Statistical Arbitrage

Pure arbitrage cannot encode a drift. It is invariant across an equivalence class of measures, so any equivalent martingale measure annihilates it. Real strategies lose on many paths yet cannot lose on average, a statement about the physical measure $\P$ rather than the pricing measure $\Q$ . This post makes that statement precise.

#Primitives

Fix $(\Omega,\Filt,(\Filt_t)_{t\ge 0},\P)$ under the usual conditions. Prices carry a positive numeraire $B_t$ , the money market $B_t=e^{rt}$ , and a semimartingale $S_t$ , with discounted price $\tilde S_t = S_t/B_t$ . A strategy is a predictable, $\tilde S$ -integrable process $H$ ; its discounted gain is the stochastic integral $v(t)=\int_0^t H_u\,d\tilde S_u$ , so $v(0)=0$ . Every expectation, variance, and probability is taken under $\P$ unless a measure is named.

#Arbitrage belongs to the equivalence class

Definition1

An arbitrage is a strategy with $v(0)=0$ , $v(T)\ge 0$ almost surely, and $\P(v(T)>0)>0$ .

Lemma2

If $\Q\sim\P$ , then $v$ is an arbitrage under $\P$ if and only if it is an arbitrage under $\Q$ .

Proof

Equivalence preserves the semimartingale property and the stochastic integral up to indistinguishability, so $v(T)$ is the same random variable under $\P$ and $\Q$ . Equivalence also gives identical null sets. The event $\{v(T)<0\}$ is $\P$ -null exactly when it is $\Q$ -null, so $v(T)\ge 0$ holds almost surely under either measure simultaneously, and $\{v(T)>0\}$ carries positive mass under one measure exactly when it does under the other. Both clauses of Definition 1 are preserved, and the drift of $S$ never appears.

By Lemma 2 arbitrage lives in the equivalence class $[\P]$ and is blind to expected return. The construction below exploits this blindness.

#The fundamental theorem, proved

Let $\Omega=\{\omega_1,\dots,\omega_K\}$ be finite with full support $\P(\omega_k)>0$ for every $k$ , and a single trading period spanning the dates $0$ and $T$ ; the multiperiod case iterates over the filtration. Collect discounted price increments in $G\in\R^{K\times d}$ , where $G_{kj}=\tilde S^{\,j}_T(\omega_k)-\tilde S^{\,j}_0$ . A portfolio $h\in\R^d$ has discounted payoff $Gh\in\R^K$ , and an arbitrage is an $h$ with $Gh\ge 0$ componentwise and $Gh\ne 0$ . No arbitrage is therefore exactly

\range(G)\cap\R^K_{\ge 0}=\{0\}. \tag{1}

Theorem3

(First Fundamental Theorem, finite $\Omega$ .) No arbitrage holds if and only if there exists $q\in\R^K$ with $q_k>0$ , $\sum_k q_k=1$ , and $q^\top G=0$ . The measure $\Q=(q_k)$ is then equivalent to $\P$ and renders discounted prices martingales, $\E_\Q[\tilde S_T-\tilde S_0]=0$ .

Proof

$(\Leftarrow)$ Given such $q>0$ , suppose $h$ is an arbitrage. Then $Gh\ge 0$ with one strict component, so $q^\top(Gh)>0$ , while $q^\top(Gh)=(q^\top G)h=0$ , a contradiction.

$(\Rightarrow)$ Assume no arbitrage. The range $C:=\range(G)$ is a linear subspace, hence closed and convex, and the unit simplex $\Delta=\{p\in\R^K_{\ge0}:\sum_k p_k=1\}$ is compact and convex. By Equation (1) the two are disjoint, since $C\cap\R^K_{\ge0}=\{0\}$ and $0\notin\Delta$ . Strict separation yields $\psi\neq 0$ and $\alpha$ with

\psi^\top c\;\le\;\alpha\;<\;\psi^\top p \qquad\text{for all } c\in C,\ p\in\Delta. \tag{2}

Since $C$ is a subspace, $\psi^\top c$ is linear and bounded above on $C$ , which forces $\psi^\top c=0$ for every $c\in C$ and hence $\alpha\ge 0$ . Evaluating the right inequality of Equation (2) at each vertex $p=e_k$ gives $\psi_k>0$ . Put $q=\psi/\sum_k\psi_k$ , so $q>0$ and $\sum_k q_k=1$ . Since $\P$ has full support, $q_k>0$ for every $k$ makes the null sets of $\Q$ and $\P$ coincide, so $\Q\sim\P$ . Orthogonality $\psi\perp C$ reads $\psi^\top(Gh)=0$ for all $h$ , that is $\psi^\top G=0$ , whence $q^\top G=0$ and $\E_\Q[\tilde S_T-\tilde S_0]=0$ [1], [2].

