Computing deltas without derivatives

Abstract

A well-known application of Malliavin calculus in mathematical finance is the probabilistic representation of option price sensitivities, the so-called Greeks, as expectation functionals that do not involve the derivative of the payoff function. This allows numerically tractable computation of the Greeks even for discontinuous payoff functions. However, while the payoff function is allowed to be irregular, the coefficients of the underlying diffusion are required to be smooth in the existing literature, which for example already excludes simple regime-switching diffusion models. The aim of this article is to generalise this application of Malliavin calculus to Itô diffusions with irregular drift coefficients, where we focus here on the computation of the delta, which is the option price sensitivity with respect to the initial value of the underlying. To this end, we first show existence, Malliavin differentiability and (Sobolev) differentiability in the initial condition for strong solutions of Itô diffusions with drift coefficients that can be decomposed into the sum of a bounded, but merely measurable, and a Lipschitz part. Furthermore, we give explicit expressions for the corresponding Malliavin and Sobolev derivatives in terms of the local time of the diffusion, respectively. We then turn to the main objective of this article and analyse the existence and probabilistic representation of the corresponding deltas for European and path-dependent options. We conclude with a small simulation study of several regime-switching examples.

Appendix: Proofs of results in Sect. 3

In this appendix, we collect the proofs of the results in Sect. 3.

1.1 A.1 Proof of Theorem 3.1

We start by giving two estimates that will be used throughout some proofs in the sequel.

Remark A.1

From Lemma 2.6, it follows immediately that if the approximating functions $b_{n}$, $n\ge1$, are as in (3.1), then for any compact subset $K\subset\mathbb{R}$, one can find an $\varepsilon>0$ such that

$$\sup_{x\in K} \sup_{n\geq0}E\bigg[ \mathcal{E}\bigg( \int_{0}^{T} b_{n}(u,B_{u}^{x})dB_{u}\bigg)^{1+\varepsilon} \bigg] < \infty, $$

where we recall that $b_{0}:=b$.

Lemma A.2

Let $f:[0,T]\times\mathbb{R}\rightarrow\mathbb{R}$ be a bounded measurable function. Then for every $t\in[0,T]$, $\lambda\in\mathbb {R}$ and compact subset $K\subset\mathbb{R}$, we have

$$\sup_{x\in K} E\bigg[\exp\bigg(\lambda\int_{0}^{t} \int_{\mathbb {R}} f(s,y) L^{B^{x}} (ds,dy)\bigg)\bigg] < \infty, $$

where $L^{B^{x}}(ds,dy)$ denotes integration with respect to the local time of the Brownian motion $B^{x}$ in both time and space; see Sect. 2.2.

Proof

By using the decomposition (2.5) from Sect. 2 and the Cauchy–Schwarz inequality twice, we have

$$\begin{aligned} &E\bigg[\exp\bigg(\lambda\int_{0}^{t} \int_{\mathbb{R}} f(s,y) L^{B^{x}} (ds,dy)\bigg)\bigg]\\ &\leq\, E\bigg[\exp\bigg(2\lambda\int_{0}^{t} f(s,B_{s}^{x}) dB_{s}\bigg)\bigg]^{1/2}\\ &\phantom{=:}\times E\bigg[\exp\bigg(4\lambda\int_{T-t}^{T} f(T-s,B_{T-s}^{x}) dW_{s}\bigg)\bigg]^{1/4}\\ &\phantom{=:}\times E\bigg[\exp\bigg(-4\lambda\int_{T-t}^{T} f(T-s,B_{T-s}^{x}) \frac{B_{T-s}}{T-s} ds\bigg)\bigg]^{1/4}\\ &=:\, I\times \mathit{II} \times \mathit{III}, \end{aligned}$$

where $W_{t} := \int_{0}^{t} \frac{B_{T-s}}{T-s}ds + B_{T-t}-B_{T}$ is a Brownian motion. Since $f$ is bounded, the factors $I$ and $\mathit{II}$ are bounded uniformly in $x$ by Bernstein’s inequality for exponential martingales. Finally, the boundedness of $\mathit{III}$ follows from

$$\begin{aligned} E\bigg[\exp\bigg(k\int_{0}^{T} \frac{|B_{T-s}|}{T-s} ds\bigg)\bigg]< \infty \end{aligned}$$

(A.1)

for any $k\in\mathbb{R}$. To see (A.1), using (2.5) with $f(s,y)=\mathrm{sgn}(y)$, $x=0$ and $t=T$ in connection with Tanaka’s formula, we find that

$$\begin{aligned} \int_{0}^{T} \frac{|\hat{B}_{s}|}{T-s}ds = 2|B_{T}|-\int_{0}^{T} \mathrm {sgn}(B_{s})dB_{s} +\int_{0}^{T} \mathrm{sgn}(\hat{B}_{s}) dW_{s}. \end{aligned}$$

Then using Hölder’s inequality, we obtain

$$\begin{aligned} E\bigg[\exp\bigg(k \int_{0}^{T} \frac{|B_{T-s}|}{T-s}ds\bigg)\bigg] &\leq E[\exp(6k |B_{T}|)]^{1/3} \\ &\phantom{=:}\times E\bigg[\exp\bigg(-3k\int_{0}^{T} \mathrm {sgn}(B_{s}) dB_{s} \bigg)\bigg]^{1/3}\\ &\phantom{=:} \times E\bigg[\exp\bigg(3k \int_{0}^{T}\mathrm {sgn}(\widehat{B}_{s})dW_{s}\bigg)\bigg]^{1/3}, \end{aligned}$$

and the result follows. □

We now develop the proof of Theorem 3.1 according to the four-step scheme outlined in Sect. 3. In order to construct a weak solution of (1.4) in the first step, let $(\varOmega, \mathcal{F}, \widetilde{P})$ be some given probability space which carries a Brownian motion $\widetilde{B}$, and put $X^{x}_{t}:=\widetilde{B}_{t} + x$, $t\in[0,T]$. By Remark 2.5, $\frac{dP}{d\tilde{P}} := \mathcal{E}( \int_{0}^{T} b(u,X^{x}_{u})d\widetilde{B}_{u})$ defines an equivalent probability measure $P$, under which the process

$$B_{t} := X^{x}_{t} - x - \int_{0}^{t} b(s,X^{x}_{s})ds, \quad t\in[0,T], $$

is a Brownian motion on $(\varOmega, \mathcal{F},P)$. Hence the pair $(X^{x},B)$ is a weak solution of (1.4) on $(\varOmega, \mathcal{F},P)$. The stochastic basis that we operate on in the following is now given by the filtered probability space $(\varOmega, \mathcal{F}, P,(\mathcal{F}_{t})_{t\in[0,T]})$ which carries the weak solution $(X^{x},B)$ of (1.4), where $(\mathcal{F}_{t})_{t\in [0,T]}$ denotes the filtration generated by $B_{t}$, $t\in[0,T]$, augmented by the $P$-nullsets.

