My definition of a Martingale Difference Sequence (MDS) is: A sequence $(Y_n)_{n \geq 1}$ of integrable random variables, adapted to the filtration $(\mathcal{F}_n)_{n \geq 1}$, is an MDS w.r.t $(\mathcal{F}_n)_{n \geq 1}$ if $\mathbb{E}[Y_{n+1}|\mathcal{F}_n] = 0$ a.s.
I am stuck showing the following statement, which it seems should be trivial but for some reason I can't figure out.
$(X_n)_{n \geq 0}$ is a martingale w.r.t $(\mathcal{F}_n)_{n \geq 0}$ iff $X_0$ is integrable and $\mathcal{F}_0$-measurable, and $(X_n - X_{n-1})_{n \geq 1}$ is a MDS w.r.t $(\mathcal{F}_n)$.
Right now I have:
$\mathbb{E}[X_{n+1} - X_n | \mathcal{F}_n] = \mathbb{E}[X_{n+1}| \mathcal{F}_n] - \mathbb{E}[X_{n}| \mathcal{F}_n] = \mathbb{E}[X_{n+1}] - X_n = \mathbb{E}[X_n]-X_n$.
I'm not sure how to get this expression to equal $0$ as required.
For the reverse direction I'm trying to use the fact that $X_{n} = X_0 + \sum_{i=1}^n (X_{i}-X_{i-1})$. I get
$\mathbb{E}[X_{n+1}|\mathcal{F}_n] = \mathbb{E}[X_0 + \sum_{i=1}^{n+1}(X_i - X_{i-1})|\mathcal{F_n}] = \mathbb{E}[X_0|\mathcal{F}_n] + \sum_{i=1}^{n+1}\mathbb{E}[X_i - X_{i-1}|\mathcal{F_n}] = \mathbb{E}[X_0|\mathcal{F}_n] + \sum_{i=1}^{n}\mathbb{E}[X_i - X_{i-1}|\mathcal{F_n}] + \mathbb{E}[X_{n+1}-X_n|\mathcal{F_n}] = \mathbb{E}[X_0|\mathcal{F}_n] + \sum_{i=1}^{n}\mathbb{E}[X_i - X_{i-1}|\mathcal{F_n}] = \mathbb{E}[X_n|\mathcal{F}_n]$
But again I'm not sure how to finish this. We know that $(X_n - X_{n-1})_{n \geq 1}$ is adapted to the filtration. But to complete this proof I need $X_n$ to be $\mathcal{F}_n$-measurable.
Thanks
Why would you think that $X_{n+1}$ is independent of $\mathcal{F}_{n}=\sigma(X_{n},X_{n-1},..,X_{1})$? It's just that $X_{n+1}$ is not "necessarily" measurable wrt $\mathcal{F}_{n}$. For example, take $S_{n}$ to be the simple symmetric random walk on $\Bbb{Z}$ . Then $S_{n}$ is a martingale. Does that mean that $S_{n+1}$ which is the ($n+1$ -th state) independent of the $n$-th state?. No, ofcourse not. It just means that we have don't have a full idea(we do have some idea) about what the $n+1$-th state will be. That's what measurability means. Intuitively, $X$ is $\mathcal{F}$ measurable should mean to you that if all information of $\mathcal{F}$ is known then $X$ is known.
Now, as far as the problem is concerned, first let $X_{n}$ be a martingale.
Then $E(X_{n+1}-X_{n}|\mathcal{F}_{n})=E(X_{n+1}|\mathcal{F}_{n})-E(X_{n}|\mathcal{F}_{n})=E(X_{n+1}|\mathcal{F}_{n})-X_{n}$ which is $0$ as by definition of martingale $E(X_{n+1}|\mathcal{F}_{n})=X_{n}$ for each $n$. So $(X_{n+1}-X_{n})_{n}$ is a martingale difference sequence. (Sometimes also called "fair sequence"). Also, by definition of martingales, $X_{0}$ is $\mathcal{F}_{0}$ measurable as $(X_{n})$ is an adapted sequence.
Conversely le $(Y_{n})_{n}=(X_{n}-X_{n-1})_{n}$ be a martingale difference sequence. Then $Y_{n}$ is $\mathcal{F}_{n}$ measurable and by definition $E(Y_{n}|\mathcal{F}_{n})=0$
Then $\sum_{k=1}^{n}Y_{n}=X_{n}-X_{0}$ is $\mathcal{F_{n}}$ measurable. As $X_{0}$ is $\mathcal{F}_{0}$ measurable and $\mathcal{F}_{0}\subseteq \mathcal{F}_{n}$ for all $n\geq 1$, we have $X_{n}=X_{n}-X_{0}+X_{0}$ is $\mathcal{F}_{n}$ measurable.
Then \begin{align}E(X_{n}|\mathcal{F}_{n-1})& =E(\sum_{k=1}^{n}Y_{n}+X_{0}|\mathcal{F}_{n-1})\\ &=E(\sum_{k=1}^{n-1}Y_{k}|\mathcal{F_{n-1}})+E(X_{0}|\mathcal{F}_{n-1})+E(Y_{n}|\mathcal{F}_{n-1})\\ &=E(X_{n-1}-X_{0}|\mathcal{F}_{n-1})+E(X_{0}|\mathcal{F}_{n-1})+E(Y_{n+1}|\mathcal{F}_{n-1})\\ &=E(X_{n-1}|\mathcal{F}_{n-1})+E(Y_{n}|\mathcal{F}_{n-1})\\ &=X_{n-1} \end{align}
for each $n\geq 1$. This shows that $X_{n}$ is a martingale sequence.