1. Thanks, yeKcim. I can't draw to save my life

2. Originally Posted by bluesblues
thanks this was the reasoning i was trying to do in my head without going over old noted it seems to be correct what you have put
although should not the line be y=x nvm,,,but having done this integration a hundred times i still keep getting ... as an answer

$\displaystyle \int_o^\infty \int_0^x \frac{1}{y}exp(-\frac{x}{y})exp(-y)dy dx =-1$

hehehehe

line $\displaystyle x=y$ is the same as line $\displaystyle y=x$ (you know that but i'm just pointing anyway )

as for integral ... put some work that you have done... and we'll see where is the problem

P.S. your result should be $\displaystyle \displaystyle \frac {e-1}{e} = 1 - \frac {1}{e} = 0.63212$

3. really confused actually,,,,what determines the order of integration i would want to start with,,oh one sec ill see this last post.(ignore) order is naturally x first

4. yeKciM THANKYOU but are you 100% positive this is correct answer,,,i just want like one other peron to confirm it. i fully understand now that my original thinking is correct due to the diagram drawn above everything makes sense i think,,,,although still trying to do the other question i found

5. Originally Posted by bluesblues
really confused actually,,,,what determines the order of integration i would want to start with,,oh one sec ill see this last post.
let's go like this

$\displaystyle \displaystyle \int _0 ^{\infty} dy \int _0 ^y \frac {e^{-{\frac {x}{y}-y}}}{y}\; dx$

and solve first :

$\displaystyle \displaystyle \int _0 ^y \frac {e^{-{\frac {x}{y}-y}}}{y}\; dx = (e-1)e^{-y-1}$

after that it's much easier

6. rv's have joind density fn exp(-x-y) for x,y>0

find p(x>y) answer given in 1/2 : O

using same method you dont get 1/2 :s

7. yes u'll get the 1/2... go and check it out there (on that thread)

just pay attention to the how u can write down "-e^(-x-y)" and you will have no problems

8. btw i lost my password y day,,,,,,,,,,,,,,,,,,,,created new account,,,,ive confused myslef again,,i cant believe it,,why did we choose to put dydx and then do the dx integral first i thought you cant do that? is the reason we put dy first because x>y?????

furthermore by doing the x integral first do we not have to change the limits in which case it doesnt work youd get x from y to infinity and y from 0 to infinty thanks?

]i.e you cannot do dx first when dy came fist$\displaystyle \int^{\infty}_{0} \int^{x}_{0}\frac {e^{-{\frac {x}{y}-y}}}{y} dy dx = \int _0 ^{\infty} dy \int _0 ^y \frac {e^{-{\frac {x}{y}-y}}}{y}\; dx$

9. if y>x i can understand y=$\displaystyle \infty$ but this is x>y so it has to be x=$\displaystyle \infty$ can someone please explain

10. can someone confirm the answer is simply $\displaystyle \int_0^\infty \int_o^y f(x,y) dx dy$ with that working and there is no reason to put dy before dx

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