Approximation of $\int_0^tF_x(s,X_s)Φ_0dW_s$ where $dX_s=φ_sds+Φ_sdW_s$ and $F_x$ is the Fréchet derivative of some $F:[0,t]×H→\mathbb R$

Question

Let

• $U$ and $H$ be Hilbert spaces
• $Q\in\mathfrak L(U)$ be nonnegative and symmetric with finite trace
• $U_0:=Q^{1/2}U$ be equipped with the usual inner product
• $(\Omega,\mathcal A,\operatorname P)$ be a probability space
• $(\mathcal F_t)_{t\ge 0}$ be a filtration of $\mathcal A$
• $(W_t)_{t\ge 0}$ be a $Q$-Wiener process on $(\Omega,\mathcal A,\operatorname P)$ with respect to $\mathcal F$
• $X_0$ be a $\mathcal F_0$-measurable random variable on $(\Omega,\mathcal A,\operatorname P)$
• $(\varphi_t)_{t\ge 0}$ be a $H$-valued $\mathcal F$-adapted stochastic process on $(\Omega,\mathcal A,\operatorname P)$
• $(\Phi_t)_{t\ge 0}$ be a $\operatorname{HS}(U_0,H)$-valued $\mathcal F$-adapted stochastic process on $(\Omega,\mathcal A,\operatorname P)$

Now, let

• $t\ge 0$ and $0=t_0<\cdots<t_n=t$ for some $n\in\mathbb N_0$
• $\Delta W_i:=W_{t_i}-W_{t_{i-1}}$
• $L_1,\ldots,L_n\in\mathfrak L(H,\mathbb R)$

We can show (see Da Prato, Proposition 4.30) that if $\Phi$ is $\mathcal F$-predictable, $$\operatorname P\left[\int_0^t\left\|\Phi_s\right\|_{\operatorname{HS}(U_0,H)}{\rm d}s<\infty\right]=1\tag 1$$ and $$\operatorname P\left[\int_{t_{i-1}}^{t_i}\left\|L_i\Phi_s\right\|_{\operatorname{HS}(U_0,\mathbb R)}{\rm d}s<\infty\right]=1\;\;\;\text{for all }i\in\left\{1,\ldots,n\right\}\;,\tag 2$$ then $$L_i\int_{t_{i-1}}^{t_i}\Phi_s\;{\rm d}W_s=\int_{t_{i-1}}^{t_i}L_i\Phi_s\;{\rm d}W_s\;\;\;\text{for all }i\in\left\{1,\ldots,n\right\}\;.\tag 3$$

Now, assume that $\Phi_s=\Phi_0$ for all $s\in[0,t]$ and $$L_i=F_x(t_{i-1},X_{t_{i-1}})$$ for some $F:[0,t]\times H\to\mathbb R$ with partial Fréchet derivative $F_x$ and $$X_s=X_0+\int_0^s\varphi_r\;{\rm d}r+\int_0^s\Phi_r\;{\rm d}W_r\;\;\;\text{for all }s\in[0,t]\;.$$ Then, $(1)$ and $(2)$ are satisfied such that we can conclude that $$S_n:=\sum_{i=1}^nL_i(\Phi_0\Delta W_i)=\sum_{i=1}^nL_i\left(\int_{t_{i-1}}^{t_i}\Phi_0\;{\rm d}W_s\right)=\sum_{i=1}^n\int_{t_{i-1}}^{t_i}L_i\Phi_0\;{\rm d}W_s$$ by $(3)$.

Question: Can we show that $$\lim_{n\to\infty}S_n=\int_0^tF_x(s,X_s)\Phi_0\;{\rm d}W_s$$ $\operatorname P$-almost surely?