#The definition

Definition4

A statistical arbitrage is a strategy with $v(0)=0$ whose discounted cumulative profit, indexed by the completed trade count $n$ , satisfies [3]

\begin{aligned} &\text{(i)} &&\liminf_{n\to\infty}\E[v(n)]>0, \\[2pt] &\text{(ii)} &&\lim_{n\to\infty}\P\!\big(v(n)<0\big)=0, \\[2pt] &\text{(iii)} &&\lim_{n\to\infty}\frac{\Var[v(n)]}{n^{2}}=0. \end{aligned} \tag{3}

The index $n$ runs over the disjoint trade-completion times, so the conditions are limits along that discrete sequence.

Condition (iii) says the variance of the time-averaged profit $v(n)/n$ vanishes. Condition (i) is an expectation under $\P$ , a real-world drift rather than a $\Q$ -mispricing, and that is the source of the word statistical.

#Sufficiency

Proposition5

Let disjoint trades produce discounted increments $Y_1,Y_2,\dots$ i.i.d. with $\E[Y_i]=\mu>0$ and $\Var(Y_i)=\sigma^2\in(0,\infty)$ . Then $v(n)=\sum_{i=1}^n Y_i$ is a statistical arbitrage.

Proof

Write $\bar Y_n=v(n)/n$ , with mean $\mu$ and variance $\sigma^2/n$ . Condition (i) holds since $\E[v(n)]=n\mu\to\infty$ , so $\liminf_n\E[v(n)]=\infty>0$ . For condition (ii), Chebyshev gives

\P\!\big(v(n)<0\big)=\P\!\big(\bar Y_n-\mu<-\mu\big)\le\P\!\big(\lvert\bar Y_n-\mu\rvert\ge\mu\big)\le\frac{\sigma^2}{n\mu^2}, \tag{4}

whose right side vanishes. Condition (iii) holds since $\Var[v(n)]/n^2=\sigma^2/n$ vanishes. Every clause of Definition 4 is met.

The bound Equation (4) is the entire engine. Any increments with positive mean and finite variance inherit the conclusion of Proposition 5.

#The hierarchy is strict

Proposition6

A repeatable, square-integrable arbitrage run on disjoint windows is a statistical arbitrage, and the converse fails.

Proof

A repeatable, square-integrable arbitrage furnishes i.i.d. increments with $Y_i\ge 0$ , $\E[Y_i]>0$ , and $\Var(Y_i)<\infty$ . Since $v(n)\ge 0$ surely, condition (ii) holds at once; condition (i) holds because $\E[v(n)]=n\,\E[Y_1]>0$ ; and condition (iii) holds because $\Var[v(n)]/n^2=\Var(Y_1)/n$ vanishes. For the converse take $Y_i\sim\mathcal N(\mu,\sigma^2)$ with $\mu>0$ . Here $\P(Y_i<0)>0$ , so $v$ fails the almost-sure nonnegativity of Definition 1 and is no arbitrage, yet Proposition 5 certifies it as a statistical arbitrage.

#The word statistical and the role of the physical measure

Remark

Statistical arbitrage, unlike Lemma 2, is not invariant under an equivalent change of measure. Condition (i) demands $\mu=\E_\P[Y_i]>0$ , and passing to the martingale measure $\Q$ of Theorem 3 alters the drift. Assume $H$ is $\Q$ -admissible, so that $v$ is a true $\Q$ -martingale rather than a strict local one; then under $\Q$ the discounted profit satisfies $\E_\Q[v(t)]=\E_\Q[v(0)]=0$ , and condition (i) fails. An arbitrage-free market, one carrying an equivalent martingale measure, may therefore still admit statistical arbitrage under $\P$ . The drift the pricing measure discards is precisely what separates the two notions.

