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In a homogeneous Poisson process with rate $\lambda$, what is the probability of observing an event in an "instant," that is, an infinitesimally small interval of length dt? I have read that the Poisson rate function $\lambda(t)$ can be defined as the "instantaneous probability of observing a spike at each point in time." (http://www.stat.columbia.edu/~liam/teaching/neurostat-spr11/uri-eden-point-process-notes.pdf) But for a homogeneous process with $\lambda(t) = \lambda$, how can this be when it is possible that $\lambda > 1$?

Eli Stein
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1 Answers1

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The instantaneous probability of observing a spike between $t$ and $t + dt$ is $\lambda(t)dt$ (mind the $dt$ term). This can be noticed directly from the definition. For example with the homogenous Poisson process: $$ P [(N(t+ \tau) - N(t)) = k] = \frac{e^{-\lambda \tau} (\lambda \tau)^k}{k!} $$

looking at $k = 1$ and $\tau = dt$ gives $P[dN(t)] = \lambda dt$.

ThePawn
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  • Thank you! But is there a well-defined limit for when dt approaches zero? – Eli Stein Jan 12 '13 at 14:10
  • I am not sure I understand your interrogation. Could you have a look at http://stats.stackexchange.com/questions/11956/why-is-the-first-postulate-of-the-poisson-process-that-lambda-dt-is-the-proba?rq=1 and see if it answers your question? – ThePawn Jan 12 '13 at 14:15
  • Thank you for asking me to clarify (and thank you for your help!) I am trying to define $X(t)$ that takes the value $1$ if there is an event at $t$, and 0 else. My goal is to find the expectation of $X(t)$. Is this question well-posed? This goes a bit beyond my undergraduate training in statistics. – Eli Stein Jan 12 '13 at 14:21
  • If you consider a poisson process, it makes no sense to consider an event at exactly $t$ because the probability will be zero. Similarly, if you have a gaussian variable $G$, the probability that $G$ be exactly equal to some real $x$ is zero. It is indeed more interesting to consider the amount $P[G \in [x, x + dx]] = f(x)dx$ where we define $f$ as the density function of $G$. – ThePawn Jan 12 '13 at 14:30