Admissible Strategies and NFLVR

Why Trading Rules Need a Loss Bound

A self-financing strategy produces gains by trading in assets without adding or withdrawing capital. In a continuous-time model, the class of permitted strategies matters: if traders may take positions whose possible losses have no lower bound, a formally attractive strategy can hide an enormous downside behind a small probability of failure. Admissibility rules out this behavior by requiring each strategy’s cumulative gains to stay above a fixed finite loss level. No free lunch with vanishing risk, abbreviated NFLVR, then asks whether admissible opportunities can approximate a nonzero, nonnegative payoff arbitrarily well.

Work on a fixed finite horizon $[0,T]$ and a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$. Let $S=(S^1,\ldots,S^d)$ be the vector of discounted traded asset prices, assumed to be semimartingales. Discounting uses a chosen numeraire, such as a bank account; all gains and prices below are in those discounted units. A strategy $H=(H^1,\ldots,H^d)$ is predictable and integrable with respect to $S$. Its cumulative gains, starting from zero, are $G_t(H)=\sum_{i=1}^d\int_0^t H_s^i\,dS_s^i$. For general semimartingale prices this is semimartingale integrability, not necessarily the Brownian $L^2$ condition. In Brownian examples below, integrands are progressively measurable and satisfy $\mathbb{E}\int_0^T H_s^2\,ds<\infty$.

Definition: A strategy $H$ is admissible on $[0,T]$ if there is a finite constant $a\ge0$ such that $G_t(H)\ge-a$ for every $t\in[0,T]$, almost surely. The bound may depend on $H$, but it must hold across the entire horizon, not just at terminal time. An arbitrage is an admissible strategy with $G_T(H)\ge0$ almost surely and $\mathbb{P}(G_T(H)>0)>0$.

The lower bound is on cumulative gains, so it constrains interim losses as well as the final payoff. It is not a requirement that the gain process be bounded above, nor does it require the same constant $a$ for every strategy. The admissible class is deliberately broad enough to include many useful strategies, while excluding strategies whose wealth can fall arbitrarily far below initial wealth.

Defining NFLVR Precisely

Let $L^\infty=L^\infty(\Omega,\mathcal{F}_T,\mathbb{P})$ denote the essentially bounded terminal random variables, identified when equal almost surely. Its norm is $\|g\|_\infty=\operatorname{ess\,sup}|g|$. Define the set of bounded claims dominated by admissible terminal gains by

\mathcal{C}=\{g\in L^\infty:\text{ there is an admissible }H\text{ with }g\le G_T(H)\ \mathbb{P}\text{-almost surely}\}.

Definition: The market satisfies no free lunch with vanishing risk (NFLVR) if the closure of $\mathcal{C}$ in the $L^\infty$ norm has no nonzero nonnegative element: $$ \overline{\mathcal{C}}^{\,\|\cdot\|_\infty}\cap L^\infty_+=\{0\}, $$ where $L^\infty_+=\{g\in L^\infty:g\ge0\ \mathbb{P}\text{-almost surely}\}$. The zero claim belongs to $\mathcal{C}$, so the intersection always contains zero.

The norm closure means that a bounded payoff $g$ is in $\overline{\mathcal{C}}^{\,\|\cdot\|_\infty}$ precisely when claims in $\mathcal{C}$ can approximate it with errors uniformly small outside null sets. Thus NFLVR excludes not only a directly available nonnegative payoff with a positive chance of gain, but also a nonzero nonnegative payoff that can be approached with uniformly vanishing error.

Worked Example: A Nonnegative Claim in the Closure

Suppose a market has admissible strategies $H^n$ such that their terminal gains are the deterministic values $G_T(H^n)=1-\frac1n$, for every integer $n\ge1$. Determine what this implies for NFLVR.

Set $g_n=1-\frac1n$. Since $g_n=G_T(H^n)$, we have $g_n\le G_T(H^n)$, so $g_n\in\mathcal{C}$; each $g_n$ is bounded. Let $g=1$, the constant-one claim. Then

\|g_n-g\|_\infty=\left|\left(1-\frac1n\right)-1\right|=\frac1n\longrightarrow0.

