# Probability of first time to an event

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?

• 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 $t$"?
– whuber
Oct 17, 2018 at 19:13
• 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. Oct 17, 2018 at 19:28
• @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.
– KRL
Oct 17, 2018 at 20:14
• @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 $f_t$
– KRL
Oct 17, 2018 at 20:16
• 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?
– whuber
Oct 17, 2018 at 21:20

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

• 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?
– KRL
Oct 19, 2018 at 7:02
• @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. Oct 19, 2018 at 13:44
• @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 $a+\frac1\lambda$ Oct 19, 2018 at 13:47