I am supposed to find $$f_{X \mid X+Y=c}, \quad \mbox{ for }c>0$$ given that $X,Y \sim \exp(1)$ and independent. I have worked back and forth with convolution formula and Bayes theorem but without any progress.
We are looking for ; $\frac {f_{X,X+Y}(x)}{f_{X+Y}(c)}$ given $f_{X}=e^{-x} \chi_{\mathbb{R}_{+}}(x)$ and $f_{X+Y}(c)$ = $\int_{\mathbb{R}}e^{-x}\chi_{\mathbb{R}_{+}}(x)e^{-c+x}\chi_{\mathbb{R}_{+}}(c-x)dx$
I cant make sense of either numerator or denominator..
Let $U=X$ and $V=X+Y$ with $0< U< V$. Note that $V\sim \mbox{Erlang}(2,1)$ as sum of two iid $\exp(1)$ random variables. Then $X=U$ and $Y=V-U$ and $$J=\begin{vmatrix}\frac{\partial x}{\partial u}&\frac{\partial x}{\partial v}\\\frac{\partial y}{\partial u}&\frac{\partial y}{\partial v}\end{vmatrix}=\begin{vmatrix}1&0\\-1&1\end{vmatrix}=1$$ Hence $$f_{U,V}(u,v)=f_{X,Y}(u,v-u)\cdot1=e^{-u}e^{-(v-u)}=e^{-v}$$ with $0< u<v$. Thus $$f_{X\mid X+Y=c}=f_{U\mid V=c}(u\mid c)=\frac{f_{U,V}(u,c)}{f_V(c)}=\frac{e^{-c}}{ce^{-c}}=\frac{1}{c}$$ for all $0<u<c$. Hence $X\mid X+Y=c \sim \mbox{Uniform}(0,c)$.