Math 300 , Spring 2004, Day37, M, May 3 Hit reload ..After class (corrected)

Continuing with Two random variables X, Y "jointly distributed".  Ash Ch. 5.  First 5.1& 5.2 Next 7.1 again, then 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.

Probability Handout problem #17:  Density f(x,y) = Cx(1+ y), 0<x<1, 0<y<2.
    C = 1/2.  f_X(x) = 2x, 0<x<1.   f_Y(x) = (1+y)/4, 0<y<2.  See X and Y are independent.

Back to Expected value (Ash ch. 7):  (law of the unconscious statistician again:)
 You can find E(X) just from f(x), or from f(x,y).  You can find E(XY) only from f(x,y) (unless X and Y are indep.)
  You can find E(X+Y) from f(x,y), or by finding E(X) +E(Y).  These things work because of the rules of iterated integration, where the variable not being integrated acts like a constant for the moment.  Reread Ch. 7-1, the parts with integration.

Find E(XY) for f(x,y) = y/18,  0<x<4, 0<y<3. E(XY)=  _oS³_oS⁴xyy/18 dxdy = (1/18) _oS ³ [(1/2) x²y² _x=o]⁴dy
          = (1/18) _oS ³ [8y²  - 0]dy = (1/18)  [8y²/3 _y=o]³ = (1/18)  [127 8· 27/3] = 4
Note E(XY) = _oS³_oS⁴xyy/18 dxdy = _oS³2y²/9 [_oS⁴x/4 dx]dy =[_oS⁴ x/4 dx][ _oS³2y²/9dy] = EX EY

Now begin E(XY) for the density f(x,y) = Cx(1+y), 0<x<1, 0<y<2.  You are integrating xyf(x,y), which = (x²y + xy²)/2 , or = x²y(1+y)/2.
  In the second form, we can see it's [x²][y(1+y)/2.], and as the things held constant pull out through the integrals, you can see it setting up to make E(X)E(Y) (or maybe E(Y)E(X)).
HW questions on Expected value?

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 (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.
    Go over  handout: Conditional distributions: Discrete.
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 .
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
Next (of course) On Wednesday
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)

Handout: Multivariate distributions of the Continuous Type: 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.
But first--Find the conditional densities and the conditional expectations for the two functions above:
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.
--- --- --- --- --- --- --- ---
HW:  Ash 8.1 covers conditional for continuous only.  Does a good job.
Discrete
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.

B. For the chips above, Find f_Y|x(y|x_o) when x_o = 3. Find f_X|y(x|y_o) when y_o = 4  (numerically.  Just to fix concept)
Conditional Distributions (discrete) handout:
  problem 2.10-1
  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)
Read Continuous handout and Ash 8.1;
Postpone continuous problems:
MultivariateContinuous handout:
  problem 3.7-6
  B. (this is checking the equation in the middle of the second page of the handout.)
  C. (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.)