# Integral over a Markov process

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.

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.

• 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 $\mathcal{F}_t = \sigma(X_s ; s \leq t)$ Sep 25, 2021 at 11:05
• 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 $G(T-t,Z_t) := E_{Z_t}[\Phi_{T-t}].$ Sep 25, 2021 at 11:06
• So my solution 2 is equivalent to yours using the shift operator Sep 25, 2021 at 11:09
• 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. Sep 25, 2021 at 12:34
• 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. Sep 25, 2021 at 12:38