Let $(X(t))_{t \geqslant 0}$ be a continuous-time Markov process with state space $I=\{1, \ldots, n\}$ and generator $Q=\left(q_{i j}\right)_{i, j \in I}$ satisfying $q_{i j}=q_{j i}$ for all $i, j \in I$. The local time up to time $t>0$ of $X$ is the random vector $L(t)=\left(L_{i}(t)\right)_{i \in I} \in \mathbb{R}^{n}$ defined by

$$L_{i}(t)=\int_{0}^{t} 1_{X(s)=i} d s \quad(i \in I)$$

(a) Let $f: I \times \mathbb{R}^{n} \rightarrow \mathbb{R}$ be any function that is differentiable with respect to its second argument, and set

$$f_{t}(i, \ell)=\mathbb{E}_{i} f(X(t), \ell+L(t)), \quad\left(i \in I, \ell \in \mathbb{R}^{n}\right)$$

Show that

$$\frac{\partial}{\partial t} f_{t}(i, \ell)=M f_{t}(i, \ell)$$

where

$$M f(i, \ell)=\sum_{j \in I} q_{i j} f(j, \ell)+\frac{\partial}{\partial \ell_{i}} f(i, \ell)$$

(b) For $y \in \mathbb{R}^{n}$, write $y^{2}=\left(y_{i}^{2}\right)_{i \in I} \in[0, \infty)^{n}$ for the vector of squares of the components of $y$. Let $f: I \times \mathbb{R}^{n} \rightarrow \mathbb{R}$ be a function such that $f(i, \ell)=0$ whenever $\sum_{j}\left|\ell_{j}\right| \geqslant T$ for some fixed $T$. Using integration by parts, or otherwise, show that for all $i$

$$-\int_{\mathbb{R}^{n}} \exp \left(\frac{1}{2} y^{T} Q y\right) y_{i} \sum_{j=1}^{n} y_{j} M f\left(j, \frac{1}{2} y^{2}\right) d y=\int_{\mathbb{R}^{n}} \exp \left(\frac{1}{2} y^{T} Q y\right) f\left(i, \frac{1}{2} y^{2}\right) d y,$$

where $y^{T} Q y$ denotes $\sum_{k, m \in I} y_{k} q_{k m} y_{m}$.

(c) Let $g: \mathbb{R}^{n} \rightarrow \mathbb{R}$ be a function with $g(\ell)=0$ whenever $\sum_{j}\left|\ell_{j}\right| \geqslant T$ for some fixed $T$. Given $t>0, j \in I$, now let

$$f(i, \ell)=\mathbb{E}_{i}\left[g(\ell+L(t)) 1_{X(t)=j}\right]$$

in part (b) and deduce, using part (a), that

$$\begin{aligned} \int_{\mathbb{R}^{n}} \exp \left(\frac{1}{2} y^{T} Q y\right) y_{i} y_{j} g\left(\frac{1}{2} y^{2}\right) d y \\ &=\int_{\mathbb{R}^{n}} \exp \left(\frac{1}{2} y^{T} Q y\right)\left(\int_{0}^{\infty} \mathbb{E}_{i}\left[1_{X(t)=j} g\left(\frac{1}{2} y^{2}+L(t)\right)\right] d t\right) d y \end{aligned}$$

[You may exchange the order of integrals and derivatives without justification.]