f_Y|x(y|x) = f_Y(y); that is, knowing X gives no info about Y)
Continue with continuous Mon.
Read Continuous (conditional) handout and Ash 8.1;
On the handout:
  problem 3.7-6
  B. (this is checking the equation in the middle of the second page of the handout.)
  C. (x+y. I may have started or even completed this in class.  Write it up.)
Ash, p. 249, #2, #5a, #3.  (These don't need to be handed in--but DO understand them.)

= = = = = = = = = = = = = = = = = = = = = = = = =
Continuing with Two random variables X, Y "jointly distributed".  Ash Ch. 5.  First 5.1& 5.2 Next 7.1 again, now 8.1
In practice: Two random variables X, Y  measured on the same experiment.
Sample space:  points in x-y space.
   Probability of a region:  sum or integrate over the region.
"Marginal" probability/density function:
      Function for just X (not "looking at" y):   p_X(x) or f_X(x)
                    or just Y (not "looking at" y):   p_Y(x) or f_Y(y) .
X and Y INDEPENDENT:  p(x, y) =  p_X(x)p_Y(y),  f(x,y) = f_X(x) f_Y(y) for every pair (x,y) 

Continuous (p. 171 ff + Probability Handout) Joint density function f(x, y)--a surface above the base x-y space.
    Probability of a region R  in x-y space= area under f(x,y) and above the region.
The total probability (area) has to = 1.
 

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Conditional distributions: Assuming a particular x value is true/known (x_o),  what is the distribution of Y?
   If you have 4 chips, red on one side (X) and blue on the other (Y),
x red/y blue   1/4  2/2  2/4  3/4   and you choose a chip & see that the red side is 2, what is now the probability distribution of the numbers on the blue side?
Put in usual 2-dimensional x-y grid.
Assuming x_o known, what is the distribution of Y?. Look just at the column for x_o.
Can do the other way: assuming y_o known, what is the distribution of X?.Look just at the row for y_o.
  Discrete: P(Y= y | X = x_o) = P(Y = y & X = x_o)_/P(X = x_o)   (Old conditional probability, in new clothes)
     Changing to distribution language:  The conditional probability ( or density) of Y given x_o:
           P(Y= y | X = x_o) =  f_Y|x(y|x_o) = f(x_o, y)/f_X(x_o)
   For a single particular number x, you can plug in that number, and then everything is in y.
        For a particular number x, sum over all the y's.  The sum over the y's of f(x_o,y) = f_X(x_o).
             So the sum over all the y's of f_Y|x(y|x_o) = f_X(x_o)/f_X(x_o) = 1.
    Look at handout: Conditional distributions: Discrete.
Pick up here Monday
  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 sum of all   y f_Y|x(y|x_o) terms.
     "Average" of the y-column  at x_o .
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, E(Y|x_o)  can play that role.  We find it for "all" x_o's, and graph it on the x-y plane.

Continuous:  same formulas, 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)

  Continuous (conditional) handout "Multivariate distributions of the Continuous Type--Conditional:
Example 3.7-2 ff.
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, E(Y|x_o)  can play that role.  We find it for "all" x_o's, and graph it on the x-y plane.
                (It may not be a straight line)
But first--Find the conditional densities and the conditional expectations for two functions:
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.
f(x,y) = x+y, 0<x<1, 0<y<1.  f_X(x) = x + 1/2, 0<x<1
Another? f(x,y) = 2e^-x-y , y>x, x>0.
     Caution: if support of f(x,y) isn't rectangular, you have to keep track of possible values of x and y.