Math 300 , Spring 2002, Day34, M, April 22 Hit
reload to get most current version
Ch. 4, continuing 4.6.
How do we find the distribution of a function of a continuous
random variable? continued.
Last time: Derivative of F(x) when it's an
integral with a function of x in the limit of the integral.
Y = g(X). Two methods:
For both: a) If the transformation
function is not 1-1, you'll have two (more?) x-chunks mapping onto the
same y. Be sure to account for both.
b) If X does not have the same formula everywhere,
you'll have to follow each piecewise formula in the transformation.
1) CDF method: Find FY(y0) = P(Y <
y0) = P(g(X) < y0)
Do what you need to, to rewrite this as P(X <?
g-1?(y0) ) (If
g doesn't have an inverse over the whole range this could be more complicated.
And the inequality might reverse.)
This can (usually) be written as an integral with upper limit x= g-1
(y0), f(x)dx. NOW WHAT? Different choices.
a) Change variables to y, inside the integral and
in the
limits, until you get an integral with y0 as the upper limit.
That's your FY(y0).
b) Take the derivative of the formula for FY(y), getting
fY(y). Usually involves chain rule with fundamental th.
of calc.
c) If the integral you got is something you can integrate,
do so and plug in the limits.
2) Density method: Suppose X has density fX(x). We
want the density of Y, fY(y). A thin rectangle at xo
of width dx and height fX(xo)gets mapped by y = g(x)into
a rectangle at yo of width dy and height fY(yo).
(If g is one-to-one, that's all that gets mapped into the new rectangle.)
Set the rectangles equal: fY(yo)dy = fX(xo)
dx
Solve: fY(yo) = fX(xo)
dx/dy. You have to find dx/dy, and change variables to get everything
in terms of y.
Since y = g(x), dy/dx = g'(x). You can put it under 1 to get dx/dy.
Or find x = g-1 (y) and take the derivative of x with respect
to y.
Note: If the transformation function "flips"
things (x1< x2 , but y1 > y2),
dy/dx will be negative. Just throw away the negative (take the absolute
value.)
= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
= = = = =
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.
Integrals:
In the inner integral, the "other" variable acts
like a constant. Try it with f(x,y) = y
Practice here: http://www.math.temple.edu/~cow/
Calculus Book III > 5.Integration > 2.Double Integrals >
Modules 1 and 2 Double integrals over rectangles (These
appear to be the same material--I don't know why so many problems.
Don't worry if you can't compute all the indefinite integrals, do enough
to understand the pattern.)
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.) Don't worry about ones where you can't
guess what the indefinite integrals are.
Do as many as you need to. 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.
--- --- --- --- --- --- --- --- ---
HW: The problems on transformation of one variable on the
green sheet, including
Ash p. 156 #1, whatever method you like. Also
Sketch the density of X and the density of Y, and show where x = 0, 2,
and 5 map to.
The problems on iterated integrals sheet (From Stewart, Calculus),
3, 11, 15, 20 (recall ea+b = eaeb)
This page belongs to Sally Sievers who is solely responsible
for its content. Please see our statement
of responsibility.