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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 /> | ||
| + | <math> E(g(x))=\int g(x)f_{x}(x) dx</math> and <br /> | ||
| + | <math>var(x) = E(x^{2})-E(x)^{2}</math><br /> | ||
| + | Because integration is a linear operation you can split the integral into two parts, i.e.<br /> | ||
| + | <math> E(x)=\int_0^{0.5} 4x^{2}dx +\int_{0.5}^1 (4x-4x^{2}) dx </math> <br /><br /> | ||
| + | |||
| + | You can follow the same rules for finding the mean and variance of y. | ||
| + | --[[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)
