Càdlàg semimartingale strategies for optimal trade execution in stochastic order book models

Abstract

We analyse an optimal trade execution problem in a financial market with stochastic liquidity. To this end, we set up a limit order book model in continuous time. Both order book depth and resilience are allowed to evolve randomly in time. We allow trading in both directions and for càdlàg semimartingales as execution strategies. We derive a quadratic BSDE that under appropriate assumptions characterises minimal execution costs, and we identify conditions under which an optimal execution strategy exists. We also investigate qualitative aspects of optimal strategies such as e.g. appearance of strategies with infinite variation or existence of block trades, and we discuss connections with the discrete-time formulation of the problem. Our findings are illustrated in several examples.

Appendices

Appendix A: Once again on the cost functional and the dynamics of the deviation process

Here, we motivate the dynamics (1.2) of the deviation process and the cost functional (1.3) via a limiting procedure from a discrete-time setting.

Without loss of generality, we consider the starting time $t=0$. We fix an initial position $x\in \mathbb{R}$ and an initial deviation $d\in \mathbb{R}$ and consider a continuous-time execution strategy $X \in \mathcal{A}_{0}(x,d)$. For any (large) $N\in \mathbb{N}$, we set $h=\frac{T}{N}$ and consider trading in discrete time at points of the grid $\{kh\colon k=0,\ldots ,N\}$. More precisely, the continuous-time strategy $X$ is approximated by the discrete-time strategy that consists of trades $\xi _{kh}$, $k\in \{0,\ldots ,N\}$, at the grid points, where

$$ \xi _{0}=X_{0}-x,\qquad \xi _{kh}=X_{kh}-X_{(k-1)h},\quad k\in \{1, \ldots ,N\}. $$

Notice that $\xi _{kh}$ is $\mathcal{F}_{kh}$-measurable, $k=0,\ldots ,N$. Further, for $k\in \{1,\ldots ,N\}$, we introduce $\beta _{kh} = \exp ( -\int _{(k-1)h}^{kh} \rho _{s} d[M]_{s} )$ and the notations $\eta _{r} = \exp ( - \int _{0}^{r} \rho _{s} d[M]_{s} )$ and $\nu _{r}=\gamma _{r}\exp ( \int _{0}^{r} \rho _{s} d[M]_{s} )$, $r \in [0,T]$.

In the discrete-time setting of Ackermann et al. [1], the deviation process (now denoted by $\widetilde{D}^{(h)}$) is defined by

$$ \widetilde{D}^{(h)}_{0-}=d, \qquad \widetilde{D}^{(h)}_{(kh)-} = \big( \widetilde{D}^{(h)}_{((k-1)h)-} + \gamma _{(k-1)h} \xi _{(k-1)h} \big)\beta _{kh}, \quad k\in \{1,\ldots ,N\}. $$

The minus in the subscript of $\widetilde{D}^{(h)}_{(kh)-}$ is purely notational (this is a discrete-time process); the meaning of $\widetilde{D}^{(h)}_{(kh)-}$ is that this is the deviation at time $kh$ directly prior to the trade $\xi _{kh}$ at time $kh$, and we preserve the minus sign in order to make the notation consistent with [1]. A straightforward calculation shows that

$$ \widetilde{D}^{(h)}_{(kh)-} = d \prod _{l=1}^{k} \beta _{lh} + \sum _{i=1}^{k} \gamma _{(i-1)h} \xi _{(i-1)h} \prod _{l=i}^{k} \beta _{lh}, \qquad k \in \{1,\ldots ,N\} . $$

Substituting the definition of $\beta _{kh}$, we obtain that for all $k\in \{1,\ldots ,N\}$,

$$\begin{aligned} \widetilde{D}_{(kh)-}^{(h)} & = e^{ ( - \int _{0}^{kh} \rho _{s} d[M]_{s} )} d + \sum _{i=1}^{k} \gamma _{(i-1)h} \xi _{(i-1)h} e^{ ( - \int _{(i-1)h}^{kh} \rho _{s} d[M]_{s} )} \\ & = \eta _{kh} \bigg( d + \sum _{i=1}^{k} \nu _{(i-1)h} \xi _{(i-1)h} \bigg) =\eta _{kh}L^{(h)}_{(k-1)h}, \end{aligned}$$

(A.1)

where for $k\in \{0,\ldots ,N\}$, we set

$$ \begin{aligned} L^{(h)}_{kh} &=d + \sum _{j=0}^{k} \nu _{jh} \xi _{jh} = d + \gamma _{0} ( X_{0} - x ) + \sum _{j=1}^{k} \nu _{jh} ( X_{jh} - X_{(j-1)h} ) \\ &=d + \gamma _{0} (X_{0}-x) + \sum _{j=1}^{k} \nu _{(j-1)h} ( X_{jh} - X_{(j-1)h} ) \\ &\phantom{=:} + \sum _{j=1}^{k} ( \nu _{jh} - \nu _{(j-1)h} ) ( X_{jh} - X_{(j-1)h} ). \end{aligned} $$

