Example of 2 random variables s.t. $(X+Y)$ ~ $U(0,2)$

Question

In my book I found:

Can you give an example of 2 random variables $X,Y$ S.T $(X+Y)$ ~ $U(0,2)$ and $X,Y$ are not independent.

Any ideas of how I can find such 2 random variabes?

I would prefer if those random variables tell a story so I can relate to the real world, for example selecting number in $[0,1]$ has uniform disturbution of $(0,1)$.

Answer 1

Set $X \sim U(0,1)$ and let $Y = X$. Then I claim that $X+Y \sim U(0,2)$ and that $X$ and $Y$ are not independent.

If they were independent, $E(XY) = E(X)E(Y)$; instead, we have: $$E(XY) = E(X^2) = \int_0^1 x^2 dx = 1/3 \neq 1/4 = E(X)E(Y)$$

To see that $X+Y \sim U(0,2)$, notice that $$P(X+Y \leq z) = P(2X \leq z) = P(X \leq z/2) = \begin{cases} 0 & z < 0 \\ z/2 & 0 \leq z \leq 2 \\ 1 & z > 2 \end{cases}$$ which is the CDF of a $U(0,2)$ random variable.

Answer 2

Per Peter O's suggestion, I am posting an answer to tie together the comments posted below this question.

Quick/intuitive answer:

Define $Z:=X+Y$ and let $X=Y$ and $Y\sim U(0,1)$.

We can define two events $X \in B, Y \in A$ for $A,B \subset (0,1)$

If $X$ and $Y$ are independent then $P(X\in B | Y \in A) = P(X\in B) \;\;\forall A,B \subset (0,1)$

However, this is not true:

$$A\cap B = \emptyset \implies P(X\in B| Y \in A) = 0 \neq P(X\in B)=\text{leb}(B)$$

That addresses independence.

Forthe distribution, I've pointed out that $2X$ is simply a location-scale transformation of $U(0,1)$ and will result in stretching the dentisty to $U(0,2)$