$$ M_t = M_0 + \sum_{s=1}^tH_s(X_s-E(X_s)) $$ where $M_0=0,$ $H$ is a square integrable predictable process and $(X_t)$ are a sequence of square integrable iid random variables.

a) Show $M$ is a martingale.

b) Show $E(M_t^2) = \sum_{s=1}^\infty H_s^2(E(X_s^2)-E(X_s)^2) $.

So, since $H$ is square integrable, then

$$\sum_{s=1}^t E(H_s^2)(E(X_s^2)-E(X_s)^2) <\infty$$

For standard procedures of finding martingales, we find $E(M_t|\mathcal{F}_s)=M_s,$ but here it seems a little difficult to do, and I think for square integrable processes, we need to show something else? I found online that they possess the following property:

$$ \mathbb{E}((X_u - X_t)X_s)=0 \quad \text{and} \quad \mathbb{E}((X_t-X_s)^2|\mathcal{F}_s) = \mathbb{E}(X_t^2|\mathcal{F}_s)-X_s^2$$

for $s\le t \le u$. Is this what I need to prove?

I suppose predictability of $H$ will help with this question where it is $\mathcal{F}_{t-1}-$measurable. Also, a sidenote, why is $M_0$ defined here if it $M_0=0$..?

b) I guess i need to use since $H$ is square integrable then we can use $\sum_{s=1}^\infty E(H_s^2)(E(X_s^2)-E(X_s)^2)$, and then the only difference is that it suggests $E(H_s^2) = H_s^2$ because of predictability, is this right?

  • $\begingroup$ Presumably the filtration is the one generated by the iid sequence, i.e. $\mathcal{F}_t=\sigma(X_j:j\leq t)$. If so you may want to check that if $Y$ and $W$ are independent random variables then $E[f(Y)|\sigma(W)]=E[f(Y)]$ whenever $f(Y)$ is integrable. $\endgroup$
    – Mittens
    Commented Jun 12, 2020 at 15:25
  • $\begingroup$ Do you have a link for a theorem of this? But, yes, it is generated by $X$. So, since $H$ is integrable, do you mean that if $f(Y) = H$, we prove $\mathbb{E}[H|\mathcal{F}_t]=\mathbb{E}[H]$? Since it is predictable then this is true? $\endgroup$
    – Jia
    Commented Jun 12, 2020 at 15:48
  • $\begingroup$ $H$ being predictable only tells you that $H_t$ is $\mathcal{F}_{t-1}$ measurable, it doesn't tell you whether it's independent of anything. $\endgroup$ Commented Jun 12, 2020 at 15:53
  • $\begingroup$ It is best if you try to prove that just following the definition of conditional expectation and the definition of independence. Many books that covers conditional expectation and martingale will prove that. Breiman’s book ir Durret’s book on probability will do $\endgroup$
    – Mittens
    Commented Jun 12, 2020 at 16:03
  • $\begingroup$ That's fine... I wasn't sure whether to use the standard way which or to use the properties above due to square integrability $\endgroup$
    – Jia
    Commented Jun 12, 2020 at 16:04

1 Answer 1


For a), we only need the square integrability of $H$ and $X$ to make sure that $M$ is integrable. To check the martingale property for discrete time processes, it's enough to show $\mathbb{E}[M_{t+1}|\mathcal{F}_{t}] = M_t$. We compute

\begin{align*} \mathbb{E}[M_{t+1}|\mathcal{F}_{t}] &= \mathbb{E}[M_{t} + H_{t+1}(X_{t+1} - \mathbb{E}[X_{t+1}])|\mathcal{F}_{t}] \\ &= M_t + \mathbb{E}[H_{t+1}(X_{t+1} - \mathbb{E}[X_{t+1}])|\mathcal{F}_{t}] \\ &= M_t + H_{t+1} \mathbb{E}[(X_{t+1} - \mathbb{E}[X_{t+1}])|\mathcal{F}_{t}] \\ &= M_t + H_{t+1}(\mathbb{E}[X_{t+1}|\mathcal{F}_t] - \mathbb{E}[X_{t+1}]) \\ &= M_t \end{align*}

so $M$ is a martingale.

You're right that it doesn't make much sense to include $M_0$ in the definition only to set it to $0$ immediately afterwards. The only reason I can think of for that is to slightly simplify the problem? But it really isn't that much different regardless.

For b), I'm fairly sure there's a typo in the question. Even though $H$ is predictable, there is no reason that $\mathbb{E}[H_s^2] = H_s^2$. Predictable processes can still be random, so the only way this would hold true is if $H$ is deterministic. You can show $\mathbb{E}[M_t^2] = \sum_{s=1}^t\mathbb{E}[H_s^2](\mathbb{E}[X_s^2]-\mathbb{E}[X_s]^2)$ but what's stated in the problem doesn't work because $\mathbb{E}[M_t^2]$ is non-random while the right hand side is random.

