This question is regarding the convolution theorem. So far, I have only used it as a tool to solve some recurrence relations. A nice example that I am familiar with is the renewal formula for finding first passage probabilities $-$ consider a continuous time Markov process on a discrete state space $\mathcal{S}$. We denote by $P(C,t|C\_0)$ (where $C,C\_0\in\mathcal{S}$) the probability density that the process is in state $C$ at time $t$, given that initially it was in state $C\_0$. Then, we have $$P(C,t|C\_0) = \int\_0^t dt'~F(C,t'|C\_0)P(C,t-t'|C)$$  

where $F(C,t|C\_0)$ denotes the probability density that, starting from $C\_0$, the first visit to configuration $C$ happens at time $t$. We can define Laplace transforms of $P$ and $F$, and using convolution theorem, we get $$F(C,s|C\_0)=\frac{P(C,s|C\_0)}{P(C,s|C)}.$$ where $s$ is the Laplace variable.


Thinking more along these lines, consider the following equation $$P(C,t|C\_0) = \sum\_{C'\in\mathcal{S}} P(C',t^\*|C\_0)P(C,t-t^\*|C')$$ where $t^\*$ is some intermediate time between $0$ and $t$. Could the above equation be classified as a convolution in some abstract configuration space? If yes, could a convolution theorem like result be used to *simplify* the equation (bring it into a product form in the abstract space)? I understand that if we were considering a $1D$ lattice, then going from $C\_0=x\_0$ to $C'=x'$ would be equivalent to a displacement of $x'-x\_0$. And similarly going from $C'$ to $C=x$ is equivalent to covering up the rest of the displacement $x-x'$ needed to finally reach $x$. This allows us to view the problem as a conventional convolution in space. But what if the state space isn't *ordered* like a lattice, and instead has an arbitrary graph structure?


I would appreciate any comments/suggestions that you may have regarding the question.



Convolution in state space of a Markov process?
$\newcommand\S{\mathcal S}\newcommand\V{\mathcal V}\newcommand\tC{\tilde C}\newcommand\tD{\tilde D}$
The answer is: in general, no. Indeed, suppose the contrary: that you have an imbedding $f$ of your state space $\S$ into a vector space $\V$ such that for
$$Q(D',t|D):=
\left\{
\begin{aligned}
P(f^{-1}(D'),t|f^{-1}(D))\quad&\text{if}\quad f^{-1}(\{D'\})\ne\emptyset \text{ and } f^{-1}(\{D\})\ne\emptyset, \\
0\quad&\text{if}\quad f^{-1}(\{D'\})=\emptyset \text{ and } f^{-1}(\{D\})\ne\emptyset, \\
1(D'=0)\quad&\text{if}\quad f^{-1}(\{D\})=\emptyset, 
\end{aligned}
\right.
\tag{0}$$
we have
$$Q(D'+d,t|D+d)=Q(D',t|D)\tag{1}$$
for all $D',D,d$ in $\V$.


Suppose now that there are $C',C,\tC$ in $\S$ such that
$$0<P(C',t|C)\ne P(\tC',t|\tC)\tag{2}$$
for all $\tC'\in\S$. Let
$$D':=f(C'),\quad D:=f(C),\quad \tD:=f(\tC),\quad d:=\tD-D,\quad \tD':=D'+d.\tag{3}$$One of the following two cases takes place:


*Case 1: $f^{-1}(\{\tD'\})=\emptyset$*. Then, by (3), (0), and (2),
$$Q(D'+d,t|D+d)=Q(\tD',t|D+d)=0\ne P(C',t|C)=Q(D',t|D),$$
so that (1) fails to hold.


*Case 2: $f^{-1}(\{\tD'\})\ne\emptyset$*, so that $f(\tC')=\tD'$ for some $\tC'\in\S$. Then, again by (3), (0), and (2),
$$Q(D'+d,t|D+d)=Q(\tD',t|\tD)=P(\tC',t|\tC)\ne P(C',t|C)=Q(D',t|D),$$
so that (1) fails to hold in Case 2 either.


Thus, you cannot have a desired imbedding whenever condition (2) is satisfied for some $C',C,\tC$ in $\S$ and all $\tC'\in\S$ -- which should be generically so. $\Box$


