(Brian Thomas rhea edit hw10 (sorry if this is late :) )) |
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I'm not sure if this is right | I'm not sure if this is right | ||
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| + | I agree with your proposal, Arie. Let's finish this section of the problem: | ||
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| + | <math>f_{Y|X}(y|a)= \frac {f_{X,Y}(a,y)}{f_X(a)}</math> | ||
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| + | We know <math>f_X (a) = \frac{2}{3}</math> from part a, and from the graph, <math>f_{XY} (a,y) = \{\frac{4}{3}, \ \ 0.5 \leq y \leq 1; \ \ \ 0, \ \ \text{else}\}</math>. | ||
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| + | So, dividing, <math>f_{Y|X}(y|a)= = \{2, \ \ 0.5 \leq y \leq 1; \ \ \ 0, \ \ \text{else}\}</math>. | ||
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| + | <math>E(Y|X=a) = \frac{3}{4}</math>, and <math>Var(Y|X=a) = \frac{1}{48}</math>, since <math>f_{Y|X}(y|a) \frac{}{}</math> has a uniform PDF. | ||
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| + | -Brian ([[User:Thomas34|Thomas34]] 00:23, 9 December 2008 (UTC)) | ||
Revision as of 19:23, 8 December 2008
$ f_{y|x}(y|a)= \frac {f_{X,Y}(x,y)}{f_X(a)}\qquad \mathrm{Where}\ f_X(a) = f_X(x)\ \mathrm{for}\ 0<x<0.5 $
I'm not sure if this is right
I agree with your proposal, Arie. Let's finish this section of the problem:
$ f_{Y|X}(y|a)= \frac {f_{X,Y}(a,y)}{f_X(a)} $
We know $ f_X (a) = \frac{2}{3} $ from part a, and from the graph, $ f_{XY} (a,y) = \{\frac{4}{3}, \ \ 0.5 \leq y \leq 1; \ \ \ 0, \ \ \text{else}\} $.
So, dividing, $ f_{Y|X}(y|a)= = \{2, \ \ 0.5 \leq y \leq 1; \ \ \ 0, \ \ \text{else}\} $.
$ E(Y|X=a) = \frac{3}{4} $, and $ Var(Y|X=a) = \frac{1}{48} $, since $ f_{Y|X}(y|a) \frac{}{} $ has a uniform PDF.
-Brian (Thomas34 00:23, 9 December 2008 (UTC))
