Showing a property of a martingale

Question

Let $(Z_t)_{t\in \mathbb{N}}$ be bounded and r.v. and assume $(Z_t)_{t\in\mathbb{N}}$ is a martingale with $\mathcal{F}_{n}=\sigma(Z_1,\dots ,Z_n)$. Show that $$\mathbb{E}((Z_{n+1}-Z_n)(Z_{l+1}-Z_l))=0$$

I'm struggling with this exercise. I made multiple attempts such as:

Let $n>l$ then

$$\mathbb{E}((Z_{n+1}-Z_n)(Z_{l+1}-Z_l))=\mathbb{E}((Z_{n+1}-Z_n)(\mathbb{E}(Z_{n+1}\mid \mathcal{F}_{l+1})-\mathbb{E}(Z_{n+1}\mid \mathcal{F}_{l})))$$ $$=\mathbb{E}((Z_{n+1}\mathbb{E}(Z_{n+1}\mid \mathcal{F}_{l+1})-Z_n\mathbb{E}(Z_{n+1}\mid \mathcal{F}_{l+1})-Z_{n+1}\mathbb{E}(Z_{n+1}\mid \mathcal{F}_{l})-Z_n\mathbb{E}(Z_{n+1}\mid \mathcal{F}_{l}))$$

But this is always where I get stucked.

Answer 1

If $n>l$, then $$ \mathbb{E}[(Z_{n+1}-Z_n)(Z_{l+1}-Z_l)]=\mathbb{E}[\mathbb{E}[(Z_{n+1}-Z_n)(Z_{l+1}-Z_l)\mid\mathcal{F}_{l+1}]]=\mathbb{E}[(Z_{l+1}-Z_l)\mathbb{E}[Z_{n+1}-Z_n\mid \mathcal{F}_{l+1}]]$$ since $Z_{l+1}-Z_l$ is $\mathcal{F}_{l+1}$-measurable. And since $\{Z_n\}$ is a martingale and $n\geq l+1$, we have $$ \mathbb{E}[Z_{n+1}-Z_n\mid \mathcal{F}_{l+1}]=\mathbb{E}[Z_{n+1}\mid \mathcal{F}_{l+1}]-\mathbb{E}[Z_{n}\mid \mathcal{F}_{l+1}]=Z_{l+1}-Z_{l+1}=0$$ hence $\mathbb{E}[(Z_{n+1}-Z_n)(Z_{l+1}-Z_l)]=0$.

Answer 2

Let $n>l$ then $E\left[(Z_{n+1}-Z_n)\big|\mathcal{F}_{l+1}\right] = Z_{l+1} - Z_{l+1} = 0$,so

$$\begin{align*} E\left[(Z_{n+1}-Z_n)(Z_{l+1}-Z_l)\right] &= E\left[E\left[(Z_{n+1}-Z_n)(Z_{l+1}-Z_l)\big|\mathcal{F}_{l+1}\right]\right] \\ &= E\left[(Z_{l+1}-Z_l)E\left[(Z_{n+1}-Z_n)\big|\mathcal{F}_{l+1}\right]\right] \\ &=E[Z_{l+1}-Z_l]\end{align*}$$

Conditioning on $\mathcal{F}_l$ gives your result.

Answer 3

I think you forgot FOIL. Continuing where you left off with one change:

$$=E[Z_{n+1}E(Z_{n+1}\mid \mathcal{F}_{l+1})-Z_nE(Z_{n+1}\mid \mathcal{F}_{l+1})-Z_{n+1}E(Z_{n+1}\mid \mathcal{F}_{l})\color{red}{+}Z_nE(Z_{n+1}\mid \mathcal{F}_{l})]$$

By iterated expectations,

$$=E[E[Z_{n+1}E(Z_{n+1}\mid \mathcal{F}_{l+1})-Z_nE(Z_{n+1}\mid \mathcal{F}_{l+1})|\mathcal F_{l+1}]$$$$-E[Z_{n+1}E(Z_{n+1}\mid \mathcal{F}_{l})+Z_nE(Z_{n+1}\mid \mathcal{F}_{l})|\mathcal F_l]$$

$$=E[E[Z_{n+1}E(Z_{n+1}\mid \mathcal{F}_{l+1})|\mathcal F_{l+1}]]-E[E[Z_nE(Z_{n+1}\mid \mathcal{F}_{l+1})|\mathcal F_{l+1}]]$$$$-E[Z_{n+1}E(Z_{n+1}\mid \mathcal{F}_{l})|\mathcal F_{l}]+E[Z_nE(Z_{n+1}\mid \mathcal{F}_{l})]|\mathcal F_l]$$

$$=E[E(Z_{n+1}\mid \mathcal{F}_{l+1})E[Z_{n+1}|\mathcal F_{l+1}]]-E[E(Z_{n+1}\mid \mathcal{F}_{l+1})E[Z_n|\mathcal F_{l+1}]]$$$$-E(Z_{n+1}\mid \mathcal{F}_{l})E[Z_{n+1}|\mathcal F_{l}]+E(Z_{n+1}\mid \mathcal{F}_{l})]E[Z_n|\mathcal F_l]$$

$$=E[Z_{l+1}Z_{l+1}]-E[Z_{l+1}E[Z_n|\mathcal F_{l+1}]]$$$$-Z_lZ_l+Z_lZ_l$$

$$=E[Z_{l+1}Z_{l+1}]-E[Z_{l+1}E[Z_n|\mathcal F_{l+1}]]$$

$$=E[Z_{l+1}Z_{l+1}]-E[Z_{l+1}Z_{l+1}] \tag{*}$$

$$= 0$$

QED

$(*)$ we're not sure if $n > l+1$, but we have $n \ge l+1$ and if $n = l+1$,

$$E[Z_n | \mathcal F_{l+1}] = E[Z_{l+1} | \mathcal F_{l+1}] = Z_{l+1}$$

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I think something more efficient is instead of turning $Z_{l \ / \ l+1}$ into a conditional expectation, we use iterated expectation from the very beginning:

$$\mathbb{E}[(Z_{n+1}-Z_n)(Z_{l+1}-Z_l)]$$

$$= \mathbb{E}[(Z_{n+1}-Z_n) Z_{l+1}] - \mathbb{E}[(Z_{n+1}-Z_n)Z_{l}]$$

$$= \mathbb{E}[(Z_{n+1}-Z_n)(Z_{l+1})| \mathscr F_{l+1}] - \mathbb{E}[(Z_{n+1}-Z_n)(Z_{l})| \mathscr F_l]$$

$$= Z_{l+1}\mathbb{E}[Z_{n+1}-Z_n| \mathscr F_{l+1}] - Z_{l}\mathbb{E}[Z_{n+1}-Z_n| \mathscr F_l]$$

$$= Z_{l+1}(0) - Z_{l}(0) = 0$$

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