Let $X$, and $Y$ be two i.i.d exponential r.v's with rate $1$. I'm supposed to find the distribution of $Y|X+Y = c$, where $c>0$ is some constant. I can prove this by using properties of arrival times of the Poisson process, but not by just using standard conditional probability tricks (if any?!).
This has been my approach so far, but I don't like it much because it looks handwavy and I needed to draw the boundaries on paper:
Given the conditions the joint pdf of $X$, and $Y$ is $f_{XY}(x,y)=e^{-(x+x)}$ and we have $$P(X<x|X+Y=c) = \lim_{\delta\rightarrow 0}\frac{P(X<x,c\leq X+Y<c+\delta)}{P(c\leq X+Y<c+\delta)} $$ $$= \lim_{\delta\rightarrow 0}\frac{\int_0^x\int_c^{c+\sqrt{2}\delta}e^{-(u+v)}dudv}{\int_0^{c-\sqrt{2}\delta}\int_c^{c+\sqrt{2}\delta}e^{-(u+v)}dudv+o(\delta)} $$ $$=\lim_{\delta\rightarrow0}(\text{calculate the integrals})= \frac{x}{c}$$
Is there a more elegant way to do this? There must be..
You can compute the joint pdf of $X$ and $Z=X+Y$ as $$ f_{X,Z}(x,z) =f_{X,Y}(x,z-x) = e^{-z}I_{x<z}$$ (where the Jacobian factor is just one).
Then the pdf for $Z$ is $$ f_Z(z)=\int_0^z e^{-z}dx = ze^{-z}$$ (which should probably be familiar... the sum of two exponentials is a Gamma). So the conditional is $$ f_{X\mid Z}(x\mid z) = \frac{f_{X,Z}(x,z)}{f_Z(z)}=\frac{I_{x<z}}{z}.$$ In other words, $X$ is conditionally uniformly distributed between $0$ and $z.$ (Or $0$ and $c$ in your notation.)