No-arbitrage conditions and pricing from discrete-time to continuous-time strategies

Abstract

In this paper, a general framework is developed for continuous-time financial market models defined from simple strategies through conditional topologies that avoid stochastic calculus and do not necessitate semimartingale models. We then compare the usual no-arbitrage conditions of the literature, e.g. the usual no-arbitrage conditions NFL, NFLVR and NUPBR and the recent AIP condition. With appropriate pseudo-distance topologies, we show that they hold in continuous time if and only if they hold in discrete time. Moreover, the super-hedging prices in continuous time coincide with the discrete-time super-hedging prices, even without any no-arbitrage condition.

Appendices

Appendix A: Proof of Proposition 10

The proof of Proposition 10 is deduced from Lemma 37. To get it, we first show intermediate steps such as the following Lemmas 33,  34,  35 and  36.

Lemma 33

Let $(M_t)_{t\in [0,T]}$ be a sub-maxingale. Let $\tau$ be a stopping time such that $\tau (\Omega )= \{ t_1, t_2, \cdots ,t_n \}$ where $(t_i)_{i=1}^n$ is an increasing sequence of discrete dates. Then, for all $i=1,\cdots ,n$, we have ${{\,\mathrm{ess\;sup}\,}}_{{\mathcal {F}}_{t_i}}(M_{\tau }) \ge M_{\tau \wedge t_i}$.

Proof

We have:

$$\begin{aligned} {{\,\mathrm{ess\;sup}\,}}_{{\mathcal {F}}_{t_i}}(M_{\tau \wedge t_{i+1}})= & {} {{\,\mathrm{ess\;sup}\,}}_{{\mathcal {F}}_{t_i}}(M_{\tau \wedge t_{i+1}} 1_{\{\tau \le t_i \}}) +{{\,\mathrm{ess\;sup}\,}}_{{\mathcal {F}}_{t_i}}(M_{\tau \wedge t_{i+1}} 1_{\{\tau>t_i \}}),\\= & {} 1_{\{\tau>t_i \}}{{\,\mathrm{ess\;sup}\,}}_{{\mathcal {F}}_{t_i}}(M_{ t_{i+1}}) + 1_{\{\tau \le t_i \}} {{\,\mathrm{ess\;sup}\,}}_{{\mathcal {F}}_{t_i}}(M_{\tau \wedge t_{i}} ),\\\ge & {} 1_{\{\tau >t_i \}}M_{ t_{i}} + 1_{\{\tau \le t_i \}} M_{\tau \wedge t_{i}}= M_{\tau \wedge t_i}. \end{aligned}$$

If $j >i+1$, argue by induction. By the tower property, we first have ${{\,\mathrm{ess\;sup}\,}}_{{\mathcal {F}}_{t_i}}(M_{\tau \wedge t_{j}}) ={{\,\mathrm{ess\;sup}\,}}_{{\mathcal {F}}_{t_i}}({{\,\mathrm{ess\;sup}\,}}_{{\mathcal {F}}_{t_{j-1}}}(M_{\tau \wedge t_{j}}))$. Therefore, by the first step above, ${{\,\mathrm{ess\;sup}\,}}_{{\mathcal {F}}_{t_i}}(M_{\tau \wedge t_{j}})\ge {{\,\mathrm{ess\;sup}\,}}_{{\mathcal {F}}_{t_i}}(M_{\tau \wedge t_{j-1}})$ and we conclude by induction. $\square$

Lemma 34

Let $\tau$ be a stopping time such that $\tau (\Omega )= \{ t_1, t_2, \cdots ,t_n \}$ where $(t_i)_{i=1}^n$ is an increasing sequence of discrete dates. Then, for any random variable X, we have

$$\begin{aligned} {{\,\mathrm{ess\;sup}\,}}_{{\mathcal {F}}_{\tau }}(X 1_{\{\tau = t_i\}})= {{\,\mathrm{ess\;sup}\,}}_{{\mathcal {F}}_{t_i}} (X)1_{\{\tau = t_i\}}. \end{aligned}$$

