Calls are received at a company according to a Poisson process at the rate of 5 calls per minute. Find the probability that $25$ calls are received in the first $5$ minutes and six of those calls occur during the first minute.

Denote the number of calls with $N_t$ at time $t$. We have that $N_t\sim\text{Poi}(\lambda t),$ where $\lambda=5$. We are looking for

$$\mathbb{P}(N_5=25\ | \ N_1=6 )=\frac{\mathbb{P}(N_1=6,N_5-N_1=19)}{\mathbb{P}(N_1=6)}=\frac{\mathbb{P}(N_1=6,\tilde{N_4}=19)}{\mathbb{P}(N_1=6)}=...$$

by stationary increments. Independent icrements also give that we can proceed with

$$...=\frac{\mathbb{P}(N_1=6)\mathbb{P}(\tilde{N_4}=19)}{\mathbb{P}(N_1=6)}=\mathbb{P}(\tilde{N_4}=19)=\frac{(5\cdot 4)^{19}e^{-5\cdot 4}}{19!}\approx0.0888.$$

Which is incorrect. However I get the correct answer if I, with the same method using increments, calculate $\mathbb{P}(N_5=25\ , \ N_1=6 ).$


Why is it wrong to calculate $\mathbb{P}(N_5=25\ | \ N_1=6 )$? To me this seems intuitive: We want to find the probability that $25$ calls are received given that $6$ calls already have happened in the first minute.

  • $\begingroup$ $25$ calls occuring during the first five minutes, with $6$ of those occurring during the first minute, means how many calls occurred during minutes two through five? $\endgroup$ – Math1000 Jan 15 at 17:30

We weren't given a conditional probability question. It isn't previously known or given that $6$ calls happened in the first minute. Had it said something along those lines then your conditional probability approach would have been correct.

Instead we have two events $A$ and $B$, and we want the probability of $A\cap B$, and because events over disjoint time intervals are independent in the Poisson process, we can find $\mathbb{P}(A\cap B) = \mathbb{P}(A) \mathbb{P}(B)$

  • $\begingroup$ I thought the exponential distribution was memorylessa not the Poisson? That formula holds if $A$ and $B$ are disjoint and/or independent. I don't see how memorylessness implies it. Could you elaborate on this? $\endgroup$ – Parseval Jan 15 at 18:13
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    $\begingroup$ Both distributions have this property (in fact the exponential distribution and Poisson distribution are closely related). If we let event $A$ be that we get $6$ calls in the first minute and let $B$ be the event that we get $19$ calls in minutes 2 through 5 (as Math1000 points out) then event $B$ will have no memory of event $A$, thus making them independent events $\endgroup$ – WaveX Jan 15 at 18:17
  • $\begingroup$ Perhaps calling the Poisson distribution "memoryless" may have been somewhat of a stretch; hopefully my comment and edited answer makes more sense $\endgroup$ – WaveX Jan 15 at 18:28

The key word is highlighted below:

Find the probability that 25 calls are received in the first 5 minutes and six of those calls occur during the first minute.

Since they used the word and, you want $P(N_5=25 \cap N_1=6)$, and not $P(N_5=25 \mid N_1=6)$.


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