Novikov's Condition and Its Direct Tests

Why an Exponential-Integrability Test Is Useful

A stochastic exponential is always a local martingale under suitable local square-integrability assumptions, but a local martingale need not be a true martingale. Novikov's condition gives a convenient sufficient test: it controls the exponential of the accumulated squared integrand, rather than requiring a direct computation of the stochastic exponential's expectation. The test is useful precisely because that direct computation is often difficult.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a Brownian motion $B_t$, whose increments after time $s$ are independent of $\mathcal{F}_s$. For a progressively measurable process $H_t$, the Itô integral $\int_0^t H_s\,dB_s$ is an $L^2$ integral when $\mathbb{E}\int_0^T H_s^2\,ds<\infty$; more generally, it is defined locally when $\int_0^T H_s^2\,ds<\infty$ almost surely. The latter assumption will be made whenever we define a stochastic exponential. The stochastic integral in its differential need not be globally $L^2$.

Definition: Given a progressively measurable $H$ with $\int_0^T H_s^2\,ds<\infty$ almost surely, its Itô stochastic exponential on $[0,T]$ is $$ Z_t=\exp\left(\int_0^t H_s\,dB_s-\frac12\int_0^t H_s^2\,ds\right),\qquad 0\le t\le T. $$ It is a positive local martingale and satisfies $dZ_t=Z_tH_t\,dB_t$ in the local Itô sense. In this expression, the integral defining $Z$ uses $H$; the differential uses $Z H$, which need only be locally square-integrable. The stochastic exponential is an Itô object: its half-quadratic-variation term reflects $dB_t\,dB_t=dt$, rather than an ordinary chain rule or a Stratonovich integral.

Novikov's Condition

Theorem (Novikov's criterion): Suppose $H$ is progressively measurable and $\int_0^T H_s^2\,ds<\infty$ almost surely. If $$ \mathbb{E}\left[\exp\left(\frac12\int_0^T H_s^2\,ds\right)\right]<\infty, $$ then $Z_t$ is a true martingale on $[0,T]$. In particular, $Z_t$ is integrable and $\mathbb{E}[Z_t\mid\mathcal{F}_s]=Z_s$ for $0\le s\le t\le T$, so $\mathbb{E}[Z_t]=1$. This is a standard result; its proof by localization is given in the subsequent treatment of the criterion.

The condition is imposed under $\mathbb{P}$, and the expectation in it is also under $\mathbb{P}$. The criterion is sufficient, not necessary: if the exponential moment is infinite, the theorem gives no conclusion. In particular, failure of the test does not by itself show that $Z$ is a strict local martingale. Nor does the criterion alone make a change to a risk-neutral measure $\mathbb{Q}$; measure changes and the resulting Brownian motion are separate consequences of Girsanov's theorem.

Direct Tests

Worked Example: A Uniform Bound on the Integrand

Suppose $T>0$, $H$ is progressively measurable, and $|H_s|\le c$ for almost every $s\in[0,T]$, almost surely, where $c\ge0$ is a deterministic constant. Check Novikov's condition.

For almost every outcome, the pointwise bound gives $$ \int_0^T H_s^2\,ds\le \int_0^T c^2\,ds=c^2T. $$ Exponentiating and taking expectations yields $$ \mathbb{E}\left[\exp\left(\frac12\int_0^T H_s^2\,ds\right)\right]\le \exp\left(\frac12c^2T\right)<\infty. $$ Thus Novikov's criterion applies, and $Z$ is a true martingale with $\mathbb{E}[Z_t]=1$ and the stated conditional-expectation property. This includes the degenerate case $c=0$: then $H=0$ almost everywhere, $Z_t=1$, and the bound equals $1$.

Check: the estimate controls the accumulated squared integrand, not the random variable $\int_0^T H_s\,dB_s$ by a pointwise bound. For every allowed $H$, the conclusion follows from the same deterministic upper bound $c^2T$.

Bounded-integrand consequence. Under the assumptions of the example, the stochastic exponential is a true martingale. This conclusion follows from the displayed bound and Novikov's criterion; the criterion supplies both integrability and the conditional-expectation identity, not merely the local-martingale property.

Worked Example: A Deterministic Integrand

Let $T=2$ and $H_t=t$. Compute the Novikov expectation and verify the martingale property directly.

The accumulated squared integrand up to time $t$ is $$ A_t=\int_0^t s^2\,ds=\frac{t^3}{3}. $$ In particular, $\frac12 A_2=\frac43$, so the Novikov expectation is the finite deterministic number $$ \mathbb{E}\left[\exp\left(\frac12\int_0^2 s^2\,ds\right)\right]=e^{4/3}. $$ The criterion therefore applies. We can also verify the martingale property directly. For $0\le s\le t\le2$, write $$ Z_t=Z_s\exp\left(\int_s^t u\,dB_u-\frac12\int_s^t u^2\,du\right). $$ The future integral is independent of $\mathcal{F}_s$ and is normal with mean $0$ and variance $v=\int_s^t u^2\,du=(t^3-s^3)/3$. If $G$ is normal with mean $0$ and variance $v$, its moment-generating function gives $\mathbb{E}[e^{G-v/2}]=e^{v/2-v/2}=1$. Hence $$ \mathbb{E}[Z_t\mid\mathcal{F}_s]=Z_s. $$ This conditional expectation is legitimate: $Z_s$ is integrable, since the same Gaussian calculation gives $\mathbb{E}[Z_s]=1$, and the independent future factor is integrable with mean $1$. Thus $Z$ is a true martingale, verified independently of the criterion.

Check: at $t=2$, the variance of $\int_0^2 u\,dB_u$ is $8/3$, and the exponential subtracts half this variance, $4/3$. Its expectation is therefore $e^{(8/3)/2-4/3}=1$, in agreement with the conditional calculation.

