Line 17: Line 17:
 
Can anyone verify this?
 
Can anyone verify this?
 
--[[User:Leedj|Leedj]] 21:46, 29 March 2009 (UTC)
 
--[[User:Leedj|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:
 +
<math>f_{x,y}(x,y)=f_{x}(x)*f_{y}(y)</math>
 +
Using this identity we can see that X and Y are not independent because:
 +
<math>f_{x}(x)*f_{y}(y)=8*x , 0\leq x \leq 0.5,  0\leq y \leq x</math>
 +
<math>f_{x}(x)*f_{y}(y)=8*(1-x) , 0.5 \leq x \leq 1,  x \leq y \leq 1</math>
 +
--[[User:Kklacik|Kklacik]] 09:48, 2 April 2009 (UTC)
  
 
I thought that the part b is
 
I thought that the part b is

Revision as of 05:48, 2 April 2009



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)=8*x , 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)

Alumni Liaison

Questions/answers with a recent ECE grad

Ryne Rayburn