From Osher Doctorow
The example in the last post fails because F(x, y) = R(x, y) goes
against the following facts in probability theory:
1) lim F(x, y) (as x --> infinity) = FY(y)
2) lim F(x, y) (as y --> infinity) = FX(x)
3) lim F(x, y) (as x --> -infinity) = 0
4) lim F(x, y) (as y --> -infinity) = 0
5) lim F(x, y) (as x and y --> infinity) = 1
6) lim F(x, y) (as x and y --> -infinity) = 0
Notice that if X and Y are nonnegative, for example defined on [0,
infinity) or (0, infinity), then the limits as x and/or y --> -infinity
become limits as x and/or y --> 0+ (approach 0 from the right, i.e.,
from x and/or y > 0).
It is rather easy to see that, on the contrary, R(x, y) --> 0 as x and
y --> infinity, while R(x, y) --> 1 as x and y --> -infinity. If one
of x or y but not both --> infinity, R(x, y) still --> 0, and if one
but not both of x or y --> -infinity then since R(x, y) = I f(x, y) dy
dx where the integral I...dydx is the multiple or iterated integral
from x to infinity and from y to infinity, and I f(x, y)dx = fY(y) and
I f(x, y)dy = fX(x), the approach of only y to -infinity "integrates
out" y and leaves 1 - FX(x) and similarly for only x--> -infinity
integrating out x and leaving 1 - FY(y).
To actually evaluate R(x, y) in the example, do the homework assignment
of calculating the integral:
7) R(x, y) = I1[I2[f(x,y)dy]dx, I1 from x to infinity, I2 from y to
infinity.
Osher Doctorow
.
|