The next lemma provides the second step of the proof.

Lemma A.3

Let $b_{n}:[0,T]\times\mathbb{R}\rightarrow\mathbb{R}$ be a sequence of functions approximating $b$ a.e. as in (3.1) and $X^{n,x}$, $n\geq1$, the corresponding strong solutions to (3.2). Then for every $t\in[0,T]$ and function $\varphi\in L^{2p} (\mathbb{R};w_{t})$, where the space $L^{2p} (\mathbb{R};w_{t})$ is defined as in (4.6) with $p$ being the conjugate exponent of $1+\varepsilon$, $\varepsilon>0$ from Lemma 2.6, we have

$$\begin{aligned} \varphi(X_{t}^{n,x}) \longrightarrow E[\varphi (X_{t}^{x})|\mathcal{F}_{t}] \qquad \textit{as } n\to\infty \end{aligned}$$

weakly in $L^{2}(\varOmega;\mathcal{F}_{t})$.

Proof

First note that $\varphi(X_{t}^{n,x})$ and $E[\varphi(X_{t}^{x})|\mathcal {F}_{t}]$ are in $L^{2}(\varOmega;\mathcal{F}_{t})$, $n\geq0$. Indeed, Girsanov’s theorem, Remark A.1 and the fact that $\varphi \in L^{2p}(\mathbb{R};w_{t})$ imply that for some constant $C_{\varepsilon}>0$ with $\varepsilon>0$ small enough, we have

$$\begin{aligned} \begin{aligned} \sup_{n\geq0} E\big[|\varphi(X_{t}^{n,x})|^{2}\big] &\leq C_{\varepsilon} E\big[|\varphi(x+B_{t})|^{2\frac{1+\varepsilon }{\varepsilon}}\big]^{\frac{\varepsilon}{1+\varepsilon}} \\ &= C_{\varepsilon}\frac{1}{\sqrt{2\pi t}}\int_{\mathbb{R}} |\varphi(x+z)|^{2\frac{1+\varepsilon}{\varepsilon}} e^{-\frac {|z|^{2}}{2t}}dz < \infty. \end{aligned} \end{aligned}$$

(A.2)

To show the weak convergence, it suffices to show

$$\begin{aligned} \mathcal{W}\big(\varphi(X_{t}^{n,x})\big)(f) \longrightarrow\mathcal {W}\big(E[\varphi(X_{t}^{x})|\mathcal{F}_{t}]\big)(f)\qquad \text{as } n \to\infty \end{aligned}$$

for every $f\in L^{2}([0,T])$. Indeed, by Girsanov’s theorem and (1.8), we can write

$$\begin{aligned} &\mathcal{W}\big(\varphi(X_{t}^{n,x})\big)(f) - \mathcal{W}\big(E[\varphi(X_{t}^{x})|\mathcal{F}_{t}]\big)(f) \\ &= E\bigg[ \varphi(B_{t}^{x}) \bigg( \mathcal{E}\Big(\int_{0}^{T} \big(b_{n}(u,B_{u}^{x})+f(u)\big)dB_{u}\Big) \\ &\phantom{=:E\bigg[ \varphi(B_{t}^{x}) \Big(}{}- \mathcal{E}\Big(\int _{0}^{T} \big(b(u,B_{u}^{x})+f(u)\big)dB_{u}\Big)\bigg) \bigg]. \end{aligned}$$

Using the inequality $|e^{x}-e^{y}|\leq|x-y|(e^{x}+e^{y})$ for $x,y\in\mathbb {R}$, we get

$$\begin{aligned} &E\bigg[ \big(\varphi(X_{t}^{n,x}) - E[\varphi(X_{t}^{x})|\mathcal {F}_{t}]\big) \mathcal{E}\bigg(\int_{0}^{T} f(u) dB_{u}\bigg) \bigg]\\ &\leq E\bigg[|\varphi(B_{t}^{x})| |U_{n}| \bigg(\mathcal{E}\Big(\int _{0}^{T} \big(b_{n}(u,B_{u}^{x})+f(u)\big)dB_{u}\Big)\\ &\phantom{=:E\bigg[|\varphi(B_{t}^{x})| |U_{n}| \Big(}{}+\mathcal{E}\Big(\int_{0}^{T} \big(b(u,B_{u}^{x})+f(u)\big)dB_{u}\Big)\bigg)\bigg], \end{aligned}$$

where

$$\begin{aligned} U_{n} &:= \int_{0}^{T} \big(b_{n}(u,B_{u}^{x})-b(u,B_{u}^{x})\big)dB_{u}\\ &\phantom{=:} - \frac{1}{2}\int_{0}^{T} \Big(\big(b_{n}(u,B_{u}^{x})+f(u)\big)^{2} - \big(b(u,B_{u}^{x})+f(u)\big)^{2}\Big)du. \end{aligned}$$

Using Hölder’s inequality with exponents $p=\frac{1+\varepsilon }{\varepsilon}$ and $q=1+\varepsilon$ and then the Cauchy–Schwarz inequality yields

$$\begin{aligned} &E\bigg[ \big(\varphi(X_{t}^{n,x}) - E[\varphi(X_{t}^{x})|\mathcal {F}_{t}]\big) \mathcal{E}\bigg(\int_{0}^{T} f(u) dB_{u}\bigg) \bigg] \\ &\leq E\big[ | \varphi(B_{t}^{x})|^{2\frac{1+\varepsilon}{\varepsilon }}\big]^{\frac{\varepsilon}{2(1+\varepsilon)}} E\big[ | U_{n}|^{2\frac{1+\varepsilon}{\varepsilon}}\big]^{\frac{\varepsilon }{2(1+\varepsilon)}}\\ &\phantom{=:}\times E\bigg[\bigg(\mathcal{E}\Big(\int_{0}^{T} \big(b_{n}(u,B_{u}^{x})+f(u)\big)dB_{u}\Big)\\ &\phantom{=:\times E\bigg[\bigg(}{}+\mathcal{E}\Big(\int_{0}^{T} \big(b(u,B_{u}^{x})+f(u)\big)dB_{u}\Big)\bigg)^{1+\varepsilon}\bigg]^{\frac {1}{1+\varepsilon}}\\ &=: I^{1} \times I_{n}^{2} \times I_{n}^{3}, \end{aligned}$$

where $\varepsilon>0$ is such that $I_{n}^{3}$ is bounded uniformly in $n\geq0$ (see Remark A.1). The first factor $I^{1}$ is controlled as shown in (A.2). For the second factor $I_{n}^{2}$, using the Minkowski, Burkholder–Davis–Gundy and Hölder inequalities, we can write

$$\begin{aligned} (I_{n}^{2})^{2p} &\lesssim\int_{0}^{T} E\big[|b_{n}(u,B_{u}^{x})-b(u,B_{u}^{x})|^{p} \big] du \\ &\phantom{=:}{}+ \int_{0}^{T} E\big[ \big|\big(b_{n}(u,B_{u}^{x})+f(u)\big)^{2}-\big(b(u,B_{u}^{x})+f(u)\big)^{2}\big|^{2p} \big]du, \end{aligned}$$

and by dominated convergence we obtain $I_{n}^{2} \to0$ as $n\to\infty$. □

We now turn to the third step of our scheme to prove Theorem 3.1.

