Let $_$ and $Y_t$ be two independent Poisson Process with rate parameters $\lambda_1$ and $\lambda_2$ respectively, measuring the number of customers arriving in stores 1 and 2. What is the probability that a customer arrives in store 1 before any customers arrive in store 2?
I have found some solutions for this question that I don't understand, and would like some explanation for. Firstly let me explain my thinking. Let $T_1, T_2$ be the first times a customer arrives at stores 1 and 2 respectively. Fix any $t$, then we can consider $P(T_2 > t) \cdot P(T_1 = t).$ Then we can integrate this from $t = 0$ to $\infty$ and that should give the answer. I think this is similar to the method outline in this question where the integral
$$\int_0^\infty P(T_2>t)dP(T_1=t)=\int_0^\infty e^{-\lambda_2 t}\lambda_1 e^{-\lambda_1 t} dt$$
is calculated. I think this is just me being bad with integration/continuous distributions, but I don't see how my intuition leads to the integral on the LHS. Specifically, what is the $dP(T_1=t)$ term doing? Secondly, how does this translate to the integral on the right? The derivative of $P(T_1 > t) = \lambda_1 e^{-\lambda_1 t}$ so how does this arise from $P(T_1 = t)$ which should be 0?
The integral on the LHS is essentially considering all possible times $T_1$ can happen and integrates over $t$. It can be re-written as
$$ \int_0^\infty P(T_2>t)f_{T_1}(t)dt $$
where $f_{T_1}(t)$ is the density of $T_1$. Some manipulation yields
$$ = \int_0^\infty S_{T_2}(t) \cdot f_{T_1}(t)dt = \int_0^\infty e^{-t\lambda_2} \cdot \lambda_1 e^{-t\lambda_1}dt $$
where $S_{T_2}(t) = 1 - F_{T_2}(t)$ is the survival function of $T_2$.