Suppose X,Y are both Uniform on [0,1] and indpendent. I am interested in the distribution of X/Y conditional on X+Y=constant. (Obviously, the constant must be <2.)
My hunch is that that distribution is essentially the same as the unconditional distribution of X/Y, but I have a hard time proving it.
On
Truncated of uniform is uniform so, $(X,Y)|X+Y<1$ is a uniform vector in the triangle of vertices $(0,0)$, $(0,1)$ and $(1,0)$.
Let $W=\frac{X}{Y}$, we want to compute its conditional cdf: $$\mathbf{P}(W\leq w|X+Y<1)=\mathbf{P}(X\leq wY|X+Y<1)$$
For $w<0$ this probability is zero. For $w>0$, if you draw the areas you can compute the probability with the area ratio; i.e. $$\mathbf{P}(W\leq w|X+Y<1)=\frac{w}{1+w}\mathbf{1}\{w>0\}$$
So the pdf is
$$f_{W|X+Y<1}(w)=\frac{1}{(1+w)^2}\mathbf{1}\{w>0\}$$
I don't think they are identically distributed, so it's a good thing you're having trouble proving your conjecture. :)
Consider the full, x-y domain, the unit square in the x-y plane with each point equally likely (uniform density). Your conditional then splits this domain along the diagonal from (0,1) to (1,0).
When you condition your distribution of X/Y on the lower triangle (x+y<1), you will generate large values of X/Y all along (near) the x-axis (when y is small, x/y blows up).
When you look at the upper triangle that you see you have a much lower density of these points that generate large values: just in the corner near (1,0) as opposed to all along the x-axis.
So, to summarize, your conditional (x+y<1) probability will have a distribution more heavily weighted to higher values than the overall distribution (which includes the upper triangular domain where the density is weighted to lower values of x/y).