Theorem A.4

Let $b_{n}: [0,T] \times\mathbb{R}\rightarrow\mathbb{R}$, $n\geq1$, be as in (3.1) and $X^{n,x}$ the corresponding strong solutions to (3.2). Then for each $t\in[0,T]$,

$$\begin{aligned} X_{t}^{n,x} \longrightarrow E[X_{t}^{x} |\mathcal{F}_{t}] \qquad \textit{in } {L^{2}(\varOmega)} \end{aligned}$$

(A.3)

as $n\to\infty$. Moreover, the right-hand side of (A.3) is Malliavin-differentiable.

Proof

The main step is to show relative compactness of $(X_{t}^{n,x})_{n\geq 1}$ by applying Proposition 2.3. Let $t\in[0,T]$, $0\leq s\leq s'\leq t$ and a compact set $K\subset\mathbb{R}$ be given. Using successively the explicit representation introduced in (3.3), Girsanov’s theorem, the mean value theorem, Hölder’s inequality with exponent $1+\varepsilon$ for a sufficiently small $\varepsilon>0$ and the Cauchy–Schwarz inequality, we obtain

$$\begin{aligned} &E\big[(D_{s} X_{t}^{n,x} - D_{s'}X_{t}^{n,x})^{2}\big]\\ &= E\bigg[\exp\bigg(2\int_{s'}^{t} b_{n}'(u, B_{u}^{x})du\bigg) \bigg(\exp\Big(\int_{s}^{s'} b_{n}'(u, B_{u}^{x})du\Big) -1\bigg)^{2} \mathcal {E}(b_{n})_{T}\bigg]\\ &\leq E\bigg[\exp\bigg(2\int_{s'}^{t} b_{n}'(u, B_{u}^{x})du\bigg) \bigg(\sup_{0\leq\theta\leq1}\exp\Big(\theta\int_{s}^{s'} b_{n}'(u, B_{u}^{x})du\Big)\bigg)^{2}\\ &\phantom{=:E\bigg[}\times\bigg(\int_{s}^{s'} b_{n}'(u, B_{u}^{x})du\bigg)^{2} \mathcal{E}(b_{n})_{T}\bigg]\\ &\leq E\bigg[\exp\bigg(2\frac{1+\varepsilon}{\varepsilon}\int _{s'}^{t} b_{n}'(u, B_{u}^{x})du\bigg) \sup_{0\leq\theta\leq1}\exp\bigg(2\frac{1+\varepsilon}{\varepsilon}\theta\int_{s}^{s'} b_{n}'(u, B_{u}^{x})du\bigg)\\ &\phantom{=:E\bigg[}\times\bigg|\int_{s}^{s'} b_{n}'(u, B_{u}^{x})du\bigg|^{2\frac{1+\varepsilon}{\varepsilon}}\bigg]^{\frac{\varepsilon }{1+\varepsilon}} E\big[ \mathcal{E}(b_{n})_{T}^{1+\varepsilon}\big]^{\frac{1}{1+\varepsilon}}\\ &\leq E\bigg[\exp\bigg(4\frac{1+\varepsilon}{\varepsilon}\int _{s'}^{t} (\tilde{b}_{n}'(u, B_{u}^{x})+\hat{b}'(u,B_{u}^{x}))du\bigg)\bigg]^{\frac{\varepsilon}{2(1+\varepsilon)}}\\ &\phantom{=:}\times E\bigg[ \sup_{0\leq\theta\leq1}\exp\bigg( 8\frac{1+\varepsilon}{\varepsilon}\theta\int_{s}^{s'} \big(\tilde {b}_{n}'(u, B_{u}^{x})+\hat{b}'(u,B_{u}^{x})\big)du\bigg)\bigg]^{\frac {\varepsilon}{4(1+\varepsilon)}}\\ &\phantom{=:}\times E\bigg[\bigg|\int_{s}^{s'} \big(\tilde{b}_{n}'(u, B_{u}^{x})+\hat{b}'(u,B_{u}^{x})\big)du\bigg|^{8\frac{1+\varepsilon }{\varepsilon}}\bigg]^{\frac{\varepsilon}{4(1+\varepsilon)}} E\big[ \mathcal{E}(b_{n})_{T}^{1+\varepsilon}\big]^{\frac{1}{1+\varepsilon }}\\ &=: I_{n}^{1} \times I_{n}^{2} \times I_{n}^{3} \times I_{n}^{4}. \end{aligned}$$

Here, by Remark A.1, $\varepsilon>0$ is chosen such that $\sup_{x\in K}\sup_{n\geq0}I_{n}^{4} <\infty$. For $I_{n}^{1}$ and $I_{n}^{2}$, we use the Cauchy–Schwarz inequality and the fact that $\hat{b}'$ is bounded to get

$$\begin{aligned} I_{n}^{1} \lesssim E\bigg[\exp\bigg(4\frac {1+\varepsilon}{\varepsilon}\int_{s'}^{t} \tilde{b}_{n}'(u, B_{u}^{x})du\bigg)\bigg]^{\frac{\varepsilon}{2(1+\varepsilon)}}=:\mathit{II}_{n}^{1} \end{aligned}$$

and

$$\begin{aligned} I_{n}^{2} \lesssim E\bigg[ \sup_{0\leq\theta\leq 1}\exp\bigg( 8\frac{1+\varepsilon}{\varepsilon}\theta\int_{s}^{s'} \tilde{b}_{n}'(u, B_{u}^{x})du\bigg)\bigg]^{\frac{\varepsilon }{4(1+\varepsilon)}}=:\mathit{II}_{n}^{2}. \end{aligned}$$