Consequently $g\in\overline{\mathcal{C}}^{\,\|\cdot\|_\infty}$. It is nonnegative and nonzero, so the market does not satisfy NFLVR. Check: for $n=4$, $g_4=1-\frac14=\frac34$, and $\|g_4-g\|_\infty=|\,\frac34-1\,|=\frac14$, matching $1/n$. This example describes the closure test; it does not assert that such strategies exist in any particular market.

NFLVR Rules Out Ordinary Arbitrage

NFLVR is stronger than the absence of arbitrage: it controls limits of bounded claims, not just individual strategies. The implication from NFLVR to no arbitrage follows directly from the definitions.

Proposition: If a market satisfies NFLVR, it has no admissible arbitrage.

Proof. Suppose, to the contrary, that an admissible strategy $H$ is an arbitrage. Then $G_T(H)\ge0$ almost surely and $\mathbb{P}(G_T(H)>0)>0$. Define $g=\min(G_T(H),1)$. This random variable is bounded, is nonnegative almost surely, and is strictly positive on the event $\{G_T(H)>0\}$, so it is nonzero as an element of $L^\infty$. Also $g\le G_T(H)$, hence $g\in\mathcal{C}$. Since every element of $\mathcal{C}$ belongs to its norm closure, the closure contains a nonzero element of $L^\infty_+$, contradicting NFLVR. Therefore no such $H$ exists. $\square$

The converse is not part of the definition: a market can lack a directly available arbitrage and still fail the stronger closure condition. The uniform norm is important here. Merely having gains that are positive with probability tending to one does not show convergence in $L^\infty$; a rare loss of large magnitude remains visible to this norm.

Worked Examples: Admissibility in Continuous-Time Models

Worked Example: A Long Position in a Positive Martingale

Under a probability measure $\mathbb{Q}$, let a discounted asset be $S_t=\exp(B_t-\tfrac12t)$ on $[0,T]$, where $B$ is Brownian motion. Take $S_0=1$ and hold one share, $H_t=1$. Verify the gain, admissibility and martingale property.

First, the Ito integral in the gain is well-defined: its Brownian integrand is $S_t$, which is progressively measurable, and $\mathbb{E}^{\mathbb{Q}}\int_0^T S_t^2\,dt=\int_0^T e^t\,dt=e^T-1<\infty$. Ito’s formula applied to $f(t,x)=e^{x-t/2}$ gives $f_t=-\tfrac12f$, $f_x=f$, and $f_{xx}=f$. Therefore

dS_t=f_t\,dt+f_x\,dB_t+\tfrac12f_{xx}\,dt =-\tfrac12S_t\,dt+S_t\,dB_t+\tfrac12S_t\,dt =S_t\,dB_t.

The gain process is $G_t(H)=\int_0^t1\,dS_s=S_t-S_0=S_t-1$. Since $S_t>0$, $G_t(H)>-1$ at every time; thus $H$ is admissible with $a=1$. For the martingale check, if $s\le t$, then $S_t=S_s\exp(B_t-B_s-\tfrac12(t-s))$, and the Brownian increment is independent of $\mathcal{F}_s$. Its exponential factor has expectation $1$, so $\mathbb{E}^{\mathbb{Q}}[S_t\mid\mathcal{F}_s]=S_s$. Also $\mathbb{E}^{\mathbb{Q}}S_t=1<\infty$, so this is a true martingale, not merely a local martingale. The gain has conditional mean zero, and its terminal value cannot be a strictly positive arbitrage payoff.

Check: $\mathbb{E}^{\mathbb{Q}}[G_T(H)]=\mathbb{E}^{\mathbb{Q}}[S_T]-1=1-1=0$, consistent with the conditional martingale calculation. The lower bound is $-1$, not a lower bound on the size of the asset itself.

Worked Example: A Stop-Loss Rule Gives a Lower-Bounded Gain

Use the same discounted asset $S_t=\exp(B_t-\tfrac12t)$, with $S_0=1$. Hold one share until it first reaches $\tfrac12$, or until the finite horizon $T$, whichever comes first. Show that this stopped strategy is admissible.

