What is the PDF for this uniform distribution? Suppose we wish to select $X$ and $Y$ uniformly such that $X+Y=1$ and $X,Y \in (0,1)$
I know the following integration result:
$\int_{S_K}\Pi_{i=1}^n\theta_i^{a_i-1}d\theta_1d\theta_2...d\theta_n = \frac{\Pi_{i=1}^n\Gamma(a_i)}{\Gamma(\sum\limits_{i=1}^na_i)}$
Where $S_k = \{ (\theta_1,\theta_2,...,\theta_n) \in (0,1)^n : \sum\limits_{i=1}^n\theta_i=1\}$
Hence taking $a_i = 1$ for all $i$ yields $\int_{S_K}d\theta_1d\theta_2...d\theta_n = \frac{\Pi_{i=1}^n\Gamma(1)}{\Gamma(n)} = \frac{1}{(n-1)!}$
And so our PDF should be: $f_{X,Y}(x,y) = \begin{cases} 
      1 & (x,y) \in S_2 \\
     0 & o.w \\ 
   \end{cases}
$
However the age old rule about uniform pdfs is just $\frac{1}{area}$ and looking at the following graph: 
We see the area/length is just $\sqrt2$ not $1$ ?
And so the pdf should just be $f_{X,Y}(x,y) = \begin{cases} 
      \frac{1}{\sqrt2} & (x,y) \in S_2 \\
     0 & o.w \\ 
   \end{cases}
$
Which of these (if any) answers is correct?
 A: There seems to be some confusion as to what is meant by "select $X$ and $Y$ uniformly such that $X+Y=1$ and $X,Y \in (0,1)$." I am interpreting this as: select $X$ and $Y$ such that the vector $(X,Y)$ has a uniform distribution on the set $S_2 := \{ (x,y) \in (0,1) \times (0,1) \: | \: x+y = 1 \}$.
This begs the question - how does one define a uniform distribution on a given set? And is it even possible to give any set a uniform distribution? If your definition of uniform distribution is "a distribution whose PDF is constant", then it is not possible to create a uniform distribution on $S_2$, this is because
$$\int_{S_2} c \: dx dy = c \cdot \text{area}(S_2) = 0$$
for any constant $c$. So both of your suggested PDF's are incorrect.
A possible way to work around this is to notice that $S_2$ is contained in a 1-dimensional affine subspace of $\mathbb{R}^2$, so we should really consider $S_2$ as a subspace of $\mathbb{R}$ rather than $\mathbb{R}^2$. We can do this by considering the isometry
\begin{align*}
f:(0,\sqrt{2}) &\rightarrow S_2 \\
  x &\mapsto (\frac{x}{\sqrt{2}} , 1- \frac{x}{\sqrt{2}})
\end{align*}
Using this mapping it is natural to say that $(X,Y)$ is uniform on $S_2$ if and only if $f^{-1}((X,Y)) = \sqrt{2}X$ has a uniform distribution on $(0,\sqrt{2})$, which is equvalent to saying that $X \sim \operatorname{Uniform}(0,1)$.
Note also that this gives an explicit way to sample such a random variable. Sample $X\sim \operatorname{Uniform}(0,1)$ and let $Y=1-X$.
A: You can replace variables $X$ and $Y$ with variables $U = X+Y$ and $V= X-Y$.
Then find the conditional distribution $$f_V|U(v|u) = \frac{f_{U,V}(u,v)}{f_U(u)}$$ and plug in $u=1$ to get the density you want.
