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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.

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