For $I_{n}^{3}$, Minkowski’s inequality and the boundedness of $\hat{b}'$ give

$$\begin{aligned} I_{n}^{3} &\leq E\bigg[ \bigg|\int_{s}^{s'} \tilde{b}_{n}'(u,B_{u}^{x})du\bigg|^{8\frac{1+\varepsilon}{\varepsilon}}+ \bigg|\int_{s}^{s'} \hat {b}'(u,B_{u}^{x})du\bigg|^{8\frac{1+\varepsilon}{\varepsilon}}\bigg]^{\frac{\varepsilon}{4(1+\varepsilon)}}\\ &\lesssim E\bigg[ \bigg|\int_{s}^{s'} \tilde{b}_{n}'(u,B_{u}^{x})du \bigg|^{8\frac{1+\varepsilon}{\varepsilon}}\bigg]^{\frac{\varepsilon }{4(1+\varepsilon)}}+ \|\hat{b}'\|_{\infty}^{2} T |s'-s|\\ &=: \mathit{II}_{n}^{3} + \|\hat{b}'\|_{\infty}^{2} T |s'-s|. \end{aligned}$$

To get rid of the derivatives $\tilde{b}_{n}'$ in $\mathit{II}_{n}^{1},\mathit{II}_{n}^{2}$ and $\mathit{II}_{n}^{3}$, we use integration with respect to the local time of the Brownian motion (see Theorem 2.10) and obtain

$$\begin{aligned} &E\big[(D_{s} X_{t}^{n,x} - D_{s'}X_{t}^{n,x})^{2}\big] \\ & \lesssim E\bigg[\exp\bigg(-4\frac{1+\varepsilon}{\varepsilon }\int_{s'}^{t}\int_{\mathbb{R}} \tilde{b}_{n}(u,y)L^{B^{x}}(du,dy)\bigg)\bigg]^{\frac{\varepsilon}{2(1+\varepsilon)}} \\ &\phantom{=:}\times E\bigg[ \sup_{0\leq\theta\leq1}\exp\bigg(- 8\frac{1+\varepsilon}{\varepsilon}\theta\int_{s}^{s'}\int_{\mathbb {R}} \tilde{b}_{n}(u, x)L^{B^{x}}(du,dy)\bigg)\bigg]^{\frac{\varepsilon }{4(1+\varepsilon)}} \\ &\phantom{=:}\times\bigg(E\bigg[\bigg|\int_{s}^{s'}\int_{\mathbb {R}} \tilde{b}_{n}(u, x)L^{B^{x}}(du,dy)\bigg|^{8\frac{1+\varepsilon }{\varepsilon}}\bigg]^{\frac{\varepsilon}{4(1+\varepsilon)}}+ \| \hat{b}'\| |s'-s|\bigg). \end{aligned}$$

(A.4)

Observe that the first two factors in (A.4) can be controlled uniformly in $n\geq1$ and $x\in K$ by virtue of Lemma A.2. Denote $p_{\varepsilon} := 4\frac{1+\varepsilon}{\varepsilon}$, and recall from above that the first summand in the third factor in (A.4) was denoted by $\mathit{II}_{n}^{3}$. Then we use (2.6) in connection with (2.5) and apply Minkowski’s inequality, the Burkholder–Davis–Gundy inequality on the stochastic integrals and Hölder’s inequality to obtain

$$\begin{aligned} \mathit{II}_{n}^{3} & \leq E\bigg[\bigg|-\int_{s}^{s'} \tilde{b}_{n}(u,B_{u}^{x}) dB_{u} - \int_{T-s'}^{T-s} \tilde{b}_{n}(T-u, \hat{B}_{u}^{x}) dW_{u}\\ &\phantom{=:E\bigg[\bigg|}{}+ \int_{T-s'}^{T-s} \tilde{b}_{n}(T-u, \hat{B}_{u}^{x}) \frac{\hat{B}_{u}}{T-u}du\bigg|^{2p_{\varepsilon}} \bigg]^{1/p_{\varepsilon}}\\ &\lesssim E\bigg[\bigg(\int_{s}^{s'} |\tilde{b}_{n}(u,B_{u}^{x})|^{2} du\bigg)^{p_{\varepsilon}}\bigg]^{1/p_{\varepsilon}} \\ &\phantom{=:}{}+ E\bigg[\bigg(\int_{T-s'}^{T-s} |\tilde{b}_{n}(T-u, \hat{B}_{u}^{x})|^{2} du\bigg)^{p_{\varepsilon}} \bigg]^{1/p_{\varepsilon }}\\ &\phantom{=:}{}+ E\bigg[\bigg|\int_{T-s'}^{T-s} \tilde{b}_{n}(T-u, \hat{B}_{u}^{x}) \frac{\hat{B}_{u}}{T-u}du\bigg|^{2p_{\varepsilon}} \bigg]^{1/p_{\varepsilon}}. \end{aligned}$$

Since $\tilde{b}_{n}$ is uniformly bounded, we have for any $\alpha\in (0,1)$ that

$$\begin{aligned} \mathit{II}_{n}^{3} &\leq\|\tilde{b}_{n}\|_{\infty}^{2}|s'-s| + \|\tilde{b}_{n}\| _{\infty}^{2} \bigg(\int_{T-s'}^{T-s} \bigg\| \frac{\hat {B}_{u}}{T-u}\bigg\| _{L^{4p_{\varepsilon}}(\varOmega)} du \bigg)^{2}\\ &\leq \|\tilde{b}_{n}\|_{\infty}^{2}|s'-s| + \|\tilde{b}_{n}\|_{\infty }^{2} |s'-s|^{\alpha} . \end{aligned}$$

Altogether, we can find a constant $C>0$ depending on $T$ and $\varepsilon>0$ such that

$$\begin{aligned} \sup_{x\in K}\sup_{n\geq1} E\big[(D_{s'} X_{t}^{n,x} - D_{s} X_{t}^{n,x})^{2}\big] \leq C |s'-s|^{\alpha} \end{aligned}$$

(A.5)

for $0\leq s'\leq s\leq t$ and $\alpha\in(0,1)$. Similarly, one also obtains

$$\begin{aligned} \sup_{x\in K} \sup_{0\leq s\leq t} \sup_{n\geq1} E\big[(D_{s} X_{t}^{n,x})^{2}\big] \leq C \end{aligned}$$

(A.6)

for a constant $C>0$. Then (A.2) with $\varphi= \mathrm{Id}$, (A.5) and (A.6) together with Proposition 2.3 imply that the set $(X_{t}^{n,x})_{n\geq1}$ is relatively compact in $L^{2}(\varOmega)$. Since the sequence of solutions $X_{t}^{n,x}$ also converges weakly to $E[X_{t}^{x}|\mathcal{F}_{t}]$ due to Lemma A.3 with $\varphi= \mathrm{Id}$, we get by uniqueness of the limit that

$$\begin{aligned} X_{t}^{n_{k},x} \longrightarrow E[X_{t}^{x} |\mathcal{F}_{t}] \qquad \text{in } L^{2}(\varOmega) \end{aligned}$$

for a subsequence $n_{k}$, $k\geq0$. Further, the $L^{2}(\varOmega)$-convergence even holds for the whole sequence because by the same reasoning as above, every subsequence has a further subsequence that converges to $E[X_{t}^{x}|\mathcal{F}_{t}]$.