Proof

As $1_{\{\tau = t_i\}}$ is ${\mathcal {F}}_{\tau }$-mesurable, then we get that

$$\begin{aligned} {{\,\mathrm{ess\;sup}\,}}_{{\mathcal {F}}_{\tau }}(X1_{\{\tau = t_i\}})={{\,\mathrm{ess\;sup}\,}}_{{\mathcal {F}}_{\tau }}(X) 1_{\{\tau = t_i\}}. \end{aligned}$$

Since $X 1_{\{\tau = t_i\}}\le {{\,\mathrm{ess\;sup}\,}}_{{\mathcal {F}}_{t_i}}(X )1_{\{\tau =t_i\}}$, we deduce that

$$\begin{aligned} {{\,\mathrm{ess\;sup}\,}}_{{\mathcal {F}}_{\tau }}(X 1_{\{\tau = t_i\}})\le \ {{\,\mathrm{ess\;sup}\,}}_{{\mathcal {F}}_{\tau }} ({{\,\mathrm{ess\;sup}\,}}_{{\mathcal {F}}_{t_i}}(X )1_{\{\tau = t_i\}}). \end{aligned}$$

We claim that $Z={{\,\mathrm{ess\;sup}\,}}_{{\mathcal {F}}_{t_i}}(X )1_{\{\tau = t_i\}}$ is ${\mathcal {F}}_{\tau }$-mesurable. For any $k \in \textbf{R}$,

$$\begin{aligned} \{Z \le k \} = \{0 \le k \} \cap \{ \tau \ne t_i \} \cup \{ \tau = t_i \} \cap \{ {{\,\mathrm{ess\;sup}\,}}_{{\mathcal {F}}_{t_i}}(X) \le k \}. \end{aligned}$$

Note that $\{0 \le k \} = \emptyset$ or $\Omega$ and $\{ \tau \ne t_i \} \in {\mathcal {F}}_{\tau }$ hence $\{0 \le k \} \cap \{ \tau \ne t_i \}\in {\mathcal {F}}_{\tau }$. Now let us show that $B = \{ \tau = t_i \} \cap \{ {{\,\mathrm{ess\;sup}\,}}_{{\mathcal {F}}_{t_i}}(X) \le k \} \in {\mathcal {F}}_{\tau }$. To do so, we evaluate $B \cap \{ \tau \le t \}$ for $t\ge 0$. Note that $t_j \le t < t_{j+1}$ for some $t_j \in \{t_0,\cdots ,t_n,t_{n+1}\}$, where $t_{n+1}=\infty$. So, we deduce that $B \cap \{ \tau \le t \}$ coincides with $B \cap \{ \tau \le t_j \}= \emptyset$ if $t_j < t_i$. Otherwise, we obtain that $B \cap \{ \tau \le t \}=B \in F_{t_i} \subseteq F_{t_j} \subseteq F_t.$ Therefore, $B \cap \{ \tau \le t \}\in {\mathcal {F}}_t$, for all $t\in \textbf{R}$, hence $B \in {\mathcal {F}}_{\tau }$. Finally, Z is ${\mathcal {F}}_{\tau }$-mesurable and the inequality ${{\,\mathrm{ess\;sup}\,}}_{{\mathcal {F}}_{\tau }}(X 1_{\{\tau = t_i\}})\le \ {{\,\mathrm{ess\;sup}\,}}_{{\mathcal {F}}_{t_i}}(X )1_{\{\tau = t_i\}}$ holds. For the reverse inequality it suffices to show that $Y={{\,\mathrm{ess\;sup}\,}}_{{\mathcal {F}}_{\tau }}(X 1_{\{\tau = t_i\}} )$ is ${\mathcal {F}}_{t_i}$-measurable. Since $\{\tau \ne t_i\}\in {\mathcal {F}}_{\tau }$, we get that $Y 1_{\{\tau \ne t_i\}}=0$ and

$$\begin{aligned} \{ Y \le k \} = (\{0 \le k \} \cap \{\tau \ne t_i \}) \cup (A \cap \{\tau = t_i \}), \end{aligned}$$