The last expression shows that the continuous-time limit, for $N\to \infty$ (and hence $h=\frac{T}{N}\to 0$) of the processes $(L^{(h)}_{kh})_{k\in \{0,\ldots ,N\}}$, is the process $(L_{s})_{s\in [0,T]}$ given by

$$ L_{s} = d+ \int _{[0,s]} \nu _{r}\,dX_{r} + \int _{[0,s]} d[\nu ,X]_{r}, \qquad s\in [0,T] $$

(apply Jacod and Shiryaev [25, Proposition I.4.44 and Theorem I.4.47]). Combining this with (A.1) and the definition of $\nu _{r}$, $r \in [0,T]$, recovers that the continuous-time limit of the processes $(\widetilde{D}^{(h)}_{(kh)-})_{k\in \{0,\ldots ,N\}}$ is the process $(D_{s})_{s\in [0,T]}$ given by

$$ D_{s}=\eta _{s} L_{s},\qquad s\in [0,T] $$

(and $D_{0-}=d$), which is nothing else but (2.2) or, equivalently, (1.2).

We now turn to the cost functional. In the discrete-time setting, the cost is given by $\sum _{j=0}^{N} ( \widetilde{D}^{(h)}_{(jh)-} + \frac{\gamma _{jh}}{2} \xi _{jh} ) \xi _{jh}$. Set $X_{-h}=X_{0-}\;(=x)$. Then it holds that

$$\begin{aligned} &\sum _{j=0}^{N} \bigg( \widetilde{D}^{(h)}_{(jh)-} + \frac{\gamma _{jh}}{2} \xi _{jh} \bigg) \xi _{jh} \\ &= \sum _{j=0}^{N} \widetilde{D}^{(h)}_{(jh)-} ( X_{jh} - X_{(j-1)h} ) + \sum _{j=0}^{N} \frac{\gamma _{(j-1)h}}{2} ( X_{jh} - X_{(j-1)h} )^{2} \\ & \phantom{=:}+ \sum _{j=0}^{N} \frac{1}{2} ( \gamma _{jh} - \gamma _{(j-1)h} ) ( X_{jh} - X_{(j-1)h} )^{2} . \end{aligned}$$

(A.2)

For the first term on the right-hand side of (A.2), we have

$$ \begin{aligned} \sum _{j=0}^{N} \widetilde{D}^{(h)}_{(jh)-} ( X_{jh} - X_{(j-1)h} ) &= \sum _{j=0}^{N} \eta _{jh} L^{(h)}_{(j-1)h} ( X_{jh} - X_{(j-1)h} ) \\ & = \sum _{j=0}^{N} \eta _{(j-1)h} L^{(h)}_{(j-1)h} ( X_{jh} - X_{(j-1)h} ) \\ &\phantom{=:} + \sum _{j=0}^{N} L^{(h)}_{(j-1)h} ( \eta _{jh} - \eta _{(j-1)h} ) ( X_{jh} - X_{(j-1)h} ), \end{aligned} $$

which has the continuous-time limit

$$ \int _{[0,T]}\eta _{s}L_{s-}\,dX_{s}+\int _{[0,T]}L_{s-}\,d[\eta ,X]_{s} =\int _{[0,T]}D_{s-}\,dX_{s}, $$

as $\eta$ is a continuous process of finite variation. Further, the second term on the right-hand side of (A.2) tends to $\int _{[0,T]}\frac{\gamma _{s}}{2}\,d[X]_{s}$, and the third term to $\frac{1}{2}[\gamma ,[X]]_{T}=0$ because $\gamma$ is continuous. As the continuous-time limit of the discrete-time cost, we thus obtain

$$ \int _{[0,T]}D_{s-}\,dX_{s}+\int _{[0,T]}\frac{\gamma _{s}}{2}\,d[X]_{s}, $$

which motivates our form of the cost functional in continuous time.

Appendix B: Heuristic derivation of the BSDE (3.2)

We have seen that the BSDE (3.2) plays a central role both in our results and in the proofs. But where does it come from? In this appendix, we motivate the BSDE (3.2) via a heuristic limiting procedure from discrete time.

