Novikov's Criterion for Exponential Martingales

From a Local Martingale to a True Martingale

Exponentiating a Brownian stochastic integral produces a process that is always a local martingale under suitable integrability assumptions, but a local martingale need not be a true martingale. The distinction matters whenever an argument uses expectations or conditional expectations: a nonnegative local martingale has expectation at most its initial value, and equality is not automatic. Novikov’s criterion supplies a useful sufficient condition for equality.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a standard Brownian motion $B_t$, with the filtration satisfying the usual conditions. Fix $T>0$. Let $\theta_s$ be progressively measurable and suppose $\mathbb{E}\int_0^T\theta_s^2\,ds<\infty$. Thus $\theta$ belongs to the class $\mathcal{L}^2$ on $[0,T]$, and its Ito integral is well-defined and square-integrable.

Definition: Set $M_t=\int_0^t\theta_s\,dB_s$ and $A_t=\int_0^t\theta_s^2\,ds$. The stochastic exponential associated with $\theta$ is $Z_t=\exp(M_t-\tfrac12 A_t)$, for $0\le t\le T$. The process $A_t$ is the quadratic variation $[M]_t$. Ito’s formula, with $f(x)=e^x$, $f'(x)=f''(x)=e^x$, and the correction $\tfrac12 f''(x)\,dt$, gives $dZ_t=Z_t\theta_t\,dB_t$; the $dt$ terms cancel because $dM_t\,dM_t=\theta_t^2dt$.

This stochastic differential shows that $Z$ is a nonnegative local martingale with $Z_0=1$. It does not by itself show that $Z$ is a martingale. Here “local martingale” means that there are stopping times increasing to infinity (or to the time horizon) for which the stopped processes are martingales. A nonnegative local martingale is a supermartingale, so its expectation can decrease. The technical issue is whether localization loses expectation in the limit.

Novikov’s Criterion

Theorem (Novikov’s criterion): Suppose $\theta$ is progressively measurable and $\mathbb{E}\int_0^T\theta_s^2\,ds<\infty$. If $$ \mathbb{E}\!\left[\exp\!\left(\frac12\int_0^T\theta_s^2\,ds\right)\right]<\infty, $$ then $Z_t=\exp(\int_0^t\theta_s\,dB_s-\tfrac12\int_0^t\theta_s^2\,ds)$ is a true martingale on $[0,T]$. In particular, $\mathbb{E}[Z_t]=1$ for every $t\in[0,T]$.

Proof status. Novikov’s criterion is a standard theorem, and its proof relies on a technical localization estimate that controls the expectation of the stopped exponentials using the exponential moment of $A_T/2$. That estimate is not derived here. In particular, the theorem above is being invoked as a named standard result, not deduced from the claim that a stopped exponential has expectation one. The latter claim is precisely the issue that the criterion resolves.

We can prove directly the martingale consequence once the criterion gives the constant expectations. A nonnegative local martingale is a supermartingale: apply conditional Fatou’s lemma to a localizing sequence of stopped martingales to obtain $\mathbb{E}[Z_t\mid\mathcal{F}_s]\le Z_s$ for $s\le t$. Novikov’s criterion applies on every interval $[0,t]$, since $A_t\le A_T$, so $\mathbb{E}[Z_t]=1=\mathbb{E}[Z_s]$. The nonnegative random variable $Z_s-\mathbb{E}[Z_t\mid\mathcal{F}_s]$ therefore has expectation zero, and hence is zero almost surely. Thus $\mathbb{E}[Z_t\mid\mathcal{F}_s]=Z_s$, which is the martingale property. $\square$

The criterion is sufficient, not necessary. Its hypothesis is an exponential-moment bound on the accumulated squared integrand, rather than merely a bound on its mean. In applications, one checks the stated exponential expectation and then applies the theorem; one must not infer the result just from the local-martingale differential.

Worked Example: A Bounded Integrand

Suppose $|\theta_s|\le 3$ almost surely for all $s\in[0,2]$, with $\theta$ progressively measurable. Verify Novikov’s condition and conclude that the associated stochastic exponential is a martingale.

For every outcome, $$ A_2=\int_0^2\theta_s^2\,ds\le\int_0^2 9\,ds=18. $$ Therefore $\exp(A_2/2)\le e^9$, and taking expectations gives $\mathbb{E}[\exp(A_2/2)]\le e^9<\infty$. The bound also gives $\mathbb{E}\int_0^2\theta_s^2\,ds\le18$, so the Ito integral is in $\mathcal{L}^2$. Novikov’s criterion applies, and $Z_t$ is a true martingale for $0\le t\le2$.

Check: at the largest permitted squared-integrand bound, $A_2=18$, the criterion’s exponential is $e^{18/2}=e^9$, exactly the finite upper bound used above. No distributional assumption on $\theta$ is needed.

Worked Example: An Unbounded Integrand Near the Initial Time

Let $T=1$ and $\theta_s=s^{-1/4}$ for $0<s\le1$, with any value assigned at $s=0$. This deterministic integrand is unbounded near zero. Verify the required integrability and Novikov’s condition.

Since $\theta_s^2=s^{-1/2}$, $$ A_1=\int_0^1s^{-1/2}\,ds=\left[2s^{1/2}\right]_0^1=2. $$ In particular, $\mathbb{E}\int_0^1\theta_s^2\,ds=2<\infty$, so $\theta\in\mathcal{L}^2$. The accumulated squared integrand is deterministic, and $$ \mathbb{E}\!\left[e^{A_1/2}\right]=e^{2/2}=e<\infty. $$ Novikov’s criterion therefore implies that $Z_t$ is a true martingale on $[0,1]$.

