# Proving martingale properties

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?

• $2.\Longrightarrow 3.$ Can be simplified (and fixed): $$E\left[\Delta X(t+1)|\mathcal{F}_t\right]=E\left[X(t+1)-X(t)|\mathcal{F}_t\right]=E\left[X(t+1)|\mathcal{F}_t\right]-\color{red}{E\left[X(t)|\mathcal{F}_{t}\right]}=0\,.$$ Otherwise, Problem 2 looks good. Mar 14 at 20:06
• $$E\left[X(t+1)|\mathcal{F}_t\right]-E\left[X(t)|\mathcal{F}_{t}\right]=X(t)-X(t)=0\,.$$ Let's get to Problem 1 once that is understood. Mar 14 at 20:15
• Ah. Finally you are letting us know what $\mathbb T$ is. Acceptable but I would write it as in my comment further above using $t\le T\,.$ Mar 14 at 20:36
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.