Q:

Solving inequality of two independent exponentially distributed RVs

I have huge problems solving following excersice:
There are two molecules. The decay of the molecules is exponentially distributed with $\alpha_1 = 1$ (for molecule 1) and $\alpha_2 = 2$ (for molecule 2). Molecule 1 is in the system since $t = 0$ and molecule 2 enters the system at $t = \Delta t$. What is the probability, that molecule one decays before molecule 2?
I came so far:
Let $X_1$ be the decay of molecule 1 and $X_2$ be the decay of molecule 2
$P(X_1 > X_2)$
$=P(X_1 > X_2|X_1 <\Delta t)*P(X_1 <\Delta t) + P(X_1 > X_2|X_1 >\Delta t)*P(X_1 >\Delta t)$
$= P(X_1 > X_2|X_1 <\Delta t)*P(X_1 <\Delta t) + \frac{P(X_1 > X_2,X_1 >\Delta t)}{P(X_1 >\Delta t)} *  P(X_1 >\Delta t)$
$= P(X_1 > X_2|X_1 <\Delta t)*P(X_1 <\Delta t) + P(X_1 > X_2,X_1 >\Delta t)$
I know the answer for 
$P(X_1 > X_2|X_1 <\Delta t)*P(X_1 <\Delta t)$ but I have problems with $P(X_1 > X_2,X_1 >\Delta t)$....or is there something completely wrong with my approach?
Thanks for the help! :)

A:

For simplicity, let $T_1$ be the shelflife of molecule 1 after it enters the system, at time $0$, and $T_2$ be that of molecule 2 after it enters the system, at time $\Delta t$.   Then the event of molecule 1 decaying before molecule 2 is represented by: $\{T_1< T_2+\Delta t\}$
Noting that the supports of the exponential distribution asserts that $T_1\geq 0, T_2\geq 0$ , we then partition the space at $\Delta t$.
$$\mathsf P(T_1<T_2+\Delta t) = \mathsf P(T_1\leqslant \Delta t)+\mathsf P(T_1-\Delta t<T_2\mid T_1\gt \Delta t)~\mathsf P(T_1\gt \Delta t)$$
We make use of the memoryless property of exponential distributions.   Where $T_1'$ is $T_1-\Delta t$ when given that $T_1\gt \Delta t$, then $T_1'$ has identical distribution to $T_1$. [ That is $T_1'\sim\mathcal{Exp}(\alpha_1)$ ]
$$\mathsf P(T_1 < T_2+\Delta t) = \mathsf P(T_1 \leqslant \Delta t)+\mathsf P(T_1'<T_2)~\mathsf P(T_1\gt \Delta t)$$
Then as $T_1\sim\mathcal{Exp}(\alpha_1)$
$$\begin{align}\mathsf P(T_1 < T_2+\Delta t) =&~ 1-\mathsf e^{-\alpha_1\Delta t}+\mathsf P(T_1'<T_2)~\mathsf e^{-\alpha_1\Delta t}\\[1ex] =&~ 1-\mathsf e^{-\alpha_1 \Delta t}\mathsf P(T_1'\geqslant T_2)\end{align}$$
Leaving you to determine: $\mathsf P(T_1'\geqslant T_2)$
(This is a somewhat well known result, but easy enough to resolve by integration if you don't already know it. Do you need to do so? Can you?)
Then we have$$\begin{align}\mathsf P(T_1'\geqslant T_2) =&~ \int_0^\infty f_{T_2}(\tau)\,\mathsf P(T_1'\geqslant \tau\mid T_2=\tau)\operatorname d \tau\\[1ex] =&~ \int_0^\infty f_{T_2}(\tau)\,\mathsf P(T_1'\geqslant \tau)\operatorname d \tau &\text{because independence}
\\[5ex] &\vdots  \end{align}$$

 $$\begin{align}\mathsf P(T_1'\geqslant T_2) =&~ \dfrac{\alpha_2}{\alpha_1+\alpha_2} \end{align}$$