Check: the singularity at zero does not make the square integral diverge: the antiderivative $2s^{1/2}$ tends to zero as $s$ decreases to zero. The criterion depends on the integral of $\theta_s^2$, not on boundedness of $\theta$ itself.

A Random Unbounded Integrand

Worked Example: A Gaussian Initial Value

Let $Y$ be $\mathcal{F}_0$-measurable with distribution $N(0,1/4)$, independent of the Brownian motion, and set $\theta_s=Y$ for $0\le s\le1$. Verify Novikov’s condition. The filtration contains $Y$ at time zero and is such that $B_t$ remains Brownian motion relative to it.

Here $A_1=Y^2$, so $\mathbb{E}\int_0^1\theta_s^2\,ds=\mathbb{E}[Y^2]=1/4<\infty$. For a centered normal random variable with variance $v$, direct integration of its density gives $$ \mathbb{E}[e^{cY^2}]=\frac{1}{\sqrt{1-2vc}}\quad\text{when }2vc<1. $$ Indeed, the density is proportional to $e^{-y^2/(2v)}$; multiplying by $e^{cy^2}$ leaves a Gaussian integral with positive quadratic coefficient $1/(2v)-c$, which is finite exactly when $2vc<1$. In this example, $v=1/4$ and $c=1/2$, so $2vc=1/4<1$, and $$ \mathbb{E}[e^{A_1/2}]=\mathbb{E}[e^{Y^2/2}]=\frac{1}{\sqrt{1-1/4}}=\frac{2}{\sqrt{3}}<\infty. $$ Thus Novikov’s criterion applies. The integrand is unbounded as a random variable, but its exponential square-integral moment is finite.

Check: substituting $v=1/4$ and $c=1/2$ gives $2vc=2(1/4)(1/2)=1/4$, so the denominator is $\sqrt{1-1/4}=\sqrt{3}/2$, and its reciprocal is $2/\sqrt{3}$.

Why the Criterion Is Not Necessary

The previous example also illustrates why Novikov’s condition is a sufficient test rather than a characterization. Keep $Y$ independent of $B$, but instead let $Y\sim N(0,2)$ and again set $\theta_s=Y$ on $[0,1]$. The integrand is still in $\mathcal{L}^2$, since $\mathbb{E}\int_0^1\theta_s^2\,ds=\mathbb{E}[Y^2]=2$. However, the exponential moment $\mathbb{E}[e^{Y^2/2}]$ is infinite: the Gaussian integral above has $2vc=2$, which is not less than $1$.

Nevertheless, in this particular example the stochastic exponential is a true martingale. Conditional on $\mathcal{F}_0$, the value $Y$ is fixed and the Brownian increment $B_t$ is independent of $\mathcal{F}_0$. For $0\le s\le t\le1$, $$ Z_t=Z_s\exp\!\left(Y(B_t-B_s)-\frac12Y^2(t-s)\right). $$ The conditional normal moment-generating function gives $$ \mathbb{E}\!\left[\left.\exp\!\left(Y(B_t-B_s)-\frac12Y^2(t-s)\right)\right|\mathcal{F}_s\right]=1. $$ Consequently $\mathbb{E}[Z_t\mid\mathcal{F}_s]=Z_s$. Also $Z_t$ is integrable: conditioning first on $Y$, the normal moment-generating function gives $\mathbb{E}[Z_t\mid\mathcal{F}_0]=1$, and hence $\mathbb{E}[Z_t]=1$. This direct verification does not use Novikov’s criterion.

A Common Mistake: Replacing an Exponential Moment by a Mean

A tempting but incorrect shortcut is to check only that $\mathbb{E}[A_T]$ is finite and then treat this as if it implied $\mathbb{E}[e^{A_T/2}]<\infty$. Finiteness of the mean does not control an exponential moment. For the $Y\sim N(0,2)$ example, $\mathbb{E}[A_1]=2$ is finite, while $\mathbb{E}[e^{A_1/2}]=\infty$. The correct calculation tests the expectation of the exponential itself.

A related error is to write $\mathbb{E}[e^X]=e^{\mathbb{E}[X]}$. In general this identity is false. For $X=Y^2/2$ in the last example, $\mathbb{E}[X]=1$, so $e^{\mathbb{E}[X]}=e$ is finite, but $\mathbb{E}[e^X]=\infty$. The criterion requires the latter quantity, computed under $\mathbb{P}$, to be finite.

Takeaway: First establish that $\theta$ is progressively measurable and square-integrable on the stated horizon. Then check the precise condition $\mathbb{E}[\exp(A_T/2)]<\infty$. Novikov’s theorem implies that the stochastic exponential is a true martingale; failure of the condition means only that this sufficient test does not apply.

Check Your Understanding

For each question, distinguish the exponential-moment hypothesis from ordinary square-integrability.

If $|\theta_s|\le C$ on $[0,T]$ almost surely, give an explicit finite upper bound for $\mathbb{E}[\exp(A_T/2)]$.
For deterministic $\theta_s=s^{-\alpha}$ on $(0,1]$, which values of $\alpha$ make $\int_0^1\theta_s^2\,ds$ finite?
For $Y\sim N(0,v)$, state the condition on $v$ under which $\mathbb{E}[e^{Y^2/2}]$ is finite, and evaluate the expectation when $v=1/3$.
Why does a nonnegative local martingale not automatically have constant expectation?
Does failure of Novikov’s condition prove that the stochastic exponential is a strict local martingale? Explain.