I'm trying to solve a joint distribution problem with $X$ and $Y$ being distributed uniformly on $[0,1.]$ Then, $U=X$ and $V=X+Y$.
I know that $f_{(U,V)}(u,v)= \text{Jacobian} \cdot f_X(u)\cdot f_Y(v-u)$
My question is how to evaluate $f_Y(v-u)$. For a normal, exponential, poisson distribution the intuition on where to plug in is clear. I'm guessing for the uniform, it doesn't matter what value is plugged into $f_Y$, because the pdf is flat?
EDIT: I've figured out the domain of the joint distribution, being $0{\leq}u{\leq}v{\leq}u+1{\leq}2$
This gives rise to a parallelogram on $(U,V)$ bounded by $0{\leq}v{\leq}2$ and $0{\leq}u{\leq}1$, and more specifically bounded by lines $v{\leq}u+1$ and $u{\leq}v$.
I worked out the Jacobian to be 1.
But still, for evaluating $f_Y(v-u)$ and $f_X(u)$, I don't see the connection. I know that in the end, $f_{(U,V)}(u,v)= 1$ on the domain I defined above, but still not quite sure how to get there.
In the problem statement, it seems that the independence of $X$ and $Y$ is assumed. Therefore, the joint density $f_{(X,Y)}$ of $(X,Y)$ is the product of the marginal densities $f_{X}$ and $f_{Y}$, i.e. \begin{aligned} f_{(X,Y)}(x,y) &= f_{X}(x)\, f_{Y}(y) \\ &= \boldsymbol{1}_{[0,1]}(x) \, \boldsymbol{1}_{[0,1]}(y) \, . \end{aligned} The random variable $(X,Y)$ is uniformly distributed over the square $[0,1]\times [0,1]$. Since the Jacobian of the transformation $(X,Y) \mapsto (X,X+Y)$ is equal to one, the change of variable formula writes \begin{aligned} f_{(U,V)}(u,v) &= f_{(X,Y)}(u,v-u)\\ &= \boldsymbol{1}_{[0,1]}(u) \, \boldsymbol{1}_{[0,1]}(v-u) \, . \end{aligned} The random variable $(U,V)$ is uniformly distributed over a parallelogram $(\mathcal {P})$, defined by $$ (\mathcal{P}):\qquad 0\leq u\leq 1 \quad\text{and} \quad u\leq v \leq 1+u \, . $$