I have the following questions: Let $Z$ be a continuous one-dimensional Markov process on some probability space $(\Omega,\mathcal{F},\mathbb{P})$ and $\mathcal{F}_t = \sigma(Z_s,s \leq t)$. Then show that for all $T > 0$:

for all $t \leq T$ there exists a function $F(t,T,\cdot)$ such that

$$E[e^{-\int_{t}^T R(Z_s) ds} | \mathcal{F}_t] = F(t,T,Z_t),$$

where $R$ is some function. If $Z$ is time-homogeneous then for all $t\leq T$ there exists a function $G$ such that $$F(t,T,z) = G(T-t,z).$$

So for question (1) I intuitively know that $E[e^{-\int_{t}^T R(Z_s) ds} | \mathcal{F}_t] = E[e^{-\int_{t}^T R(Z_s) ds} | Z_t] $ but how do I prove it? For question (2) I was thinking about expressing it in terms of $Q_{T-t}f(Z_t)$ where $(Q_{t})_{t \geq 0}$ is the transition semi-group, but since there is an integral expression I have no clue how to do this.


1 Answer 1


Regarding question (1): There are many ways to phrase the Markov property. One of the more convenient ones is as follows: if $\Phi$ is a non-negative function measurable with respect to $\mathcal F_{t,\infty} := \sigma\{X_s : s \ge t\}$, then $$ E[\Phi | \mathcal F_t] = E[\Phi | X_t] \text{ a.s.} $$ Furthermore, by the Doob–Dynkin lemma, the right-hand side is equal a.s. to a Borel function of $X_t$.

Clearly, in our case $\Phi_{t,T} = \exp(-\int_t^T R(Z_s) ds)$ is $\mathcal F_{t,\infty}$-measurable (thanks to continuity of $Z_s$; I assume that $R$ is nice enough — say: locally bounded from below), so clearly $$ E[\Phi_{t,T} | \mathcal F_t] = F_{t,T}(X_t) \text{ a.s.} $$ for a Borel function $F_{t,T}$.

The true question is whether we can choose $F_{t,T}$ in such a way that $(t,T,x) \mapsto F_{t,T}(x)$ is jointly measurable. I expect that under reasonable assumptions (say: the function $R$ is bounded from below), this is standard, but tiresome; however, I did not attempt to write a detailed proof.

(What is easy is that $\Phi_{t,T}$ is uniformly bounded and almost surely continuous with respect to $t, T$, and hence, by the dominated convergence theorem, $(t,T) \mapsto \Phi_{t,T}$ is a continuous map into $L^1(\Omega, P)$; in particular, $(t,T) \mapsto F_{t,T}(X_t)$ is a continuous map into $L^1(\Omega, P)$.)

The answer to question (2) really depends on your favourite definition of a time-homogeneous Markov process. If we are allowed to use the shift operators $\theta_t$ and expectation $E^x$ for the process started at $x$, then things are pretty clear: we have $\Phi_{t,T} = \Phi_{0,T-t} \circ \theta_t$, and hence $$ E[\Phi_{t,T} | \mathcal F_t] = E[\Phi_{0,T-t} \circ \theta_t | \mathcal F_t] = G_{T-t}(X_t) , $$ where $$ G_s(x) = E^x \Phi_{0,s} $$ is a Borel function. Noteworthy, here joint measurability of $(s,x) \mapsto G_s(x)$ presents no difficulties: $s \mapsto \Phi_{0,s}$ is a continuous map into $L^1(\Omega, P)$, and so $s \mapsto G_s(x)$ is continuous for every $x$.

I should also add that there is no simple expression for the conditional expectation of the Feynman–Kac functional $\Phi_{t,T}$ in terms of the transition semigroup of the process.

  • $\begingroup$ thank you very much for your quick answer. Regarding (1), I have never seen this definition to be honest, the definition I know is actually the other way around: $$E[f(X_t) | \mathcal{F}_s] = E[f(X_t) | Z_s].$$ where $\mathcal{F}_t = \sigma(X_s ; s \leq t)$ $\endgroup$
    – Oli Bernet
    Sep 25, 2021 at 11:05
  • $\begingroup$ For (2), your notation for the function $\Phi$ got me thinking about the simple Markov property, and I came up with the following solution: Let $\mathbb{D}(\mathbb{R})$ be set of cadlag functions $t \mapsto f(t)$ Define $\Phi_{T-t}:\mathbb{D}(\mathbb{R}) \mapsto [0,\infty)$ by $\Phi_{T-t}(f) = e^{-\int_0^{T-t} f(s) ds}.$ by simple Markov property $E_{Z_t}[\Phi_{T-t}] = E[\phi((R(Z_{t+s}))_{s \geq 0}) | \mathcal{F}_t] = E[e^{-\int_0^{T-t} R(Z_{s+t}) ds} | \mathcal{F}_t] = E[e^{-\int_t^{T} R(Z_s) ds} | \mathcal{F}_t],$ thus define $G(T-t,Z_t) := E_{Z_t}[\Phi_{T-t}].$ $\endgroup$
    – Oli Bernet
    Sep 25, 2021 at 11:06
  • $\begingroup$ So my solution 2 is equivalent to yours using the shift operator $\endgroup$
    – Oli Bernet
    Sep 25, 2021 at 11:09
  • $\begingroup$ Regarding your first comment: this is equivalent. Indeed, the $\sigma$-algebra generated by $f(X_t)$, where $t \geqslant s$ and $f$ is a Borel function, is precisely the $\sigma$-algebra $\mathcal F_{s,\infty}$ in my answer. $\endgroup$ Sep 25, 2021 at 12:34
  • $\begingroup$ In your second comment, you either implicitly use the shift operators (as suggested by the notation $\phi((R(Z_{t+s}))_{s\geqslant 0})$; I guess you meant $\Phi_{T-t}$ rather than $\phi$) or you work with the canonical realisation of the process with $\Omega$ being simply $\mathbb D(\mathbb R)$ (as suggested by $E_{Z_t}[\Phi_{T-t}]$. Both are fine, but mixing them is somewhat inconsistent. $\endgroup$ Sep 25, 2021 at 12:38

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