with $A=\{{{\,\mathrm{ess\;sup}\,}}_{{\mathcal {F}}_{\tau }}(X 1_{\{\tau = t_i\}}) \le k \}$. As $A \in {\mathcal {F}}_{\tau }$, $A \cap \{\tau \le t_i \} \in {\mathcal {F}}_{t_i}$ and, finally, $A \cap \{\tau = t_i \}= A \cap \{\tau = t_i \} \cap \{\tau \le t_i \}\in {\mathcal {F}}_{t_i}$. Therefore, for all $k\in \textbf{R}$, $\{ Y \le k \}\in {\mathcal {F}}_{t_i}$, i.e. Y is ${\mathcal {F}}_{t_i}$-measurable. At last, notice that ${{\,\mathrm{ess\;sup}\,}}_{{\mathcal {F}}_{\tau }}(X 1_{\{\tau = t_i\}} )\ge X 1_{\{\tau = t_i\}}$ and, since Y is ${\mathcal {F}}_{t_i}$-measurable, we get that ${{\,\mathrm{ess\;sup}\,}}_{{\mathcal {F}}_{\tau }}(X 1_{\{\tau = t_i\}} )\ge {{\,\mathrm{ess\;sup}\,}}_{{\mathcal {F}}_{t_i}}(X 1_{\{\tau = t_i\}})$. The conclusion follows. $\square$

Lemma 35

Let $(M_t)_{t\in [0,T]}$ be a sub-maxingale. Let $\tau$, S be two stopping times. Suppose that $S(\Omega )= \{ t_1, t_2, \cdots ,t_n \}$ where $(t_i)_{i=1}^n$ is an increasing sequence of discrete dates and suppose that $\tau (\Omega )$ is also a finite set. Then ${{\,\mathrm{ess\;sup}\,}}_{{\mathcal {F}}_{S}}(M_{\tau })\ge M_{\tau \wedge s}$.

Proof

By Lemma 34, we obtain ${{\,\mathrm{ess\;sup}\,}}_{{\mathcal {F}}_{S}}(M_{\tau }) =\sum \nolimits _{i=1}^n {{\,\mathrm{ess\;sup}\,}}_{{\mathcal {F}}_{t_i}}(M_{\tau }) 1_{\{S=t_i\}}$. By Lemma 33, we deduce that

$$\begin{aligned} {{\,\mathrm{ess\;sup}\,}}_{{\mathcal {F}}_{S}}(M_{\tau }) \ge \sum _{i=1}^n M_{\tau \wedge t_i} 1_{\{S=t_i\}}= \sum _{i=1}^n M_{ \tau \wedge S} 1_{\{S=t_i\}} = M_{ \tau \wedge S}. \end{aligned}$$

$\square$

Lemma 36

Let $\tau \in [0,T]$ be a stopping time. Suppose that the filtration $({\mathcal {F}}_t)_{t\in [0,T]}$ is right-continuous. There exists a non increasing sequence $(\tau _n)_n$ of stopping times converging to $\tau$ such that, for any $X\in L^0(\textbf{R},{\mathcal {F}}_T)$,

$$\begin{aligned} {{\,\mathrm{ess\;sup}\,}}_{{\mathcal {F}}_{\tau }}(X)= \lim _n \uparrow {{\,\mathrm{ess\;sup}\,}}_{{\mathcal {F}}_{\tau _n}}(X). \end{aligned}$$

Moreover, $\tau ^n(\Omega )$ is finite for all $n\ge 1$.

Proof

Let $\tau$ be a stopping time taking values in [0, T]. For any $n\ge 1$, we define $\tau ^n(\omega ) = T(i+1)/2^n$ where $i=i(\omega )$ is uniquely defined such that $Ti/2^n< \tau (\omega ) \le T(i+1)/2^n$ for $i\ge 1$ or $0\le \tau (\omega ) \le T/2^n$ when $i=0$. Note that $\tau ^n(\Omega )$ is finite and $\tau ^n \ge \tau$. It is easily seen that $(\tau _n)_n$ is non increasing, positive and $\lim _n \tau _n=\tau$. Moreover, $\tau ^n$ is a stopping time. Indeed, for any fixed $t \in [0,T)$, there exists $i \in {\mathbb {N}}$ such that $Ti/2^n\le t < T(i+1)/2^n$. Then $\{ \tau ^n \le t \}= \{ \tau \le Ti/2^n \} \in {\mathcal {F}}_{Ti/2^n} \subset {\mathcal {F}}_t$ and the conclusion follows.