Answer 1

Well, \begin{align} S_n&:=\sum_{i=1}^n\int_{t_{i-1}}^{t_i}L_i\Phi_0\;{\rm d}W_s\\ &=\sum_{i=1}^n\int_{t_{i-1}}^{t_i}F_x(t_{i-1},X_{t_{i-1}})\Phi_0\;{\rm d}W_s\\ &=\int_0^t\bigg[\sum_{i=1}^nF_x(t_{i-1},X_{t_{i-1}})\Phi_0\mathbf{1}_{(t_{i-1},t_‌​i]}(s)\bigg]\;{\rm d}W_s. \end{align} Now if $F(s,x)$ is continuous for $(s,x)\in[0,t]\times H$ and also $X_s$ is a continuous process on $[0,t]$, then the integrand in the last integral converges surely (not merely almost surely) and thereby converges with respect to norm $|||\cdot|||$ (in p96 of Prato's book) to $F_x(s,X_s)\Phi_0$. Hence by the isometry property of stochastic integral, $$S_n\to\int_0^tF_x(s,X_s)\Phi_0\;{\rm d}W_s$$ in $L^2$ and thereby almost surely, as $n\to\infty$.

Answer 2

Let's try to prove this rigorously.

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Lemma 1 Let

• $E$ be a topological space
• $t\ge 0$ and $t_0^{(n)},\ldots,t_n^{(n)}\ge 0$ with $$0=t_0^{(n)}<\cdots<t_n^{(n)}=t$$ for some $n\in\mathbb N$
• $F:[0,t]\times H\to E$ be continuous
• $(X_t)_{t\ge 0}$ be a left-continuous $H$-valued stochastic process on $(\Omega,\mathcal A,\operatorname P)$

Then, $$Y_s^{(n)}:=\sum_{i=1}^nF\left(t_{i-1}^{(n)},X_{t_{i-1}^{(n)}}\right)1_{\left(t_{i-1}^{(n)},t_i^{(n)}\right]}(s)\stackrel{n\to\infty}\to F(s,X_s)=:Y_s\;\;\;\text{for all }s\in (0,t]\;.$$

Proof:

• We can assume that $$t_i^{(n)}=\frac int\;\;\;\text{for all }i\in\left\{1,\ldots,n\right\}$$
• Let $s\in (0,t]$ $\Rightarrow$ $\exists!i\in\left\{1,\ldots,n\right\}$ with $$s\in\left(t_{i-1}^{(n)},t_i^{(n)}\right]$$
• Since $$t_{i-1}^{(n)}\uparrow s\;\;\;\text{for }n\to\infty\;,$$ we obtain $$\left(t_{i-1}^{(n)},X_{t_{i-1}^{(n)}}\right)\stackrel{n\to\infty}\to (s,X_s)$$ by left-continuity of $X$ and hence $$Y_s^{(n)}=F\left(t_{i-1}^{(n)},X_{t_{i-1}^{(n)}}\right)\stackrel{n\to\infty}\to F(s,X_s)=Y_s$$ by continuity of $F$

Lemma 2 Let

• $U$ and $H$ be Hilbert spaces
• $L\in\mathfrak L(H,\mathbb R)$
• $Q\in\operatorname{HS}(U,H)$

Then, $$L\circ Q\in\operatorname{HS}(U,\mathbb R)\;.$$

Proof:

• $L$ is bounded $\Rightarrow$ $\exists C>0$ with $$|Lx|\le C\left\|x\right\|_H\;\;\;\text{for all }x\in H\tag 5$$
• Let $(e_n)_{n\in\mathbb N}$ be an orthonormal basis of $U$ $\Rightarrow$ $$\left\|L\circ Q\right\|_{\operatorname{HS}(U,\mathbb R)}^2\stackrel{\text{def}}=\sum_{n\in\mathbb N}\left|L(Qe_n)\right|^2\stackrel{(5)}\le C^2\sum_{n\in\mathbb N}\left\|Qe_n\right\|_H^2\stackrel{\text{def}}=C^2\left\|Q\right\|_{\operatorname{HS}(U,H)}^2<\infty\tag 6$$

Lemma 3 In the situation of (Lemma 2), let $(L_n)_{n\in\mathbb N}\subseteq\mathfrak L(H,\mathbb R)$ with $$C_n:=\left\|L_n\right\|_{\mathfrak L(H,\mathbb R)}\stackrel{n\to\infty}\to 0\;.\tag 7$$ Then, $$\left\|L_n\circ Q\right\|_{\operatorname{HS}(U,\mathbb R)}\stackrel{n\to\infty}\to 0\;.$$

