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Let's say there is an induction loop in a road capable of counting the number of cars passing over it. By keeping a list of moments for when a car passed the detection loop, I am able to determine the average interval $\mu$ and variance $\sigma^2$ there over. I also know the amount of intervals $N$ measured and the last moment a car passed the induction loop.

I wish to know the chance of no cars passing for a time interval since the last moment a car passed the loop. This question seems to ask for a Poisson-distribution, but these seem no to keep into account the variance of time intervals, nor do they seem to use the amount of measurements. Which method should I use to determine the probability of no car passing the detection loop within a certain interval?

known values:

  • $\mu$: The average interval between two cars passing the induction loop
  • $\sigma^2$: The variance in these intervals
  • $N$: the amount of interval measurements

unknown:

  • $P(t,c)$: The chance of $c$ cars passing the induction loop in interval $t$

where:

  • $t$: the interval for which the probability is sought
  • $c$: the amount of cars which should pass within the interval. 0 in my case.
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You refer to the mean $\mu$ and the variance $\sigma^2$. Do you mean the mean and variance of the times between cars? Or maybe of the number of cars passing in some specified time? Or something else? –  Michael Hardy Apr 26 '13 at 17:22
    
With $\mu$ I mean the mean time in between cars. With $\sigma^2$ I mean the variance in time between cars. Given that I keep a list of each car passing, this could be calculated. If you find it helpful to extract other data from this set of time stamps, please do. –  Roelof van den Berg Apr 26 '13 at 17:28

1 Answer 1

Variance is not needed, and it is perfectly fine to use Poisson formula. So your answer would be just

$$P(X=0)=e^{-\mu}$$

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Why is the variance not needed? And on that same note, why is the amount of measurements $N$ for the interval-estimates not needed? Could you please clarify why nothing is lost when putting this information aside? –  Roelof van den Berg Apr 26 '13 at 16:58
    
Beware of "gambler's fallacy". To take an analogy, if you throw a fair dice, the probability of getting a four is always 1/6. It doesn't depend on how many times you have tossed. As for variance, it is better interpreted as a measure of a group of data instead of a predictor of probability, although we know variance is equal in value to the mean of a Poisson distribution. Of course, the ultimate reason Poisson distribution doesn't depend on variance is because of its derivation. –  Neoh Apr 26 '13 at 17:14

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