Let $f\left(x,y\right)=e^{-x}$ for $0<y<1$, $x>0$. We assume that $Z=X-Y$ and $V=X+Y$. What's $P\left(Z>1\:|\:V=2\right)$?
As far as I know:
$P\left(Z>1\:|\:V=2\right)=\frac{P\left(Z>1,\:V=2\right)}{P\left(V=2\right)}=\frac{P\left(X-Y>1,\:X+Y=2\right)}{P\left(X+Y=2\right)}=\frac{P\left(X-Y>1,\:Y=2-X\right)}{P\left(X=2-Y\right)}=\frac{P\left(X-\left(2-X\right)>1\right)}{P\left(1<X<2\right)}=\frac{P\left(X>\frac{3}{2}\right)}{P\left(1<X<2\right)}=\frac{1-P\left(X\le \frac{3}{2}\right)}{P\left(1<X<2\right)}$
$f_X\left(x\right)=e^{-x}$ which is easily checked.
$P\left(X\le \frac{3}{2}\right)=\int _0^{\frac{3}{2}}\:e^{-x}dx=\left(1-e^{\left(-\frac{3}{2}\right)}\right)$
$P\left(1<X<2\right)=\int _1^2\:e^{-x}dx=e^{-1}-e^{-2}$
Thus $\frac{e^{-\frac{3}{2}}}{e^{-1}-e^{-2}}=\frac{\sqrt{e}}{e-1}$
Is this correct?
First we notice: ${Z=X-Y, V=X+Y}$ iff ${X=(V+Z)/2, Y=(V-Z)/2}$
And the domain $D_{\small X,Y}=\{\langle x,y\rangle: 0<x, 0<y<1\}$ maps to $D_{\small V,Z}=\{\langle v, z\rangle: 0<v+z, 0<v-z<2\}$
Then we may find the conditional distribution of $V,Z$ using the Jacobian Transformation:
$$\begin{align}f_{\small V,Z}(v,z)&=f_{\small X,Y}((v+z)/2,(v-z)/2)\begin{Vmatrix} \dfrac{\partial((v+z)/2)}{\partial v} & \dfrac{\partial((v-z)/2)}{\partial v}\\\dfrac{\partial((v+z)/2)}{\partial z} & \dfrac{\partial((v-z)/2)}{\partial z}\end{Vmatrix}\\[2ex]&= \dfrac{\mathrm e^{-(v+z)/2}}8 \mathbf 1_{\langle v,z\rangle\in D_{\tiny V,Z}}\end{align}$$