Math 300 Day34

Math 300 , Spring 2004, Day34, M, April 26 Hit reload to get most current version

Two random variables X, Y "jointly distributed". Ash Ch. 5. First 5.1 and 5.2
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. Day 33
Only got to here, almost...
"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(x) .
X and Y INDEPENDENT: p(x, y) = p_X(x)p_Y(y), f(x,y) = f_X(x) f_Y(x) for every pair (x,y)
Discrete (p. 180+ Discrete Handout): joint probability function p(x, y) = P(X = x and Y= y): lump of probability sitting on that spot.
    Probability of a set: Sum the probabilities of the points in the set. P(X+Y < 4) = 10/20

Marginal probability function for X: p_X(x) = sum down--sum over the y's.
Marginal probability function for Y: p_Y(x) = sum across--sum over the x's.
X and Y independent? Need only one point to show not independent:
p(1,1) = 1/20 p_X(x)p_Y(y)= (5/20)(7/20) = 35/400=7/80. Not independent.

When the discrete distribution is represented by a formula, sometimes summing over the x's (or y's) can result in a tidy formula as well. Handout ex. 2.8.3 (see handwritten page for pictures, computations). For that distribution, X and Y are independent.

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.

Integrals:

In the inner integral, the "other" variable acts like a constant. Try it with g(x,y) = y, 0<x<4, 0<y<3

Day 33: Find area under the curve: Dividing g by this will give a function with area 1.DID THIS today.
Let f(x,y) = Cg(x,y) = Cy as above.C = 1/18
HW: Find P( X+Y<4) by integration.

Find the marginal distributions:
f_Y(y) = 2y/9, 0<y<3
Are X and Y independent? Check.
NOTE: X and Y canNOT be independent unless their joint support (region where p or f is Not 0) is rectangular!
Because if it isn't a rectangle, there will be a pair (x,y) with X and Y having positive probabilities/densities, but the pair having 0 prob/density.
--- --- --- --- --- --- --- --- ---
HW: Reorganized: Do Boldface for tonight:
Finish review of double integrals with COW as needed. With Ash 164-170. Day 33:
Find P( X+Y<4) by integration. where f(x,y) = y/18 on the rectangle 0<x<4, 0<y<3.

Get 2 handouts from outside my door.
Handout on discrete:
For tonight, read the handout, study the handwritten examples. Do problem B on handwritten page. On the Exercises, last page , Sketch and label the space for 5.1-1, and do c, d, e, f. Think about adding across, then down (or vice versa.) Next: Finish 5.1-1, do 5.1e,f (read the solutions to a-d).

Postpone to next: A) Find f_X(x) for the example above, where f(x,y) = y/18 on the rectangle 0<x<4, 0<y<3.
DO Handout on Continuous "Probability", #17 and #18 (p. 2): For this assignment, Sketch the region of positive density of f, and do parts a.
Next: Also determine their marginal densities and whether X and Y are indep.for 17,18,19, and finish 17, do 18b, 19b.
Ash p. 181, #2a For this assignment, graph the region of f, and the region you want the prob. of. Next time finish it. (Hint--p. 164 has pictures for min and max--but in my book the text refers to the wrong pictures (the captions are right),
#7. Do it by graphing the regions, and since all pairs x,y are equally likely, find the prob's by proportional areas (cf. pp. 31-2, and #7 p. 35.)

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