Conditional Distribution of Sum of two Independent Exponential Random Variables $X$ and $X'$ are independent exponentially distributed random variables with parameters $\alpha$ and $\beta$. Now $Z = X+ X'$. 
We need to find: $P(Z \mid X> \tau, X' > \tau)$.
I am writing this as: 
\begin{equation}
f_{Z \mid X > \tau, {X'} > \tau}(z) = \frac{f_{Z}(z)}{f_{X}(x) f_{{X'}}({x'})} \\
   =   \frac{ \int_\tau^z f(x) f(z - {x}) d x  }  {\int_\tau^z f(x) dx \int_\tau^z f({x'}) d{x'} }
\end{equation}
Am I doing it right? Thank you.
 A: $\text{“}P(Z \mid X> \tau, X' > \tau)\text{''}$ is not actually proper use of notation, since $Z$ is a random variable, not an event. I take it you want the conditional probability distribution of $Z$ given that event.
The part where you equate a quotient of density functions with a quotient involving integrals is nonsense. It looks like something that would be written only by someone who does not understand what those expressions mean. In particular, the expression involving integrals is a function only of the variable $z,$ but $z$ does not even appear in the quotient of density functions. If you're attempting to learn the subject—any mathematical subject—the question of what the symbols mean, rather than just how to manipulate them, should always be there.
The conditional distribution of $X-\tau$ given the event $X>\tau$ is $\alpha e^{-\alpha t} \, dt \text{ for } t>0.$ That is the memorylessness of the exponential distribution. And if we condition on the event that both $X>\tau$ and $X'>\tau,$ the conditional distribution of $X-\tau$ is still the same, since $X'$ and $X$ are independent. Similarly the conditional distribution of $X'-\tau$ given that same event is $\beta e^{-\beta t}\, dt \text{ for } t>0.$
So
\begin{align}
& \Pr(Z>\sigma \mid X>\tau\ \&\ X'>\tau) \\[8pt]
= {} & \Pr((X-\tau) + (X'-\tau) > \sigma-2\tau\mid (X-\tau)>0\ \&\ (X'-\tau)>0)
\end{align}
and this is just the probability that the sum of two exponentially (but not identically) distributed random variables is greater than a specified quantity. This is
$$
\iint\limits_{\begin{smallmatrix} u\,>\,0 \\ v\,>\,0\\ u\,+\,v\,>\,\sigma\,-\,2\tau \end{smallmatrix}} \alpha\beta e^{-\alpha u - \beta v} \, d(u,v)
$$
The value of this expression is well known in the case $\alpha=\beta,$ but we don't have that assumption, so we can plod through it in a pedestrian manner:
\begin{align}
& \int_0^{\sigma-2\tau} \left( \int_{\sigma-2\tau-u}^\infty \alpha\beta e^{-\alpha u - \beta v} \, dv \right) \, du \\[10pt]
& {} + \int_{\sigma-2\tau}^\infty \left( \int_0^\infty \alpha\beta e^{-\alpha u - \beta v} \, \, dv \right) \, du
\end{align}
Turn the crank.
