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


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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.

  • $\begingroup$ $2.\Longrightarrow 3.$ 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. $\endgroup$
    – Kurt G.
    Mar 14 at 20:06
  • $\begingroup$ How did you get that the last difference equals 0? You somehow used 2.? What about problem 1? $\endgroup$
    – eMathHelp
    Mar 14 at 20:12
  • $\begingroup$ $$E\left[X(t+1)|\mathcal{F}_t\right]-E\left[X(t)|\mathcal{F}_{t}\right]=X(t)-X(t)=0\,.$$ Let's get to Problem 1 once that is understood. $\endgroup$
    – Kurt G.
    Mar 14 at 20:15
  • $\begingroup$ I understood this. You've used the martingale property 1. As for me, you've proved 1.->3., not 2.->3. Am I wrong? $\endgroup$
    – eMathHelp
    Mar 14 at 20:19
  • 1
    $\begingroup$ Ah. Finally you are letting us know what $\mathbb T$ is. Acceptable but I would write it as in my comment further above using $t\le T\,.$ $\endgroup$
    – Kurt G.
    Mar 14 at 20:36

1 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.


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