Where does the second line "p(r,theta)d(r)d(theta)..." come from?
kc1
@ ChrisZzh I agree that there's some explanation missing here. p(r,theta) is the probability density function. If we integrate this over the circle it should equal 1. Equating the LHS (Left-hand side) to RHS in line 2 is just saying exactly this. Note the RHS integrated over unit circle is equal to 1.
However, usually the integral p(r,theta)d(r)d(theta) on the LHS should have an extra r term in there ie. p(r,theta) r d(r)d(theta) since integrating over a circle we have dA = r dr dtheta. There looks to be some explanation missing/being swept under the rug as to why this r term is being moved into p(r,theta). I think it is being done because we are trying to figure out how to draw random samples from r and theta variables agnostic of geometry. So we integrate over the variables r and theta agnostic of geometric meaning ie. integrate over dr dtheta instead of dA = r d(r)d(theta).
But just my thoughts, would need insight from prof.
ChrisZzh
@kc1 Thanks for this great stuff! I think the reason why we don't have an r term on the left-hand side is that we are integrating a probability density function. If we just ignore the context here and treat r and theta as two random variables without special meanings, this LHS is how the integration of a pdf looks like.
The one thing I'm not sure about is: since the integral of RHS equals to 1, LHS and RHS should be equal?
cou
I think it would be good to mention that the styled capital E is a random number from [0,1], since it is not clear looking from the slides
Where does the second line "p(r,theta)d(r)d(theta)..." come from?
@ ChrisZzh I agree that there's some explanation missing here. p(r,theta) is the probability density function. If we integrate this over the circle it should equal 1. Equating the LHS (Left-hand side) to RHS in line 2 is just saying exactly this. Note the RHS integrated over unit circle is equal to 1.
However, usually the integral p(r,theta)d(r)d(theta) on the LHS should have an extra r term in there ie. p(r,theta) r d(r)d(theta) since integrating over a circle we have dA = r dr dtheta. There looks to be some explanation missing/being swept under the rug as to why this r term is being moved into p(r,theta). I think it is being done because we are trying to figure out how to draw random samples from r and theta variables agnostic of geometry. So we integrate over the variables r and theta agnostic of geometric meaning ie. integrate over dr dtheta instead of dA = r d(r)d(theta).
But just my thoughts, would need insight from prof.
@kc1 Thanks for this great stuff! I think the reason why we don't have an r term on the left-hand side is that we are integrating a probability density function. If we just ignore the context here and treat r and theta as two random variables without special meanings, this LHS is how the integration of a pdf looks like.
The one thing I'm not sure about is: since the integral of RHS equals to 1, LHS and RHS should be equal?
I think it would be good to mention that the styled capital E is a random number from [0,1], since it is not clear looking from the slides