Actex 8.12

[vishusmartishus]

f(x,y) = 15y, x^2<y<x.

What is fy(y)?

Well, the book is giving some ridiculous method and claims that the answer is 15y^(3/2)*(1-y)^1/2

Clearly x&y are independent so why is it not the case that f(x,y) = fy(y)?

[Kaner3339]

draw out x^2= y and y = x. Integrate the joint density for all x. it's really not that ridiculous.

[Actuarialsuck]

Originally Posted by vishusmartishus

f(x,y) = 15y, x^2<y<x.

What is fy(y)?

Well, the book is giving some ridiculous method and claims that the answer is 15y^(3/2)*(1-y)^1/2

Clearly x&y are independent so why is it not the case that f(x,y) = fy(y)?

Wait, why is X and Y ind? Clearly x^2 < y < x would suggest otherwise...

[vishusmartishus]

Yeah, x is from 0 to 1, so after integrating what am I left with? 15y. which again suggests x&y are independent. I always thought if f(x,y) doesn't even have x in it, then they are automatically independent? The explanation in the back of the book is frustrating and provides no explanation of the work done.

[Kaner3339]

if you draw it out, x doesn't go from 0 to 1. x is stuck between x = y^1/2 and x = y.

[vishusmartishus]

I suppose.... I thought you could just integrate over all the values where x > x^2? Thats what I've been doing the past 5000 problems and getting them all right

[alexsadowski]

Nope you have to integrate from x= y^1/2 to y, you'd only go from 0 to 1 on the second integral if you were doing one (to find expected value of Y, probability, etc.). If Y has bounds defined by x, or vice versa, then i believe it can't be independent.

[vishusmartishus]

Originally Posted by alexsadowski

Nope you have to integrate from x= y^1/2 to y, you'd only go from 0 to 1 on the second integral if you were doing one (to find expected value of Y, probability, etc.). If Y has bounds defined by x, or vice versa, then i believe it can't be independent.

Since y has bounds defined by x, then it should have been a tip-off that they were dependent and there was more to the problem.... Thanks for the useful reply!

[joni308]

Originally Posted by vishusmartishus

f(x,y) = 15y, x^2<y<x.

What is fy(y)?

Well, the book is giving some ridiculous method and claims that the answer is 15y^(3/2)*(1-y)^1/2

Clearly x&y are independent so why is it not the case that f(x,y) = fy(y)?

How about you ease up on the usage of words such as 'obviously', 'clearly' etc until you actually get to that level. Feeling overconfident while there is no justification at all for that is wreckless and quite dangerous.

[vishusmartishus]

I was in a bad mood; could you go away forever? Thanks.