# probability

This figure below describes the joint PDF of the random variables X and Y. These random variables take values over the unit disk. Over the upper half of the disk (i.e., when y>0), the value of the joint PDF is 3a, and over the lower half of the disk (i.e., when y<0), the value of the joint PDF is a (for some positive constant a which we will solve for in the problem).
In each part below, express your answer using standard notation . Enter 'pi' for π.
Helpful fact: the area of a circle of radius r is πr2.

1)Find a.
2)Compute E[XY].
Hint: once you setup your integral in terms of dx and dy, you may want to use polar coordinates (take x=rcos⁡(θ), y=rsin⁡(θ), and dxdy=rdrdθ). Alternatively, you can find the solution with very little calculation through a symmetry argument.
3)Compute the marginal PDFs fX(x) and fY(y).
If −1<x<1, fX(x)=
If −1<y<0, fY(y)=
If 0<y<1, fY(y)=
4)Compute the expected values E[X] and E[Y].
Hint: In evaluating integrals, it may be useful to write 2ydy as d(y2) and make a change of variables.
5)Compute E[X|Y>0] and var(X|Y>0).

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1. question 2 and the first part of 4 and 5 =0

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2. question 3:

If −1<x<1, fX(x)= 2/(pi) * sqrt(1-x^2)
If −1<y<0, fY(y)= 1/pi * sqrt(1 - y^2)
If 0<y<1, fY(y)= 3/pi * sqrt(1 - y^2)

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3. so what is the answer for 1) and 4b) + 5b)?Thanks for helping out

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4. 1: a=0.1592

4b: 0.2122

5b is a pain. Did not find the correct answer so far.

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5. For part 5 (though this only affects 5b), ensure that you have normalized the conditional PDF.

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6. 5b: 1/4 (one fourth) or 0.25 if you want

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