Suppose we have an increasing, non-negative sequence of random variables $0 = A_0 \leq A_1 \leq ... \leq A_n \leq A_{n+1} \leq ...$ with all $A_n$ integrable. The question I am attempting to solve states that $$\{A_n\}_{n \in \mathbb{N}} \text{ is predictable} \quad \iff \quad E(M_nA_n) = E\left(\sum_{k=1}^n M_{k-1}(A_k-A_{k-1}) \right) $$

for every bounded martingale $M_n$.

One of the directions is ok/easy, but I cannot do the converse:

$(\implies)$ If $A_n$ is predictable, then for each bounded martingale $M$ we have $$E\left(\sum_{k=1}^n M_{k-1}(A_k-A_{k-1}) \right) = \sum_{k=1}^n E\left( E(M_n|\mathcal{F}_{k-1})(A_k-A_{k-1})\right) \\ = \sum_{k=1}^n E\left( E(M_n(A_k-A_{k-1})|\mathcal{F}_{k-1})\right) = E\left( M_n\sum_{k=1}^n(A_k-A_{k-1})\right) \\ = E(M_nA_n)$$

where the second equality follows from the $\mathcal{F}_{k-1}$ measurability of the $A_k$ and the last equality follows as the sum telescopes and the first term $A_0 = 0$.

$(\impliedby)$ For this direction, I am very unsure about how to proceed. I know that if the expectations are equal then we have $$E\left(\sum_{k=1}^n M_{k-1}(A_k-A_{k-1}) \right) = E\left(\sum_{k=1}^n M_{k}(A_k-A_{k-1}) \right) $$

since $E(M_nA_n) = E\left(M_n \sum_{k=1}^n (A_k - A_{k-1}) \right) = \sum_{k=1}^n E\left(E(M_n (A_k - A_{k-1}) | \mathcal{F}_k) \right)$, but I can't figure anything from here. I figured maybe fixing $N \in \mathbb{N}$ and considering the event $E_N = \{A_{N+1} > E(A_{N+1} | \mathcal{F}_N\}$ with associated martingale $M_n \equiv E(1_{E_N}|\mathcal{F}_n)$ might yield something useful but I can't come up with something from there either.

I'd be grateful for any help! Thanks for taking the time to read.


1 Answer 1


I have a solution based on $E_N$ as defined in the question but I'm not sure if it's enough to make the conclusion because I obtain that $A_{N+1} = E(A_{N+1}|\mathcal{F}_N)$ almost surely - thus I'm not sure if I can conclude that $A_{N+1}$ is indeed $\mathcal{F}_N$ measurable.

(1) Notice that $M_{N+1} = 1_E$ as $E$ is $\mathcal{F}_{N+1}$ measurable and that $$E(M_{N+1}A_{N+1}) - E(M_NA_N) = E(M_N A_{N+1}) - E(M_N A_N)$$ by the hypothesis we are assuming.

(2) Also since $Y \equiv E(A_{N+1}|\mathcal{F}_N)$ is trivially $\mathcal{F}_N$ measurable, we have $E(Y(M_{N+1}-M_N)) = 0$

Combining (1) and (2) yields $$E(1_{E_N} (A_{N+1} - E(A_{N+1}|\mathcal{F}_N))) - \underbrace{E(M_N(A_{N+1} - E(A_{N+1}|\mathcal{F}_N)))}_{=0} = 0$$

As $1_{E_N} (A_{N+1} - E(A_{N+1}|\mathcal{F}_N)) \ge 0$ by definition of the set $E_N$, this quantity must be $=0$ almost surely. Hence $A_{N+1} \leq E(A_{N+1}|\mathcal{F}_N)$ almost surely. Replacing $E_N$ with the reversed strict inequality yields that $$ A_{N+1} = E(A_{N+1}|\mathcal{F}_N) \quad \text{almost surely for every } N \in \mathbb{N}$$

Is this enough to conclude? i.e. is almost sure equivalence to an $\mathcal{F}_N$ measurable random variable imply that $A_{N+1}$ is $\mathcal{F}_N$ measurable? I know it is in the case where the filtration is complete, but is this necessarily the case in general?

  • $\begingroup$ I don't think you can do better. whatever proof you do, since you will go through expectations, and conditional expectation is defined almost surely, you will run into the same issue. (it is in general not true, since you can fudge measurability on your 0 sets (just take a trivial sigma algebra for your F_n) I think this proof is fine (I would have done a different proof where I would look for the set that violates that conditional expectation inequality, and constructed that martingale, but this proof is great) $\endgroup$
    – E-A
    Aug 14, 2020 at 21:44
  • $\begingroup$ That makes sense, thanks. I'm guessing the creator of the question meant it to be a complete filtration but just didn't write it explicitly since it's technically one of the "usual assumptions" $\endgroup$
    – qp212223
    Aug 16, 2020 at 0:55

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