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Let

  • $(\Omega,\mathcal A,\operatorname P)$ be a probability space
  • $(\mathcal F_t)_{t\ge0}$ be a filtration on $(\Omega,\mathcal A)$
  • $M_{c,\:\text{loc}}(\mathcal F,\operatorname P)$ denote the set of continuous local $\mathcal F$-martingales on $(\Omega,\mathcal A,\operatorname P)$

Is $M_{c,\:\text{loc}}(\mathcal F,\operatorname P)$ equipped with the topology of uniform convergence on compact sets$^1$ complete?

$^1$ i.e. If $(M^n)_{n\in\mathbb N}\subseteq M_{c,\:\text{loc}}(\mathcal F,\operatorname P)$ and $M\in M_{c,\:\text{loc}}(\mathcal F,\operatorname P)$, then $M_n\xrightarrow{n\to\infty}M$ in $M_{c,\:\text{loc}}(\mathcal F,\operatorname P)$ if and only if $$\sup_{0\le s\le t}\left|M^n_s-M_s\right|\xrightarrow{n\to\infty}0\;\;\;\text{in probability for all }t\ge 0\tag1.$$

0xbadf00d
  • 14,208

1 Answers1

1

Yes, the space of continuous local martingales equipped with ucp convergence is complete.

Let $(M^n)_{n \in \mathbb{N}} \subseteq M_{c,\text{loc}}(\mathcal{F},P)$ be a Cauchy sequence with respect to the ucp convergence. Then we can choose for any fixed $T>0$ numbers $n_k \in \mathbb{N}$, $n_1 < n_2 < \ldots$, such that

$$\mathbb{P} \left( \sup_{t \leq T} |M^{n_k}_s-M_s^{n_{k-1}}|>2^{-k} \right) \leq 2^{-k}.$$

By the Borel Cantelli lemma, this implies that

$$\mathbb{P} \left( \sup_{t \leq T} |M^{n_k}_t-M_t^{n_{k-1}}|>2^{-k} \, \, \text{for infinitely many $k$} \right)=0.$$

Using that

$$M_t^n - M_t^m = \sum_{k=m+1}^n (M_t^k-M_t^{k-1}), \qquad m<n,$$

it is not difficult to see that this implies that $(M_t^{n_k}(\omega))_{t \leq T}$ is a Cauchy sequence in $C([0,T],\|\cdot\|_{\infty})$ for $\mathbb{P}$-almost all $\omega$. By the completeness of $C[0,T]$, we find that

$$M_t(\omega) := \lim_{k \to \infty} M_t^{n_k}(\omega)$$

exists almost surely and defines a process with continuous sample paths. Since $(M^n)_{n \in \mathbb{N}}$ is a Cauchy sequence with respec to the ucp convergence and $M^{n_k} \to M$ uniformly on $[0,T]$ with probability $1$, it follows that

$$\forall \epsilon>0: \quad \lim_{n \to \infty} \mathbb{P} \left( \sup_{t \leq T} |M_t^n-M_t|> \epsilon \right)=0,$$

and so $M^n \to M$ in ucp. Consequently, $M$ is the ucp limit of a sequence of continuous local martingales, and so $M$ is itself a local martingale (see Theorem 5 here).

