I'm interested in the ordering of arrivals of types of 2 independent Poisson processes, denoted type 1 and 2. Given interarrival times are exponentially distributed, I know that if the processes have rate $\alpha_1,\alpha_2$, then the probability of either occurring first in any sequence is given by $$\frac{\alpha_1}{\alpha_1\cdot\alpha_2}~\text{or}~\frac{\alpha_2}{\alpha_1\cdot\alpha_2}$$ By the memorylessness property, you simply start fresh after new arrivals. However, I am unsure how to go about calculating the probability that a sequence starts with a type 1 arrival and is simultaneously followed by a type 2 arrival $d$ time units after. That is, if $X_1,X_2$ denote the interarrival times of the first arrival of each type, I want to calculate $\mathbb{P}[X_1<X_2<X_1+d]$.
My thoughts are that because these arrival times are memoryless, I can break it up into: $$ \mathbb{P}[X_1<X_2<X_1+d]=\mathbb{P}[X_1<X_2]\cdot\mathbb{P}[X_2<d] $$ Is this logic sound?
Could you not make this problem into something like
$$ \mathbb{P}[X_1<X_2<X_1+d]= \mathbb{P}[0<X_2-X_1<d] = \mathbb{P}[0<Z<d] $$ where defining a new random variable $Z = X_1-X_2$
This should be solvable by the above transformation.
Let $Z = X_2-X_1$. First we will find the cumulative distribution function $F_Z(z)$ of $Z$, that is, the probability that $Z\le z$.
Consider $z$ fixed and positive, and draw the line $x_2-x_1=z$. We want to find the probability that the ordered pair $(X_1,X_2)$ ends up below that line or on it. The only relevant region is in the first quadrant. So let $D$ be the part of the first quadrant that lies below or on the line $x_2=x_1+z$. Then $$P(Z \le z)=\iint_D \alpha_1\alpha_2e^{-(\alpha_1x_1+\alpha_2x_2)}dx_1\,dx_2.$$
We will evaluate this integral, by using an iterated integral. First we will integrate with respect to $x_2$, and then with respect to $x_1$. Note that $x_2$ travels from $0$ to $x_1+z$, and then $x_1$ travels from $0$ to infinity. Thus $$P(Z\le z)=\int_0^\infty \alpha_1 e^{-\alpha_1 x_1}\left(\int_{y=0}^{x_1+z} \alpha_2 e^{-\alpha_2 x_2}\,dx_2\right)dx_1.$$
The inner integral turns out to be $1-e^{-\alpha_2(x_1+z)}$. So now we need to find $$\int_0^\infty \left(\alpha_1 e^{-\alpha_1 x}-\alpha_1 e^{-\alpha_2 z} e^{-(\alpha_1+\alpha_2)x_1}\right)dx_1.$$ We end up with $$P(Z \le z)=1-\frac{\alpha_1}{\alpha_1+\alpha_2}e^{-\alpha_2 z}.$$
$$P(Z \le d)=1-\frac{\alpha_1}{\alpha_1+\alpha_2}e^{-\alpha_2 d}.$$