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Homework Statement:

For the following joint density
f(x,y)=3/4(2xy) for 0<x<2,0<y<2,x+y<2 and f(x,y) = 0 otherwise
1) What is the conditional probability P(X<1Y<1)?
2) What is E[XY<1]?
Relevant Equations:

1) P(AB) = P(A and B)/P(B) or ∫_0^1▒∫_0^1▒〖f(x│y)dydx〗
for
2) E[X│Y]=∫_0^(2y)▒〖x∙(f(x,y))/(f_Y (y))〗 dx
For 1) I found two ways but I get difference results.
The first way is I use P(AB) = P(A and B)/P(B).
I get P(X<1Y<1)=(∫_0^1▒∫_0^1▒〖3/4 (2xy)dydx〗)/(∫_0^1▒∫_0^1▒〖3/4 (2xy)dydx〗+∫_1^2▒∫_0^(2x)▒〖3/4 (2xy)dydx〗)=6/7
The 2nd method is I use is
f(x│y)=f(x,y)/(f_X (x) )=(3/4(2xy))/(∫_0^(2x)▒〖3/4(2xy)dy〗)=(2(2xy))/〖(2x)〗^2 for 0<x<2
So P(X<1│Y<1)=∫_0^1▒∫_0^1▒〖2(2xy)/(2x)^2 dydx〗=ln(4)1/2
For 2) I know that E[X│Y]=∫_0^(2y)▒〖x∙(f(x,y))/(f_Y (y))〗 dx but I have no idea how to start E[X│Y<1]=∫_0^(2y)▒〖x∙(f(x,y))/(f_Y (y))〗 dx
Im thinking of a double integral maybe.
Thanks!
The first way is I use P(AB) = P(A and B)/P(B).
I get P(X<1Y<1)=(∫_0^1▒∫_0^1▒〖3/4 (2xy)dydx〗)/(∫_0^1▒∫_0^1▒〖3/4 (2xy)dydx〗+∫_1^2▒∫_0^(2x)▒〖3/4 (2xy)dydx〗)=6/7
The 2nd method is I use is
f(x│y)=f(x,y)/(f_X (x) )=(3/4(2xy))/(∫_0^(2x)▒〖3/4(2xy)dy〗)=(2(2xy))/〖(2x)〗^2 for 0<x<2
So P(X<1│Y<1)=∫_0^1▒∫_0^1▒〖2(2xy)/(2x)^2 dydx〗=ln(4)1/2
For 2) I know that E[X│Y]=∫_0^(2y)▒〖x∙(f(x,y))/(f_Y (y))〗 dx but I have no idea how to start E[X│Y<1]=∫_0^(2y)▒〖x∙(f(x,y))/(f_Y (y))〗 dx
Im thinking of a double integral maybe.
Thanks!
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