Martingales in Banach spaces
1 Three line summary
Conditional expectations exist in a natural way for simple functions, by taking extensions they also exist for integrable functions to a Banach space \(L^1(\Omega\to E)\).
Using conditional expectations we can define what a martingale is just like in the real case.
The space of continuous \(p\)-integrable martingales is a Banach space.
2 Why should I care?
Banach valued martingales form the basis of SPDEs. This is because analogously to Itô integration of real-valued processes. Integrating against a Wiener process valued in a Banach space the same will produce a square integrable continuous martingale.
3 Conditional expectation
In graduate-level probability courses, given a \(\sigma\)-algebra \(\Gg\) one shows that by applying Radon–Nikodym’s theorem, for any real-valued random variable \(X\in L^1(\Omega\to \R)\) there exists a conditional expectation \(\E_{\Gg}[X]\) verifying that
\[ \int_A \E_{\Gg}[X]=\int_A X,\quad \forall A\in\Gg. \]
Of course, now that we have created an integral for integral random variables to a Banach space \(L^1(\Omega\to E)\) we would like to see whether such a conditional expectation also exists for these functions. If we are given a simple function
\[ X=\sum_{k=1}^n x_k 1_{A_k}, \quad x_k\in E, A_k\in\Gg. \]
It is a simple calculation to show that, since \(1_{A_k}\) are real-valued and thus \(\E_{\Gg}[1_{A_k}]\) are well defined, then
\[ \E_{\Gg}[X]=\sum_{k=1}^n x_k \E_{\Gg}[1_{A_k}], \]
verifies the desired formula. Furthermore, we have that \(\mathbb{E}_\mathcal{F}\) is a linear, and pointwise continuous operator with
\[ \left\|\E_{\Gg}[X]\right\|\leq \sum_{k=1}^n \|x_k\| \E_{\Gg}[1_{A_k}]=\E_{\Gg}\left[\sum_{k=1}^n \|x_k\|1_{A_k}\right]=\E_{\Gg}\left[\|X\|\right]. \]
This allows us to show the following:
Theorem 1 (Existence and uniqueness of conditional expectation) Let \(X\in L^1(\Omega\to E)\) for some Banach space \(E\). Then \(X\) has a conditional expectation satisfying
\[ \left\|\E_{\Gg}[X]\right\|\leq\E_{\Gg}\left[\|X\|\right]. \]
Proof. We have already proved the above inequality for simple processes. By our previous post on the Bochner integral, we can take \(X_n\) converging to \(X\) in \(L^1(\Omega\to E)\) to obtain that
\[ \begin{aligned} \left\|\E_{\Gg}[X_n-X_m]\right\|&\leq\E_{\Gg}\left[\|X_n-X_m\|\right]\\&\implies \E[\left\|\E_{\Gg}[X_n]-\E_{\Gg}[X_m]\right\|]\leq \E\left[\|X_n-X_m\|\right]\to 0 \end{aligned} \]
As a result, \(\E_{\Gg}[X_n]\) is a Cauchy sequence in \(L^1(\Omega\to E)\) and converges to some function \(Y\), passing to the limit in the defining equation for the conditional expectation shows that \(Z=\E_\Gg[X]\). Finally, to prove uniqueness we have that if both \(Z_1,Z_2\) satisfy
\[ \int_A Z_1=\int_A X=\int_A Z_2,\quad \forall A\in\Gg. \]
Then using the linearity of the integral we obtain that \(w(Z_1)=w(Z_2)\) for all linear functions \(w \in E^*\), so \(Z_1=Z_2\).
4 Martingales
Okay, so we leveraged some inequalities to prove the existence of a conditional expectation. This done, the following definition mimicking the real case is quite natural. Given measurable spaces \((F,\Ff)\) and \((G,\Gg)\), we use the notation \(f: \Ff \to \Gg\) to mean \(f:(F,\Ff)\to (G,\Gg)\) is measurable.