Proof:

• By definition of operator boundedness and $(6)$, we obtain $$\left\|L_n\circ Q\right\|_{\operatorname{HS}(U,\mathbb R)}^2\le C_n^2\left\|Q\right\|_{\operatorname{HS}(U,H)}^2\tag 8$$
• Thus, the claim follows by continuity of $$\mathbb R\to\mathbb R\;,\;\;\;x\mapsto x^2$$ and $(7)$

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Using (Lemma 1) with $E=\mathfrak L(H,\mathbb R)$ and $F$ replaced by $F_x$ from the question, we obtain $$\sqrt{Z_s^{(n)}}:=\left\|\left(Y_s^{(n)}-Y_s\right)\Phi_0\right\|_{\operatorname{HS}(U_0,\mathbb R)}\stackrel{n\to\infty}\to 0\;\;\;\text{for all }s\in (0,t]\tag 9$$ by (Lemma 3), if we assume that $F$ (from the question) is continuously partially Fréchet differentiable with respect to the second argument. Notice that by (Lemma 2) the $\left(Y^{(n)}-Y\right)\Phi_0$ are indeed $\operatorname{HS}(U_0,\mathbb R)$-valued.

Since $\Phi_0=\Phi_s$ for all $s\in [0,t]$, $\Phi_0$ is $(\mathcal F_s)_{s\in[0,t]}$-adapted. Thus, since $X$ is $\mathcal F$-adapted, each $Y^{(n)}\Phi_0$ is an $(\mathcal F_s)_{s\in[0,t]}$-elementary stochastic process on $(\Omega,\mathcal A,\operatorname P)$.

Claim $\exists g\in\mathcal L^1\left(\left.\lambda\right|_{[0,t]}\otimes\operatorname P\right)$ with $$Z^{(n)}\le g\;\;\;\left(\left.\lambda\right|_{[0,t]}\otimes\operatorname P\right)\text{-almost everywhere}\;.\tag{10}$$

Using the claim and $(9)$, Lebesgue's Dominated Convergence Theorem (combined with Fubini's theorem and calling the relationship of the Lebesgue and the Riemann integral to mind) yields $$\operatorname E\left[\int_0^tZ_s^{(n)}\;{\rm d}s\right]\stackrel{n\to\infty}\to 0\;.$$ Thus, by definition of the Itō integral with respect to $W$ (see Da Prato, Proposition 4.22 (ii)) $$\int_0^tY_s^{(n)}\Phi_0\;{\rm d}W_s\stackrel{n\to\infty}\to\int_0^tY_s\Phi_0\;{\rm d}W_s$$ in $L^2(\operatorname P;\operatorname{HS}(U_0,\mathbb R))$.

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However, I've got some problems in showing the claim. Let $$L_s^{(n)}(\omega):=Y_s^{(n)}(\omega)-Y(\omega)\;\;\;\text{for }s\in(0,t]\text{ and }\omega\in\Omega\;.$$ Then, $$C_s^{(n)}:=\left\|L_s^{(n)}(\omega)\right\|_{\mathfrak L(H,\mathbb R)}\stackrel{n\to\infty}\to 0\;\;\;\text{for all }s\in(0,t]\text{ and }\omega\in\Omega\;.$$ Thus, $\exists C:\Omega\times(0,t]$ with $$C_s^{(n)}(\omega)\le C_s(\omega)\;\;\;\text{for all }s\in(0,t]\text{ and }\omega\in\Omega\;.$$ By $(8)$, we can conclude that $$Z_s^{(n)}(\omega)\stackrel{\text{def}}=\left\|L_s^{(n)}(\omega)\Phi_0(\omega)\right\|^2_{\operatorname{HS}(U_0,\mathbb R)}\le\left[C_s(\omega)\right]^2\left\|\Phi_0(\omega)\right\|^2_{\operatorname{HS}(U_0,H)}=:\tilde g(s,\omega)$$ for all $s\in(0,t]$ and $\omega\in\Omega$.

However, I'm failing to show that $\tilde g\le g$ for some $g\in\mathcal L^1\left(\left.\lambda\right|_{[0,t]}\otimes\operatorname P\right)$.