Define $\tau=\inf\{t\ge0:S_t=\tfrac12\}$, with $\inf\varnothing=\infty$, and set $H_t=\mathbf{1}_{\{t\le\tau\}}$ on $[0,T]$. This predictable stopped-position convention holds the share up to the stopping time and zero thereafter; changing its value at the single time $\tau$ does not change the integral. The integrand for the Brownian representation of the gain is $H_tS_t$. It is progressively measurable and satisfies $\mathbb{E}^{\mathbb{Q}}\int_0^T(H_tS_t)^2\,dt\le\int_0^T\mathbb{E}^{\mathbb{Q}}[S_t^2]\,dt=e^T-1<\infty$. By the stopped-integral identity, $G_t(H)=S_{t\wedge\tau}-1$. Before the stopping time the price is above $\tfrac12$; at the stopping time it equals $\tfrac12$. If the level is never reached by $t$, the price is also above $\tfrac12$. Hence $S_{t\wedge\tau}\ge\tfrac12$ and $G_t(H)\ge-\tfrac12$ for every $t\le T$. The strategy is admissible with $a=\tfrac12$.

Check: at a time when the stop-loss has been triggered, $G_t(H)=\tfrac12-1=-\tfrac12$, exactly the permitted lower bound. If the level has not been reached, $G_t(H)=S_t-1>-\tfrac12$. This argument is a pathwise bound and does not require the stopping time to occur before the finite horizon.

Worked Example: A Brownian Exit Gamble on an Infinite Horizon

This example is explicitly on an infinite horizon, to separate an eventual exit payoff from a payoff at a fixed finite date. Let $B$ be Brownian motion under $\mathbb{P}$. For an integer $n\ge1$, set $\tau=\inf\{t\ge0:B_t\notin(-n,1)\}$, and consider the gain process $G_t=B_{t\wedge\tau}$, corresponding to holding one unit of the Brownian gain process until exit. Its stochastic integral has integrand $\mathbf{1}_{\{s\le\tau\}}$, which is progressively measurable and satisfies $\mathbb{E}\int_0^T\mathbf{1}_{\{s\le\tau\}}^2\,ds\le T$ on every finite interval $[0,T]$. By continuity, $G_t\in[-n,1]$, so the strategy is admissible with lower bound $-n$.

The exit time is finite almost surely. Indeed, while the process remains in $(-n,1)$, an increment over the next unit interval greater than $n+1$ forces its value above $1$. This increment has probability $p_0=\mathbb{P}(N(0,1)>n+1)>0$, independently of the past. Thus the probability of remaining in the interval for $k$ successive unit intervals is at most $(1-p_0)^k$, which tends to zero. At exit, continuity gives $B_\tau\in\{-n,1\}$.

For each finite $t$, $B_{t\wedge\tau}$ is a martingale: it is the Brownian motion stopped at $\tau$, and it is bounded between $-n$ and $1$, hence integrable. Optional stopping for this bounded stopped process gives $\mathbb{E}[B_{t\wedge\tau}]=0$. As $t\to\infty$, $B_{t\wedge\tau}\to B_\tau$ almost surely and the variables remain bounded by $n+1$ in absolute value. Dominated convergence therefore yields $\mathbb{E}[B_\tau]=0$. If $q=\mathbb{P}(B_\tau=1)$, then

0=\mathbb{E}[B_\tau]=q\cdot1+(1-q)(-n), \qquad q=\frac{n}{n+1}.

For $n=9$, the probability of winning $1$ is $\frac9{10}$, and the probability of losing $9$ is $\frac1{10}$; the expected exit payoff is $\frac9{10}\cdot1+\frac1{10}\cdot(-9)=0$. This is not an arbitrage. On a fixed finite horizon $T$, the payoff is $B_{T\wedge\tau}$, and on $\{\tau>T\}$ it equals $B_T$, not either exit value. The two-outcome calculation applies to the payoff at the infinite-horizon exit time $\tau$, not automatically to a finite-horizon terminal gain.

Why a Lower Bound Makes the Martingale Argument Work

A key fact is that a local martingale bounded below cannot have a positive expected gain. A local martingale is a process that becomes a martingale when stopped at a suitable increasing sequence of stopping times. It need not itself be a true martingale, because its expectations may fail to be preserved without sufficient integrability. The lower bound supplies the control needed for the following result.

Theorem: Every nonnegative local martingale $Y$ is a supermartingale. In particular, it is integrable at each time and, for $s\le t$, $\mathbb{E}[Y_t\mid\mathcal{F}_s]\le Y_s$ almost surely.

