Cdf of continous RV transformation

[Laura1321412]

Homework Statement

Let f(x)= 2x , 0<x<1 , zero elsewhere, be the pdf of X.
Compute the cdf of Y=1/X

Homework Equations

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
= 1/y^2

so the cdf is 1/y^2 for 1<y<infinty

however i don't think this is right... the textbook answers state the cdf as 1-1/y^2


so I am confused. thanks for any help!

 

[Ray Vickson]

Since 2 < 4, does this imply that 1/2 < 1/4?

RGV