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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.

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$\newcommand\S{\mathcal S}\newcommand\V{\mathcal V}\newcommand\tC{\tilde C}\newcommand\tD{\tilde D}$ 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. $\Box$

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    $\begingroup$ Thanks for a detailed answer. Condition (2) is sufficient to show that an imbedding doesn't exist in general. However, could we also write a set of necessary conditions for this? I would be really interested in knowing under what circumstances is such an imbedding possible, so if you could point me to any references, that would be very helpful. $\endgroup$ Nov 13, 2020 at 22:34
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    $\begingroup$ @StatisticalMechanic : As I wrote, condition (2) should be satisfied generically. However, at the moment I don't have a good idea about a nontrivial necessary and sufficient condition for the existence of such an imbedding. You may want to ask this question in a separate post. Also, I don't think that such questions have been considered in the existing literature. $\endgroup$ Nov 15, 2020 at 0:44

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