To this end, we consider a discrete-time version of the stochastic control problem (1.4). For $h>0$ such that $h=\frac{T}{N}$ for some $N\in \mathbb{N}$, $t\in [0,T]$ and $x, d \in \mathbb{R}$, let $\mathcal{A}_{t}^{h}(x,d)$ be the subset of all processes $X = (X_{s})_{s\in [t,T]} \in \mathcal{A}_{t}(x,d)$ of the form $X_{s}= \sum _{k=0}^{N} X_{(kh) \vee t} 1_{[kh,(k+1)h)}(s)$ for all $s \in [t,T]$. Moreover, let

$$ V_{t}^{h}(x,d) = \mathop {\mathrm {ess\,inf}}_{X \in \mathcal{A}_{t}^{h}(x,d)}J_{t}(x,d,X) $$

for all $x,d \in \mathbb{R}$, $t\in [0,T]$, $h>0$ with $h=\frac{T}{N}$ for some $N\in \mathbb{N}$. Then it follows from Ackermann et al. [1] that for each $h>0$ with $h=\frac{T}{N}$ for some $N\in \mathbb{N}$, there exists a process $Y^{h}=(Y_{t}^{h})_{t\in \{0,h,\ldots ,T\}}$ such that $V_{t}^{h}(x,d) = \frac{Y_{t}^{h}}{\gamma _{t}} ( d-\gamma _{t} x)^{2} - \frac{d^{2}}{2\gamma _{t}}$, $x,d \in \mathbb{R}$, $t\in \{0,h,\ldots ,T\}$. The discrete-time process $Y^{h}$ is given by the backward recursion $Y_{T}^{h}=\frac{1}{2}$ and, for $t \in \{0,h,\ldots ,T-h\}$,

$$\begin{aligned} Y_{t}^{h} =& E_{t}\bigg[ \frac{\gamma _{t+h}}{\gamma _{t}} Y_{t+h}^{h} \bigg] \\ &- \frac{ ( E_{t} [ Y_{t+h}^{h} ( e^{-\int _{t}^{t+h} \rho _{s} d[M]_{s}} - \frac{\gamma _{t+h}}{\gamma _{t}} ) ] )^{2}}{E_{t} [ Y_{t+h}^{h} \frac{\gamma _{t}}{\gamma _{t+h}} ( e^{-\int _{t}^{t+h} \rho _{s} d[M]_{s}} - \frac{\gamma _{t+h}}{\gamma _{t}} ) ^{2} + \frac{1}{2} ( 1- \frac{\gamma _{t}}{\gamma _{t+h}} e^{-2\int _{t}^{t+h} \rho _{s} d[M]_{s}} ) ]} . \end{aligned}$$

(B.1)

We aim at deriving — at least heuristically — the dynamics of the continuous-time limit $Y=(Y_{t})_{t\in [0,T]}$ of $Y^{h}$. To this end, we suppose that $Y$ can be decomposed as

$$ dY_{t} = a_{t} d[M]_{t} + Z_{t}dM_{t} + dM^{\perp }_{t}, \qquad t \in [0,T], $$

(B.2)

where $(a_{t})_{t\in [0,T]}$, $(Z_{t})_{t\in [0,T]}$ are progressively measurable processes ($(a_{t})_{t\in [0,T]}$ is still to be determined) and $M^{\perp }= (M^{\perp }_{t})_{t\in [0,T]}$ is a local martingale orthogonal to $M$. From (B.2), we deduce that $(a_{t})_{t\in [0,T]}$ should be identified as the limit

$$ \begin{aligned} a_{t} & = \lim _{h \to 0} \frac{E_{t}[Y_{t+h}]-Y_{t}}{E_{t} [ [M]_{t+h} ] - [M]_{t}}, \qquad t \in [0,T]. \end{aligned} $$

Assume that replacing $Y^{h}$ with $Y$ in (B.1) introduces an error only of the magnitude $o(E_{t}[ [M]_{t+h} ] - [M]_{t})$. Then for all $t\in [0,T]$, we can get the expression for $a_{t}$ by evaluating the limit, for $h \to 0$, of

$$\begin{aligned} &\frac{1}{E_{t} [ [M]_{t+h} ] - [M]_{t}} \\ &\times \bigg( E_{t} [ Y_{t+h} ] - E_{t}\bigg[ \frac{\gamma _{t+h}}{\gamma _{t}} Y_{t+h} \bigg] \\ & \quad \,\,\, + \frac{ ( E_{t} [ Y_{t+h} ( e^{-\int _{t}^{t+h} \rho _{s} d[M]_{s}} - \frac{\gamma _{t+h}}{\gamma _{t}} ) ] )^{2}}{E_{t} [ Y_{t+h} \frac{\gamma _{t}}{\gamma _{t+h}} ( e^{-\int _{t}^{t+h} \rho _{s} d[M]_{s}} - \frac{\gamma _{t+h}}{\gamma _{t}} ) ^{2} + \frac{1}{2} ( 1- \frac{\gamma _{t}}{\gamma _{t+h}} e^{-2\int _{t}^{t+h} \rho _{s} d[M]_{s}} ) ]} \bigg). \end{aligned}$$

(B.3)

