Math 300 , Spring 2008, Day 32, W, April 16 Hit
reload After class
Lecture, David Aldous: Coincidences happen. People are not very good
at estimating probability/risk in real world situations.
Read Ash 5.1,
Handouts: "MultDisc",
"MultCont"
HW
Practice
here: http://www.math.temple.edu/~cow/
Calculus Book III > 5.Integration > 2.Double Integrals >
Modules 1 and 2 Double integrals over rectangles (Module
1 tells you the limits of integration and steps through. Module 2
has you set the limits and do the whole thing.
Don't worry if you can't compute all the indefinite integrals, do
enough
to understand the pattern, and skip sin and cos if you want.) Do at
least Module 1:
1, 10, 12, (recall ea+b = eaeb),
Module 2: 1, 9
Module 3 Limits of double integrals--Figuring out the limits
of integration. Caution--some problems can be done easily both
ways,
and the package will allow you to do either. But if there is only
one way to do it without breaking it into 2 double integrals, it won't
tell you you've chosen an impossible way--it'll just keep
rejecting
your answers. (e.g. Problem 1--if you (foolishly) try to integrate over
x first, you need to break the area into 2 pieces, at the line y =
1.
The integration limits are "between y/2 and y for
0<y<1,
and between y/2 and 1 for 1<y<2 ", which the program has no way
of
reading or understanding.) Copy the sketch at the first step, to
help keep track.
Don't worry about ones where you can't
guess what the indefinite integrals are.
Do as many as you need to. At least 1, 5, 12. Ash pp. 165-69 is a good
followup here.
Module 4 Changing the order of integration. This requires
figuring out from the integrals what the shape of the region is and the
functions defining the borders. Then switching, using the methods
of module 3. A good test for your understanding. Try a few,
at least 1, 4, 12.
Work on the above, and try to finish the Transformations (Ch. 4)
HW. We'll do at most one more in class, so make it count!
Postpone the rest.
HW
Multivariate Discrete; "MultDisc" p. 3, all those
exercises. In 5.1.3c,f: There's not much of a
double
sum to write when the probability wanted is only one point. What
I'm looking for is that you try to write double sums that parallel what
you do in writing double integrals, but in the sum case you can
actually
check your work by brute force. Do what makes sense.
With them: p.4: B) P (3 < X+Y < 4) for the
two distributions on that page (those of examples 2.8.2 and 3)
Then A) Find P( 0 < X-Y<1) for f(x,y) = xy2/30
, x = 1,2,3,4, y = 1,2 (dist. of example 2.8.2 in handout)
= = = = = = = = = = = = = = = = = = = = = = = = =
Ch. 4, continuing 4.6.
How do we find the distribution of a function of a continuous
random variable? continued.
Y=g(X)
As the values on the x-axis get mapped
to the y-axis, the probability has to follow along. To find the
distribution of Y, we have to track back and see where the probability
came from, when it was attached to the x-axis.
Density method: fY(y)
dy =
Area = fX(x) dx. Solve for fY(y)
CDF method: FY(Y
< yo) = P(Y < yo)
= P(g(X)
< yo). Restate so it is a statement about P
(? X ?). Use the distribution of X, and whatever
means is easiest to get this into a formula, either for FY or for fY.
Homework questions? Most notes: Day 29
HW Day 30 Notes & links Day 31
Class was spent working examples of the
above.
= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
= = = = =
Next: Two random variables "jointly
distributed".
Ash Ch. 5. First 5.1 and 5.2
In practice: Two random variables X, Y) measured on the same
experiment.
We've looked at discrete already (balls and chips in urns, tables in
two rooms in a restaurant, two dice).
Sample space: points in x-y space.
Discrete (p. 180): 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.
Continuous (p. 171 ff) 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.
Need calculus for that:
The
dA
is a little region in x-y space--you chop the x-y plane into little
dA's
and sum the volumes over all the little dA's that are in R.
How to calculate? Chop into rectangular dA's by chopping
the x axis into dx's and the y axis into dy's and making the grid. Then
you can sum them by first summing in (say) the x direction, and then
summing
those values in the y direction.
Discrete
analog:
To find the probability of the whole space:
First sum across the x direction, get P(Y=1) = 7/20, P(Y=2) =
6/20, P(Y=3) = 7/20,for each y.. Then sum over the y values, to get
(7+6+7)/20
=1.
To find P(X + Y < 4): In the yellow region, Sum
across
the x direction, get 4/20, 3/20, 3/20 for y = 1,2,3. Then sum
over
the y values to get 10/20.
Handout: "MultDisc"
"Marginal" probability/density function:
Function for just X (not
"looking
at" y): pX(x) or fX(x)
or just Y (not "looking at" y): pY(x)
or fY(x) .
X and Y INDEPENDENT: p(x, y) = pX(x)pY(y),
f(x,y) = fX(x) fY(x)
for every pair (x,y)
Discrete (Ash p. 180+ "MultDisc"):
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: pX(x)
= sum down--sum over the y's.
Marginal probability function for Y: pY(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 pX(x)pY(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 example. 2.8.3 (see page 4 for pictures,
computations).
For that distribution, X and Y are independent.
Continous: Integrals:
First the total volume, then P(X + Y < 4).
In the inner integral, the "other" variable
acts
like a constant. Try it with g(x,y) = y, 0<x<4,
0<y<3. Volume =?
Handout: "MultCont"
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