Finally, since $(D_{s} X_{t}^{n,x})_{n\geq1}$ is bounded in the $L^{2}([0,T]\times\varOmega)$-norm uniformly in $n$ because of (A.6), we also have that the limit $E[X_{t}^{x}|\mathcal{F}_{t}]$ is Malliavin-differentiable; see e.g. [31, Lemma 1.2.3]. □

Remark A.5

Note that we have proved the estimates (A.5) and (A.6) uniformly in $x \in K$ for a compact set $K$ even though this is not needed to apply Proposition 2.3. However, we use these uniform bounds later on in the proofs of Lemma A.6 and Theorem 3.4.

We are now ready to complete the proof of Theorem 3.1 by use of the previous steps.

Proof of Theorem 3.1

To prove that $X^{x}$ is a strong solution, it only remains to prove that $X_{t}^{x}$ is $\mathcal{F}_{t}$-measurable for every $t\in[0,T]$. Indeed, let $\varphi$ be a continuous bounded function. Then by Theorem A.4, we have for a subsequence $n_{k}$, $k\geq0$, that

$$\begin{aligned} \varphi(X_{t}^{n_{k},x}) \longrightarrow\varphi (E[X_{t}^{x}|\mathcal{F}_{t}]) \quad P\mbox{-a.s. as } k\to\infty. \end{aligned}$$

On the other hand, by Lemma A.3 we also have

$$\begin{aligned} \varphi(X_{t}^{n,x}) \longrightarrow E[ \varphi (X_{t}^{x})|\mathcal{F}_{t}] \quad \text{weakly in } L^{2}(\varOmega). \end{aligned}$$

By the uniqueness of the limit, we get

$$\begin{aligned} \varphi( E[X_{t}^{x}|\mathcal{F}_{t}] ) = E[ \varphi (X_{t}^{x})|\mathcal{F}_{t}] \quad P\mbox{-a.s.} \end{aligned}$$

for all continuous bounded functions $\varphi$, which implies that $X_{t}^{x}$ is $\mathcal{F}_{t}$-measurable, for every $t\in[0,T]$.

To show uniqueness, suppose we have two strong solutions $X^{x}$ and $Y^{x}$ to the SDE (1.4). Then $\mathcal{W}(X_{t}^{x})(h)=E[X_{t}^{x}(h)]$ for $h\in L^{2}([0,T])$, where $X_{t}^{x}(h),0\leq t\leq T$, satisfies the SDE

$$ dX_{t}^{x}(h)=\Big(b\big(t,X_{t}^{x}(h)\big)+h(t)\Big)dt+d\widehat {B}_{t}, \qquad X_{0}^{x}(h)=x $$

(A.7)

for a Brownian motion $\widehat{B}_{t},0\leq t\leq T$. Analogously, $\mathcal{W}(Y_{t}^{x})(h)=E[Y_{t}^{x}(h)]$ with $Y_{t}^{x}(h),0\leq t\leq T$, solving (A.7). From the linear growth condition of the drift coefficient $b$, it follows that $X_{t}^{x}(h)$ and $Y_{t}^{x}(h),0\leq t\leq T$, are unique in law (see e.g. [19, Proposition 3.3.10]). Thus $\mathcal {W(}X_{t}^{x})(h)=\mathcal{W(}Y_{t}^{x})(h)$ for all $t,h$, and therefore $X^{x}$ and $Y^{x}$ are indistinguishable. □

1.2 A.2 Proof of Proposition 3.2

By (3.3) and (2.6), we can write for regular coefficients $b_{n}$ that

$$\begin{aligned} D_{s} X_{t}^{n,x} = \exp\bigg(-\int_{s}^{t} \int_{\mathbb {R}}b_{n}(u,y) L^{X^{n,x}}(du,dy)\bigg). \end{aligned}$$

Since $X_{t}^{n,x}$, $n\geq1$, is relatively compact in $L^{2}(\varOmega )$ and $\|D_{s} X_{t}^{n,x}\|_{L^{2} ([0,T]\times\varOmega)}$ is bounded uniformly in $n\geq0$ due to the proof of Theorem A.4, we know that the sequence $D_{s} X_{t}^{n,x}$, $n\geq1$, converges weakly to $D_{s} X_{t}^{x}$ in $L^{2}([0,T]\times\varOmega)$; see [31, Lemma 1.2.3]. Therefore it is enough to check that our candidate is the weak limit, i.e., that

$$\begin{aligned} &\bigg\langle \mathcal{W} \bigg(\exp\Big(-\int_{\cdot}^{t} \int _{\mathbb{R}}b_{n}(u,y) L^{X^{n,x}}(du,dy)\Big)\\ &\phantom{\bigg\langle \mathcal{W} \bigg(}{}- \exp\Big(-\int _{\cdot}^{t} \int_{\mathbb{R}}b(u,y) L^{X^{x}}(du,dy)\Big) \bigg)(f), g\bigg\rangle _{L^{2}([0,T])} \longrightarrow0\quad \text{as }n\to \infty \end{aligned}$$

for every $f\in L^{2}([0,T])$ and $g\in C_{0}^{\infty}([0,T])$. It suffices to show that the Wiener transform goes to zero. As in Lemma A.3, we obtain by Girsanov’s theorem that

