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What is the relation between the infinitesimal generator $\mathcal{A}$ of an Ito diffusion and the transition density function of the process?

More specifically, let's consider this subcase. Suppose I have the transition density $p_t(x|y)$ of a process $X_t$ with infinitesimal generator $\mathcal{A}$. Let $\kappa\in\mathbb{R}$ be fixed. What is the transition density of the process generated by $\kappa\mathcal{A}$? Is other information needed?

My thoughts: from PDEs, scaling the heat equation essentially scales time in the fundamental solution. So I think we should get something like $\tilde{p}_{\kappa t}(x|y)$ for the density of the new process. Not sure how to show it though.

user3658307
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  • Is your SDE autonomous, so that the generator does not depend explicitly on time? (It's not necessary, just a little bit simpler to explain.) – Ian Apr 04 '17 at 00:50
  • @Ian Yes, the diffusion process does not depend on time directly (i.e. it's $dX_t = \mu(X_t),dt + \sigma(X_t),dB_t$). – user3658307 Apr 04 '17 at 01:06

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By definition, the (strong) generator of a time-homogeneous (or "autonomous" if you like) Markov process is

$$\mathcal{A}[u](x)=\left. \frac{d}{dt} [E[u(X_t) \mid X_0=x] \right |_{t=0}.$$

So

$$\kappa \mathcal{A}[u](x) = \kappa \left. \frac{d}{dt} E[u(X_t) \mid X_0=x] \right |_{t=0} \\ = \left. \frac{d}{d(t/\kappa)} E[u(X_t) \mid X_0=x] \right |_{t=0} \\ = \left. \frac{d}{dt} E[u(X_{\kappa t}) \mid X_0=x] \right |_{t=0}. $$

That is, $\kappa \mathcal{A}$ is just the generator of $X_{\kappa t}$. So its transition density is just $p_{\kappa t}$. In effect you have, as you said, just accelerated time by a factor of $\kappa$.

Notably, doing so multiplies $\mu$ by $\kappa$ and multiplies $\sigma^2$, not $\sigma$, by $\kappa$. That is, multiplying through by $\kappa$ in the generator does not do the same thing as multiplying through by $\kappa$ in the SDE, because $\kappa dB_t$ is not the same as $dB_{\kappa t}$.

Ian
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  • Thanks very much! Small question: it multiplies $\mu$ by $\kappa$, but does it multiply $\sigma$ by $\sqrt{\kappa}$ if $\sigma$ is symmetric positive definite? – user3658307 Apr 04 '17 at 01:23
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    @user3658307 So this is a slightly tricky matter: $\sigma$ is not uniquely defined from the generator. Thus, $\sigma^2$ becomes $\kappa \sigma^2$, and given an admissible SPD matrix $\sigma$ for the old generator, $\sqrt{\kappa} \sigma$ is an admissible SPD matrix for the new generator. But there are other such SPD matrices. What is going on here is the fact that if $Q$ is an orthogonal matrix then $QB_t$ is another Brownian motion, which from the distributional perspective we cannot distinguish from $B_t$. – Ian Apr 04 '17 at 01:25
  • Very cool; thank you! Hmm, so, given that the SPD matrix root of a real SPD matrix is unique, if we assume instead that $\sigma^2$ is real SPD, I suppose demanding $\sigma$ be SPD then yields a unique $\sigma$? – user3658307 Apr 04 '17 at 02:03
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    @user3658307 Yes, that's right. Also, I made an error in my last comment: I should not have said that my $\sigma$s were required to be SPD, but only that they are solutions to $\sigma \sigma^T = \sigma^2$. The point of my comment is that these are not unique, they are only uniquely determined up to an orthogonal matrix. But if you require them to be SPD then they are unique. Relatedly, if you require them to be triangular then they are also unique. – Ian Apr 04 '17 at 02:05