Memoryless property for Poisson process

Question

Suppose that arrivals of a certain Poisson process occur once every $4$ seconds on average. Given that there are no arrivals during the first $10$ seconds, what is the probability that there will be 6 arrivals during the subsequent $10$ seconds?

My solution

Suppose $N(t) =$ number of arrivals at time $t$. Then we need to find $$P(N(20) = 6 \mid N(10) = 0) = P(N(20) = 6) = e^{-\lambda t}\sum_{x=0}^6\frac{(\lambda t)^x}{x!} = e^{-80}\sum_{x=0}^6\frac{80^x}{x!}.$$

Since the fact that nothing occurs during the first $10$ seconds gives us no additional information.

Poisson *process* is not memoryless. Exponential *distribution* is memoryless. PP has independent *increments*. — A.S., Mar 22 '16 at 23:44

score 4 · Accepted Answer · answered Mar 22 '16 at 23:45

4

$$ P(N(20) = 6 \mid N(10) = 0) \ne P(N(20) = 6) $$ $$ P(N(20) = 6 \mid N(10) = 0) = P(N(20) -N(10) = 6) = P(N(10) = 6) = \frac{e^{-\lambda\cdot 10} (\lambda\cdot10)^6}{6!} $$ where $\lambda = \dfrac 1 {\text{4 seconds}} $.

It is not the Poisson distribution that is memoryless; it is the distribution of the waiting times in the Poisson process that is memoryless. And that is an exponential distribution.

answered Mar 22 '16 at 23:45

Michael Hardy

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Do you mind explaining why P(N(20)=6∣N(10)=0)=P(N(20)−N(10)=6) is true? – An Ignorant Wanderer Oct 28 '19 at 16:43
@user10796158 : $$\begin{align} & \Pr(N(20) = 6\mid N(10)=0) \\ = {} & \Pr(N(20) - N(10)= 6\mid N(10)=0) \\ = {} & \Pr(N(20)-N(10)=6). \end{align}$$ The reason for the "equals" sign at the beginning of the last line is that the numbers of arrivals in disjoint intervals of time are independent. The reason for the "equals" sign at the beginning of the second line is that, given that $N(10)=0,$ the events $N(20)=6$ and $N(20)-N(10) = 6$ are both the same event. $\qquad$ – Michael Hardy Oct 28 '19 at 18:56
So I have 2 followup questions: 1-"The reason for the 'equals' sign at the beginning of the last line is that the numbers of arrivals in disjoint intervals of time are independent. " If that's the case, then why P(N(20)=6∣N(10)=0)≠P(N(20)=6)? 2- After that, how did you get from P(N(20)−N(10)=6) to P(N(10)=6)? – An Ignorant Wanderer Oct 28 '19 at 21:06
@user10796158 : I said _disjoint_ intervals. The intervals $[0,20]$ and $[0,10]$ are not disjoint. – Michael Hardy Oct 29 '19 at 00:44
@user10796158 : I never said $N(10)= 6;$ I said $$ N(20) - N(10) = 6. $$ – Michael Hardy Oct 29 '19 at 00:45
For your last comment, I'm talking about what you wrote in your answer to this post (P(N(20)−N(10)=6)=P(N(10)=6) – An Ignorant Wanderer Oct 29 '19 at 02:34
@user10796158 : The reason why $$ \Pr\big(N(20) - N(10)=6\big) = \Pr\big(N(10)= 6\big) $$ is that both intervals are of the same length. Each of them is the number of arrivals during a period of $10$ second, so each has the same probability of seeing $6$ arrivals. – Michael Hardy Oct 29 '19 at 02:41

score 1 · Answer 2 · edited Oct 28 '19 at 18:51

By the no-memory property of exponential inter-arrival times, you need to start over at the end of the first 10 seconds. What did or did not happen in the first 10 seconds is irrelevant to what happens in the second 10 seconds. So your answer should be based on 10 sec. remaining. Also, you need exactly 6 arrivals, not at least 6 arrivals.

The rate for a 10 second period of time is $\lambda = 10/4 = 2.5.$ If $X \sim \operatorname{Pois}(\lambda),$ then $P(X = 6) = e^{-\lambda} \lambda^6/6!.$

Related computations in R:

 lam = 10/4;  exp(-lam)*lam^6/factorial(6)
 ## 0.02783373
 dpois(6, lam)
 ## 0.02783373

Memoryless property for Poisson process

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