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We have a stream of events over time. Suppose that $f_t$ is the probability density that an event happens at time $t$. For example, $f_t$ can be the probability density that any bus arrives at time $t$. Note that:

  1. $f_t$ is not a probability because the probability that an event happens exactly at time $t$ is zero.

  2. $f_t$ is not a probability distribution over $t$. My understanding is events over time form a random process rather than a random variable. However, we can have a probability distribution over $t$ for the arrival time of the third bus.

For my problem, I have computed $f_t$. My questions are:

  1. Is $f_t$ a known mathematical measure? It seems to me $f_t$ is a probability density over with parameter $t$.

  2. Suppose that we know no bus has come until time $t$ and we know that probability that a bus comes at $t+\Delta t$ is independent of the past. We want to compute the probability density over the arrival time of the first bus, denoted by $h(t)$, using $f_t$. We have:

$h(t) = \lim_{\Delta t \rightarrow 0} [(1-f_{0})\Delta t] [(1-f_{\Delta t})\Delta t] ... [(1-f_{t - \Delta t})\Delta t][f_t\Delta t]$

In other words, we want no bus to arrive before $t$ and a bus arrives at $t$. Is this calculation correct? Can we simplify the product of the limit in this equation?

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    $\begingroup$ Could you clarify how you arrived at your first conclusion (2)? If we define a random variable $X$ to be the time when the bus first arrives, why isn't the distribution of $X$ a "probability distribution over $t$"? $\endgroup$
    – whuber
    Oct 17, 2018 at 19:13
  • $\begingroup$ The probability that an event happens exactly at time t tends to 0, but is not equal 0. I will prove it by contradiction. Let's assume that the probability of that event to happen on time t is zero, for all times. Therefore, the definite integral from 0 to t' is zero. However, the event need to occur until time t'. Thus, the probability of an event to happen on time t should not be equal to 0. $\endgroup$ Oct 17, 2018 at 19:28
  • $\begingroup$ @whuber I agree arrival time of the first bus has a probability distribution over $t$. However, the probability that "a bus" arrives at $t$ does not a probability distribution over $t$ because it doesn't sum to one. $\endgroup$
    – KRL
    Oct 17, 2018 at 20:14
  • $\begingroup$ @IagoAugusto I agree. The probability tends to zero at any given time $t$. However, probability density of an event happening at different times is different and denoted by $f_t$ $\endgroup$
    – KRL
    Oct 17, 2018 at 20:16
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    $\begingroup$ It looks like you might be confusing yourself with informal language. The event "the bus arrives at time $t$" is the event $X=t.$ All random variables have distributions; ergo, $X$ has one too. Since its possible values are times, we might say it "has a distribution over $t.$" If that's not what you mean by this phrase, exactly what do you mean? $\endgroup$
    – whuber
    Oct 17, 2018 at 21:20

1 Answer 1

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I would call $f_t$ the instantaneous rate of the process at time $t$, or perhaps the hazard function

So, for example, you can find the expected number of arrivals between time $a$ and time $b$, which would be $\int\limits_a^b f_t\, dt$

The expected number of arrivals in the short interval between $t$ and $t+\Delta t$ is $\int\limits_t^{t+\Delta t} f_s\, ds$ which for small $\Delta t$ and well-behaved $f_t$ is about $f_t \Delta t$, and this is then also approximately the probability for finding at least one arrival, making the probability of no arrivals in that short interval about $1-f_t \Delta t\approx \exp\left(-f_t \Delta t\right)$.

Taking products (which turn in a sum inside the $\exp$) and then the limit, this then makes the probability of no arrivals in the long interval from $a$ to $t$ be $\exp\left(-\int\limits_a^t f_s\, ds\right)$ which is a survival function, but it would be more useful to have the cumulative distribution function for $T$ being the first arrival after time $a$, which is $$F(t)= \mathbb P(T \le t \mid T \gt a)= 1 -\exp\left(-\int\limits_a^t f_s\, ds\right)$$ which is a probability when $t \ge a$

The density for the first arrival time after $a$ is then the derivative of this, which is $$f_t \exp\left(-\int\limits_a^t f_s\, ds\right) = f_t \left(1-F(t)\right)$$

If $f_t$ is in fact a constant over time, say $\lambda$, then you have a Poisson process with that parameter, making $F(t)=1-e^{-\lambda(t-a)}$ and the density $\lambda e^{-\lambda(t-a)}$ , i.e. an exponential distribution starting at $a$, much as you might expect

With a variable rate, I believe you have what is called an inhomogeneous Poisson process

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  • $\begingroup$ If we want to know what is the expected time to the first event, can we say it is when integral of $f_t$ becomes 1? $\endgroup$
    – KRL
    Oct 19, 2018 at 7:02
  • $\begingroup$ @Paris - Not for variable $f_t$ - the point at which the integral of $f_t$ is $1$ would be the point at which the point at which the cumulative probability is $1-e^{-1}\approx 0.632$; if the rate were constant then this would indeed be the mean but not in general. On the other hand, if you took the point at which the integral of $f_t$ was $\log_e(2) \approx 0.693$ then the cumulative probability would be $\frac12=0.5$ and so you would have found the median. $\endgroup$
    – Henry
    Oct 19, 2018 at 13:44
  • $\begingroup$ @Paris: the expected first arrival time given it has not happened by time $a$ can be found from a double integral $$a+\int_a^\infty \exp\left(-\int\limits_a^t f_s\, ds\right)\, dt$$ which with constant $f_t=\lambda$ would give $a+\frac1\lambda$ $\endgroup$
    – Henry
    Oct 19, 2018 at 13:47

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