#Numerical illustration

Simulating i.i.d. Gaussian increments with positive mean makes Equation (3) visible. The mean grows linearly, the loss probability collapses inside the Equation (4) envelope, and the averaged variance vanishes.

import numpy as np
from numpy.random import Generator


def loss_probability_envelope(mu: float, sigma: float, n: int) -> np.ndarray:
    """Chebyshev upper bound on the loss probability of cumulative profit.

    Args:
        mu: Per-trade expected discounted profit, strictly positive.
        sigma: Per-trade profit standard deviation.
        n: Horizon in number of trades.

    Returns:
        The bound sigma**2 / (k * mu**2) for k = 1, ..., n.
    """
    horizon = np.arange(1, n + 1)
    return sigma**2 / (horizon * mu**2)


def simulate_diagnostics(
    mu: float, sigma: float, n: int, paths: int, rng: Generator
) -> dict[str, np.ndarray]:
    """Monte Carlo diagnostics for the three limiting conditions.

    Args:
        mu: Per-trade expected discounted profit, strictly positive.
        sigma: Per-trade profit standard deviation.
        n: Horizon in number of trades.
        paths: Number of independent profit trajectories.
        rng: Seeded generator for reproducibility.

    Returns:
        Mapping of the cumulative-profit mean, empirical loss probability, and
        time-averaged variance against the horizon.
    """
    increments = rng.normal(mu, sigma, size=(paths, n))
    v = np.cumsum(increments, axis=1)
    horizon = np.arange(1, n + 1)
    return {
        "mean": v.mean(axis=0),
        "loss_probability": (v < 0).mean(axis=0),
        "averaged_variance": v.var(axis=0) / horizon**2,
    }


rng = np.random.default_rng(0)
diagnostics = simulate_diagnostics(mu=0.05, sigma=1.0, n=2_000, paths=20_000, rng=rng)
envelope = loss_probability_envelope(mu=0.05, sigma=1.0, n=2_000)

The construction is complete. A market may satisfy the fundamental theorem yet still admit a strategy meeting Definition 4, separated from arbitrage by a drift invisible to every equivalent martingale measure.

[1]

J. M. Harrison and S. R. Pliska, “Martingales and stochastic integrals in the theory of continuous trading,” Stochastic Processes and their Applications, vol. 11, no. 3, pp. 215–260, 1981.

[2]

H. Föllmer and A. Schied, Stochastic Finance: An Introduction in Discrete Time, 4th ed. De Gruyter, 2016.

[3]

S. Hogan, R. Jarrow, M. Teo, and M. Warachka, “Testing market efficiency using statistical arbitrage with applications to momentum and value strategies,” Journal of Financial Economics, vol. 73, no. 3, pp. 525–565, 2004.

Explore connections

see in the atlas

referenced by (1)

The Mean-Variance Portfolio

cite

@misc{statistical-arbitrage,
  author = {Zac Kienzle},
  title  = {Statistical Arbitrage},
  year   = {2026},
  month  = {06},
  url    = {https://zackienzle.com/blog/statistical-arbitrage}
}

#Primitives

#Arbitrage belongs to the equivalence class

Definition1

An arbitrage is a strategy with $v(0)=0$ , $v(T)\ge 0$ almost surely, and $\P(v(T)>0)>0$ .

Lemma2

If $\Q\sim\P$ , then $v$ is an arbitrage under $\P$ if and only if it is an arbitrage under $\Q$ .

Proof

By Lemma 2 arbitrage lives in the equivalence class $[\P]$ and is blind to expected return. The construction below exploits this blindness.

#The fundamental theorem, proved

\range(G)\cap\R^K_{\ge 0}=\{0\}. \tag{1}

Theorem3

Proof

$(\Leftarrow)$ Given such $q>0$ , suppose $h$ is an arbitrage. Then $Gh\ge 0$ with one strict component, so $q^\top(Gh)>0$ , while $q^\top(Gh)=(q^\top G)h=0$ , a contradiction.

