Simple conditional probability problem
Not without some independence. (If I understand the question.)
Say X and Y have joint density m(x,y) = x+y on
[0,1] x [0,1]. That has double integral 1, right?
Let g(x) = x. Then
P( X + Y <= 1 | X = 2/3)
= (integral(y=0 to 1/3) m(2/3,y) dy)/(integral(y=0 to 1) m(2/3,y) dy)
= 5/21 , but
P( 2/3 + Y <= 1)
= doubleintegral((x=0 to 1)(y=0 to 1/3) m(x,y) dx dy) = 2/9 .
If X and Y are independent, I think it is correct.
--
G. A. Edgar http://www.math.ohio-state.edu/~edgar/
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