# (Poisson process) probability of least one arrival of $N_2$ and $N_3$, between 2 successive arrivals of $N_1$.

Let $N_1(t),~N_2(t),~N_3(t)$ be independent Poisson processes of rates $\lambda_1,~\lambda_2,~\lambda_3>0$, respectively. Evaluate the probability between two successive arrivals of $N_1$ we get at least one arrival of $N_2$ and at least one arrival of $N_3$.

Attempt. The probability of getting an arrival of $N_i$ (and not of $N_j,~j\neq i$) is $\frac{\lambda_i}{\lambda_1+\lambda_2+\lambda_3}.$ If we already have an arrival of $N_1$, then for

$E_i$= the event we get at least one arrival of $N_i$,

we seek $P(E_2\cap E_3)=P(E_2)P(E_3).$ So:

$$P(E_2)=1-P(E_2^c)=1-\sum_{k=0}^{\infty}\bigg(\frac{\lambda_3}{\lambda_1+\lambda_2+\lambda_3}\bigg)^k\frac{\lambda_1}{\lambda_1+\lambda_2+\lambda_3}$$

and similar we work for $E_3.$ Am I on the right path?

• I think that the problem can be rephrased as: find $P(X_1<X_3\text{ and }X_2<X_3)$ where $X_i$ has exponential distribution with parameter $\lambda_i$. Here $X_i$ corresponds with the first arrival time connected with process $N_i(t)$. – drhab Sep 5 '18 at 9:38
• @drhab well, wouldn't it be $\mathbb{P}(X_2 < X_1 \text{ and } X_3 < X_1)$ ? – P. Quinton Sep 5 '18 at 10:21
• I don't think $E_2$ and $E_3$ are independent. They both make early arrivals of $N_1$ less likely, so they should correlate positively? – joriki Sep 5 '18 at 10:26
• @P.Quinton Yes, that is what I meant to write of course. – drhab Sep 5 '18 at 11:33

After the initial $N_1$ you need to either first get $N_2$ and then $N_3$ before $N_1$, or first $N_3$ and then $N_2$ before $N_1$. Thus the probability is
$$\frac{\lambda_2}{\lambda_1+\lambda_2+\lambda_3}\frac{\lambda_3}{\lambda_1+\lambda_3}+\frac{\lambda_3}{\lambda_1+\lambda_2+\lambda_3}\frac{\lambda_2}{\lambda_1+\lambda_2}=\frac{\lambda_2\lambda_3}{\lambda_1+\lambda_2+\lambda_3}\left(\frac1{\lambda_1+\lambda_3}+\frac1{\lambda_1+\lambda_2}\right)\;.$$
• @Henry: I'm not sure whether you'd count this as "using the memoryless property". I'm using the fact that Poisson processes with rates $\lambda_i$ can be combined into a single Poisson process (with rate $\sum_k\lambda_k$) in which each arrival is independently of type $i$ with probability $\lambda_i/\sum_k\lambda_k$. This is stated in the question, so I thought it would be OK to use it. – joriki Sep 5 '18 at 12:23
• I would say that your $\frac{\lambda_3}{\lambda_1+\lambda_3}$ term uses the memoryless property, since you are saying once $N_2$ has happened first, the probabilities in the subsequent race between $N_3$ and $N_1$ are the same as they would have been at the start – Henry Sep 5 '18 at 12:56