\psi^\top c\;\le\;\alpha\;<\;\psi^\top p \qquad\text{for all } c\in C,\ p\in\Delta. \tag{2}

#The definition

Definition4

A statistical arbitrage is a strategy with $v(0)=0$ whose discounted cumulative profit, indexed by the completed trade count $n$ , satisfies [3]

\begin{aligned} &\text{(i)} &&\liminf_{n\to\infty}\E[v(n)]>0, \\[2pt] &\text{(ii)} &&\lim_{n\to\infty}\P\!\big(v(n)<0\big)=0, \\[2pt] &\text{(iii)} &&\lim_{n\to\infty}\frac{\Var[v(n)]}{n^{2}}=0. \end{aligned} \tag{3}

The index $n$ runs over the disjoint trade-completion times, so the conditions are limits along that discrete sequence.

#Sufficiency

Proposition5

Let disjoint trades produce discounted increments $Y_1,Y_2,\dots$ i.i.d. with $\E[Y_i]=\mu>0$ and $\Var(Y_i)=\sigma^2\in(0,\infty)$ . Then $v(n)=\sum_{i=1}^n Y_i$ is a statistical arbitrage.

Proof

Write $\bar Y_n=v(n)/n$ , with mean $\mu$ and variance $\sigma^2/n$ . Condition (i) holds since $\E[v(n)]=n\mu\to\infty$ , so $\liminf_n\E[v(n)]=\infty>0$ . For condition (ii), Chebyshev gives

\P\!\big(v(n)<0\big)=\P\!\big(\bar Y_n-\mu<-\mu\big)\le\P\!\big(\lvert\bar Y_n-\mu\rvert\ge\mu\big)\le\frac{\sigma^2}{n\mu^2}, \tag{4}

whose right side vanishes. Condition (iii) holds since $\Var[v(n)]/n^2=\sigma^2/n$ vanishes. Every clause of Definition 4 is met.

The bound Equation (4) is the entire engine. Any increments with positive mean and finite variance inherit the conclusion of Proposition 5.

#The hierarchy is strict

Proposition6

A repeatable, square-integrable arbitrage run on disjoint windows is a statistical arbitrage, and the converse fails.

Proof

#The word statistical and the role of the physical measure

Remark

#Numerical illustration

import numpy as np
from numpy.random import Generator


def loss_probability_envelope(mu: float, sigma: float, n: int) -> np.ndarray:
    """Chebyshev upper bound on the loss probability of cumulative profit.

    Args:
        mu: Per-trade expected discounted profit, strictly positive.
        sigma: Per-trade profit standard deviation.
        n: Horizon in number of trades.

    Returns:
        The bound sigma**2 / (k * mu**2) for k = 1, ..., n.
    """
    horizon = np.arange(1, n + 1)
    return sigma**2 / (horizon * mu**2)


def simulate_diagnostics(
    mu: float, sigma: float, n: int, paths: int, rng: Generator
) -> dict[str, np.ndarray]:
    """Monte Carlo diagnostics for the three limiting conditions.

    Args:
        mu: Per-trade expected discounted profit, strictly positive.
        sigma: Per-trade profit standard deviation.
        n: Horizon in number of trades.
        paths: Number of independent profit trajectories.
        rng: Seeded generator for reproducibility.

    Returns:
        Mapping of the cumulative-profit mean, empirical loss probability, and
        time-averaged variance against the horizon.
    """
    increments = rng.normal(mu, sigma, size=(paths, n))
    v = np.cumsum(increments, axis=1)
    horizon = np.arange(1, n + 1)
    return {
        "mean": v.mean(axis=0),
        "loss_probability": (v < 0).mean(axis=0),
        "averaged_variance": v.var(axis=0) / horizon**2,
    }


rng = np.random.default_rng(0)
diagnostics = simulate_diagnostics(mu=0.05, sigma=1.0, n=2_000, paths=20_000, rng=rng)
envelope = loss_probability_envelope(mu=0.05, sigma=1.0, n=2_000)

[1]

J. M. Harrison and S. R. Pliska, “Martingales and stochastic integrals in the theory of continuous trading,” Stochastic Processes and their Applications, vol. 11, no. 3, pp. 215–260, 1981.

[2]

H. Föllmer and A. Schied, Stochastic Finance: An Introduction in Discrete Time, 4th ed. De Gruyter, 2016.

[3]

Explore connections

see in the atlas

referenced by (1)

The Mean-Variance Portfolio

cite

@misc{statistical-arbitrage,
  author = {Zac Kienzle},
  title  = {Statistical Arbitrage},
  year   = {2026},
  month  = {06},
  url    = {https://zackienzle.com/blog/statistical-arbitrage}
}