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#1
08-03-2011, 11:58 AM
 vishusmartishus Member Join Date: Jun 2011 College: Capital University Posts: 77
Actex 8.12

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)?
#2
08-03-2011, 12:02 PM
 Kaner3339 Member SOA Join Date: Jun 2011 Location: Canada Studying for MLC & MFE Posts: 907

draw out x^2= y and y = x. Integrate the joint density for all x. it's really not that ridiculous.
#3
08-03-2011, 12:04 PM
 Actuarialsuck Member Join Date: Sep 2007 Posts: 5,324

Quote:
 Originally Posted by vishusmartishus f(x,y) = 15y, x^2
Wait, why is X and Y ind? Clearly x^2 < y < x would suggest otherwise...
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#4
08-03-2011, 12:08 PM
 vishusmartishus Member Join Date: Jun 2011 College: Capital University Posts: 77

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.
#5
08-03-2011, 12:12 PM
 Kaner3339 Member SOA Join Date: Jun 2011 Location: Canada Studying for MLC & MFE Posts: 907

if you draw it out, x doesn't go from 0 to 1. x is stuck between x = y^1/2 and x = y.
#6
08-03-2011, 12:19 PM
 vishusmartishus Member Join Date: Jun 2011 College: Capital University Posts: 77

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
#7
08-03-2011, 01:15 PM
 alexsadowski Member SOA Join Date: Jun 2011 Studying for Exam 3/MLC College: Auburn University Favorite beer: Leinenkugel Posts: 85

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.
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#8
08-03-2011, 01:21 PM
 vishusmartishus Member Join Date: Jun 2011 College: Capital University Posts: 77

Quote:
 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!
#9
08-03-2011, 01:31 PM
 joni308 Member CAS SOA Join Date: Jun 2011 Posts: 2,122

Quote:
 Originally Posted by vishusmartishus f(x,y) = 15y, x^2
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.
#10
08-03-2011, 01:37 PM
 vishusmartishus Member Join Date: Jun 2011 College: Capital University Posts: 77

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

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