Likelihood of multiple event times modeled as independent Poisson processes

Question

I am modeling three events A, B, and C as Poisson processes with rates $\lambda_A$, $\lambda_B$, and $\lambda_C$ and I would like to calculate the likelihood of observing some data given my model.

A distinction that I have come across is the difference between an event happening at a specific time $t$, $P\left(T\left(E\right) = t \right)$, versus by time $t$, $P\left(T\left(E\right) \leq t \right)$, where $T\left(E\right) = E_n - E_{n-1}$ refers to the inter-arrival time between two consecutive occurrences of the event. For a single event with rate $\lambda$, the probability of these scenarios are the exponential likelihood function

$P\left(T\left(E\right) = t \right) = \lambda e^{-\lambda t}$

and an integration over it

$P\left(T\left(E\right) \leq t \right) = \int_0^t \lambda e^{-\lambda \tau} \, \textrm{d}\tau = 1 - e^{-\lambda t}$.

It follows that the odds of the event not occurring by $t$ is

$P\left(T\left(E\right)\gt t\right) = 1 - \left(1 - e^{-\lambda t}\right) = e^{-\lambda t}$.

Returning to my three-event model, the probability of observing event A at time $t$ but not events B nor C by that time is

$P\left(T\left(A\right) = t \cap T\left(B\right) \gt t \cap T\left(C\right) \gt t \right) = \lambda_A e^{-\lambda_A t} e^{-\lambda_B t} e^{-\lambda_C t} = \lambda_A e^{-\lambda t}$

where $\lambda = \lambda_A + \lambda_B + \lambda_C$, the overall rate of events. (It is worth noting that this calculation corresponds with the steps for simulating under multiple Poisson processes.)

However, I see two non-equivalent approaches to calculate the probability of observing event A by time $t$ but not events B nor C by that time, $P\left(T\left(A\right) \leq t \cap T\left(B\right) \gt t \cap T\left(C\right) \gt t \right)$.

The first method I see is to explicitly evaluate each expression, as I did previously; that is,

$P\left(T\left(A\right) \leq t \cap T\left(B\right) \gt t \cap T\left(C\right) \gt t \right) = \left(1 - e^{-\lambda_A t}\right) e^{-\lambda_B t} e^{-\lambda_C t} = e^{-(\lambda-\lambda_A)t} - e^{-\lambda t}$.

The second is to integrate over all possible times:

$P\left(T\left(A\right) \leq t \cap T\left(B\right) \gt t \cap T\left(C\right) \gt t \right) = \int_0^t \lambda_A e^{-\lambda \tau} \, \textrm{d}\tau = \frac{\lambda_A - \lambda_A e^{-\lambda t}}{\lambda}$.

Both approaches seem equally valid to me. Which calculation should I be using, if either, and why? Thanks in advance for your help; I really appreciate it.

Answer 1

Let's begin by observing that

$$\Pr\left(T\left(E\right) = t \right) = 0.$$

The probability density function $\lambda e^{-\lambda t}$, which is not a probability, does become a probability when multiplied by an (infinitesimal) time difference $dt$, equal to $\lambda e^{-\lambda t}dt$ (and the $dt$ is an essential part of it).

With this in mind, consider the final expression

$$\Pr\left(T\left(A\right) \leq t \cap T\left(B\right) \gt t \cap T\left(C\right) \gt t \right).$$

Assuming, as the question seems to do, that the three events $T(A)$, $T(B)$, and $T(C)$ are independent, this means

$$\eqalign{ \Pr\left(T\left(A\right) \leq t \cap T\left(B\right) \gt t \cap T\left(C\right) \gt t \right) &=\Pr\left(T\left(A\right)\le t\right)\Pr\left(T\left(B\right)\gt t\right)\Pr\left(T\left(C\right)\gt t\right) \\ &=\left(\int_0^t \lambda_A e^{-\lambda_A u}e^{-\lambda_B t} e^{-\lambda_C t} du\right) \\ &=\left(\int_0^t \lambda_A e^{-\lambda_A u}du\right)e^{-\lambda_B t} e^{-\lambda_C t} \\ &= \left(1 - e^{-\lambda_A t}\right) e^{-\lambda_B t} e^{-\lambda_C t} }$$

which agrees with the alternative (penultimate) calculation in the question. The point evidently is that you should not be integrating the "$t$"s that appear in the events $T\left(B\right)\gt t$ and $T\left(C\right)\gt t$. It is clear that $e^{-\lambda_B t}$ and $e^{-\lambda_C t}$ do not belong inside the integral because they are not accompanied by the "$du$" that is needed for integration to make sense and, indeed, they are independent of the variable of integration $u$ altogether.

At this juncture one might protest and ask why shouldn't all three of the probability density functions be multiplied in the integral? Remembering the essential infinitesimal part, this would cause us to write

$$\int_0^t \left(\lambda_A e^{-\lambda_A u} du\right) \left(-\lambda_B e^{-\lambda_b u} du\right) \left(-\lambda_C e^{-\lambda_c u} du\right) = \int_0^t \lambda_A \lambda_B \lambda_C e^{-\lambda u} (du)^3$$

which, containing the triple infinitesimal $(du)^3$, makes no sense: that would be a nice automatic warning against making this natural mistake.

Answer 2

I was initially puzzled by why the 2nd expression is wrong, because the integrand is the probability, $\lambda_A/\lambda$, A occurs given an event occurs, multiplied by the probability density, $\lambda e^{-\lambda \tau}$, the 1st event occurs. Thus, the integrand is the probability density the 1st event occurs, and the event is A. How is this different from the probability density A occurs, while B and C do not?

Further thought led me to realize the 2nd expression gives the probability the 1st event occurs before $t$, and is A, instead of the probability A occurs, while B and C do not, because B and C can still occur before $t$ after the 1st A event. What we really want is the probability at least 1 event occurs before $t$, and all are A:

$$ P(T(A)\leq t \cap T(B)>t \cap T(C)>t)=\sum_{n=1}^\infty P_{Poi}(n;\lambda t)(\frac{\lambda_A}{\lambda})^n=e^{-(\lambda-\lambda_A)t}-e^{-\lambda t} $$

$dP/dt$ can be interpreted as the probability density B and C do not occur, and A has occurred.