$$\begin{aligned} &\bigg| E\bigg[\mathcal{E}\bigg(\int_{0}^{T} f(u) dB_{u}\bigg)\bigg(\exp\Big(-\int_{s}^{t} \int_{\mathbb{R}}b_{n}(u,y) L^{X^{n,x}}(du,dy)\Big)\\ &\phantom{\bigg| E\bigg[\mathcal{E}\bigg(\int_{0}^{T} f(u) dB_{u}\bigg)\bigg(}{}- \exp\Big(-\int_{s}^{t} \int_{\mathbb{R}}b(u,y) L^{X^{x}}(du,dy)\Big)\Bigg)\Bigg]\Bigg| \\ &= \bigg| E\bigg[\exp\bigg(-\int_{s}^{t}\int_{\mathbb{R}} b_{n}(u,y)L^{B^{x}}(du,dy)\bigg)\mathcal{E}\bigg(\int_{0}^{T} \big(b_{n}(u,B_{u}^{x})+f(u)\big) dB_{u}\bigg)\\ &\phantom{=:\bigg| E\bigg[}{}- \exp\bigg(-\int_{s}^{t} \int_{\mathbb {R}}b(u,y) L^{B^{x}}(du,dy)\bigg)\mathcal{E}\bigg(\int_{0}^{T} \big(b(u,B_{u}^{x})+f(u)\big) dB_{u}\bigg)\bigg]\bigg|\\ &\leq\bigg| E\bigg[\bigg(\exp\Big(-\int_{s}^{t}\int_{\mathbb{R}} \tilde{b}_{n}(u,y)L^{B^{x}}(du,dy)\Big)\\ &\phantom{=:\bigg| E\bigg[\bigg(}{}-\exp\Big(-\int_{s}^{t}\int _{\mathbb{R}} \tilde{b}(u,y)L^{B^{x}}(du,dy)\Big) \Bigg)\\ &\phantom{=:\bigg| E\bigg[}\times\exp\bigg(\int_{s}^{t}\hat {b}'(u,B_{u}^{x})du\bigg) \mathcal{E}\bigg(\int_{0}^{T} \big(b(u,B_{u}^{x})+f(u)\big) dB_{u}\bigg)\bigg]\Bigg|\\ &\phantom{=:}{}+\bigg| E\bigg[\bigg( \mathcal{E}\Big(\int_{0}^{T} \big(b_{n}(u,B_{u}^{x})+f(u)\big) dB_{u}\Big)-\mathcal{E}\Big(\int_{0}^{T} \big(b(u,B_{u}^{x})+f(u)\big) dB_{u}\Big)\bigg)\\ &\phantom{=:+\bigg| E\bigg[}\times\exp\bigg(-\int_{s}^{t} \int _{\mathbb{R}}\tilde{b}_{n}(u,y) L^{B^{x}}(du,dy)\bigg)\exp\bigg(\int _{s}^{t}\hat{b}'(u,B_{u}^{x})du\bigg)\bigg]\bigg|\\ &=: I_{n} + \mathit{II}_{n}. \end{aligned}$$

For term $I_{n}$, we define $p:=\frac{1+\varepsilon}{\varepsilon}$ for a suitable small $\varepsilon>0$ such that the stochastic exponential is bounded due to Lemma 2.6, and then apply Hölder’s inequality with exponent $1+\varepsilon$ on the stochastic exponential. Then we apply the Cauchy–Schwarz inequality and bound the factor with $\|\hat{b}'\|_{\infty}$, and finally we use the inequality $|e^{x} - 1| \leq|x|( e^{x} +1)$. As a result, we obtain

$$\begin{aligned} I_{n}& = \bigg| E\bigg[ \exp\bigg(-\int_{s}^{t} \int_{\mathbb {R}}\tilde{b}(u,y) L^{B^{x}}(du,dy)\bigg) \\ &\phantom{=:\bigg| E\bigg[}\times\bigg(\exp\Big(-\int_{s}^{t} \int _{\mathbb{R}}\big(\tilde{b}_{n}(u,y)-\tilde{b}(u,y)\big) L^{B^{x}}(du,dy)\Big) -1 \bigg) \\ &\phantom{=:\bigg| E\bigg[}\times\exp\bigg(\int_{s}^{t}\hat {b}'(u,B_{u}^{x})du\bigg) \mathcal{E}\bigg(\int_{0}^{T} \big(b(u,B_{u}^{x})+f(u)\big) dB_{u}\bigg)\bigg]\bigg| \\ &\lesssim E\bigg[ \exp\bigg(-2p \int_{s}^{t} \int_{\mathbb{R}}\tilde {b}(u,y) L^{B^{x}}(du,dy)\bigg)\bigg| \\ &\phantom{=:\lesssim E\bigg[}\times\bigg(\exp\Big(-\int_{s}^{t} \int _{\mathbb{R}}\big(\tilde{b}_{n}(u,y)-\tilde{b}(u,y)\big) L^{B^{x}}(du,dy)\Big) -1 \bigg)^{2p}\bigg|\bigg]^{1/(2p)} \\ &\phantom{=:\lesssim E\bigg[}\times E\bigg[\mathcal{E}\bigg(\int _{0}^{T} \big(b(u,B_{u}^{x})+f(u)\big) dB_{u}\bigg)^{1+\varepsilon}\bigg]^{1/(1+\varepsilon)} \\ &\lesssim E\bigg[ | V_{n}|^{2p} \bigg(\exp\Big(-\int_{s}^{t} \int _{\mathbb{R}}\tilde{b}_{n}(u,y) L^{B^{x}}(du,dy)\Big) \\ &\phantom{=:E\bigg[ | V_{n}|^{2p} \bigg(}{}+\exp\Big(-\int_{s}^{t} \int _{\mathbb{R}}\tilde{b}(u,y) L^{B^{x}}(du,dy)\Big) \bigg)^{2p}\bigg]^{1/(2p)} \\ &\lesssim E\bigg[ | V_{n}|^{2p} \exp\bigg(-2p\int_{s}^{t} \int_{\mathbb {R}}\tilde{b}_{n}(u,y) L^{B^{x}}(du,dy)\bigg)\bigg]^{\frac {1}{2p}} \\ &\phantom{=:}{}+ E\bigg[ | V_{n}|^{2p}\exp\bigg(-2p\int_{s}^{t} \int _{\mathbb{R}}\tilde{b}(u,y) L^{B^{x}}(du,dy)\bigg) \bigg]^{\frac{1}{2p}}, \end{aligned}$$

(A.8)

where

$$\begin{aligned} V_{n} := \int_{s}^{t} \int_{\mathbb{R}}\big(\tilde {b}_{n}(u,y)-\tilde{b}(u,y)\big)L^{B^{x}}(du,dy). \end{aligned}$$

The Cauchy–Schwarz inequality and Lemma A.2 give

$$\begin{aligned} &E\bigg[ | V_{n}|^{2p}\exp\bigg(-2p\int_{s}^{t} \int_{\mathbb{R}}\tilde {b}_{n}(u,y) L^{B^{x}}(du,dy)\bigg)\bigg] \\ &\leq E[ | V_{n}|^{4p}]^{1/2}E\bigg[\exp\bigg(-4p\int_{s}^{t} \int _{\mathbb{R}}\tilde{b}_{n}(u,y) L^{B^{x}}(du,dy)\bigg)\bigg]^{1/2} \\ &\lesssim E[ | V_{n}|^{4p}]^{1/2}. \end{aligned}$$

