# How to generally express a shifted PDF ?

1. Dec 16, 2013

### nikozm

Hello,

i am trying to solve the following.

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

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

Any help would be useful.

2. Dec 16, 2013

### Office_Shredder

Staff Emeritus
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.