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For a (index-)homogeneous Markov process $X_t$, its infinitesimal generator A is defined to act on suitable functions $f : \mathbb R^n → \mathbb R$ by $$ A f (x) = \lim_{t \downarrow 0} \frac{\mathbf{E}^{x} [f(X_{t})] - f(x)}{t}. $$

  1. In the case of a inhomogeneous Markov process, how is its generator defined?

  2. How can we go from the generator of a (inhomogeneous or homogeneous) Markov process to the Markov process (or its transition kernels/probabilities)?

    There are two steps here, if I am correct:

    • first go from the generator to the one-parameter semigroup of operators defined by the cauchy problem, and
    • secondly go from the semigroup of operators to the transition kernels, by letting the integrand to be the indicator function of a measurable subset.

Thanks and regards!

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Tim
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    As far as I'm concerned the generator doesn't make (great) sense in the inhomogeneous case. In the non-homogeneous case there is no transition semigroup, just a two-parameter Markovian family: $$ P^{s,t} f(x) = E(f(X_t) , | , X_s = x)$$ When the process is homogeneous the above family only depends on the difference $s-t$ and you write $$ P^{s,t} f(x) = E(f(X_t) , | , X_s = x) = E(f(X_{t-s} , | , X_0 = x)) = E^x(f(X_{t-s}))$$ – Bunder Apr 08 '13 at 12:39
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    The generator $A$ makes sense in the homogeneous case since it generates the semigroup of transition probabilities (at least for a suitable subfamily of functions $f$): $$ E^x[f(X_t)] = P^t f(x)= f(x) + Af(x) + o(t) $$ If there is a generalization, it will would be something like: $$ A(s)f(x) = \lim_{t \to s} \frac{E(f(X_t) , | , X_s = x) - f(x)}{t-s} $$ But I don't know if this object can be exploited as well as the generator in the homogeneous case. – Bunder Apr 08 '13 at 12:39
  • @Bunder: Thanks! In $E^x[f(X_t)] = P^t f(x)= f(x) + Af(x) + o(t)$, do you miss a $t$ that multiplies $Af(x)$? I.e. do you mean $E^x[f(X_t)] = P^t f(x)= f(x) + [Af(x)], t + o(t)$ instead? – Tim Apr 08 '13 at 13:19
  • indeed :). I can't correct it though. – Bunder Apr 08 '13 at 14:28
  • @Bunder:Thanks! Regarding the second part of my post, are the two steps correct for going from the generator of a (inhomogeneous or homogeneous) Markov process to the Markov process (or its transition kernels/probabilities)? – Tim Apr 08 '13 at 18:14

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