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--[[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:<br /><br />
 
<math>f_{x,y}(x,y)=f_{x}(x)f_{y}(y)</math><br />
 
Using this identity we can see that X and Y are not independent because:<br /><br />
 
<math>f_{x}(x)f_{y}(y)=8x , 0\leq x \leq 0.5,  0\leq y \leq x</math><br />
 
<math>f_{x}(x)f_{y}(y)=8(1-x) , 0.5 \leq x \leq 1,  x \leq y \leq 1</math><br /><br />
 
--[[User:Kklacik|Kklacik]] 09:48, 2 April 2009 (UTC)
 
  
 
I thought that the part b is
 
I thought that the part b is
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part d is confused for me. Could anyone tell me how to do for part d?--[[User:Kim415|Kim415]] 03:54, 30 March 2009 (UTC)
 
part d is confused for me. Could anyone tell me how to do for part d?--[[User:Kim415|Kim415]] 03:54, 30 March 2009 (UTC)
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'''For part D''' you will just use the expected value functions for <math>E(x)</math> and <math>E(x^2)</math>  The formulas are as follows:<br /><br />
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<math> E(g(x))=\int g(x)f_{x}(x) dx</math> and <br />
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<math>var(x) = E(x^{2})-E(x)^{2}</math><br />
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Because integration is a linear operation you can split the integral into two parts, i.e.<br />
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<math> E(x)=\int_0^{0.5} 4x^{2}dx +\int_{0.5}^1 (4x-4x^{2}) dx </math>  <br /><br />
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You can follow the same rules for finding the mean and variance of y.
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--[[User:Kklacik|Kklacik]] 10:18, 2 April 2009 (UTC)

Latest revision as of 05:18, 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)


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)

For part D you will just use the expected value functions for $ E(x) $ and $ E(x^2) $ The formulas are as follows:

$ E(g(x))=\int g(x)f_{x}(x) dx $ and
$ var(x) = E(x^{2})-E(x)^{2} $
Because integration is a linear operation you can split the integral into two parts, i.e.
$ E(x)=\int_0^{0.5} 4x^{2}dx +\int_{0.5}^1 (4x-4x^{2}) dx $

You can follow the same rules for finding the mean and variance of y. --Kklacik 10:18, 2 April 2009 (UTC)

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