Finally, using (2.5) and the Minkowski and Burkholder–Davis–Gundy inequalities, we have

$$\begin{aligned} E[|V_{n}|^{4p}] &= E\bigg[\bigg|\int_{s}^{t} \big(\tilde {b}_{n}(u,B_{u}^{x})-\tilde{b}(u,B_{u}^{x})\big)dB_{u}\\ &\phantom{=:E\bigg[}{}+ \int_{T-t}^{T-s} \big(\tilde{b}_{n}(T-u,\hat {B}_{u}^{x})-\tilde{b}(T-u,\hat{B}_{u}^{x})\big)dW_{u}\\ &\phantom{=:E\bigg[}{}-\int_{T-t}^{T-s} \big(\tilde{b}_{n}(T-u,\hat {B}_{u}^{x})-\tilde{b}(T-u,\hat{B}_{u}^{x})\big)\frac{\hat {B}_{u}}{T-u}du\bigg|^{4p}\bigg]\\ &\leq E\bigg[\left|\int_{s}^{t} \big(\tilde{b}_{n}(u,B_{u}^{x})-\tilde {b}(u,B_{u}^{x})\big)dB_{u}\right|^{4p} \bigg]\\ &\phantom{=:}{}+ E\bigg[\bigg|\int_{T-t}^{T-s}\big(\tilde {b}_{n}(T-u,\hat{B}_{u}^{x})-\tilde{b}(T-u,\hat{B}_{u}^{x})\big)dW_{u} \bigg|^{4p}\bigg]\\ &\phantom{=:}{}+E\bigg[\bigg|\int_{T-t}^{T-s} \big(\tilde {b}_{n}(T-u,\hat{B}_{u}^{x})-\tilde{b}(T-u,\hat{B}_{u}^{x})\big)\frac{\hat {B}_{u}}{T-u}du\bigg|^{4p}\bigg]\\ &\leq E\bigg[\bigg(\int_{s}^{t} |\tilde{b}_{n}(u,B_{u}^{x})-\tilde {b}(u,B_{u}^{x})|^{2} du\bigg)^{2p}\bigg]\\ &\phantom{=:}{}+ E\bigg[\bigg(\int_{T-t}^{T-s}|\tilde{b}_{n}(T-u,\hat {B}_{u}^{x})-\tilde{b}(T-u,\hat{B}_{u}^{x})|^{2} du \bigg)^{2p}\bigg]\\ &\phantom{=:}{}+\bigg(\int_{T-t}^{T-s} \|\tilde{b}_{n}(T-u,\hat {B}_{u}^{x})-\tilde{b}(T-u,\hat{B}_{u}^{x})\|_{L^{8p}(\varOmega)}\frac {1}{\sqrt{T-u}}du \bigg)^{4p}. \end{aligned}$$

By dominated convergence, all terms converge to zero as $n\to\infty$. The second term in (A.8) is estimated in the same way. Similarly, one can also bound $\mathit{II}_{n}$.  □

1.3 A.3 Proof of Proposition 3.3

We first introduce the following auxiliary lemma.

Lemma A.6

Let $b_{n}: [0,T] \times\mathbb{R}\rightarrow\mathbb{R}$, $n\geq0$, be as in (3.1) and $X^{n,x}$ the corresponding strong solutions with drift coefficients $b_{n}$. Then for any compact subset $K\subset \mathbb{R}$ and $p\geq1$,

$$ \sup_{n\geq1}\sup_{x\in K} \sup_{t\in[0,T]} E\bigg[\bigg(\frac {\partial}{\partial x} X_{t}^{n,x}\bigg)^{p}\bigg] \leq C_{K,p} $$

(A.9)

for a constant $C_{K,p}>0$ depending on $K$ and $p$. Here, $\frac{\partial}{\partial x} X^{n,x}$ is the first variation process of $X^{n,x}$, $n\geq1$ (see Proposition 2.4).

Proof

The proof of this result relies on the proof of (A.6) in Theorem A.4 by observing that $\frac{\partial}{\partial x}X_{t}^{n,x} = D_{0} X_{t}^{n,x}$ by setting $s=0$ in (2.3) of Proposition 2.4. Following the same steps as in Theorem A.4, we see that all computations can be done for an arbitrary power $p\geq1$. Finally, (A.9) is obtained from the term $\mathit{II}_{n}^{1}$ in the proof of Theorem A.4. □

Proof of Proposition 3.3

For any $p\geq1$, the linear growth of $b_{n}$ implies that

$$ E[ |X_{t}^{n,x}|^{p} ] \lesssim |x|^{p} + |t| + \int_{0}^{t} E[ |X_{u}^{n,x}|^{p}] du + |t|^{p/2}. $$

Hence, Gronwall’s inequality gives

$$ \sup_{n\geq1}E[ |X_{t}^{n,x}|^{p}] \leq C. $$

(A.10)

Let $0\leq s < t \leq T$. Then

$$\begin{aligned} X_{t}^{n,x} - X_{s}^{n,y} &= x-y+ \int_{s}^{t} b_{n} (u, X_{u}^{n,x})du\\ &\phantom{=:}{}+ \int_{0}^{s} \big(b_{n}(u,X_{u}^{n,x}) - b_{n}(u, X_{u}^{n,y}) \big)du + B_{t} - B_{s}. \end{aligned}$$

The linear growth of $b_{n}$ together with (A.10) and Itô’s isometry then yield

$$\begin{aligned} E\big[ |X_{t}^{n,x} - X_{s}^{n,y}|^{2}\big] &\lesssim|x-y|^{2} + |t-s|\\ & \phantom{=:}{}+ E\bigg[ \bigg| \int_{0}^{s} \big(b_{n}(u,X_{u}^{n,x}) - b_{n}(u, X_{u}^{n,y}) \big)du\bigg|^{2}\bigg]. \end{aligned}$$

Then we use the fact that $X_{t}^{n,\cdot}$ is a stochastic flow of diffeomorphisms (see e.g. [22, Theorem 4.6.5]), the mean value theorem and Lemma A.6 in order to obtain

$$\begin{aligned} &E\bigg[\bigg| \int_{0}^{s} \big(b_{n}(u,X_{u}^{n,x}) - b_{n}(u, X_{u}^{n,y}) \big)du \bigg|^{2}\bigg]\\ &= |x-y|^{2} E\bigg[\bigg|\int_{0}^{s}\int_{0}^{1} b_{n}'(u,X_{u}^{n,x+\tau (y-x)}) \frac{\partial}{\partial x}X_{u}^{n,x+\tau(y-x)}d\tau du \bigg|^{2}\bigg] \\ &\lesssim|x-y|^{2} \int_{0}^{1} E\bigg[ \bigg|\int _{0}^{s}b_{n}'(u,X_{u}^{n,x+\tau(y-x)}) \frac{\partial}{\partial x}X_{u}^{n,x+\tau(y-x)}du \bigg|^{2}\bigg] d\tau\\ &= |x-y|^{2}\int_{0}^{1} E\bigg[ \bigg|\frac{\partial}{\partial x}X_{s}^{n,x+\tau(y-x)}{}-(1-\tau)\bigg|^{2}\bigg] d\tau\\ &\lesssim|x-y|^{2} \sup_{\substack{s\in[0,T]\\ x\in K}}E\bigg[\bigg| \frac{\partial}{\partial x}X_{s}^{n,x} \bigg|^{2}\bigg] \\ &\lesssim|x-y|^{2}. \end{aligned}$$

