How to generally express a shifted PDF ?

[nikozm]

Hello,

i am trying to solve the following.

Given a general PDF (i.e., f_x(x), where x ≥ 0), how can i express the PDF of y = c x in terms of f_x(x)?

I think that goes like this: f_y(y) = f_x(y/c)/c, but i 'm not sure.

Any help would be useful.

Thanks in advance

 

[Office_Shredder]

nikozm, it's best to work with the CDF and then convert to the PDF. Let
F(x) = \int_{0}^{x} f_x(t) dt

Then
P(y<M) = P(cx < M) = P(x < M/c) = F(M/c)
So
P(y<M) = \int_{0}^{M/c} f_x(t) dt
To find the pdf you just need to do some manipulations to the integral so that you have an \int_{0}^{M}, alternatively you can differentiate with respect to M to get the PDF.