Definition 1 Let \(\{M(t)\}_{t\in I}\) be a stochastic process on \((\Omega, \Ff, \PP)\) with a filtration \(\{\Ff_{t}\}_{t \in I}\). The process \(M\) is called an \(\Ff_{t}\)-martingale, if:
\(M(t)\in L^1(\Omega\to E)\) for all \(t\in I\),
\(M(t):\Ff_{t} \to \Bb(E)\) for all \(t\in I\),
\(\E_{\Ff_{s}}\left[M(t)\right]=M(s)\) for all \(s \leq t\).
The concept of submartingale is defined by replacing the equality in 3. with a \(\geq\). Let us abbreviate \(\E_{\Ff_t}\) by \(\E_t\). Then, as in the real case, we have the following.
Lemma 1 (Norm: martingale \(\to\) submartingale) Let \(M(t)\) be a martingale, then \(\norm{M(t)}\) is a submartingale.
Proof. We recall that, by the Hahn Banach theorem, it holds for any metric space that given \(y\in E\)
\[ \norm{z}=\sup_{\ell\in E^*:\|\ell\|=1}\ell(z) \]
As a result, by the linearity of the integral and abbreviating the supremum to just \(\sup_\ell\),
\[ \begin{aligned} \|M(s)\|&=\|\E_{s}[M(t)]\|= \sup_\ell \ell\left(\E_{s}[M(t)]\right)=\sup_\ell\|{\E_{s}\left[\ell(M(t))\right]\|}\\ &\leq \E_{s}\left[\sup_\ell\ell(M(t))\right]=\E_{s}\left[\|M(t)\|\right] \end{aligned} \]
Let us recall the following result for real-valued martingales (Çinlar 2011, Theorem 3.2.6).
Lemma 2 (Doob’s maximal Martingale inequality) Let \(\{X_k\}_{k=1}^\infty\) be a real-valued submartingale. Then it holds that
\[ \norm{\max_{k\in\{1,...,n\}}\abs{X_k}}_{L^p(\Omega)}\leq \frac{p}{p-1}\norm{X_n}_{L^p(\Omega)} \]
As a consequence, if \(X_t,t\in[0,T]\) is left (or right) continuous then
\[ \norm{\max_{t\in[0,T]}X_k}_{L^p(\Omega)}\leq \frac{p}{p-1}\norm{X_T}_{L^p(\Omega)}. \]
The idea of the above result is that, since \(X_k\) is a submartingale, \(X_k\lesssim X_{k+1}\lesssim...\lesssim X_n\). Getting from the continuous to the discrete case is possible by using the continuity of \(X\) and approximating it on some finer and finer mesh \(t_0,...,t_n\). This said, applying Doob’s maximal martingale inequality together with Lemma 1 gives that
Theorem 2 (Maximal Inequality) Let \(p>1\) and let \(E\) be a separable Banach space. If \(M(t)\), is a right-continuous \(E\)-valued \(\mathcal{F}_{t}\)-martingale, then
\[ \begin{aligned} \left(E\left(\sup _{t \in[0, T]}\|M(t)\|^{p}\right)\right)^{\frac{1}{p}} &\leq \frac{p}{p-1} \sup _{t \in[0, T]}\left(E\left(\|M(t)\|^{p}\right)\right)^{\frac{1}{p}} \\ &=\frac{p}{p-1}\left(E\left(\|M(T)\|^{p}\right)\right)^{\frac{1}{p}} \end{aligned} \]
Proof. This follows by using that \(\norm{M(t)}\) is a submartingale and Doob’s maximal inequality.
Doob’s inequality is essentially an equality between different function norms we can place on the space of continuous Martingales and will provide a very powerful tool later on.
Corollary 1 Let \(M\) be a (left or right) continuous martingale to a separable Banach space \(E\). Then the following are equivalent
\(M\in \hat{L}^\infty([0,T]\to \hat{L}^2(\Omega\to E))\)
\(M\in \hat{L}^2(\Omega\to \hat{L}^\infty([0,T]\to E))\)
\(\mathbb{E}[\norm{M(T)}^2]<\infty\)
Where we recall from the previous post that \(\hat{L}^p\) symbolizes that \(M\) may not be separately valued and only have an integrable norm. That said, the same reasoning shows that the above result also holds for the integrable \(L^p\) spaces.
