Definition: $X_n$ is a martingale difference sequence (MDS) with respect to a filtration $\mathcal{F}_n$ if it is adapted, integrable, and $E[X_n\mid \mathcal{F}_{n-1}]=0$.
Let $Y_n:=X_n+Y_{n-1}$. If $X_n$ is an MDS, does that mean $Y_n$ is a martingale?
I saw this question that claims that it does, but I did not understand the proof. It says: $Y_n=\sum\limits_{k=0}^{n} X_k$, so that $E[Y_n\mid \mathcal{F}_{n-1}]= Y_{n-1}+E[X_n\mid \mathcal{F}_{n-1}]=Y_{n-1}$.
I think they mean: $E[Y_n\mid \mathcal{F}_{n-1}] = E[X_n + Y_{n-1}\mid\mathcal{F}_{n-1}] = 0 + E[Y_{n-1}\mid\mathcal{F}_{n-1}]$
But I don't get why $E[Y_{n-1}\mid\mathcal{F}_{n-1}]=Y_{n-1}$.
Using induction, assuming $E[Y_{n-1}\mid\mathcal{F}_{n-2}] = Y_{n-2}$, we need to prove $E[Y_{n}\mid\mathcal{F}_{n-1}] = Y_{n-1}$. We get that:
\begin{align}
E[Y_{n}\mid\mathcal{F}_{n-1}] &= E[Y_{n-1}\mid\mathcal{F}_{n-1}] \\
&= E[X_{n-1} + Y_{n-2}\mid\mathcal{F}_{n-1}] \\
&= E[X_{n-1}\mid\mathcal{F}_{n-1}] + E[Y_{n-2}\mid\mathcal{F}_{n-1}] \\
&= E[X_{n-1}\mid\mathcal{F}_{n-1}] + E[E[Y_{n-1}\mid\mathcal{F}_{n-2}]\mid\mathcal{F}_{n-1}]\\
&= E[X_{n-1}\mid\mathcal{F}_{n-1}] + E[Y_{n-1}\mid\mathcal{F}_{n-2}]\\
&= E[X_{n-1}\mid\mathcal{F}_{n-1}] + Y_{n-2}
\end{align}
So it's enough to prove that $E[X_{n-1}\mid\mathcal{F}_{n-1}] = X_{n-1}$. Or that $X_n$ is measurable with respect to $\mathcal F_n$. How can this be proved?
Please explain without assuming any knowledge since I'm studying this alone with google so I may not be familiar with trivial things. Thank you very much!
Best Answer
You have answered your question yourself but you did not realize it yet. You have said that the $X_n$ are adapted to the filtration $\mathcal F_n$. What does that mean? Well, that means that $X_n$ are $\mathcal F_n$-measurable.
In the other post, they should have written that $Y_n$ can be written as $$\tag{$*$}Y_n=Y_0+ \sum_{k=1}^n X_k$$ Assuming $Y_0=0$ (or constant) then we can conclude that $Y_n$ is also $\mathcal F_n$-measurable since it is a function of $\mathcal F_n$-measurable random variables.
This observation leads to $$\mathbb E[Y_{n-1}\mid \mathcal F_{n-1}] =Y_{n-1}$$ That is just a property of the conditional expectation, if a random variable is measurable with respect to the $\sigma$-algebra it is conditioned on then you get the random variable itself. This actually solves the problem.
Things can change if $Y_0$ is not constant though.
About your approach with induction: it is actually correct what you did, ending up whether we have $X_n$ is $\mathcal F_n$-measurbale which is indeed the case from the definition. But did you check the base case? Do we have $$\mathbb E[Y_1 \mid \mathcal F_0 ] = Y_0?$$ What we do have is $$\mathbb E[Y_1\mid \mathcal F_0 ] = \mathbb E[ X_1 + Y_0 \mid \mathcal F_0] = \mathbb E[Y_0 \mid \mathcal F_0]$$ It boils down to the question whether $\mathbb E[Y_0\mid \mathcal F_0] = Y_0$ holds. That does not hold in general, but the case where $Y_0$ is constant it does, that is also why I said that things might change without the assumption that $Y_0$ is constant.
Here are two questions for you to think about. Can you think of another assumption on $Y_0$ which leads to $\mathbb E[Y_0 \mid \mathcal F_0 ] =Y_0$? If you have that assumption, can you argue why $Y_n$ is (still) $\mathcal F_n$-measurbale from ($*$)?