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In https://en.wikipedia.org/wiki/It%C3%B4's_lemma

Under the section of Poisson jump processes, it is said that

We may also define functions on discontinuous stochastic processes. Let h be the jump intensity. The Poisson process model for jumps is that the probability of one jump in the interval [t, t + Δt] is hΔt plus higher order terms.

How can I know what the higher order terms are? Is jump intensity not accurate/sufficient?

For example, the higher order terms are $h(Δt)^3$, but the jump intensity hides that, so why do we still use jump intensity?

2 Answers 2

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Jump intensity is sufficient to describe a Poisson process. The intuitive reason is because the Poisson process is memoryless. For a tiny fraction of time, there is a tiny probability $h\Delta t$ of jumping, and if it doesn't jump, the next instant of time is another independent trial. So only the leading order in $\Delta t$ matters since only the limit $\Delta t\rightarrow 0$ matters.

You can say things about longer intervals of time, but they're uniquely implied by the infinitessimal description. For instance, the number of jumps in a time period $T$ is Poisson distributed with mean $hT.$

I should add that this is not only true for Poisson processes. For instance, modulated poisson processes and jump diffusion also involve an instantaneous jump intensity... it just can change or fluctuate randomly in time. The key is that it gives a momentary infinitessimal probability of jumping (i.e. an instantaneous rate).

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    Thanks. But let's say the probability of jump inside $\Delta t$ is $h(Δt)+h(Δt)^3$. Thus the intensity is also $h$. I understand that for infinitesimal $dt$ it does not matter and for long interval the mean is hT. But often we are interested in very small but finite interval. In that case we have the misleading information that the jump probability if $h(Δt)$, instead of $h(Δt)+h(Δt)^3$? – 2017-01-11
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    @peter When are we interested in a small but finite interval? – 2017-01-11
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    @peter Do you find it implausible that a Poisson process is uniquely specified by constant jump intensity in small intervals and independent increments? – 2017-01-11
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    i think i may be talking another process. Is there a random process where the probability for finite interval depends on second or higher order of interval length, so intensity is not enough to describe the actual phenomenon? – 2017-01-11
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    @peter As soon as we drop the Markov assumption, things are no longer determined by their infinitessimal behavior, but then it's unclear why we'd be seeking a description in terms of jump probabilities over a short time. – 2017-01-11
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    say i am doing a simulation the interval is finite. In that case, what I should implement is the probability is $h(Δt)+h(Δt)^3$, instead of $h(Δt)$, even with very tight error control, because the former is the "actual" probability. – 2017-01-11
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    If the probability is $h(Δt)+h(Δt)^3$ over a very small finite interval $\Delta t$, can this process be poisson? – 2017-01-11
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    Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/51572/discussion-between-spaceisdarkgreen-and-peter). – 2017-01-11
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You can actually derive the higher order terms. Say that the jump intensity is $h·Δt+C·Δt^2+O(Δt^3)$. Now subdivide into $N$ partial intervals of equal length. Then the probability in each of the small intervals is $h·Δt/N+C·Δt^2/N^2+O(Δt^3/N^3)$. The probability of no jump in all of the sub-intervals is $$ \bigl(1-h·Δt/N-C·Δt^2/N^2-O(Δt^3/N^3)\bigr)^N $$ which converges to $e^{-h·Δt}$ independent of the original higher order terms. This then just tells us that the higher order terms are those of the series expansion of the exponential, $$ 1-e^{-h·Δt}=h·Δt-\frac12h^2·Δt^2+\frac16h^3·Δt^3\mp… $$