# Rigorously showing that a process is Markov (edit made)

Let $$S_n = X_1 + \dots + X_n$$, where the increments $$X_i$$ are not i.i.d. Let $$\tau$$ be some random time (not necessarily a stopping time) and assume $$(\tau, X_1, \dots, X_\tau)\quad \text{ and } \quad (X_{\tau + 1}, X_{\tau + 2}, \dots) \qquad (1)$$ to be independent. I want to show with rigor that the process $$(S_{\tau + n}- S_\tau)_{n \geq 1}$$ is a Markov chain (satisfies the Markov property). For convenience, let us write $$\tilde{S}_n = S_{\tau + n} - S_\tau$$. As $$\tilde{S}_n$$ is independent of $$\tau$$ and $$S_\tau$$ it makes absolutely sense that the process is Markov but I would like to see the argument in more detail.

Edit: Possibly, the assumption (1) is not enough. Let us in addition to (1) assume that $$(X_1, X_2, \dots) \overset{d}{=} (X_{\tau + 1}, X_{\tau +2}, \dots).$$ Maybe this is now enough to prove the claim that $$\tilde{S}_n$$ is a Markov chain.

Edit2: I am starting to think that the assumptions are not necessary. Simply by the structure of $$S$$ and in view of $$\tilde{S}_n = X_{\tau + 1 } + \dots + X_{\tau + n}$$ we have should have that $$\tilde{S}_n$$ conditioned on the values of $$\tilde{S}_1, \dots , \tilde{S}_{n-1}$$ only depends on $$\tilde{S}_{n-1}$$. Is this true / why is it not that easy?

• Is the independence well defined? $X_{\tau+k}$ is a well defined random variable for $k\geq 0$ since it is just $X_{\tau+k}(\omega) = \sum_{i=0}^\infty X_{i+k}(\omega) \mathcal{X}_{\tau = i}(\omega)$. But what is hidden on the dots of $X_1, \cdots, X_\tau$? What is $X_{\tau-1}$ and how many variables are there? Aug 10, 2020 at 6:37
• @LucasResende Maybe first of all, how to read assumption (1): The random variables up to the random time $\tau$, that is $X_1, \dots, X_\tau$, are independent from the random variables after the random time $\tau$, that is $X_{\tau+1}, X_{\tau +2}, \dots$. If I understand your question correctly: Hidden in the dots are the random variables $X_{\tau - i}$, where it is difficult to write them out precisely, because it is only well-defined for $i \leq \tau$ and $\tau$ is as well random. So, the number of variables that are in $(X_1, \dots, X_\tau)$ is $\tau$. I hope this helps.
– MMM
Aug 10, 2020 at 14:29
• But $\tau$ is a random variable. You can't say that $(X_1, \cdots, X_k)$ is independent of $(X_{k+1}, X_{k+2}, \cdots)$ where $\tau=k$. Aug 10, 2020 at 14:58
• @LucasResende That is true. But I am afraid that I don't really understand the question. To me the independence stated in (1) makes sense.
– MMM
Aug 10, 2020 at 16:31
• My point is that your independence seems to be dependent on the value of $\tau$. You can't say that $(X_1, \cdots, X_k)$ is independent of $(X_{k+1}, \cdots)$ when $\tau=k$ and $(X_1, \cdots, X_j)$ is independent of $(X_{j+1}, \cdots)$ when $\tau=j$. You need to properly define what you mean by $X_{1}, \cdots, X_{\tau}$, because you can't change which variables are independent based on $\tau$. Aug 10, 2020 at 16:36

By assumption (1), $$S_{\tau}$$ and $$\tau$$ are independent from $$S_{\tau+ k } - S_\tau$$. Thus,
\begin{align} & P(S_{\tau + n } - S_\tau \in dx \vert S_{\tau + 1} - S_\tau \in dx_1, \dots , S_{\tau + n-1} - S_\tau \in dx_{n-1}) \\ & \overset{(1)}{=} P(S_{\tau + n } - S_\tau \in dx \vert S_{\tau + 1} - S_\tau \in dx_1, \dots , S_{\tau + n-1} - S_\tau \in dx_{n-1}, S_\tau = dy, \tau = m) \\ & = P(S_{m + n } \in dx+dy \vert S_{m + 1} \in dx_1 + dy, \dots , S_{m + n-1} \in dx_{n-1} + dy) \\ & = P(S_{m + n } \in dx+dy \vert S_{m + n-1} \in dx_{n-1} + dy) \\ & = P(S_{\tau + n } - S_\tau \in dx \vert S_{\tau + n-1} - S_\tau \in dx_{n-1}, S_\tau = dy, \tau = m) \\ & \overset{(1)}{=} P(S_{\tau + n } - S_\tau \in dx \vert S_{\tau + n-1} - S_\tau \in dx_{n-1}), \end{align} where I just used the definition of conditional probabilities when I wrote (1) over the equal sign. What do you think?
• I want to ask what does $\overset{d}{=}$ mean in your equation? Aug 16, 2020 at 18:21
• @GENIVI-LEARNER It means equal in distribution. So, $X \overset{d}{=} Y$, if $X$ and $Y$ follow the same distribution.