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If two events occur at a rate of 1.8 per hour on average, and this occurrence follows a poisson process, what is the probability that there is at least 1 hour between two events?

My approach for this question is the following:

$$\lambda = \frac{1}{E(x)} = \frac{1}{1.8} = 0.556$$

The probability of the time being under an hour (using the cumulative probability density for the poisson process):

$$P(X<1) = 1-e^{-\lambda x} = 1-e^{-0.556(1)} = 0.4265$$

Therefore the probability of the time being above an hour would be:

$$P(X\ge1) = 1-P(X<1) = 1-(1-e^{-0.556}) = e^{-0.556} = 0.573$$

The logic here seems obvious: The probability of a given wait time for independent events following a poisson process is determined by the exponential probability distribution $\lambda e^{-\lambda x}$ with $\lambda = 0.556$ (determined above), so the area under this density curve (the cumulative probability) is 1. Therefore, the probability of any time range would be the integral of this distribution function over the given time range. The integral from 0 to 1 is the probability of the time between events being under 1 hour, which is $0.4265$, and so the probability of the time between events being above 1 hour is $1-0.4265 = 0.573$.

...however, this is apparently not correct. The answer to this in the back of the book I am using is $0.1653$, and by the logic I've outlined above I am not seeing how. I am very much hoping my approach is proper and that the book is wrong, and am hoping someone can help me correct or confirm my logic.

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The Poisson distribution $\lambda$ is $1.8$. The probability of no event in an hour is $e^{-\lambda}$.

Another way: The waiting time has exponential distribution with parameter $\lambda$. Now calculate as you did, with parameter $1.8$.

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  • $\begingroup$ Ok..I might be seeing some light here. Are the $\lambda$ for the poisson process and distribution both the same? I've been seeing them as separate lambdas, where one represents the number of events per time (for the poisson distribution), and another representing time per event (for the poisson process). I've been converting between these as I did above in the first step. $\endgroup$ – Topher May 25 '15 at 3:20
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    $\begingroup$ The $\lambda$ for the exponential associated with the Poisson process is the $\lambda$ of the Poisson process. But the slightly confusing thing is that the mean waiting time, the mean of the exponential distribution, is $\\frac{1}{\lambda}$. $\endgroup$ – André Nicolas May 25 '15 at 3:43
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I think the book's answer is correct. The idea is that if there is a wait of at least an hour between two events, that means that during the hour there were $0$ event occurrences. This has probability $P(X=0)=e^{-\lambda}=e^{-1.8}$.

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  • $\begingroup$ Isn't the expected value $1/\lambda$ for the poisson process? This means it would be 1/1.8...right? $\endgroup$ – Topher May 25 '15 at 3:14
  • $\begingroup$ Yes. So the corresponding exponential wait time distribution is $1.8e^{-1.8t}$ for $t\ge 0$. You can integrate that from $1$ to $\infty$ to get the same probability of waiting at least an hour. $\endgroup$ – paw88789 May 25 '15 at 3:24
  • $\begingroup$ Ok, I think I see my problem. I've been seeing that the poisson process describing time ranges between events and the poisson distribution describing events per unit time, as having different lambda values. This is what was messing up my logic (hence the first calculation in my question). Given that lambda is the same for both, and seeing that the difference is in how the expectation calculation of both time between events and number of events is the difference, it seems to make more sense. $\endgroup$ – Topher May 25 '15 at 3:35

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