For the remainder of this section, we fix $t\in [0,T]$ and assume that all stochastic integrals with respect to $dM$ and $dM^{\perp }$ that appear are true martingales. We define the process $\Gamma =(\Gamma _{s})_{s\in [t,T]}$ by $\Gamma _{s}=\frac{\gamma _{s}}{\gamma _{t}}$. Since for all $s\in [t,T]$,

$$ d(\Gamma _{s}Y_{s}) = ( Y_{s}\Gamma _{s} \mu _{s} + \Gamma _{s} a_{s} + \Gamma _{s}\sigma _{s}Z_{s} ) d[M]_{s} + ( Y_{s}\Gamma _{s} \sigma _{s} + \Gamma _{s}Z_{s} ) dM_{s} + \Gamma _{s} dM^{\perp }_{s}, $$

it holds for all $h\in (0,T-t)$ that

$$ E_{t} [ \Gamma _{t+h} Y_{t+h} ] = Y_{t} + E_{t}\bigg[ \int _{t}^{t+h} ( Y_{s}\Gamma _{s} \mu _{s} + \Gamma _{s} a_{s} + \Gamma _{s}\sigma _{s}Z_{s} ) d[M]_{s} \bigg] . $$

(B.4)

Together with

$$ E_{t} [ Y_{t+h} ] = Y_{t} + E_{t}\bigg[ \int _{t}^{t+h} a_{s}d[M]_{s} \bigg], \qquad h\in (0,T-t), $$

we obtain heuristically that

$$\begin{aligned} \frac{E_{t} [Y_{t+h} ]-E_{t} [ \Gamma _{t+h} Y_{t+h} ]}{E_{t} [ [M]_{t+h} ] - [M]_{t}} & = \frac{E_{t} [ \int _{t}^{t+h} ( a_{s} (1-\Gamma _{s}) - Y_{s}\Gamma _{s} \mu _{s} - \Gamma _{s}\sigma _{s}Z_{s} )d[M]_{s} ]}{E_{t} [ \int _{t}^{t+h} d[M]_{s} ] } \\ & \longrightarrow -Y_{t}\mu _{t}-\sigma _{t}Z_{t} \qquad \text{as $h \to 0$}. \end{aligned}$$

(B.5)

Furthermore, it holds for all $h\in (0,T-t)$ that

$$\begin{aligned} Y_{t+h} e^{- \int _{t}^{t+h} \rho _{s} d[M]_{s}} & = Y_{t} + \int _{t}^{t+h} ( a_{s} - \rho _{s} Y_{s} ) e^{- \int _{t}^{s} \rho _{r} d[M]_{r}} d[M]_{s} \\ & \phantom{=:}+ \int _{t}^{t+h} Z_{s} e^{- \int _{t}^{s} \rho _{r} d[M]_{r}} dM_{s} + \int _{(t,t+h]} e^{- \int _{t}^{s} \rho _{r} d[M]_{r}} dM^{\perp }_{s} . \end{aligned}$$

(B.6)

From (B.4) and (B.6), we derive heuristically that

$$\begin{aligned} & \frac{E_{t} [ Y_{t+h} ( e^{-\int _{t}^{t+h} \rho _{s} d[M]_{s}} - \Gamma _{t+h} ) ]}{E_{t} [ [M]_{t+h} ] - [M]_{t}} \\ & = \frac{ E_{t} [ \int _{t}^{t+h} ( (a_{s} - \rho _{s} Y_{s} ) e^{- \int _{t}^{s} \rho _{r} d[M]_{r}} - ( Y_{s}\Gamma _{s} \mu _{s} + \Gamma _{s} a_{s} + \Gamma _{s}\sigma _{s}Z_{s} ) ) d[M]_{s} ] }{ E_{t} [ \int _{t}^{t+h} d[M]_{s} ] } \\ & \longrightarrow - \rho _{t} Y_{t} - Y_{t} \mu _{t} -\sigma _{t} Z_{t} \qquad \text{as $h \to 0$}. \end{aligned}$$

(B.7)

Recall that $\Gamma _{s}^{-1} = \frac{\alpha _{s}}{\alpha _{t}}$, $s\in [t,T]$, with

$$ d\Gamma _{s}^{-1} = \Gamma _{s}^{-1} \big( - (\mu _{s}-\sigma _{s}^{2}) d[M]_{s} - \sigma _{s} dM_{s} \big), \qquad s \in [t,T]. $$

Therefore, it holds that

$$\begin{aligned} d (Y_{s} \Gamma _{s}^{-1} ) & = \big( - Y_{s} \Gamma _{s}^{-1} (\mu _{s}- \sigma _{s}^{2}) + \Gamma _{s}^{-1} a_{s} - Z_{s} \Gamma _{s}^{-1} \sigma _{s} \big) d[M]_{s} \\ & \phantom{=:} + ( \Gamma _{s}^{-1}Z_{s} - Y_{s}\Gamma _{s}^{-1} \sigma _{s} ) dM_{s} + \Gamma _{s}^{-1} dM^{\perp }_{s} , \qquad s\in [t,T]. \end{aligned}$$

