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# Transformation of univariate random variables watch

1. 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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2. (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 stuff is coming from (but I'm sure it's becuase of the lack of context).
3. 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).
4. (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 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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