To show $\mathbb{E}[M_t^2] = \sum_{s=1}^t\mathbb{E}[H_s^2](\mathbb{E}[X_s^2]-\mathbb{E}[X_s]^2)$, we compute

\begin{align*} \mathbb{E}[M_t^2] &= \mathbb{E}\left[\left(\sum_{s=1}^t H_s(X_s-\mathbb{E}[X_s])\right)^2\right] \\ &= \mathbb{E}\left[\sum_{s=1}^t (H_s(X_s-\mathbb{E}[X_s]))^2 + 2\sum_{s=1}^t\sum_{j=1}^{s-1}H_s(X_s-\mathbb{E}[X_s])H_j(X_j-\mathbb{E}[X_j])\right] \\ &= \sum_{s=1}^t \mathbb{E}[(H_s(X_s-\mathbb{E}[X_s]))^2] + 2\sum_{s=1}^t\sum_{j=1}^{s-1}\mathbb{E}[H_s(X_s-\mathbb{E}[X_s])H_j(X_j-\mathbb{E}[X_j])]. \end{align*}

For the terms in the second sum, we can use that $j \le s-1$ and $H_s$ is $\mathcal{F}_{s-1}$ measurable to compute \begin{align*}\mathbb{E}[H_s(X_s-\mathbb{E}[X_s])H_j(X_j-\mathbb{E}[X_j])] &= \mathbb{E}\bigg[\mathbb{E}[H_s(X_s-\mathbb{E}[X_s])H_j(X_j-\mathbb{E}[X_j])|\mathcal{F}_{s-1}]\bigg]\\ &= \mathbb{E}\bigg[H_sH_j(X_j-\mathbb{E}[X_j])\mathbb{E}[(X_s-\mathbb{E}[X_s])|\mathcal{F}_{s-1}]\bigg] \\ &= \mathbb{E}\bigg[H_sH_j(X_j-\mathbb{E}[X_j])(\mathbb{E}[X_s]-\mathbb{E}[X_s])\bigg] \\ &= 0. \end{align*}

For the first sum, we use that $H_s$ is $\mathcal{F}_{s-1}$ measurable again to compute

$$\mathbb{E}[H_s^2(X_s-\mathbb{E}[X_s])^2] = \mathbb{E}[\mathbb{E}[H_s^2(X_s-\mathbb{E}[X_s])^2|\mathcal{F}_{s-1}]] = \mathbb{E}[H_s^2\mathbb{E}[(X_s-\mathbb{E}[X_s])^2|\mathcal{F}_{s-1}]] = \mathbb{E}[H_s^2\mathbb{E}[(X_s^2-\mathbb{E}[X_s]^2)]]=\mathbb{E}[H_s^2](\mathbb{E}[X_s^2]-\mathbb{E}[X_s]^2).$$

Putting everything back together, we have

\begin{align*} \mathbb{E}[M_t^2] &= \sum_{s=1}^t \mathbb{E}[(H_s(X_s-\mathbb{E}[X_s]))^2] + 2\sum_{s=1}^t\sum_{j=1}^{s-1}\mathbb{E}[H_s(X_s-\mathbb{E}[X_s])H_j(X_j-\mathbb{E}[X_j])] \\ &=\sum_{s=1}^t \mathbb{E}[H_s^2](\mathbb{E}[X_s^2]-\mathbb{E}[X_s]^2)] \end{align*}

  • $\begingroup$ OK, so for square integrable martingales, it is easier to prove the martingale property in the standard way.. got it. For b) I feel there is a trick showing $\mathbb{E}[M_t^2] = \sum_{s=1}^t\mathbb{E}[H_s^2](\mathbb{E}[X_s^2]-\mathbb{E}[X_s]^2)$because otherwise it is just taking the expectation of both sides and not doing much with it.. surely the expectation can = RHS though, I don't see the problem with it $\endgroup$
    – Jia
    Commented Jun 12, 2020 at 15:54
  • $\begingroup$ There's a little bit more to it than just taking the expectation of both sides because in general the square of a sum isn't the same as the sum of the squares and we need to show $\mathbb{E}[H_s^2X_s^2] = \mathbb{E}[H_s^2]\mathbb{E}[X_s^2]$. I'll edit the answer with those details shortly. $\endgroup$ Commented Jun 12, 2020 at 16:00
  • $\begingroup$ Thanks a lot for clarifying this, very helpful! But, is it a typo when showing the martingale, surely it should be $\mathbb{E}(M_{t+1}|\mathcal{F}_t) = \mathbb{E}(\sum_{s=1}^{t+1} H_s(X_s - E(X_s))|\mathcal{F}_t) = \mathbb{E}(H_{t+1}(X_{t+1} - E(X_{t+1}))|\mathcal{F}_t)$. I'm not sure why $M_t$ is added in? $\endgroup$
    – Jia
    Commented Jun 12, 2020 at 16:36
  • 1
    $\begingroup$ Because we can break up the sum: $M_{t+1} = \sum_{s=1}^{t+1}H_s(X_s-\mathbb{E}[X_s]) = \sum_{s=1}^{t}H_s(X_s-\mathbb{E}[X_s]) + H_{t+1}(X_{t+1}-\mathbb{E}[X_{t+1}]) = M_t + H_{t+1}(X_{t+1}-\mathbb{E}[X_{t+1}])$ $\endgroup$ Commented Jun 12, 2020 at 16:40

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