How to Obtain the conditional random variable with inequality in the conditional random variable Suppose X,Y are exponentially distributed 
I am trying to find the distribution of $X|Y>X$. I know how to find the distribution of a random variable as well as conditional random variable bit but this time, the conditional random variable is an inequality, so how should I do?
 A: How should you do? First, you should add the hypothesis you forgot, that $X$ and $Y$ are independent. Second, you should compute $u(x)=P[X\gt x\mid Y\gt X]$ for every $x\gt0$. This is a basic conditioning, no event of probability zero is involved, hence $u(x)$ is simply $u(x)=P[X\gt x,Y\gt X]/P[Y\gt X]$. Then the distribution you are looking for has  CDF $F(x)=1-u(x)$, thus $u$ gives you $F$, and you can deduce the PDF $f$ if you like.
Edit: To compute $v(x)=P[X\gt x,Y\gt X]$, note that, by definition of the distribution of $(X,Y)$,
$$
v(x)=\int_x^\infty\int_z^\infty f(z,y)\mathrm dy\mathrm dz,
$$
where $f$ is the density of the distribution of $(X,Y)$, and that, in the present case, one is given
$$
f(z,y)=\lambda^2\mathrm e^{-\lambda z}\mathrm e^{-\lambda y}\mathbf 1_{z\gt0}\mathbf 1_{y\gt0}.
$$
From this point, elementary computations yield $v(x)$, and finally, $u(x)=v(x)/v(0)$. 
A more sophisticated approach, based on the notion of conditional distribution, is to note that the independence of $(X,Y)$ and the fact that $P[Y\gt y]=\mathrm e^{-\lambda y}$ for every $y\gt0$ yield
$$
P[Y\gt X\mid X]=\mathrm e^{-\lambda X},
$$
hence
$$
v(x)=E[P[Y\gt X\mid X]\,\mathbf 1_{X\gt x}]=E[\mathrm e^{-\lambda X};X\gt x]=\int_x^\infty \mathrm e^{-\lambda z}\,\lambda\mathrm e^{-\lambda z}\mathrm dz.
$$
From here, $v(x)$ is direct and finally, $u(x)=v(x)/v(0)$.

For the interested reader, here is a generalization. Let $X$ and $(Y_i)_{1\leqslant i\leqslant n}$ denote $n+1$ i.i.d. random variables with exponential distributions of parameter $\lambda$. Then the conditional distribution of $X$ conditionally on the event
$$
[\min\{Y_i\,;\,1\leqslant i\leqslant n\}\gt X],
$$
is exponential with parameter $(n+1)\lambda$.
