Let f(x)= 2x , 0<x<1 , zero elsewhere, be the pdf of X.
Compute the cdf of Y=1/X
cdf of X = p(X< x)
The Attempt at a Solution
P(1/X <= y)
= P(X <= 1/y)
int 2x from 0 to 1/y
= x^2 eval from 0 to 1/y
so the cdf is 1/y^2 for 1<y<infinty
however i don't think this is right.... the text book answers state the cdf as 1-1/y^2
so im confused. thanks for any help!