(B.8)

Moreover, we have for all $h\in (0,T-t)$ that

$$\begin{aligned} & ( e^{-\int _{t}^{t+h} \rho _{s} d[M]_{s}} - \Gamma _{t+h} )^{2} \\ &= - 2 \int _{t}^{t+h} ( e^{-\int _{t}^{s} \rho _{r} d[M]_{r}} - \Gamma _{s} ) \Gamma _{s} \sigma _{s} dM_{s} \\ &\phantom{=:}+\int _{t}^{t+h} \big( \Gamma _{s}^{2}\sigma _{s}^{2} - 2 ( e^{-\int _{t}^{s} \rho _{r} d[M]_{r}} - \Gamma _{s} ) ( \rho _{s}e^{-\int _{t}^{s} \rho _{r} d[M]_{r}} + \Gamma _{s} \mu _{s} ) \big) d[M]_{s} . \end{aligned}$$

(B.9)

It follows from (B.8) and (B.9) that

$$ \begin{aligned} & Y_{t+h}\Gamma _{t+h}^{-1} ( e^{-\int _{t}^{t+h} \rho _{s} d[M]_{s}} - \Gamma _{t+h} )^{2} \\ & = \int _{t}^{t+h} \bigg( Y_{s} \Gamma _{s}^{-1} \big( \Gamma _{s}^{2} \sigma _{s}^{2} - 2 ( e^{-\int _{t}^{s} \rho _{r} d[M]_{r}} - \Gamma _{s} ) ( \rho _{s} e^{-\int _{t}^{s} \rho _{r} d[M]_{r}} + \Gamma _{s} \mu _{s} ) \big) \\ & \quad \quad \qquad \,+ ( e^{-\int _{t}^{s} \rho _{r} d[M]_{r}} - \Gamma _{s} )^{2} \Gamma _{s}^{-1} \big( -Y_{s} (\mu _{s}-\sigma _{s}^{2}) + a_{s} - Z_{s} \sigma _{s} \big) \\ & \quad \quad \qquad \,- 2 \sigma _{s} ( Z_{s} - Y_{s} \sigma _{s} ) ( e^{-\int _{t}^{s} \rho _{r} d[M]_{r}} - \Gamma _{s} ) \bigg) d[M]_{s} \\ & \phantom{=:} + \int _{t}^{t+h} \bigg( \Gamma _{s}^{-1} ( e^{-\int _{t}^{s} \rho _{r} d[M]_{r}} - \Gamma _{s} )^{2} ( Z_{s} - Y_{s} \sigma _{s} ) \\ & \quad \quad \quad \qquad \, - 2 Y_{s} ( e^{-\int _{t}^{s} \rho _{r} d[M]_{r}} - \Gamma _{s} ) \sigma _{s} \bigg) dM_{s} \\ & \phantom{=:} + \int _{(t,t+h]} ( e^{-\int _{t}^{s} \rho _{r} d[M]_{r}} - \Gamma _{s} )^{2} \Gamma _{s}^{-1} dM_{s}^{\perp }, \qquad h\in (0,T-t), \end{aligned} $$

and hence

$$ \begin{aligned} &E_{t} [ Y_{t+h} \Gamma _{t+h}^{-1} ( e^{-\int _{t}^{t+h} \rho _{s} d[M]_{s}} - \Gamma _{t+h} )^{2} ] \\ &= E_{t}\bigg[ \int _{t}^{t+h} \Big( Y_{s} \Gamma _{s}^{-1} \big( \Gamma _{s}^{2} \sigma _{s}^{2} - 2 ( e^{-\int _{t}^{s} \rho _{r} d[M]_{r}} - \Gamma _{s} ) ( \rho _{s} e^{-\int _{t}^{s} \rho _{r} d[M]_{r}} + \Gamma _{s} \mu _{s} ) \big) \\ & \phantom{=:} \quad \quad \qquad \quad + ( e^{-\int _{t}^{s} \rho _{r} d[M]_{r}} - \Gamma _{s} )^{2} \Gamma _{s}^{-1} \big( -Y_{s} (\mu _{s}-\sigma _{s}^{2}) + a_{s} - Z_{s} \sigma _{s} \big) \\ & \phantom{=:} \phantom{=:} \quad \quad \qquad - 2 \sigma _{s} ( Z_{s} - Y_{s} \sigma _{s} ) ( e^{- \int _{t}^{s} \rho _{r} d[M]_{r}} - \Gamma _{s} ) \Big) d[M]_{s} \bigg] , \quad \,\,\ h\in (0,T-t) . \end{aligned} $$

