Find the marginal PDF

Question

I am having a hard time trying to find the marginal of the following joint pdf.

$$f_{U,V}(u,v) =2 \frac{n!}{\left( \frac{n}{2}-1 \right)!\left( \frac{n}{2}-1 \right)!}\left[(u-\theta)(\theta + 1-2v+u)\right]^{\frac{n}{2}-1}$$ where $ u<v<\frac{u+\theta+1}{2}$ and $\theta < u <\theta+1$.

This was my attempt

$\begin{align*} f_V (v) &= \int_{\theta}^{\theta + 1} \frac{n!}{\left( \frac{n}{2}-1 \right)!\left( \frac{n}{2}-1 \right)!}\left[(u-\theta)(\theta + 1-2v+u)\right]^{\frac{n}{2}-1} \cdot 2 du\\ &= \frac{n!}{\left( \frac{n}{2}-1 \right)!\left( \frac{n}{2}-1 \right)!}\int_{\theta}^{\theta + 1}\left[(u-\theta)(\theta + 1-2v+u)\right]^{\frac{n}{2}-1} \cdot 2 du\\ \end{align*}$

Answer 1

Your support is: $\theta\lt u< \theta +1$ and $u< v< (u+\theta+1)/2$.

That is equivalently: $\theta\leq v<\theta+1$ and $\max(\theta,2v-\theta-1)\leq u<v$

Answer 2

The given joint density looks like some joint density of some order statistic but is is clearly not a nice density.

To realize that $f_{UV}$ is wrong just set $n=2$ and $\theta=1$ getting

$$f_{UV}(u,v)=2\cdot\mathbb{1}_{(1;2)}(u)\cdot\mathbb{1}_{(u;1+u/2)}(v)$$

Where the integral over all the support is $2\cdot 0.25=0.5\ne 1$