Consider the following bivariate distribution
$$ f(x,y) = \begin{cases} \frac{3}{28}(x^2 + xy) \quad &\text{for } 0 \leq x \leq 2 \text{ and } 0 \leq y \leq 2 \\ 0 &\text{otherwise} \end{cases} $$
and the following two new random variables
$$ U = X + Y, \quad V = X $$
Find the new bivariate distribution of $U$ and $V$, and find the marginal distribution of $U = X + Y$.
I'm having some trouble figuring out the second half of the question, and I would appreciate any help I could get. I think I've managed to find the new bivariate distribution. Here's what I've done so far:
First I express the original random variables in terms of the new ones
$$ x = v_1(u,v) = v, \quad y = v_2(u,v) = u - v $$
which can be inserted into the following formula to find the new bivariate distribution:
$$ g(u,v) = f(v_1(u,v),v_2(u,v)) \cdot |\det(M)| $$
$$ \det(M) = \frac{\partial v}{\partial u} \cdot \frac{\partial (u - v)}{\partial v} - \frac{\partial v}{\partial v} \cdot \frac{\partial (u - v)}{\partial u} = -1 $$
Before putting all of this together I also need to find the new region of positive density. What I found was
$$ 0 \leq x \leq 2 \implies 0 \leq v \leq 2 \\ 0 \leq y \leq 2 \implies 0 \leq u - v \leq 2 \implies v \leq u \leq v + 2 $$
and putting all of this together yields:
$$ g(u,v) = \begin{cases} \frac{3}{28} uv \quad &\text{for } v \leq u \leq v + 2 \text{ and } 0 \leq v \leq 2 \\ 0 &\text{otherwise} \end{cases} $$
As far as I understand, to find the marginal distribution of $U$ I need to evaluate
$$ f_U(u) = \int_{-\infty}^{\infty} f(u,v) dv $$
over the following region of positive density. This this looks very different to the other examples I've seen in my book, and I'm not sure how to proceed here.
Let's begin with
$$f_U(u)=\int_0^u \frac{3}{28}uv dv=\frac{3}{56}u^3\mathbb{1}_{[0;2)}(u)$$
$$f_U(u)=\int_{u-2}^2 \frac{3}{28}uv dv=\frac{3(4-u)u^2}{56}\mathbb{1}_{[2;4]}(u)$$
So the marginal U has the following density
$$f_U(u)=\frac{3}{56}u^3\mathbb{1}_{[0;2)}(u)+\frac{3(4-u)u^2} {56}\mathbb{1}_{[2;4]}(u)$$
The marginal V is easier
$$f_V(v)=\int_v^{v+2} \frac{3}{28}uv du=\frac{3v(v+1)}{56}\mathbb{1}_{[0;2]}(v)$$
This because you have
$0\leq u-v \leq 2$
with $u \in [0;4]$ and $v \in [0;2]$
So when $u \in [0;2]$ it is necessary that $v<u$ that means to integrate in $int_0^u dv$ and so on...
I other words, if you draw the two lines implied by $0\leq u-v \leq 2$ you get this graph