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I notice that every Martingale Process refers somehow to an iid. I was wondering why do we need that and what role does the iid actually play in the process.

Sometimes when the iid is not obvious to be spotted but it is there, is there a way to identify it?

For example, in the problem below, I am not sure how to relate the iid ($X_n$) to the sequence ($L_n$) even though they mentioned ($X_k$) in the definition of ($L_n$).

Thank you very much for your help

Let $(X_n,n\geqslant1)$ be i.i.d. random variables, each uniformly distributed on the set $\{-1,0,1\}$. The sequence of random variables $(L_n,n\geqslant0)$ is defined as follows: $L_0=1$ and for $n\geqslant1$: $$L_n = \begin{cases} 0,& \text{ if $X_k=0$ for some $1\leqslant k\leqslant n$}\\ \left(\frac32\right)^n,& \text{ otherwise}. \end{cases}$$

Verify that $(L_n,n\geqslant0)$ is a martingale.

site
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  • Your question is very vague. Sure, for example, a sequence of i.i.d. random variables would be a martingale, but the converse is not true. Do you have a specific example in mind? – Math1000 Oct 13 '19 at 21:38
  • Thank you very much for your answer. I have edited the post and attached a sample problem. – site Oct 13 '19 at 21:43

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In the given example, it is clear that the $L_n$ are integrable, since $\mathbb P(L_n\geqslant 0)=1$ and $$ \mathbb E[L_n] \leqslant \left(\frac32\right)^n<\infty. $$ To show the martingale condition, consider that if $X_{n+1}=0$ then $L_{n+1}=0$, and otherwise \begin{align} \mathbb E[L_{n+1}\mid L_n] &= 0\cdot\mathbb P(L_{n+1}=0) + \left(\frac32\right)^{n+1}\cdot\mathbb P(X_{n+1}\ne 0)\\ &= 0 + \left(\frac 32\right)^{n+1}\cdot \frac 23\\ &= \left(\frac 32\right)^n\\ &= L_n. \end{align}

Math1000
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  • Thank you again for your answer. I was wondering how are we getting that ℙ(+1≠0) = 2/3. Is it because that corresponds to the case of X(-1) and X(1) and each has probability 1/3. – site Oct 13 '19 at 22:18
  • Also, since [+1∣] suggests that we only know the probability distribution ℙ[+1∣], will that distribution be the same as the one of ℙ(+1) with no condition? – site Oct 13 '19 at 22:26
  • if we were to find only [], that will be 1.ℙ( =1) + 0.ℙ(=0)+(3/2)^⋅ℙ(L≠0). Did we consider only the last 2 terms of the expression while doing [+1∣] because once we know that , then the case for +1 =1 could no longer occur? – site Oct 13 '19 at 22:48