Proof. Let $(\tau_n)$ be a localizing sequence such that $Y_{t\wedge\tau_n}$ is a martingale for each $n$, and $\tau_n\to\infty$ almost surely. Since $Y\ge0$, Fatou’s lemma gives $\mathbb{E}[Y_t]\le\liminf_n\mathbb{E}[Y_{t\wedge\tau_n}]=\mathbb{E}[Y_0]<\infty$; thus $Y_t$ is integrable. Fix $s\le t$. Because $\tau_n\to\infty$ almost surely, for each fixed $u\in\{s,t\}$ we have $Y_{u\wedge\tau_n}\to Y_u$ almost surely. No path continuity is required for this fixed-time convergence. Conditional Fatou’s lemma and the martingale property of each stopped process now give \[ \mathbb{E}[Y_t\mid\mathcal{F}_s] \le\liminf_n\mathbb{E}[Y_{t\wedge\tau_n}\mid\mathcal{F}_s] =\liminf_n Y_{s\wedge\tau_n} =Y_s \quad\text{almost surely}. \] This is the supermartingale inequality, and integrability at each time was established above. $\square$

Corollary: A local martingale $M$ with $M_t\ge-a$ for all $t\le T$, for a finite constant $a$, is a supermartingale on $[0,T]$.

Proof. The process $Y=M+a$ is a nonnegative local martingale, so the Theorem makes it a supermartingale. Subtracting the constant $a$ from $\mathbb{E}[M_t+a\mid\mathcal{F}_s]\le M_s+a$ gives $\mathbb{E}[M_t\mid\mathcal{F}_s]\le M_s$. $\square$

This explains the financial role of admissibility. Suppose there is a probability measure $\mathbb{Q}$ equivalent to $\mathbb{P}$ under which every discounted traded price is a local martingale. Equivalence means the two measures have the same null events. The standard stochastic-integration result for semimartingales says that the gain process of an integrable predictable strategy against these prices is a local martingale under $\mathbb{Q}$. For an admissible strategy it is bounded below, so the corollary makes its gain a $\mathbb{Q}$-supermartingale. If the strategy were an arbitrage, then $G_T(H)\ge0$ almost surely and it would be strictly positive with positive $\mathbb{Q}$-probability, by equivalence. Its expectation would therefore be strictly positive, contradicting $\mathbb{E}^{\mathbb{Q}}[G_T(H)]\le G_0(H)=0$. Thus such a measure rules out admissible arbitrage. The lower bound is essential to this argument: without it, a local martingale need not be a supermartingale or even have an integrable terminal value.

Common Pitfalls

Checking only the terminal loss is insufficient. Admissibility requires $G_t(H)\ge-a$ at every time up to $T$. A strategy that ends safely but can incur arbitrarily large interim losses does not meet this definition.

Do not confuse likely profit with vanishing risk. In the exit gamble with $n=9$, the loss of $9$ occurs with probability $\frac1{10}$. Increasing $n$ makes the loss less likely, but also makes its magnitude larger. In particular, these payoffs do not converge uniformly to the constant-one claim: on the loss event their distance from $1$ is $n+1$, not a quantity tending to zero. A small failure probability alone is not the $L^\infty$-closure condition in NFLVR.

A local martingale is not automatically a martingale. The supermartingale conclusion above uses nonnegativity, or equivalently a fixed lower bound after adding a constant. When this lower-bound hypothesis is absent, the conditional expectation argument cannot be applied as stated.

Concept	What it requires	What it rules out
Admissibility	A finite lower bound on cumulative gains throughout the horizon	Strategies with unbounded downside during trading
No arbitrage	No admissible terminal gain is nonnegative almost surely and positive with positive probability	Directly attainable one-sided profit
NFLVR	No nonzero nonnegative bounded claim lies in the norm closure of $\mathcal{C}$	Direct arbitrage and uniformly approximable free lunches

Check Your Understanding

Use the definitions and results above to answer these questions.

Why must the lower bound in admissibility apply at every time, rather than only to $G_T(H)$?
Show directly from the definition of $\mathcal{C}$ why an admissible arbitrage contradicts NFLVR.
For the Brownian exit gamble with $n=4$, compute the probability of the upper exit and verify the expected exit payoff.
Explain why a fixed finite horizon does not generally give the two-point exit payoff in the infinite-horizon example.
In the supermartingale proof, where is nonnegativity used, and why is continuity of $Y$ unnecessary?