A Gaussian Random Integrand

Worked Example: A Brownian Value Held as the Integrand

On $[0,2]$, fix a real constant $\alpha$ and define $H_t=0$ for $0\le t\le1$, and $H_t=\alpha B_1$ for $1<t\le2$. This process is progressively measurable: after time $1$, the random variable $B_1$ is already known. Find exactly when Novikov's condition holds.

The integrated squared integrand is $$ \int_0^2 H_s^2\,ds=\alpha^2B_1^2, $$ because the interval on which the integrand is nonzero has length $1$. Since $B_1$ is standard normal, integration against its density gives $$ \mathbb{E}\left[e^{\alpha^2B_1^2/2}\right] =\frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}e^{-(1-\alpha^2)x^2/2}\,dx = \begin{cases} (1-\alpha^2)^{-1/2},&|\alpha|<1,\\ \infty,&|\alpha|\ge1. \end{cases} $$ For $|\alpha|<1$, the Gaussian integral is finite, so Novikov's criterion guarantees that $Z$ is a true martingale.

For this particular example, the martingale property can be checked even when $|\alpha|\ge1$. For $t\le1$, $Z_t=1$. For $t>1$, $$ Z_t=\exp\left(\alpha B_1(B_t-B_1)-\frac12\alpha^2B_1^2(t-1)\right). $$ Conditional on $\mathcal{F}_1$, the increment $B_t-B_1$ is normal with variance $t-1$. The Gaussian moment-generating function therefore gives $\mathbb{E}[Z_t\mid\mathcal{F}_1]=1$. If $1\le s\le t$, then, conditional on $\mathcal{F}_s$, $B_1$ and $Z_s$ are known and $B_t-B_s$ is an independent normal increment. Applying the same calculation to the factor $Z_t/Z_s$ gives $\mathbb{E}[Z_t\mid\mathcal{F}_s]=Z_s$. If $s<1<t$, the tower property and $\mathbb{E}[Z_t\mid\mathcal{F}_1]=1$ give $\mathbb{E}[Z_t\mid\mathcal{F}_s]=1=Z_s$. In particular, $Z_t$ is integrable and has expectation $1$ for every $t\le2$. Thus it is a true martingale even for $|\alpha|\ge1$; Novikov's condition is not necessary.

Check: the example satisfies the global $L^2$ condition on $H$, because $$ \mathbb{E}\int_0^2 H_s^2\,ds=\alpha^2\mathbb{E}[B_1^2]=\alpha^2<\infty. $$ That does not imply that the integrand $Z_sH_s$ in the differential $dZ_s=Z_sH_s\,dB_s$ is globally $L^2$. For $\alpha=1$ and $s=1+u$, where $0<u\le1$, conditioning on $\mathcal{F}_1$ gives $$ \mathbb{E}[Z_s^2H_s^2\mid\mathcal{F}_1] =B_1^2\,\mathbb{E}\left[e^{2B_1(B_s-B_1)-B_1^2u}\mid\mathcal{F}_1\right] =B_1^2e^{uB_1^2}. $$ The expectation of $B_1^2e^{uB_1^2}$ is infinite for $u\ge\tfrac12$, as is seen by multiplying by the normal density: the resulting integrand has a non-decaying or increasing Gaussian exponent. Tonelli's theorem then implies $\mathbb{E}\int_1^2 Z_s^2H_s^2\,ds=\infty$. The differential remains meaningful as a locally square-integrable Itô integral; it is not a globally $L^2$ Itô integral.

How to Use the Test—and a Common Mistake

For a proposed $H$, first verify progressive measurability and almost-sure finiteness of $\int_0^T H_s^2\,ds$, so the stochastic exponential is defined. Then calculate or bound the single quantity $\mathbb{E}[\exp(\tfrac12\int_0^T H_s^2\,ds)]$. If it is finite, Novikov guarantees a true martingale. If it is infinite, stop there: the criterion has not decided the martingale question. A different argument may still establish the true-martingale property, as the Gaussian-integrand example demonstrates.

Common mistake: From $\mathbb{E}\exp(\tfrac12\int_0^T H_s^2\,ds)=\infty$, one might incorrectly conclude that $Z$ is not a true martingale. For $\alpha=1$ in the preceding example, this expectation is $\mathbb{E}e^{B_1^2/2}=\infty$, but the conditional Gaussian calculation gives $\mathbb{E}[Z_t\mid\mathcal{F}_s]=Z_s$ for every $s\le t\le2$. The correction is logical as well as computational: Novikov gives “finite exponential moment implies true martingale,” not an if-and-only-if characterization.

Another error is to treat the stochastic integral in $dZ_t=Z_tH_t\,dB_t$ as automatically belonging to the global $L^2$ class merely because $H$ does. The example shows the distinction explicitly. The exponential's differential is interpreted locally unless a separate estimate proves $\mathbb{E}\int_0^T Z_s^2H_s^2\,ds<\infty$.

Check Your Understanding

In each question, distinguish what Novikov's condition proves from what would require a separate argument.

If $|H_s|\le c$ almost everywhere on $[0,T]$, what finite upper bound does this give for the Novikov expectation?
For the deterministic integrand $H_t=t$ on $[0,2]$, compute $\int_0^2 H_s^2\,ds$ and the resulting Novikov expectation.
For the Gaussian-integrand example, for which values of $\alpha$ is the Novikov expectation finite?
When Novikov's expectation is infinite, what conclusion does the criterion permit—and what conclusion does it not permit?
Why does $\mathbb{E}\int_0^T H_s^2\,ds<\infty$ not by itself establish that $Z_sH_s$ is globally square-integrable?