Let's start over.
I originally had for the marginal of X:
∫ from 0 to x of [2e^(-x-y)]dy
=-2[e^(-x-y)], evaluated from y=0 to y=x
= 2[e^(-x)-e^(-2x)] I[0,∞) (x)
To check this, I integrated this and it came out to 1, so I'm not sure why this is wrong.
Was my conditional of f(x|y) correct?
Hmm is this the MD of x then?
Integral from 0 to infinity [2e^(-x-y)dy]
= -2(e^(-x-y)] from y=0 to y=infinity
=2[1+e^(x)] I [0,infinity) (x)
I got that k=2, and checked it by seeing it integrated to 1.
For the Marginal Distributions of X and Y, I got :
f(x) = 2[e^(-x)-e^(-2x)] I [0,infinity) (x)
f(y) = 2[e^(-y)-e^(-2y)] I [0,infinity) (y)
Then, for the conditionals, I got:
f(x|y) = e^(-x) / [1-e^(-y)] I[0,infinity) (x)...
f(x,y) = ke^(-x-y), 0<x<∞ , 0<y<∞ , and x<y
Find k to make this a PDF.
So I set up: ∫(from 0 to ∞)∫(from 0 to ∞) [ke^(-x-y)]dxdy. Is this integral right?
I also need to find the marginals of X and Y but my k has to be right first.
FX(1)(x) = P(X(1) <= x)
= P(X1,...,Xn <= X)
=1-P(X1,...,Xn > X)
=1-P(X1>X)^n since the xi's are iid.
=1-[1-P(X1 <= X)]^n
=1-[1-F(x)]^n
=1-[1-∫ from 0 to x (2tdt)]^n
=1-(1-X^2)^n
For the probability, P(X1>.2) = 1-Fx(1) (0.2) = (1-(0.2)^2)^n = (0.96)^n