Therefore, we obtain heuristically that

$$ \frac{E_{t} [ Y_{t+h}\Gamma _{t+h}^{-1} ( e^{-\int _{t}^{t+h} \rho _{s} d[M]_{s}} - \Gamma _{t+h} )^{2} ] }{E_{t} [ [M]_{t+h} ] - [M]_{t} } \longrightarrow Y_{t} \sigma _{t}^{2} \qquad \text{as $h \to 0$}. $$

(B.10)

From

$$ \begin{aligned} \Gamma _{t+h}^{-1} e^{-2 \int _{t}^{t+h} \rho _{s} d[M]_{s}} & = 1 - \int _{t}^{t+h} \Gamma _{s}^{-1} e^{-2 \int _{t}^{s} \rho _{r} d[M]_{r}} ( 2 \rho _{s} + \mu _{s}-\sigma _{s}^{2} ) d[M]_{s} \\ & \phantom{=:} - \int _{t}^{t+h} e^{-2 \int _{t}^{s} \rho _{r} d[M]_{r}} \Gamma _{s}^{-1} \sigma _{s} dM_{s} , \qquad h\in (0,T-t), \end{aligned} $$

we derive heuristically that

$$ \begin{aligned} &\frac{ E_{t} [ \frac{1}{2} ( 1 - \Gamma _{t+h}^{-1} e^{-2 \int _{t}^{t+h} \rho _{s} d[M]_{s}} ) ] }{E_{t} [ [M]_{t+h} ] - [M]_{t} } \\ & = \frac{ E_{t} [ \int _{t}^{t+h} \frac{1}{2} ( \Gamma _{s}^{-1} e^{-2 \int _{t}^{s} \rho _{r} d[M]_{r}} ( 2 \rho _{s} + \mu _{s}-\sigma _{s}^{2} ) ) d[M]_{s} ] }{ E_{t} [ \int _{t}^{t+h} d[M]_{s} ] } \\ & \longrightarrow \frac{1}{2} (2 \rho _{t} + \mu _{t} - \sigma _{t}^{2} ) \qquad \text{as $h \to 0$}. \end{aligned} $$

(B.11)

We conclude from (B.5), (B.7), (B.10), and (B.11) that the limit for $h \to 0$ of (B.3) equals

$$ a_{t} = -Y_{t}\mu _{t} - \sigma _{t} Z_{t} + \frac{ ( - \rho _{t} Y_{t} - Y_{t}\mu _{t} - \sigma _{t} Z_{t} )^{2}}{ Y_{t}\sigma _{t}^{2} + \frac{1}{2} (2 \rho _{t} + \mu _{t} - \sigma _{t}^{2} )} =-f(t,Y_{t},Z_{t}) $$

with $f$ given in (3.3). Finally, the fact (which is proved in [1]) that the discrete-time processes $Y^{h}$, $h \in (0,T-t)$, are $(0,1/2]$-valued explains the requirement in (3.4) that $Y$ is $[0,1/2]$-valued.

Appendix C: Comparison argument for Sect. 7.1

Here we justify via a comparison argument that in Proposition 7.1, we get that $Y$ is $[0,1/2]$-valued.

In the following result, we are interested in a BSDE with driver $f$ and terminal value $\xi$ of the form

$$ dY_{s} = - f(s,Y_{s}) d[M]_{s} + Z_{s} dM_{s} + dM^{\perp }_{s}, \quad s \in [0,T], \qquad Y_{T}=\xi , $$

(C.1)

and denote such a BSDE by BSDE$(f,\xi )$. Recall that in Proposition 7.1, the driver does not depend on $Z$. Therefore, we do not consider a dependence on $Z$ in (C.1).

Proposition C.1

Assume (3.8). Let $f$ and $\widetilde{f}$ be progressively measurable and $f$ Lipschitz-continuous, i.e., there exists some $L \in (0,\infty )$ such that for all $y,y'\in \mathbb{R}$, it holds that $\lvert f(s,y) - f(s,y') \rvert \leq L \lvert y-y'\rvert$ $\mathcal{D}_{M}$-a.e. Moreover, let $\xi$ and $\widetilde{\xi }$ be $\mathcal{F}_{T}$-measurable random variables. Let $(Y,Z,M^{\perp })$ be a solution of the BSDE$(f,\xi )$ and $(\widetilde{Y},\widetilde{Z},\widetilde{M}^{\perp })$ a solution of the BSDE$(\widetilde{f},\widetilde{\xi })$ such that $E[ \int _{0}^{T} Z_{s}^{2} d[M]_{s} ] < \infty$, $E[ [M^{\perp }]_{T} ] < \infty$, $E[ \int _{0}^{T} \widetilde{Z}_{s}^{2} d[M]_{s} ] < \infty$ and $E[ [\widetilde{M}^{\perp }]_{T} ] < \infty$. Set $\delta Y_{t} = Y_{t}-\widetilde{Y}_{t}$ and $\delta f_{t} = f(t, \widetilde{Y}_{t}) - \widetilde{f}(t,\widetilde{Y}_{t})$ for $t\in [0,T]$. Furthermore, define

