If $X$ is a continuous random variable uniformly distributed over $[a,b]$, then is $Y=2-4X$ uniformly distributed over $[c,d]$? Why?

Question

I ran into this problem solving one of the problems on my course and if I knew that this applies and how to simply prove it, it would help me a great lot.

Answer 1

Yes, it is. For a formal proof you could apply integration by substitution, e.g., in case $Y=4X$

$$P[x_0\le Y\le x_1]=P[\frac{x_0}{4}\le X\le \frac{x_1}{4}]=\int_{x_0/4}^{x_1/4}\frac{1}{b-a}\,dx=\int_{x_0}^{x_1} \frac{1}{b-a}\frac{1}{4}\,dx$$

Then you see that $4X$ is uniformly distributed on $[4a,4b]$. I think you can solve the more general case on your own.

Answer 2

Using theorem for transformation variable, it will hold relation $$ g_Y(y)=f_X(x)\cdot|J|, $$ where $J$ is Jacobian. Since it only one variable, then the Jacobian is $$ J=\frac{dx}{dy}. $$ We have $X\sim\mathcal{U}(a,b)$, then the pdf of $X$ is $$ f_X(x)=\frac1{b-a}. $$ The transformation is $Y=2-4X\;\Rightarrow\;X=\dfrac{2-Y}{4}$, then the Jacobian is $$ J=\frac{d}{dy}\left(\dfrac{2-y}{4}\right)=-\frac14. $$ Hence, the pdf of $Y$ is $$ g_Y(y)=f_X(x)\cdot|J|=\frac1{b-a}\cdot\left|-\frac14\right|=\frac1{4b-4a}. $$ Now, we determine the region of $y$. The region of $a\le x\le b$ is corresponding to the region $a\le \dfrac{2-y}{4}\le b\;\Rightarrow\; 2-4b\le y\le2-4a$. Therefore, it can be concluded that $Y\sim\mathcal{U}(2-4b,2-4a)$. If we let $c=2-4b$ and $d=2-4a$, then $Y\sim\mathcal{U}(c,d)$.