Probability of event A reach a number of occurences before event B does. Independent poisson processes.

Question

Assume that you have 2 Independent Poisson processes with rates λ1 and λ2. The probability that the 1st event that will occur will be from the first process in a time period t is $\frac{λ1}{λ1+λ2}$ and from the second $\frac{λ2}{λ1+λ2}$. Those processes can be merged to a single process with rate $L=λ1+λ2$ therefore from poisson pmf the probability of no events will occur from either of the processes in a time period is $e^{-L}$.

• Doesn't that make that the first event will occur from the first process $\frac{λ1}{λ1+λ2}*(1- P(no\ events\ happened)) $ and similarly from the second $\frac{λ2}{λ1+λ2}*(1- P(no\ events\ happened))$?
• If so probability that the nth event that will occur will be from the first process is $$\frac{λ1}{λ1+λ2}*(1- P(no\ events\ happened)-P(1\ event\ happened)-...-P(n-1\ events\ happened))$$

If all the assumptions are true what is the probability that the first process will reach a number of event occurrences x before the second process does and what is the probability that neither of those processes will reach x?

Answer 1

I'll provide you with two different approaches to find the probability that the first processes occurs $x-$times before the second process does.

Let $T_x(j):j=1,2$ be the time at which the $x^{\text{th}}-$ event takes place for the $j^{\text{th}}$ Poisson process. You're tasked to compute $\mathbb{P}\left(T_x(1)<T_x(2)\right)$.

Since $T_x(1),T_x(2)$ are independent Erlang random variables, the joint distribution of the bivariate random vector $\left(T_x(1),T_x(2)\right)$ factors as the following: $$f_{T_x(1),T_x(2)}(u,v)=f_{T_x(1)}(u)f_{T_x(2)}(v)=\frac{(\lambda _1 \lambda _2)^{x}}{\left[(x-1)!\right]^2}(uv)^{x-1}\exp\{-(\lambda_1 u + \lambda_2 v)\}\cdot 1_{[0,\infty)^2}(u,v)$$ So, $$\mathbb{P}\left(T_x(1)<T_x(2)\right)=\int_0^{\infty} \int_0^v \frac{(\lambda _1 \lambda _2)^{x}}{\left[(x-1)!\right]^2}(uv)^{x-1}\exp\{-(\lambda_1 u + \lambda_2 v)\}\mathrm{d}u\mathrm{d}v$$

Another approach: Take $N$ as the total number of events that occur in the compound Poisson process until $x$ occurrence of the first process takes place. You're tasked to compute $\mathbb{P}\left(N=x,\dots ,2x-1\right)$.

Note $N$ is supported on $\{x,x+1,\dots\}$ and, for $n\geq x$ fixed, $$\mathbb{P}(N=n)={n-1 \choose x-1}\left(\frac{\lambda_1}{\lambda_1 + \lambda _2}\right)^{x} \left(\frac{\lambda_2}{\lambda_1 + \lambda _2}\right)^{n-x}$$ Therefore $$\mathbb{P}\left(N=x,...,2x-1\right)=\sum_{n=x}^{2x-1}{n-1 \choose x-1}\left(\frac{\lambda_1}{\lambda_1 + \lambda _2}\right)^{x} \left(\frac{\lambda_2}{\lambda_1 + \lambda _2}\right)^{n-x}$$ For the second question, take $E$ as the event that neither process reaches $x$ total occurrences and $T_x(1)$ as the arrival time of the $x^{\text{th}}$ arrival in the first process. Fix $t>0$ arbitrarily. Then $$\begin{eqnarray*}\mathbb{P}(E) &\leq & \mathbb{P}\left(T_x(1)>t\right) \\ &=& 1- \mathbb{P}(T_x(1)\leq t) \\ &=& 1-F_{T_x(1)}(t)\end{eqnarray*}$$ The above is true for any $t>0$ which implies $\mathbb{P}(E)=0$