This question is a spin on the traditional Poisson Process.

Let's assume that I have a machine that has the ability to turn on a light. It does so by following a Poisson distribution with some frequency $\lambda$. This light turns itself off after constant $0<\alpha<\infty$ units of time after it last received a signal to turn on; if the light is already on when an event happens, the light always turns itself off $\alpha$ units of time after the latest event.

Recall that a Poisson distribution has the form: $Pr(X=k)=$$\lambda^k\mathrm{e}^{-\lambda} \over k!$

For example, assume $\alpha=10$, and Event1 occurs at $t=5$ and Event2 occurs at $t=10$. The light would turn on at $t=5$ and be scheduled to turn off at $t=5+\alpha=15$. Event2 comes along and changes the time that the light will turn off to be $t=10+\alpha=20$. The light then turns itself off at $t=20$, assuming no more events occur on the interval $t \in [10, 20)$.

The first question is: If you ran the machine forever, what percent of time would you expect the light to be on?

My approach is to first calculate the probability that given any time, $t$, what is the probability that in the past $\alpha$ units of time there has not been any event? That is, what is the probability that over the interval $[t-\alpha, t]$ no event occurred?

Poisson gives us this in the form of: $\mathrm{e}^{-\lambda \alpha}$.

Now, at time $t$ what are the chances that the light is on? If the light is on then an event must have occured somewhere over the interval $[t-\alpha,t]$. Since the probability of no event happening is $\mathrm{e}^{-\lambda \alpha}$, the probability of at least one event happening must be $1-\mathrm{e}^{-\lambda \alpha}$.

Thus, the chance that at any given time $t$ that the light is on (and the percentage of time the light will be on) is $1-\mathrm{e}^{-\lambda \alpha}$.

Now for the twist: What if the machine goes into a "cooldown" period after having fired an event? That is, let $\beta$ be a unit of time for which the machine cannot attempt to turn on the light where $0\leq \beta<\infty$. That is, $\beta$ can be less than, equal to, or greater than $\alpha$ (the duration that the light is on).

To bring it back to our example, let $\beta=7$ and let Event1 happen at $t=5$. The light would turn on at $t=5$ and be scheduled to turn off at $t=5+\alpha=15$. However, the machine is now incapable of turning the light on (and thus increasing the time the light is on) until $t=5+\beta=12$. Another case would be where $\beta=15$, which then the machine could not turn the light on until $t=5+\beta=20$, in which case we have guaranteed time with the light off on the interval $t \in [15,20)$.

Now, if you ran the machine forever with this added restriction, what percent of time would you expect the light to be on?

Obviously, if $\beta=0$ then we have the original case (as in a Poisson process no two events can occur at the same instant in time).

The other two cases are where $0<\beta\leq\alpha$ and $\alpha<\beta<\infty$. I am not too sure how to proceed. I know that Poisson processes depend on independent events, and the added "cooldown" period seems to play a role in that.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.