Revision as of 04:49, 2 April 2009 by Kklacik (Talk | contribs)



For part b, I am getting
$ f_{x}(x)=\int_{0}^x 4 dy = 4x, 0\leq x \leq 0.5 $
$ f_{x}(x)=\int_{x}^1 4 dy = 4 - 4x, 0.5\leq x \leq 1.0 $
$ f_{y}(y)=\int_{y}^{0.5} 4 dx = 2 - 4y, 0\leq y \leq 0.5 $
$ f_{y}(y)=\int_{0.5}^y 4 dx = 4y - 2, 0.5\leq y \leq 1 $

For part c, they are not independent.
Can anyone verify this? --Leedj 21:46, 29 March 2009 (UTC)

We can tell if two random variables are independent by looking at their marginal density functions and joint density functions. If they are independent:

$ f_{x,y}(x,y)=f_{x}(x)f_{y}(y) $
Using this identity we can see that X and Y are not independent because:

$ f_{x}(x)f_{y}(y)=8x , 0\leq x \leq 0.5, 0\leq y \leq x $
$ f_{x}(x)f_{y}(y)=8(1-x) , 0.5 \leq x \leq 1, x \leq y \leq 1 $

--Kklacik 09:48, 2 April 2009 (UTC)

I thought that the part b is

$ f_{x}(x)=\int_{0}^x 4 dy = 4x, 0\leq x \leq 0.5 $
$ f_{x}(x)=\int_{x}^1 4 dy = 4 - 4x, 0.5\leq x \leq 1.0 $
$ f_{y}(y)=\int_{0}^{0.5} 4 dx = 2, 0\leq y \leq x $
$ f_{y}(y)=\int_{0.5}^{1} 4 dx = 2, x\leq y \leq 1 $

and part c, they are not independent because

$ f_{xy}(xy) $ is not equal to $ f_{y}(y)*f_{x}(x). $

part d is confused for me. Could anyone tell me how to do for part d?--Kim415 03:54, 30 March 2009 (UTC)

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