A useful space of Martingales is as follows
Definition 2 Let \(M(t)\) be a \(E\) valued martingale with index set \(I=[0,T]\), then we define
\[ \mathcal{M}_T^2(E):=\left\{\text{continuous martingales } M:\mathbb{E}[\norm{M(T)}^2]<\infty\right\} \]
and give it the norm
\[ \norm{M}_{\mathcal{M}_T^2(E)}:=\left(\mathbb{E}[\norm{M(T)}^2]\right)^ \frac{1}{2} \]
By Theorem 2 we have that
\[ \mathcal{M}_T^2(E)\subset \hat{L}^\infty([0,T]\to \hat{L}^2(\Omega\to E))\cap \hat{L}^2(\Omega\to \hat{L}^\infty([0,T]\to E)). \]
and that any of the norms of these spaces is equivalent to the one set on \(\mathcal{M}_T^2(E)\). This is useful in the following result
Proposition 1 Let \(E\) be a separable Banach space, then \(\mathcal{M}_T^2(E)\) is a Banach space.
Proof. By the previous observation and the completeness of the \(\hat{L}^p\) spaces proved in the previous post, \(\mathcal{M}_T^2(E)\) is a subspace of a Hilbert space. As a result, it is sufficient to show that it is closed. Let \(M_n\) converge to \(M\). Then, by the equivalence of the norms we have that \(M_n(t)\to M(t)\in \hat{L}^1(\Omega\to E)\subset \hat{L}^2(\Omega\to E)\) so that for all \(A\in\mathcal{F}_s\)
\[ \int_A M(s)d\mathbb{P}=\lim_{n\to\infty}\int_A M_n(s)d\mathbb{P}=\lim_{n\to\infty}\int_A M_n(t)d\mathbb{P}=\int_A M(t)d\mathbb{P}. \]
This shows that \(M\) is a martingale. Furthermore, as was seen in the previous post, there exists a subsequence \(M_{n_k}\) such that
\[ \lim_{n\to\infty}{M_{n_k}}(\cdot,\omega)=M(\cdot,\omega)\in \hat{L}^\infty([0,T]\to E)\quad a.e.\quad \omega\in\Omega \]
Since \(M_{n_k}(\cdot,\omega)\) are continuous and continuity is preserved by uniform limits this proves that \(M\) is continuous almost everywhere. This concludes the proof.
In future instalments, we will prove that a Wiener process valued in a Hilbert space \(H\) belongs to this space and use it to define the stochastic integral that leads to the construction of SPDEs.
Proposition 2 Let \(H\) be a separable Hilbert space and \(Q\) be a symmetric, non-negative, trace-class operator on \(H\). Let \(W(t)\) be an \(H\)-valued \(Q\)-Wiener process with respect to a filtration \(\Ff_t\). Then \(W(t) \in \mathcal{M}_T^2(H)\).
Proof. The process \(W(t)\) is adapted by definition. Let \(0 \leq s < t \leq T\). Given \(A \in \Ff_s\) and \(u \in H\), the mapping \(x \mapsto \langle x, u \rangle\) is a bounded linear functional. Since \(W(t) - W(s)\) is independent of \(\Ff_s\), we apply the properties of the Bochner integral to obtain:
\[ \begin{aligned} \br{ \int_A (W(t)-W(s)) d\mathbb{P}, u } &= \int_A \langle W(t)-W(s), u \rangle d\mathbb{P} \\ &= \mathbb{E}[1_A] \mathbb{E}[\langle W(t)-W(s), u \rangle] = 0. \end{aligned} \]
This satisfies the definition of conditional expectation, proving \(\mathbb{E}[W(t) \mid \Ff_s] = W(s)\), so \(W(t)\) is a martingale. Finally, by construction \(\mathbb{E}[\|W(t)\|^2] = t \operatorname{Tr}(Q)<\infty\) as \(Q\) is trace-class. Continuity is given by the definition of the Wiener process. This concludes the proof.
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