$$ b_{t} = 1_{\{Y_{t}\neq \widetilde{Y}_{t}\}} \big( f(t,Y_{t}) - f(t, \widetilde{Y}_{t}) \big) ( Y_{t} - \widetilde{Y}_{t} )^{-1}, \qquad t \in [0,T], $$

and introduce the process $\Gamma =(\Gamma _{t})_{t\in [0,T]}$ given by $\Gamma _{t} = \exp ( \int _{0}^{t} b_{s} d[M]_{s} )$, $t\in [0,T]$. Then $\delta Y$ admits the representation

$$ \delta Y_{t} = \Gamma _{t}^{-1} E_{t}\bigg[ \Gamma _{T} \delta Y_{T} + \int _{t}^{T} \Gamma _{s} \delta f_{s} d[M]_{s} \bigg], \qquad t \in [0,T]. $$

(C.2)

In particular:

(i) If $\xi \geq \widetilde{\xi }$ a.s. and $f(s,\widetilde{Y}_{s})\geq \widetilde{f}(s,\widetilde{Y}_{s})$ $\mathcal{D}_{M}$-a.e., then $Y_{t}\geq \widetilde{Y}_{t}$ a.s. for all $t \in [0,T]$.

(ii) If $\xi \leq \widetilde{\xi }$ a.s. and $f(s,\widetilde{Y}_{s})\leq \widetilde{f}(s,\widetilde{Y}_{s})$ $\mathcal{D}_{M}$-a.e., then $Y_{t}\leq \widetilde{Y}_{t}$ a.s. for all $t\in [0,T]$.

Proof

It holds for all $t\in [0,T]$ that

$$ \begin{aligned} \delta Y_{t} & = Y_{T} - \widetilde{Y}_{T} + \int _{t}^{T} \big( f(s,Y_{s}) - \widetilde{f}(s,\widetilde{Y}_{s}) \big) d[M]_{s} \\ &\phantom{=:} - \int _{t}^{T} Z_{s} dM_{s} - ( M^{\perp }_{T} - M^{\perp }_{t} ) + \int _{t}^{T} \widetilde{Z}_{s} dM_{s} + ( \widetilde{M}^{\perp }_{T} - \widetilde{M}^{\perp }_{t} ) . \end{aligned} $$

Since we have for all $s\in [0,T]$ that

$$ f(s,Y_{s}) - \widetilde{f}(s,\widetilde{Y}_{s}) = f(s,Y_{s}) - f(s, \widetilde{Y}_{s}) + f(s,\widetilde{Y}_{s}) - \widetilde{f}(s, \widetilde{Y}_{s}) = b_{s} \delta Y_{s} + \delta f_{s}, $$

it follows that

$$ d(\delta Y_{s}) = - ( b_{s} \delta Y_{s} + \delta f_{s} ) d[M]_{s} + Z_{s} dM_{s} - \widetilde{Z}_{s} dM_{s} + dM^{\perp }_{s} - d\widetilde{M}^{\perp }_{s}, \qquad s\in [0,T] . $$

Together with $d\Gamma _{s} = \Gamma _{s} b_{s} d[M]_{s}$, $s\in [0,T]$, we obtain by integration by parts that

$$ \begin{aligned} \Gamma _{T} \delta Y_{T} & = \Gamma _{t}\delta Y_{t} - \int _{t}^{T} \Gamma _{s} ( b_{s} \delta Y_{s} + \delta f_{s} ) d[M]_{s} + \int _{t}^{T} \Gamma _{s} Z_{s} dM_{s} - \int _{t}^{T} \Gamma _{s} \widetilde{Z}_{s} dM_{s} \\ & \phantom{=:} + \int _{(t,T]} \Gamma _{s} dM^{\perp }_{s} - \int _{(t,T]} \Gamma _{s} d \widetilde{M}^{\perp }_{s} + \int _{t}^{T} \delta Y_{s} \Gamma _{s} b_{s} d[M]_{s}, \qquad t \in [0,T]. \end{aligned} $$

If the local martingales $S=\int _{0}^{\cdot } \Gamma _{s} Z_{s} dM_{s}$, $\widetilde{S}=\int _{0}^{\cdot } \Gamma _{s} \widetilde{Z}_{s} dM_{s}$, $U=\int _{(0,\cdot ]} \Gamma _{s} dM^{\perp }_{s}$ and $\widetilde{U}=\int _{(0,\cdot ]} \Gamma _{s} d\widetilde{M}^{\perp }_{s}$ are true martingales, then it follows that

$$ \Gamma _{t} \delta Y_{t} = E_{t}\bigg[ \Gamma _{T} \delta Y_{T} + \int _{t}^{T} \Gamma _{s} \delta f_{s} d[M]_{s} \bigg], \qquad t \in [0,T], $$

which yields the representation (C.2) of $\delta Y$.

