Difference between revisions of "Question 7.4" - Rhea

Revision as of 06:17, 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.