As $(\tau ^n)_n$ is non increasing, then $({\mathcal {F}}_{\tau ^n})_n$ non increasing. As we know that ${{\,\mathrm{ess\;sup}\,}}_{{\mathcal {F}}_{\tau ^{n+1}}}(X) \ge X$ and ${{\,\mathrm{ess\;sup}\,}}_{{\mathcal {F}}_{\tau ^{n+1}}}(X)$ is ${\mathcal {F}}_{\tau ^n}$-measurable ($\tau _{n+1}\le \tau _n)$, we deduce that ${{\,\mathrm{ess\;sup}\,}}_{{\mathcal {F}}_{\tau ^n}}(X) \le {{\,\mathrm{ess\;sup}\,}}_{{\mathcal {F}}_{\tau ^{(n+1)}}}(X)$, i.e. $({{\,\mathrm{ess\;sup}\,}}_{{\mathcal {F}}_{\tau ^n}}(X))_n$ is non decreasing.

Similarly, $\tau ^n \ge \tau$ implies that ${{\,\mathrm{ess\;sup}\,}}_{{\mathcal {F}}_{\tau ^n}}(X) \le {{\,\mathrm{ess\;sup}\,}}_{{\mathcal {F}}_{\tau }}(X)$. Therefore, $\lim _n \uparrow {{\,\mathrm{ess\;sup}\,}}_{{\mathcal {F}}_{\tau ^n}}(X) \le {{\,\mathrm{ess\;sup}\,}}_{{\mathcal {F}}_{\tau }}(X)$. To obtain the reverse inequality, we consider the sequence $({{\,\mathrm{ess\;sup}\,}}_{{\mathcal {F}}_{\tau + T/n}}(X))_n.$ Since $\tau + T/n \ge \tau ^n$, then

$$\begin{aligned} \lim _n \uparrow {{\,\mathrm{ess\;sup}\,}}_{{\mathcal {F}}_{\tau + T/n}}(X) \le \lim \uparrow {{\,\mathrm{ess\;sup}\,}}_{{\mathcal {F}}_{\tau ^n}}(X) \le {{\,\mathrm{ess\;sup}\,}}_{{\mathcal {F}}_{\tau }}(X). \end{aligned}$$

It suffices to see that $Z=\lim _n \uparrow {{\,\mathrm{ess\;sup}\,}}_{{\mathcal {F}}_{\tau +T/n}}(X)$ is ${\mathcal {F}}_{\tau }$-measurable to conclude. Indeed, $Z\ge X$ hence $Z\ge {{\,\mathrm{ess\;sup}\,}}_{{\mathcal {F}}_{\tau }}(X)$ and inequalities above are equalities. For all $k\in \textbf{R}$, $t\ge 0$, and any $n_0\ge 1$

$$\begin{aligned} \{Z\le k\}\cap \{\tau \le t\}= & {} \bigcap _{n\ge 1} \{{{\,\mathrm{ess\;sup}\,}}_{{\mathcal {F}}_{\tau + T/n}}(X)\le k\}\cap \{\tau \le t\},\\= & {} \bigcap _{n\ge n_0}\{ {{\,\mathrm{ess\;sup}\,}}_{{\mathcal {F}}_{\tau + T/n}}(X)\le k\} \cap \{\tau + T/n\le t+ T/n\}. \end{aligned}$$

