I want to prove that if X is a random variable with the uniform distribution over [L, R] and Y = cX + d with c > 0, then the uniform distribution of Y is over the interval [cL + d, cR + d].

I'm using the following theorem:

$ f_Y(y) = \frac {f_X(h^{-1}(y))}{|(h'(h^{-1}(y))|} $

and I get:

$ f_Y(y) = \frac {f_X(\frac{y-d}{c})}{c} $, (*)

but I think I should be getting

$ f_Y(y) = \frac {f_X(\frac{y-d}{c})}{(R-L)c} $ (**)

to make my proof work. How do I reach (**) instead of (*)?


(*) is correct, but you need to go one more step. Since, $f_X(x) = \dfrac 1{R-L}\;\mathbf 1_{x\in[L;R]}$

$$\begin{align}f_Y(y) & = \frac {f_X(\frac{y-d}{c})}{c} \\[2ex] & = \frac{1}{c(R-L)}\;\mathbf 1_{y\in[cL+d;cR+d]}\end{align}$$

That is all.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.