Cumulative Conditional Probability Function of First Jump of Poisson Process Occurring Before Others

Question

What is the cumulative conditional probability in $(0,T)$ (i.e. $0<\tau<T$) that the first arrival of a Poisson process event of intensity $\lambda_1$ occurs before first-arrivals of all other Poisson processes $\lambda_2$, $\lambda_3$,...,$\lambda_n$

The unconditional cumulative probability function of the first arrival is then given by

$P(\tau_1<T) = (1-e^{(-\lambda_1 T)})$

For its conditional counterpart then, it feels like the below is a step in the right direction (probability of occurance of $\tau_1$ conditional on joint prob of no occurrance of $\tau_{j\neq1}$) but a more lucid & complete/correct derivation would be great

$P(\tau_1<T|\tau_1<\tau_j, \forall j=2,...,n) = \frac{(1-e^{(-\lambda_1 T)})}{e^{(-\lambda_2 T)}e^{(-\lambda_3 T)}...e^{(-\lambda_n T)}} = \frac{(1-e^{(-\lambda_1 T)}}{\prod\limits_{j=2}^n e^{(-\lambda_j T)}}$

Answer 1

Let $\tau=\min\limits_i\tau_i$ and $\nu$ the random variable such that $[\nu=i]=[\tau=\tau_i]$ for each $i$. Then $(\tau,\nu)$ is independent, $\tau$ is exponential with parameter $\lambda=\sum\limits_{i}\lambda_i$ and, for every $i$, $[\nu=i]$ has probability $\lambda_i/\lambda$. Since $[\nu=i]=[\forall j\ne i,\tau_j\gt\tau_i]$, one gets:

The conditional distribution of $\tau_i$ conditionally on the event $[\forall j\ne i,\tau_j\gt\tau_i]$ is exponential with parameter $\lambda$. In particular, this conditional distribution does not depend on $i$.

The easiest way to check the independence of $(\tau,\nu)$ asserted above might be to compute, for every $t$ and $i$, $$ P(\tau\gt t,\nu=i)=P(\tau_i\gt t,\forall j\ne i,\tau_j\gt\tau_i). $$ For every $s$, $$ P(\forall j\ne i,\tau_j\gt s)=\prod_{j\ne i}P(\tau_j\gt s)=\prod_{j\ne i}\mathrm e^{-\lambda_js}=\mathrm e^{-(\lambda-\lambda_i)s}, $$ hence $$ P(\forall j\ne i,\tau_j\gt\tau_i\mid\tau_i)=\mathrm e^{-(\lambda-\lambda_i)\tau_i}. $$ Conditioning on $\tau_i$, this implies $$ P(\tau\gt t,\nu=i)=\int_t^\infty\lambda_i\mathrm e^{-\lambda_is}\mathrm e^{-(\lambda-\lambda_i)s}\mathrm ds=(\lambda_i/\lambda)\mathrm e^{-\lambda t}. $$ This last identity is enough to identify the distributions of $\tau$ and $\nu$ and to check that $(\tau,\nu)$ is indeed independent.

Edit: For every $t\geqslant0$ and every $i$, $$ P(\tau_i\leqslant t\mid\forall j\ne i,\tau_j\gt\tau_i)=1-\mathrm e^{-\lambda t}. $$