Notice that ${{\,\mathrm{ess\;sup}\,}}_{{\mathcal {F}}_{\tau + T/n}}(X)$ is ${\mathcal {F}}_{\tau +T/n}$-measurable. We deduce that:

$$\begin{aligned}{} & {} \{{{\,\mathrm{ess\;sup}\,}}_{{\mathcal {F}}_{\tau + T/n}}(X)\le k\}\in {\mathcal {F}}_{\tau +T/n},\\{} & {} \{{{\,\mathrm{ess\;sup}\,}}_{{\mathcal {F}}_{\tau + T/n}}(X)\le k\}\cap \{\tau + T/n \le t+ T/n\}\in {\mathcal {F}}_{t+ T/n}. \end{aligned}$$

Therefore, for any $\epsilon >0$ and $n_0\ge 1$ such that $t+ T/n\le t+\epsilon$, we have ${\mathcal {F}}_{t+ T/n}\subseteq {\mathcal {F}}_{t+\epsilon }$ and, finally, $\{Z\le k\}\cap \{\tau \le t\}\in \cap _{\epsilon >0}{\mathcal {F}}_{t+\epsilon }={\mathcal {F}}_{t+}={\mathcal {F}}_t$. We deduce that $\{Z\le k\}\in {\mathcal {F}}_{\tau }$, for all $k\in \textbf{R}$, i.e. Z is ${\mathcal {F}}_{\tau }$-measurable. $\square$

Lemma 37

Suppose that the filtration $({\mathcal {F}}_t)_{t\in [0,T]}$ is right-continuous. Let $(M_t)_{t\in [0,T]}$ be a right-continuous sub-maxingale. Let $\tau$, S be two stopping times such that $\tau (\Omega )$ is a finite set. Then, we have ${{\,\mathrm{ess\;sup}\,}}_{{\mathcal {F}}_{S}}(M_{\tau })\ge M_{\tau \wedge S}$.

Proof

Let $(S_n)_n$ be a sequence of stopping times decreasing to S as given in Lemma 36. Recall that $S_n(\Omega )$ is finite for all n. Moreover, we have ${{\,\mathrm{ess\;sup}\,}}_{{\mathcal {F}}_{s}}(M_{\tau }) =\lim _n \uparrow {{\,\mathrm{ess\;sup}\,}}_{{\mathcal {F}}_{s_n}}(M_{\tau })$. By Lemma 35, we deduce that ${{\,\mathrm{ess\;sup}\,}}_{{\mathcal {F}}_{S}}(M_{\tau }) \ge \lim \uparrow M_{\tau \wedge S_n}.$ As $(\tau \wedge S_n)_n$ decreases to $\tau \wedge S$ and M is right-continuous, we conclude that ${{\,\mathrm{ess\;sup}\,}}_{{\mathcal {F}}_{s}}(M_{\tau }) \ge M_{\tau \wedge s}$. $\square$

Appendix B: Auxiliary results

Lemma 38

Suppose that, at time $t\le T$, ${\mathcal {O}}_t$ is the pseudo-distance topology defined by (8) and ${\mathcal {V}}_{t,T}^c={\mathcal {V}}_{t,T}^c({\mathcal {O}}_t)$. If ${\mathcal {V}}_{t,T}$ is ${\mathcal {F}}_t$-decomposable, then ${\mathcal {V}}_{t,T}^c$ is ${\mathcal {F}}_t$-decomposable.

Proof

Consider $V_{t,T}^{c,i}\in {\mathcal {V}}_{t,T}^c$, $i=1,2$, and $F_t\in {\mathcal {F}}_t$. By Proposition 32, $V_{t,T}^{c,i}\le V_{t,T}^{n,i}+\alpha _t^{n,i}$ where $V_{t,T}^{n,i}\in {\mathcal {V}}_{t,T}$ and $\alpha _t^{n,i}$ converges to 0 in probability as $n\rightarrow \infty$, for $i=1,2$. We set

$$\begin{aligned} V_{t,T}^{n}= & {} V_{t,T}^{n,1}1_{F_t}+V_{t,T}^{n,2}1_{\Omega \setminus F_t}, \quad \alpha _t^{n}=\alpha _t^{n,1}1_{F_t}+\alpha _t^{n,2} 1_{\Omega \setminus F_t}. \end{aligned}$$

