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Consider a recurring event for which the time periods between consecutive events is exponentially distributed. For argument's sake, I'm waiting for a taxi on a busy street. How might one calculate the likelihood of a taxi showing up in the next second?

Intuitively the chance that the event will occur in the next second should be just the integral of the PDF over the range (NOW, NOW+1). However, the fact that I am waiting for the event during this time period is not independent of the fact that the event did not occur in the time period (0, NOW]. Thus, perhaps I should consider the the integral of the PDF over the range [0, NOW+1) to get the probability of the event to occur in the next second. Though this "feels wrong" it does make sense as for each given second, the likelihood of the event occurring in that second is higher than it will be for the following second, given that the event has not already occurred.

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    $\begingroup$ if the tax arrivak process really is a poisson process, so the waiting time really is exponential, you have the memoryless propert. $\endgroup$
    – kjetil b halvorsen ♦
    Commented Mar 1, 2015 at 12:04
  • $\begingroup$ @kjetilbhalvorsen. Correct, memoryless. $\endgroup$
    – dotancohen
    Commented Mar 1, 2015 at 12:18
  • $\begingroup$ You have been waiting since NOW for a taxi. Whether any taxis arrived before NOW is irrelevant since you were not present. The probability that a taxi arrives before time NOW+1 is the integral of the pdf from $0$ to $1$, not from NOW to NOW+1. $\endgroup$
    – Dilip Sarwate
    Commented Mar 1, 2015 at 14:11
  • $\begingroup$ @DilipSarwate: Are you are suggesting that every time period has an equal likelihood of observing a taxi arriving (possibly due to the memoryless property)? If so, that would be an argument that Poisson processes are described by the uniform distribution, not the exponential distribution. $\endgroup$
    – dotancohen
    Commented Mar 1, 2015 at 14:39
  • $\begingroup$ No, the calculation $\int_0^1 \lambda e^{-\lambda x}\,dx = 1-e^{-\lambda}$ gives the probability that at least one taxi arrives in the time interval $(0,1]$ or (NOW, NOW+1] or $(t, t+1]$ etc. The total number of taxis arriving in this interval is a Poisson$(\lambda)$ random variable $N$; $P(N=0)=e^{-\lambda}$, $P(N>0)=1-e^{-\lambda}$, $P(N=1)=\lambda e^{-\lambda}$ etc. It is also true that conditioned on exactly one arrival in $(0,1]$, the time of arrival is (conditionally) uniformly distributed on $(0,1]$. $\endgroup$
    – Dilip Sarwate
    Commented Mar 1, 2015 at 14:49

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