Altogether, we get

$$\begin{aligned} E\big[| X_{t}^{n,x} - X_{s}^{n,y}|^{2}\big] \lesssim |t-s|+|x-y|^{2}. \end{aligned}$$

To conclude, we use the fact that $X_{t}^{n,x} \to X_{t}^{x}$ in $L^{2}(\varOmega)$ as $n\to\infty$ due to Theorem A.4. □

1.4 A.4 Proof of Theorem 3.4

We note that for any smooth function $\varphi\in C_{0}^{\infty}(U,\mathbb {R})$ with compact support and $t\in[0,T]$, the sequence of random variables

$$\begin{aligned} \langle X_{t}^{n} , \varphi\rangle:= \int_{U} X_{t}^{n,x}\varphi(x) dx \end{aligned}$$

converges weakly in $L^{2}(\varOmega)$ to $\langle X_{t},\varphi\rangle$; this goes by using the Wiener transform and following the same steps as in Lemma A.3. Then for all sets $A\in\mathcal{F}$, $\varphi\in C_{0}^{\infty}(\mathbb {R})$ and using the Cauchy–Schwarz inequality, we have

$$\begin{aligned} E[\textbf{1}_{A} \langle X_{t}^{n_{k},x} - X_{t}^{x},\varphi'\rangle]&\leq\| \varphi'\|_{L^{2}(U)} |U|^{1/2} \bigg(\sup_{x\in\mathrm{supp}(U)} E[\textbf{1}_{A} (X_{t}^{n_{k},x} - X_{t}^{x})^{2}]\bigg)^{1/2}\\ &< \infty, \end{aligned}$$

where the last quantity is finite by Proposition 3.3. Then by Theorem A.4, we see that

$$\begin{aligned} \lim_{k\to\infty} E[\textbf{1}_{A} \langle X_{t}^{n_{k},x} - X_{t}^{x},\varphi'\rangle] =0. \end{aligned}$$

In addition, by virtue of Lemma A.6, we have that

$$\begin{aligned} \sup_{n\geq1} E\big[\|X_{t}^{n,x}\| _{W^{1,2}(U)}^{2}\big]< \infty , \end{aligned}$$

that is, $(x\mapsto X_{t}^{n,x})_{n\ge1}$ is bounded in $L^{2}(\varOmega ,W^{1,2}(U))$. As a result, the sequence $(X_{t}^{n,x})$ is weakly relatively compact in $L^{2}(\varOmega, W^{1,2}(U))$, see e.g. [23, Theorem 10.44], and therefore there exists a subsequence $n_{k}$, $k\geq0$, such that $X_{t}^{n_{k}, x}$ converges weakly to some element $Y_{t} \in L^{2}(\varOmega, W^{1,2}(U))$ as $k\to\infty$. Let us denote by $Y_{t}'$ the weak derivative of $Y_{t}$. Then

$$\begin{aligned} E[\textbf{1}_{A} \langle X_{t}^{x},\varphi'\rangle] &= \lim_{k\to\infty } E[\textbf{1}_{A} \langle X_{t}^{n_{k},x},\varphi'\rangle]\\ &=-\lim_{k \to\infty} E\bigg[\textbf{1}_{A} \bigg\langle \frac {\partial}{\partial x}X_{t}^{n_{k},x},\varphi\bigg\rangle \bigg]\\ &= -E[\textbf{1}_{A} \langle Y_{t}',\varphi\rangle], \end{aligned}$$

and so

$$\begin{aligned} \langle X_{t}, \varphi'\rangle= - \langle Y_{t}', \varphi\rangle, \quad P\mbox{-a.s.} \end{aligned}$$

(A.11)

Finally, we need to show that there exists a measurable set $\varOmega _{0} \subset\varOmega$ with full measure such that $X_{t}^{\cdot}$ has a weak derivative on this subset. To this end, choose a sequence $( \varphi_{n} )$ in $C^{\infty}(\mathbb{R})$ dense in $W^{1,2}(U)$. Choose a measurable subset $\varOmega_{n}$ of $\varOmega$ with full measure such that (A.11) holds on $\varOmega_{n}$ with $\varphi$ replaced by $\varphi_{n}$. Then $\varOmega_{0} := \bigcap_{n \geq 1}\varOmega_{n}$ satisfies the desired property.  □

Having established the existence of $\frac{\partial}{\partial x} X_{t}^{x}$, we can now extend Lemma A.6 to the case including $n=0$.

Corollary A.7

Let $b: [0,T] \times\mathbb{R}\rightarrow\mathbb{R}$ be as in (1.5) and $X^{x}$ the corresponding strong solution of (1.4). Then for any compact subset $K\subset\mathbb{R}$ and $p\geq1$,

$$\begin{aligned} \sup_{x\in K} \sup_{t\in[0,T]} E\bigg[\bigg(\frac{\partial}{\partial x} X_{t}^{x}\bigg)^{p}\bigg] \leq C_{K,p} \end{aligned}$$

for a constant $C_{K,p}>0$ depending on $K$ and $p$. Here, $\frac{\partial}{\partial x} X^{x}$ is the first variation process of $X^{x}$ (see Proposition 3.5).

Proof

This is a direct consequence of Lemma A.6 in connection with Fatou’s lemma. □

1.5 A.5 Proof of Proposition 3.5

By Theorem 3.4, the sequence $(X_{t}^{n,x} )_{n\geq0}$ converges weakly to $X_{t}^{x}$ in $L^{2} (\varOmega{,} W^{1,2} (U))$. Therefore, it is enough to check that our candidate is the limit of $\frac{\partial}{\partial x}X_{t}^{n,x}$ in the weak topology of $L^{2}(U\times\varOmega)$ for any open bounded $U\subset\mathbb{R}$. This can be shown following exactly the same steps as in Proposition 3.2 by integrating $I_{n}$ and $\mathit{II}_{n}$ against $g(x)$ over $x\in U$. The only difference here is that we need all bounds uniformly in $x\in U$, which is obtained by employing Lemma A.2.