To show that $S$ is a martingale, note first that due to the Lipschitz-continuity of $f$, the process $b$ is bounded $\mathcal{D}_{M}$-a.e. by the corresponding Lipschitz constant. By the Cauchy–Schwarz inequality, it holds that

$$\begin{aligned} E\bigg[ \bigg( \int _{0}^{T} \Gamma _{s}^{2} Z_{s}^{2} d[M]_{s} \bigg)^{\frac{1}{2}} \bigg] & \leq E\bigg[ \Big( \sup _{t\in [0,T]} \Gamma _{t}^{2} \Big)^{\frac{1}{2}} \bigg( \int _{0}^{T} Z_{s}^{2} d[M]_{s} \bigg)^{\frac{1}{2}} \bigg] \\ & \leq \Big( E\Big[ \sup _{t\in [0,T]} \Gamma _{t}^{2} \Big] \Big)^{ \frac{1}{2}} \bigg( E\bigg[ \int _{0}^{T} Z_{s}^{2} d[M]_{s} \bigg] \bigg)^{\frac{1}{2}} . \end{aligned}$$

(C.3)

Since $b$ is bounded and (3.8) holds, we have $E[ \sup _{t\in [0,T]} \Gamma _{t}^{2} ]<\infty$. We also have by assumption that $E[ \int _{0}^{T} Z_{s}^{2} d[M]_{s} ]<\infty$. Therefore, it follows from (C.3) and the Burkholder–Davis–Gundy inequality that $E[ \sup _{t\in [0,T]} \lvert S_{t}\rvert ] < \infty$. Thus $S$ is a martingale. A similar reasoning applies also to $\widetilde{S}$, $U$ and $\widetilde{U}$.

Finally, the claims (i) and (ii) are straightforward consequences of (C.2). □

We now apply Proposition C.1 to obtain $0\leq Y\leq \frac{1}{2}$ in the proof of Proposition 7.1. Observe that $(\widetilde{Y},\widetilde{Z},\widetilde{M}^{\perp })=( \frac{1}{2},0,0 )$ is a solution of the BSDE$(0,\frac{1}{2})$, which obviously satisfies $E[ [\widetilde{M}^{\perp }]_{T}]<\infty$ and $E[ \int _{0}^{T} \widetilde{Z}_{s}^{2} d[M]_{s} ] <\infty$. Moreover, with $\overline{f}$ as defined in the proof of Proposition 7.1, it holds that

$$ \overline{f}\bigg(s,\frac{1}{2}\bigg) = \frac{-\rho _{s}^{2}}{2 ( 2\rho _{s} + \mu _{s} )} \leq 0, \qquad s \in [0,T], $$

and both BSDEs have the same terminal value $\frac{1}{2}$. Therefore Proposition C.1 applies and yields $Y\leq \widetilde{Y} = \frac{1}{2}$.

For the other bound, note that $(\widetilde{Y},\widetilde{Z},\widetilde{M}^{\perp })=( 0,0,0 )$ is a solution of the BSDE $(0,0)$ with $E[ [\widetilde{M}^{\perp }]_{T} ]<\infty$ and $E[ \int _{0}^{T} \widetilde{Z}_{s}^{2} d[M]_{s} ] <\infty$. Since $\overline{f}(s,0)=0$ for all $s\in [0,T]$ and $Y_{T} = \frac{1}{2} \geq 0 = \widetilde{Y}_{T}$, it follows from Proposition C.1 that $Y\geq \widetilde{Y}=0$.

Remark C.2

Notice that in the proof of Proposition C.1, we need (3.8) and the Lipschitz-continuity of $f$ only to show that $E[ \sup _{t\in [0,T]} \Gamma _{t}^{2} ]$ is finite. Replace these two conditions by the assumption that there exists a predictable process $R$ such that for all $y,y'\in \mathbb{R}$, $\lvert f(\omega ,s,y) - f(\omega ,s,y')\rvert \leq R_{s}(\omega ) \lvert y-y'\rvert$ $\mathcal{D}_{M}$-a.e. and for all $c\in (0,\infty )$, $E[ \exp ( c\int _{0}^{T} R_{s} d[M]_{s} ) ] <\infty$. Then we still have that

$$ \begin{aligned} E\Big[ \sup _{t\in [0,T]} \Gamma _{t}^{2} \Big] & = E \bigg[ \sup _{t\in [0,T]} \exp \bigg( 2 \int _{0}^{t} b_{s} d[M]_{s} \bigg) \bigg] \\ & \leq E\bigg[ \exp \bigg( 2 \int _{0}^{T} R_{s} d[M]_{s} \bigg) \bigg] < \infty . \end{aligned} $$

Hence the claim of Proposition C.1 also applies to the setting mentioned in Remark 7.2.