Lawrence, you are my go-to whenever I get stuck. You are so much clearer than most of what's out there. You make this stuff seem like simple arithmetic. Thank you thank you!
This is a very good video. You proove that F_Y(y)=F_X(g^-1(y)) and then you derive F_Y(y) by dy and by chain rule you first get the pdf and you multiply it then by 1/g'(y) which is the derivative of g^-1(y).
Lawrence, you are my go-to whenever I get stuck. You are so much clearer than most of what's out there. You make this stuff seem like simple arithmetic. Thank you thank you!
Thank you. You saved me from the terrible textbook I have
Thank you, Laurence.
Very helpful, thank you.
This is a very good video. You proove that F_Y(y)=F_X(g^-1(y)) and then you derive F_Y(y) by dy and by chain rule you first get the pdf and you multiply it then by 1/g'(y) which is the derivative of g^-1(y).
Should the theorem say 1-1 (meaning injective) or bijective?
bijective, or else you won't have an inverse
what if the function is non-monotonic ? and what if it's the case of two to one?
cknudson.com/StudentWork/Transforms.pdf