Let $X$ ~ $Uniform[L, R]$. Let $Y = cX + d$, where $c > 0$. Prove that $Y$ ~ Uniform$[cL + d, cR + d]$.

Question

Let $X$ ~ $Uniform[L, R]$. Let $Y = cX + d$, where $c > 0$. Prove that $Y$ ~ Uniform$[cL + d, cR + d]$.

let $h(x) = cx + d$, then $h'(x) = c$ and $h^{-1}(y) = \frac{y-d}{c}$.

Since h is strictly increasing $$f_Y(y) = \frac{f_x(\frac{y-d}{c})}{c}$$

$$= \frac{1}{(R-L)c}\text{ How did they get this part <- }$$

whenever $L \leq \frac{y-d}{c} \leq R$ which is equal to $cL + d \leq y \leq cR+d$ or $0$ otherwise.

Could someone tell me how they got $\frac{1}{(R-L)c}$?

Answer 1

$$F(y)=\mathbb P\{Y\leq y\}=\mathbb P\left\{X\leq \frac{y-d}{c}\right\}=\frac{1}{L-R}\int_L^{(y-d)/c}1_{[R,L]}(x)dx.$$

Therefore, $$f_Y(x)=\frac{1}{c(L-R)}\boldsymbol 1_{[R,L]}\left(\frac{y-d}{c}\right)=\frac{1}{(cL+d)-(cR+D)}\boldsymbol 1_{[cR+d,cL+d]}(y).$$