Note that $V_{t,T}^{n}\in {\mathcal {V}}_{t,T}$ by assumption and $\alpha _t^{n}$ converges to 0 in probability. Moreover, $V_{t,T}^{c,1}1_{F_t}+V_{t,T}^{c,2}1_{\Omega {\setminus } F_t}\le V_{t,T}^{n}+\alpha _t^{n}$. Therefore, Proposition 32 implies that $V_{t,T}^{c,1}1_{F_t}+V_{t,T}^{c,2}1_{\Omega {\setminus } F_t}\in {\mathcal {V}}_{t,T}^c$ and the conclusion follows. $\square$

Lemma 39

Let $h_T\in L^0(\textbf{R},{\mathcal {F}}_T)$ be a payoff. If ${\mathcal {V}}_{t,T}$ (resp. ${\mathcal {V}}_{t,T}^c$) is ${\mathcal {F}}_t$-decomposable (resp. infinitely ${\mathcal {F}}_t$-decomposable), then ${\mathcal {P}}_{t,T}(h_T)$ (resp. ${\mathcal {P}}_{t,T}^c(h_T)$) is ${\mathcal {F}}_t$-decomposable (resp. infinitely ${\mathcal {F}}_t$-decomposable).

Proof

Suppose that ${\mathcal {V}}_{t,T}$ is ${\mathcal {F}}_t$-decomposable and consider $p_t^1,p_t^2\in {\mathcal {P}}_{t,T}(h_T)$ and $F_t\in {\mathcal {F}}_t$. Then, $p_t^i+V_{t,T}^{i}\ge h_T$ for some $V_{t,T}^{i}\in {\mathcal {V}}_{t,T}$, $i=1,2$. By assumption, we have $V_{t,T}=V_{t,T}^{1}1_{F_t} +V_{t,T}^{2}1_{\Omega {\setminus } F_t}\in {\mathcal {V}}_{t,T}$ by assumption and $p_t^11_{F_t}+p_t^21_{\Omega \setminus F_t}+V_{t,T}\ge h_T$. We deduce that $p_t^11_{F_t}+p_t^2 1_{\Omega {\setminus } F_t}\in {\mathcal {P}}_{t,T}(h_T)$. By the same reasoning, the property holds for ${\mathcal {V}}_{t,T}^c$ and the infinite ${\mathcal {F}}_t$-decomposability is obtained similarly. The conclusion follows. $\square$

Lemma 40

Let $h_T\in L^0(\textbf{R},{\mathcal {F}}_T)$ be a payoff. If ${\mathcal {V}}_{t,T}$ is infinitely ${\mathcal {F}}_t$-decomposable, then for any $\gamma _t\in L^0(\textbf{R},{\mathcal {F}}_t)$ such that $\gamma _t>\pi _{t,T}(h_T)$, there exists a price $p_t\in {\mathcal {P}}_{t,T}(h_T)$ such that $p_t<\gamma _t$. In particular, $\gamma _t\in {\mathcal {P}}_{t,T}(h_T)$.

Proof

Since ${\mathcal {V}}_{t,T}$ is infinitely ${\mathcal {F}}_t$-decomposable, ${\mathcal {P}}_{t,T}(h_T)$ is infinitely ${\mathcal {F}}_t$-decomposable by Lemma 39. Therefore, ${\mathcal {P}}_{t,T}(h_T)$ is directed downward and we deduce that $\pi _{t,T}(h_T)\lim _n\downarrow p_t^n$ where $p_t^n\in {\mathcal {P}}_{t,T}(h_T)$, see Kabanov and Safarian (2009, Sect. 5.3.1). Then, a.s.($\omega$), there exits $n(\omega )$ such that $p_t^n(\omega )<\gamma _t(\omega )$. We then define

$$\begin{aligned} N_t= & {} \inf \{n\ge 1:~p_t^n<\gamma _t\}\in L^0({\mathbb {N}},{\mathcal {F}}_t),\\ p_t= & {} \sum _{j=1}^\infty p_t^j1_{\{N_t=j\}}. \end{aligned}$$

By assumption $p_t\in {\mathcal {P}}_{t,T}(h_T)$ and $p_t<\gamma _t$. The conclusion follows. $\square$