Show that M is a martingale

Question

Assume that $E[(HM)_N]=0$ for predictable process H and $H_0=0$. I want to prove that $M$ is a martingale.

$$(HM)_N= \sum_{j=1}^N H_j (M_j -M_{j-1})+ H_0M_0 = \sum_{j=1}^N H_j (M_j -M_{j-1})$$

I am stuck at showing that $M$ is a martinagle.

I read some theorems. (If required, I can share them) but I couldn’t not show this proof. Please help me to do this.

Many thanks.

Answer 1

Let $(\mathcal{F}_n)_{n \in \mathbb{N}}$ be the underlying filtration. Fix $n \in \mathbb{N}$ and pick $F \in \mathcal{F}_n$. The process

$$H_j := \begin{cases} 0, & \text{if $j \neq n+1$}, \\ 1_F, & \text{if $j=n+1$},\end{cases}$$ is predictable and $H_0=0$. We have

$$(H \bullet M)_{n+1} = \sum_{j=1}^{n+1} H_j (M_j-M_{j-1}) = H_{n+1} (M_{n+1}-M_n) = 1_F (M_{n+1}-M_n).$$

By assumption, the random variable has expectation zero, i.e.

$$0 = \mathbb{E}((H \bullet M)_{n+1}) = \mathbb{E}(1_F M_{n+1})-\mathbb{E}(1_F M_n),$$

and so

$$\mathbb{E}(M_{n+1} 1_F) = \mathbb{E}(M_n 1_F).$$

Since this holds for any $F \in \mathcal{F}_n$, $n \in \mathbb{N}$, it follows that $(M_n)_{n \in \mathbb{N}}$ is a martingale with respect to $(\mathcal{F}_n)_{n \in \mathbb{N}}$.

Answer 2

Put $Y=H\cdot M$. Since $H$ is predictable and $(M_j-M_{j-1})_j$ is adapted, clearly $Y$ is adapted. Now, for each positive integer $N$, \begin{align} \mathbb E[(H\cdot M)_N] &= \mathbb E\left[\sum_{j=1}^N H_j(M_j-M_{j-1}) \right]\\ &= \sum_{j=1}^N \mathbb E[H_j(M_j-M_{j-1})]\\ &=0, \end{align} which implies that $$ \sum_{j=1}^N \mathbb E[H_jM_j] = \sum_{j=1}^N \mathbb E[H_jM_{j-1}]. $$ For $N=1$ this yields $\mathbb E[H_1M_1] = \mathbb E[H_1M_0]$, and by induction $\mathbb E[H_jM_j] = \mathbb E[H_jM_{j-1}]$, $j\geqslant 1$. I am not sure how to proceed from here, but hopefully this was useful.