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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?

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  • $\begingroup$ 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? $\endgroup$
    – Lucas Resende
    Aug 10, 2020 at 6:37
  • $\begingroup$ @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. $\endgroup$
    – MMM
    Aug 10, 2020 at 14:29
  • $\begingroup$ 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 $\tau=k$. $\endgroup$
    – Lucas Resende
    Aug 10, 2020 at 14:58
  • $\begingroup$ @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. $\endgroup$
    – MMM
    Aug 10, 2020 at 16:31
  • $\begingroup$ 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 $\tau$. $\endgroup$
    – Lucas Resende
    Aug 10, 2020 at 16:36

1 Answer 1

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I will try to answer the question myself: (I would be grateful for feedback)

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?

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  • $\begingroup$ I want to ask what does $\overset{d}{=}$ mean in your equation? $\endgroup$
    – GENIVI-LEARNER
    Aug 16, 2020 at 18:21
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    $\begingroup$ @GENIVI-LEARNER It means equal in distribution. So, $X \overset{d}{=} Y$, if $X$ and $Y$ follow the same distribution. $\endgroup$
    – MMM
    Aug 18, 2020 at 20:58

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