Prove that the discrete time martingale can be represented by $E (Y_{n +1} \mid F_n) = 0$ if $Y_{n +1} = X_{n +1}-X_n$, for $n = 0,1, \ldots $

Question

Prove that the discrete time martingale can be represented by $E (Y_{n +1} \mid F_n) = 0$ if $Y_{n +1} = X_{n +1}-X_n$, for $n = 0,1, \ldots $

I want to use the sequence $(y_n)$ called "martingale difference sequence with respect to the filtration $(F_n)^n$".

$(X_{n+1}\mid F_n)=X_n$ for $n=0,1,2,\ldots$ then if $Y_{n+1}=X_{n+1}-X_n$ for $n=0,1,2,\ldots$ finally $E(Y_{n+1}\mid F_n)=0$

Is this ok?

Answer 1

I think you are a bit confused with the notation. Do you mean $\mathbb{E}[Y_{n+1}|F_n] = 0$ in your last expression?

The rest of it is true, I'll just reword your solution here: Yes, since expectation is linear, we have

$ \mathbb{E}[Y_{n+1}|F_n] = 0 \Leftrightarrow \mathbb{E}[X_{n+1}-X_n|F_n] = 0 \Leftrightarrow \mathbb{E}[X_{n+1}|F_n] - \mathbb{E}[X_n|F_n] = 0 $ $\Leftrightarrow \mathbb{E}[X_{n+1}|F_n] = \mathbb{E}[X_n|F_n] \Leftrightarrow \mathbb{E}[X_{n+1}|F_n] = X_n,$

showing that the two are equivalent (i.e. you can use one criteria to represent the other one, which is what's been asked to be proved). The last equivalence is due to $X_n$ being a $F_n$-measurable random variable.