Markov property vs. Martingale property

Question

Let $(Y_n)_{n\in\mathbb{N}_0}$ be a sequence of random variables with an arbitrary distribution $q$ and existing expectancy value.

How can I check, if the following sequences hold the markov property or the martingale property?

1) $(S_n)_{n\in\mathbb{N}_0}$, where $S_n:=Y_1+...+Y_n$

2) $(T_n)_{n\in\mathbb{N}_0}$, where $T_n:=Y_{n-1}+Y_n$

3) $(M_n)_{n\in\mathbb{N}_0}$, where $M_n:=max\{Y_1,...,Y_n\}$

Does a sequence exist, which hold the martingale property, but not the markov property?

I really hope someone can help with these questions.

Greetings

Answer 1

For i.i.d $\left\{X_{i}\right\}_{i\ge 1 }$ with $\mathbb{E}\left[X_{i}\right]=0$

\begin{eqnarray*} S_{n} &=& \sum_{i=1}^{n}{X_{i}} \end{eqnarray*} is a Martingale relative to filtration generated by the random variables $X_{n}$ (i.e., the sequence w.r.t the sequence $0,X_{1},\ldots,X_{n}$. Recall that, the definition of sequence $S_{n}$ to be a Martingale (w.r.t filtration $\mathcal{F}_{n\ge0}$ is that $\mathbb{E}\left[ S_{n+1} \lvert \mathcal{F}_{n}\right] = S_{n}, \forall n>0$.

\begin{eqnarray*} \mathbb{E}\left[ S_{n+1} \lvert \mathcal{F}_{n}\right] &=& \mathbb{E}\left[ S_{n}+X_{n+1}\lvert \mathcal{F}_{n}\right] \\ &=& \mathbb{E}\left[ S_{n} \lvert \mathcal{F}_{n}\right] + \mathbb{E}\left[ X_{n+1}\lvert \mathcal{F}_{n}\right] \\ &=& S_{n} + \mathbb{E}\left[ X_{n+1}\right] \\ &=& S_{n} \end{eqnarray*}

The 3rd one I guess is also relatively easy to establish the Markov property.

\begin{eqnarray*} M_{n+1} &=&\max \left(Y_{1},Y_{2},\ldots,Y_{n+1}\right) \\ &=& \max \left(\max\left(Y_{1},Y_{2},\ldots,Y_{n} \right),Y_{n+1}\right) \\ &=& \max \left(M_{n},Y_{n+1}\right) \\ \end{eqnarray*}

Clearly, the state $M_{n+1}$ depends only the present state $M_{n}$ and the new input $Y_{n+1}$. In other words, given $M_{n}$, the state $M_{n+1}$ is conditionally independent of the past $Y_{i}\lvert_{i\le n} $.

Answer 2

Here is a detailed answer to 1) that I hope you can use to inform your attempts for the other problems (at least for the martingale part).

A sequence of RVs $S=(S_n)$ is a martingale with respect to a filtration $\mathscr{F}_n$ if $$\mathbb{E}(S_{n+1} | \mathscr{F}_n)=S_{n}$$ for all $n$. Also $S$ must be adapted to the filtration, i.e. $S_n \in \mathscr{F}_n$ for all $n$. Intuitively this is interpreted to mean that given all the information up to time $n$, then one knows the value of $S_n$.

We assume the question wants to use the natural filtrations $\mathscr{F}_n=\sigma(Y_1,\dotsc, Y_n)$.

If $S_n=Y_1+\dotso +Y_n$ and with $Y=(Y_n)$ a sequence of independent RVs with common mean, then we first note that we can write $$S_{n+1}=Y_{n+1}+S_{n},$$ then taking the conditional expectation gives $$\mathbb{E}(S_{n+1} | \mathscr{F}_n)=\mathbb{E}(Y_{n+1}+S_{n}| \mathscr{F}_n)$$ $$=\mathbb{E}(Y_{n+1}| \mathscr{F}_n)+\mathbb{E}(S_{n} | \mathscr{F}_n),$$ using linearity. Now since each $Y$ is independent, we know that $Y_{n+1}$ is independent of $\mathscr{F}_n$. Thus $\mathbb{E}(Y_{n+1}|\mathscr{F}_n)=\mathbb{E}(Y_{n+1})$. Second, $S_{n}$ is $\mathscr{F}_n$-measurable, hence $\mathbb{E}(S_n | \mathscr{F}_n)=S_n$. Therefore, the last displayed equation becomes now, $$\mathbb{E}(S_{n+1} | \mathscr{F}_n)=\mathbb{E}(Y_{n+1})+S_n$$ but we want the RHS to just contain $S_n$ to match the above definition. This only occurs when $\mathbb{E}(Y_n)=0$ for all $n$. Therefore we have proven any sequence of mean-zero independent RVs forms a martingale w.r.t. the natural filtration. Further, this imposes no changes on $Y_n$ other than the mean being zero and hence no changes on $S_n$ other than $\mathbb{E}(S_n)=\mathbb{E}(Y_1)+\dotsc+\mathbb{E}(Y_n)=0$.

Check out Probability with Martingales by D. Williams and Markov Chains by J.R. Norris for extremely well-written presentations of these topics.