saz
  • 123,507
  • Thanks for your answer. Actually, I've figured it out in the meantime. However, I've got one remaining question: Your proof above can be generalized to the following scenario: Let $M$ be a metric space and $E$ be a $\mathbb R$-Banach space. Now, let $\mathcal B_{\text{loc}}(\mathcal A;M,E)$ denote the set of $F:\Omega\times M\to E$ with $F(;\cdot;,x)$ being $\mathcal A$-measurable for all $x\in M$ and $$\left|F(\omega,;\cdot;)\right|{B(K,:E)}:=\sup{x\in K}\left|F(\omega,x)\right|_E<\infty;;;\text{for all }\omega\in\Omega\text{ and compact }K\subseteq M.$$ – 0xbadf00d Jun 05 '18 at 11:18
  • Now, assume $(F_n){n\in\mathbb N}\subseteq\mathcal B{\text{loc}}(\mathcal A;M,E)$ with $$\left|F_m-F_n\right|{B(K,:E)}\xrightarrow{m,:n:\to:\infty}0;;;\text{in probability for all compact }K\subseteq M.$$ With the same arguments as in your proof, we can show that there is a $F:\Omega\times M\to E$ with $F(;\cdot;,x)$ being $\mathcal A$-measurable for all $x\in M$ and $$\left|F_n-F\right|{B(K,:E)}\xrightarrow{n\to\infty}0;;;\text{in probability for all compact }K\subseteq M.$$ – 0xbadf00d Jun 05 '18 at 11:19
  • However, it's not clear to me whether $F\in B_{\text{loc}}(\mathcal A;M,E)$, i.e. whether $\left|F(\omega,;\cdot;)\right|_{B(K,:E)}<\infty$ (at least for almost) all $\omega\in\Omega$ for all compact $K\subseteq M$. – 0xbadf00d Jun 05 '18 at 11:19
  • @0xbadf00d I'm somewhat puzzled about your question. If (as you say) $|F_n-F|{B(K,E)} \to 0$ in probability, then $$\mathbb{P}(\exists n: |F_n-F|{B(K,E)}<\infty) =1$$ and so $$|F|{B(K,E)} \leq |F_n-F|{B(K,E)} + |F_n|_{B(K,E)} < \infty$$ with probability $1$... or am I missing something? – saz Jun 05 '18 at 14:32
  • You're missing nothing. As you say, it's trivial. Don't know what I was thinking. However, in order to prove that $M$ is a local martingale, it seems like we need the fact that $$A_t:=\sup_{n\in\mathbb N}\sup_{s\in[0,:t]}\left|M_s^n\right|_E;;;\text{for }t\ge0$$ is continuous. But I don't see why that's the case. I've asked a separate question for that, but the downvotes indicate that it is either trivial or obviously wrong. – 0xbadf00d Jun 06 '18 at 17:33
  • @0xbadf00d I don't have much time right now to think about it (or to look at the question you linked) ,but why would we need that $A_t$ is continuous...`? – saz Jun 06 '18 at 19:33
  • It would allow us to conclude that $\inf\left{t\ge0:A_t\ge C\right}$ is a stopping time. And it's claimed here. – 0xbadf00d Jun 06 '18 at 19:58
  • There is a remaining issue in your first comment: With my definition, $F\in B_{\text{loc}}(\mathcal A;M,E)$ iff $\left|F(\omega,;\cdot;)\right|{B(K,:E)}<\infty$ for all $\omega\in\Omega$. You've shown that $\operatorname P\left[\left|F\right|{B(K,:E)}<\infty\right]=1$ for all compact $K\subseteq M$, but the null set depends on $K$. So, I don't see how we can conclude, unless $M$ is $\sigma$-compact. – 0xbadf00d Jun 10 '18 at 18:53
  • @0xbadf00d Just set it equal to some fixed element $e \in E$ for any $\omega$ for which the norm is infinite. – saz Jun 10 '18 at 19:01
  • Sorry, I don't get what you mean. Let $K\subseteq M$ be compact. There is a $N_K\in\mathcal A$ with $\operatorname P[N_K]=0$ and $\left|F(\omega)\right|_{B(K,:E)}<\infty$ for all $\omega\in\Omega\setminus N_K$. Now, if $M$ is not $\sigma$-compact, $M$ cannot be covered by countable many $K$ and the union of the $N_K$ doesn't need to be a null set. – 0xbadf00d Jun 10 '18 at 19:09
  • @0xbadf00d I see; actually I think that the $\sigma$-finiteness is already needed to construct $F$ (at least if you want to use the construction which I used in my answer). Could you, however, please consider opening a new question? It's pretty annoying to discuss this in the comments and, moreover, it is only partially related to your original question. – saz Jun 10 '18 at 19:11
  • It's not needed for the construction of $F$. $F$ is constructed as you did with the property that $F^n(x)\to F(x)$ in probability for all $x$. It's then simple to show that this already implies that $F^n\to F$ on any compact subset in probability. The $\sigma$-compactness is only needed for the selection of a null set and hence to obtain that $F$ is indistinguishable to a process belonging to $B_{\text{loc}}(\mathcal A;M,E)$. At least if I'm not missing something. – 0xbadf00d Jun 10 '18 at 19:17