Proving martingale properties

Question

I'm new to stochastic processes and have problems understanding martingales, conditional probability, $\sigma$-algebras etc.

I have two proofs that I'm now sure how to handle.

Problem 1. Prove that an integrable stochastic process $\{X(t),\mathcal{F}_t, t\in \mathbb{T}\}$ is a martingale if and only if for any bounded predictable process $\{\xi(t),\mathcal{F}_t, t\in \mathbb{T}\setminus\{0\}\}$ we have that $E\left[\sum_{k=1}^n\xi(k)\Delta X(k)\right]=0$.

Attempt on Problem 1

"$\Rightarrow$"

"$\Leftarrow$"

Am I correct? I'm not quite sure about the last step.

Problem 2. Prove the equivalence of the following statements:

1. $\{X(t),\mathcal{F}_t, t\in \mathbb{T}\}$ is a martingale;

2. $X(t)=E\left[X(T)|\mathcal{F}_t\right]$, $t\in \mathbb{T}$;

3. $E\left[\Delta X(t+1)|\mathcal{F}_t\right]=0$, $t=0,1,\ldots,T-1$.

Here $\Delta X(k)=X(k)-X(k-1)$.

Attempt on Problem 2

1. $\Rightarrow$ 2. Since $\{X(t),\mathcal{F}_t, t\in \mathbb{T}\}$ is a martingale, then $E[X(t)|\mathcal{F}_s]=X(s)$. Just take $t=T$ and $s=t\in T$ to get $$E[X(T)|\mathcal{F}_t]=X(t)$$

2. $\Rightarrow$ 3. $$E\left[\Delta X(t+1)|\mathcal{F}_t\right]=E\left[X(t+1)-X(t)|\mathcal{F}_t\right]=E\left[X(t+1)|\mathcal{F}_t\right]-E\left[X(t)|\mathcal{F}_t\right]=E\left[E\left[X(T)|\mathcal{F}_{t+1}\right]|\mathcal{F}_t\right]-E\left[X(T)|\mathcal{F}_t\right]=E\left[X(T)|\mathcal{F}_t\right]-E\left[X(T)|\mathcal{F}_t\right]=0$$ Am I correct here? The proof looks clumsy.

3. $\Rightarrow$ 1. $$E\left[\Delta X(t+1)|\mathcal{F}_t\right]=0$$ $$E\left[X(t+1)-X(t)|\mathcal{F}_t\right]=0$$ $$E\left[X(t+1)|\mathcal{F}_t\right]=X(t)$$ Am I correct here? Is it sufficient?

Thank you in advance.

Answer 1

Direction $\Rightarrow$ in Problem 1 looks good. The other direction has a bit of a gap as you noticed. I would take for fixed $k$ $$ \xi(k):=1_{\textstyle\{\mathbb E[X(k)|{\cal F}_{k-1}]> X(k-1)\}}\,. $$ and $\xi:\equiv 0$ for all other $k\,.$ Then $$ \mathbb E\Big[\xi(k)\Big(X(k)-X(k-1)\Big)\Big]=0 $$ implies $$ \mathbb E\Big[\xi(k)\Big(E[X(k)|{\cal F}_{k-1}]-X(k-1)\Big)\Big]=0 $$ and that implies $E[X(k)|{\cal F}_{k-1}]\le X(k-1)$ almost surely. The other inequality is shown similarly.