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| + | Nope. You're integrating over the box bounded by (0,0), (0,1), (1,1), (1,0), and that's not quite the shaded region :) (So, it would be wrong to substitute <math>f_{xy}(x,y) = \frac{4}{3}</math>.) | ||
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| + | I would suggest splitting the double integral up. (Think of a double integral as a nested for loop -- integrating "slowly" over the outside loop and "quickly" over the inside loop :p) | ||
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| + | <math>\int_{0}^{0.5} \int_{0.5}^{1} xy f_{XY}(x,y) dy dx + \int_{0.5}^{1} \int_{0}^{1} xy f_{XY}(x,y) dy dx</math> | ||
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| + | The first term does the box bounded by (0,0.5), (0,1), (0.5,1), (0.5, 0.5). The second term does the box bounded by (0.5,0), (0.5,1), (1,1), (1,0). | ||
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| + | (Alternatively, you could do your integration and then subtract out the space that's not shaded afterward.) | ||
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| + | -Brian ([[User:Thomas34|Thomas34]] 22:32, 9 December 2008 (UTC)) | ||
Latest revision as of 17:33, 9 December 2008
$ {y}_{\rm LMMSE}(x)=E[\theta]+\frac{COV(x,\theta)}{Var(x)}*(x-E[x]) $
Question posed by Nicholas Browdues
It's true that
$ E[XY]= \iint\limits_D xy f_{xy}(x,y)\, dx\,dy= \frac{4}{3} \int_{0}^{1}y (\int_{0}^{1} dx) dy =\frac{1}{3} $
right?
Nope. You're integrating over the box bounded by (0,0), (0,1), (1,1), (1,0), and that's not quite the shaded region :) (So, it would be wrong to substitute $ f_{xy}(x,y) = \frac{4}{3} $.)
I would suggest splitting the double integral up. (Think of a double integral as a nested for loop -- integrating "slowly" over the outside loop and "quickly" over the inside loop :p)
$ \int_{0}^{0.5} \int_{0.5}^{1} xy f_{XY}(x,y) dy dx + \int_{0.5}^{1} \int_{0}^{1} xy f_{XY}(x,y) dy dx $
The first term does the box bounded by (0,0.5), (0,1), (0.5,1), (0.5, 0.5). The second term does the box bounded by (0.5,0), (0.5,1), (1,1), (1,0).
(Alternatively, you could do your integration and then subtract out the space that's not shaded afterward.)
-Brian (Thomas34 22:32, 9 December 2008 (UTC))
