A bus arrives at a bus stop according to a Poisson process. It is given that in the last 100 hours, the bus arrived at the bus stop exactly 200 times. Predict the arrival rate for the bus at the bus stop for the next 100 hours if the bus follows the same Poisson distribution.

From the given data, we can say that the arrival rate of the bus is 200/100 =2. That means the bus arrived at the bus stop 2 times per hour. So, the arrival rate for the next 100 hours is also 2 since the bus follows the same Poisson distribution. Is it that simple or I'm missing something?

  • $\begingroup$ Yes, that is correct. $\endgroup$ – Math1000 Aug 23 '15 at 18:13
  • 1
    $\begingroup$ The poisson process has the basic property that the expected number of trials upto the next success is independent from the failures and successes which already happened. $\endgroup$ – Peter Aug 23 '15 at 18:15
  • $\begingroup$ I have a similar requirement. which prediction method should be used for predicting the possion arrivals $\endgroup$ – Amarjit Singh Nov 19 '17 at 12:28

Let $X_i \sim Pois(\lambda_1)$ be the number of arrivals in one hour, in effect, we have observations $X_1, X_2, \dots, X_{100}.$ Because $\bar X = 2$ our point estimate of the Poisson hourly rate of arrival is $\hat \lambda_1 = 2.$

We do not know the individual $X_i$ but we do know know $T = \sum_{i=1}^{100} X_i = 200.$ Because the sum of Poissons is again Poisson, we know that $T \sim Pois(\lambda_{100}),$ where $\lambda_{100}$ is the arrival rate per 100 hours.

Based on the normal approximation of the sum $T,$ a reasonable 95% confidence interval for $\lambda_{100}$ is $$T + 2 \pm 1.96\sqrt{T + 1}\; \text{or}\; 202 \pm 1.96\sqrt{201},$$ which gives the interval $(174.2,\,229.8)$ for $\lambda_{100}.$ Simply divide by 100 to get the 95% CI $(1.742, 2,298)$ for $\lambda_1.$

A Bayesian 95% probability estimate, using an 'improper' non-informative gamma prior, is based on the posterior distribution $Gamma(200, 100).$ Cutting 2.5% from each tail of the posterior distribution gives the probability interval $(1.732, 2.287)$ for $\lambda_1$, according to the computation in R below.

 qgamma(c(.025,.975), 200, 100)
 ## 1.732409 2.286527

The CI and the Bayesian interval are numerically very similar. Both give a good idea of just how accurate this estimate may be, based on the assumptions of your Problem.

I agree that the Poisson model may not be the best one, in general, to model the arrival behavior of a bus. But your Problem explicitly states that this model applies to $this$ bus route.

  • $\begingroup$ Thanks a lot for your detailed explanation. $\endgroup$ – Jenn Aug 25 '15 at 14:48

If $\lambda=2$, then in any $100$ hour period, the expected number of arrivals is indeed $200$.

However, the observation that there were $200$ arrivals in a $100$ hour period does not guarantee that $\lambda=2$: it merely provides evidence. This needs to be tested.

Predictions based on a single observation are unreliable.

Having said this, it is somewhat debatable as to whether bus arrivals could possibly follow a Poisson distribution, as any bus service does not operate on an entirely random basis, but follows a schedule.

  • $\begingroup$ $\lambda=2$ is not given. Instead we can say $\lambda_{1}=2$ and trying to estimate $\lambda_2$. The bus follows the same Poisson distribution $\lambda_1+\lambda_2=\lambda$. But I don't see any way to estimate $\lambda$ $\endgroup$ – Jenn Aug 23 '15 at 18:28

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