[Math] Cumulative distribution function (of NHPP inter-arrival time) not tending to 1

probability distributionsstochastic-processes

According to this website, for a non-homogeneous Poisson process with mean $m(t) = \int^t_0 \lambda(u) \, dt$, the cumulative distribution function (CDF) for the inter-arrival time to the first event is,

$$F_0(t) = 1 – e^{- m(t)}.$$

As $t \rightarrow 0$, $m(t) \rightarrow 0$, so $F_0(t) \rightarrow 1-e^{0} = 1- 1 = 0.$ So that is fine.

However, as $t \rightarrow \infty$, $m(t) \rightarrow \int^{\infty}_0 \lambda(u) \, du$, so $F_0(t) \rightarrow 1 – e^{-\int^{\infty}_0 \lambda(u) \, du}$ which is $1$ only if $\int^{\infty}_0 \lambda(u) \, du$ is $\infty$.

Question

Does this mean that the intensity function $\lambda(u)$ of a non-homogeneous Poisson process must be defined such that $m(t) \rightarrow \infty$ as $t \rightarrow \infty$? We cannot have an intensity function that drops to $0$ and remain $0$ after a certain point?

Best Answer

If the limit $m(+\infty)$ of $m(t)$ when $t\to+\infty$ is finite, the total number of points is Poisson with parameter $m(+\infty)$, hence there is no point at all with positive probability $\mathrm e^{-m(+\infty)}$. When this happens, there is no first event, in other words the time to wait until the first event is $+\infty$. Hence the defective distribution.

This may happen even when $\lambda(t)\gt0$ for every $t\geqslant0$, and does happen if and only if $\lambda$ is integrable, for example $\lambda(t)=\frac1{1+t^2}$ for every $t\geqslant0$.

Related Solutions

[Math] the correct inter-arrival time distribution in a Poisson process

Let $X(t)$ be a Poisson process with rate $\lambda$. For any time $t>0$, $X(t)$ follows a Poisson distribution $\mathcal{Po}(\lambda t)$. The probability that the count remains at zero at time $t$, given that is started with zero is then $$ \mathbb{P}\left(X(t)=0\right) = \mathrm{e}^{-\lambda t} $$ Let $T_1$ be the time of the first jump (a random variable). Then the event $\{X(t) = 0\}$ is the same as the event $\{T_1 > t\}$, meaning that the first jump will occur after epoch $t$, i.e. $$ \mathbb{P}\left(T_1 > t\right) = \mathbb{P}\left(X(t)=0\right) = \mathrm{e}^{-\lambda t} $$ That means that $T_1$ is an exponential random variable, whose density is $\lambda \mathrm{e}^{-\lambda t}$.

In order to establish that Poisson inter-arrival times are also exponential random variables we need to consider the values of the process at times $s$ and $t$, such that $s<t$. Since Poisson process has independent increments: $$ \mathbb{P}\left(X(s) = n, X(t) = m\right) = \mathbb{P}\left(X(s) = n, X(t) - X(s) = m -n\right) \stackrel{\text{indep. incr.}}{=} \\ \mathbb{P}\left(X(s)=n\right) \cdot \mathbb{P}\left(X(t-s)=m-n\right) = \frac{(\lambda s)^n}{n!} \mathrm{e}^{-\lambda s} \frac{(\lambda (t-s))^{m-n}}{(m-n)!} \mathrm{e}^{-\lambda (t-s)} \mathbf{1}_{ m \geqslant n \geqslant 0} = \\\frac{(\lambda t)^m}{m!} \mathrm{e}^{-\lambda t} \cdot \binom{m}{n} \left(\frac{s}{t}\right)^n \left(1-\frac{s}{t}\right)^{m-n} \mathbf{1}_{ m \geqslant n \geqslant 0} $$ This implies, for $s>0$: $$ \mathbb{P}\left(X(t+s) = n| X(t) = n\right) = \frac{\mathbb{P}\left(X(t+s) = n, X(t) = n\right)}{\mathbb{P}\left(X(t) = n\right)} = \mathrm{e}^{-\lambda s} $$ That is, given that the state of the process at times $t$, the probability that no jump occurs within $s$ seconds is $\mathrm{e}^{-\lambda s}$, i.e. the time to the next jump is an exponential random variable with rate $\lambda$.

Of course, this is all a consequence of the independence of increments and the property $X(t+s) - X(t) \stackrel{d}{=} X(s)$.

[Math] Poisson Process, Exponential inter-arrival times simulation conundrum

There does not seem to be anything wrong with your interpretation. I took your code and run it in R:

nextArrival <- 0
prevArrival <- 0
k <- 0
while (nextArrival<60) {
  nextArrival <-prevArrival +rexp(1,0.5)
  prevArrival <- nextArrival  
  k <- k+1
}
print(k);

And obtain counts that look like Poisson variable. You may want to double check your implementation.

Best Answer

Related Solutions

[Math] the correct inter-arrival time distribution in a Poisson process

[Math] Poisson Process, Exponential inter-arrival times simulation conundrum

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