How to show that $U_{2} $ is a uniform random variable?

Question

What I have done so far: $P(U_{2}<x)=P(U_{1}>1-x)=1-x$ I am not sure how I would proceed to conclude it is a uniform random variable from here?

Any help would be much appreciated!

Answer 1

$\displaystyle P(U_{2}<x)=P(1-U_{1}\le x)=P(U_{1}\ge 1-x)=\int_{1-x}^{1}1.dx=x$

$\Rightarrow P(U_{2}\le x)=x\Rightarrow F_{U_2}(x)=x \Rightarrow f_{U_2}(x)=1$

If you want to check the limits , Put $ U_2=1-0=1(@U_1=0) \text{ And } U_2=1-1=0 (@U_1=1$)

Answer 2

Hint

Let's call $F_1(x)$ and $F_2(x)$ the accumulated distribuiton for $U_1$ and $U_2$. So,

$$F_1(x)=P(U_1<x)=P(U_2>1-x)=1-P(U_2<1-x)=1-F_2(1-x)$$

The density function is then:

$$f_1(x)=\frac{dF_1(x)}{dx}=\frac{dF_2(1-x)}{dx}=f_2(1-x)$$

Can you finish?