## Multivariable Probability Distribution

 Quote by Ray Vickson It is wrong because you integrated y from 0 to x instead of from x to infinity. RGV
Ok, hopefully here's the final check:

fx(x) =
∫ from x to infinity [2e^(-x-y)]dy
=-2[e^(-x-y)] from y=x to y=∞
= 2e^(-2x) I[0,∞) (x)

fy(y) = 2[e^(-y)-e^(-2y)] I [0,infinity) (y)

f(x|y) = 2e^(-x-y) / 2[e^(-y)-e^(-2y)]
= [e^(-x)] / (1-e^(-y)) I[0,∞) (x)

f(y|x) = 2e^(-x-y) / 2e^(-2x)
= e^(-y) / e^(-x) I[0,∞) (y)

Recognitions:
Homework Help
 Quote by mathmajor23 Ok, hopefully here's the final check: fx(x) = ∫ from x to infinity [2e^(-x-y)]dy =-2[e^(-x-y)] from y=x to y=∞ = 2e^(-2x) I[0,∞) (x) fy(y) = 2[e^(-y)-e^(-2y)] I [0,infinity) (y) f(x|y) = 2e^(-x-y) / 2[e^(-y)-e^(-2y)] = [e^(-x)] / (1-e^(-y)) I[0,∞) (x) f(y|x) = 2e^(-x-y) / 2e^(-2x) = e^(-y) / e^(-x) I[0,∞) (y)
These look OK now, except you should get rid of those annoying I[0,∞)(x), etc., symbols. Just say in words that x and y are >= 0.

RGV

 Thanks a bunch! I, too, hate the annoying indicator functions, but my professor seems the need to "require" us to use them.