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Transformation of univariate random variables Watch

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    On example 2.3, shouldn't the region be divided into three intervals since it is one to one at these three intervals. These are (0,pi/2] u (pi/2,3pi/2] u (3pi/2,2pi] which would give that fY(y)= 3/[2pi(1-y^2)^1/2] for y is an element of the interval (-1,1)

    However since the region is divided into two intervals in the solution, they get that fY(y)=1/[pi(1-y^2)^1/2]
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    (Original post by Merdan)
    On example 2.3, shouldn't the region be divided into three intervals since it is one to one at these three intervals. These are (0,pi/2] u (pi/2,3pi/2] u (3pi/2,2pi] which would give that fY(y)= 3/[2pi(1-y^2)^1/2] for y is an element of the interval (-1,1)

    However since the region is divided into two intervals in the solution, they get that fY(y)=1/[pi(1-y^2)^1/2]
    It looks wrong, although I'd like to see a bit more of the context to be sure (I see them putting in a y>0 comment but without any explanation), also I'm not sure where all this \sqrt{1-y^2} stuff is coming from (but I'm sure it's becuase of the lack of context).
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    Yeah, I think it's wrong (although the error is a constant, so won't affect the PDF). I'm a little too tired/drunk/unmotivated to work out if there's an issue with different behaviour between y < 0 and y > 0 (although I'm sure they're right that a symmetry argument holds for the PDF).
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    (Original post by DFranklin)
    It looks wrong, although I'd like to see a bit more of the context to be sure (I see them putting in a y>0 comment but without any explanation), also I'm not sure where all this \sqrt{1-y^2} stuff is coming from (but I'm sure it's becuase of the lack of context).
    that is the next page of solution. the sqrt(1-y^2) comes from differentiating inverse of sine function as Y=sinX or using their method of cdf...
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