# Martingale that is not a Markov process: one technical issue in existing examples.

My question is technically not new. However, I am a bit confused by certain setups. A thread appeared here:

Martingale that is not a Markov process

The spirit of the answer provided by Did is clear. One detail is not so clear to me, though. Let me simplify to a discrete version of that argument (for instance, see https://imathworks.com/math/math-stochastic-process-that-is-martingale-but-not-markov/):

1. Pick random value for $$๐_0;$$

2. Consider a sequence of of random variables $$\{\epsilon_n: n \ge 0\}$$ that are i.i.d. with mean $$E[\epsilon_n]=0$$ AND independent of $$๐_0$$ (Note here like in Did's example independence is only assumed with respect to $$X_0;$$

3. Now we define are stochastic process: $$X_{๐+1}=๐_๐+\epsilon_{n+1}๐_0;$$

4. Then we proceed to prove that the stochastic process $$X=(X_n: n \ge 0 )$$ is a martingale. The proof goes along the lines: $$๐ธ[๐_{๐+1} | X_0,\ldots,๐_n]=๐ธ[๐_๐|๐_0 \ldots,๐_๐]+ E[\epsilon_{n+1}| X_0, \ldots, X_n]$$ $$=๐_๐ + ๐ธ[\epsilon_{๐+1}|๐_0, \ldots,๐_๐]๐ธ[๐_0|๐_0,\ldots, ๐_๐]$$ $$=๐_๐+E[\epsilon_{n+1}]X_0 = ๐_๐+ 0\cdot X_0 = X_n.$$

and here is where I am not starting to get the point, the author justifies this last step by saying that "it is key that $$\epsilon_n$$ is independent of the $$\{๐_i, 0\le i\le n\}.$$

MY QUESTION: where does the fact that $$\epsilon_n$$ is independent of the $$\{๐_i, 0\le i\le n\}$$ come from when our initial assumption was that the $$\epsilon_n$$'s are independent of $$X_0$$?

Once this is done then, yes, we have an example of a martingale that is not a Markov process.

The original example by Did, consisted in introducing a sequence $$(Z_t: t\ge 2)$$ of i.i.d. variables not identically null and with mean zero independent on $$X_0$$ (I also wonder what is the exact meaning of being independent ON $$X_0,$$ actually).

Then, Did lets $$๐_1=๐_0=1$$ and $$๐_๐ก=๐_{๐กโ1} + Z_๐ก๐_{๐กโ2}$$ for every $$t\ge 2.$$

Then, his argument uses the fact that $$E[X_t | F_{t-1}] = X_{t-1}$$ where I assume that $$F_t$$ is a member of the natural filtration for the process. He can do so using the argument that

$$๐ธ[๐_t X_{t-2}โฃ X_{t-1}] = 0$$ as a consequence of a general result that states that if $$๐$$ is independent on $$W,$$ then $$๐ธ[๐๐โฃ๐]=๐ธ[๐]๐ธ[๐โฃ๐].$$

I find myself with the same issue. $$Z_t$$ plays the role of $$\epsilon_{n+1}$$ in the previous example and $$Z_t$$ would have to be $$U$$ in applying the result for conditional expectations, $$X_{t-2}$$ is $$V$$ and $$W$$ would have to be $$X_{t-1},$$ right ? But, again, how does one say that $$Z_t$$ is independent of $$X_{t-1}$$ when we started we just saying that the $$Z_t$$'s were independent on $$X_0$$?

I think I found the answer on my own. Going back to the first version of the example it all comes down to consider the nature of how the $$X_n$$'s are constructed.
Let $$X_0$$ be a random variable with finite expectation and consider a sequence of i.i.d. random variables $$\{\epsilon_n: n \in N\}$$ such that $$E[\epsilon_n]= 0$$ for each $$n$$ and independent of $$X_0.$$ Then define the stochastic process $$X=\{X_n: n \in N_0\}$$ by letting $$X_{n+1} = X_n + \epsilon_{n+1}X_0.$$ The process $$X$$ is now a martingale. In fact, if we introduce the usual filtration $$\mathbb{F} = \{{\cal F}_n: n \in N_0\}$$ and $${\cal F}_n = \sigma(X_0, X_1, \ldots, X_n),$$ then \begin{align*} E[X_{n+1} \mid {\cal F}_n] &=E[ X_n + \epsilon_{n+1}X_0 \mid {\cal F}_n]\\ &= X_n + X_0 E[\epsilon_{n+1} \mid {\cal F}_n] = X_n + X_0 E[\epsilon_{n+1}] = X_n + 0 = X_n. \end{align*} We should notice that $$X_n = (1+\epsilon_1+ \ldots + \epsilon_n)X_0$$ and, since by assumption the $$\epsilon_n$$'s are i.i.d. and independent of $$X_0,$$ $$\epsilon_{n+1}$$ is independent of $${\cal F}_n.$$ In addition, since $$X_0$$ has finite mean, so do all the $$X_n$$'s. Thus the process $$X$$ is indeed a martingale with respect to the natural filtration.