Solution: Girsanov Failure — Novikov Condition Violation

Exercise: Girsanov Failure: Novikov Condition Violation

Part 1

With $\theta_t = \alpha W_t$ : $\int_0^T \theta_s^2\,ds = \alpha^2\int_0^T W_s^2\,ds$ . Novikov asks:

\mathbb{E}^{\mathbb{P}}\!\left[\exp\!\left(\tfrac{\alpha^2}{2}\int_0^T W_s^2\,ds\right)\right] \stackrel{?}{<} \infty.

The quantity $\int_0^T W_s^2\,ds$ is a weighted chi-squared-like sum. A classical result (via the Karhunen-Loève expansion of Brownian motion on $[0, T]$ ) is:

\mathbb{E}^{\mathbb{P}}\!\left[\exp\!\left(\lambda\int_0^T W_s^2\,ds\right)\right] = \frac{1}{\sqrt{\cos(\sqrt{2\lambda}\,T)}}, \quad \text{valid for } \sqrt{2\lambda}\,T < \pi/2.

(The formula comes from the Cameron-Martin formula / matrix-tree-type identity for quadratic forms of BM.)

Setting

\lambda = \alpha^2/2

: the Novikov integral is finite iff

\sqrt{\alpha^2}\cdot T = \alpha T < \pi/2

— i.e. $\alpha T < \pi/2$ .

Part 2

For

\alpha T \ge \pi/2

, the MGF of

\alpha^2\int_0^T W_s^2\,ds

1/2

diverges. So Novikov's condition fails when

\alpha T \ge \pi/2

Beyond this threshold, $Z_t$ is only a local martingale (not a true martingale); $\mathbb{E}^{\mathbb{P}}[Z_T] < 1$ ; the proposed $\mathbb{Q}$ is a sub-probability measure, and Girsanov's conclusion fails.

Part 3

Kazamaki's condition:

\mathbb{E}[\exp(\tfrac{1}{2}\int_0^t\theta_s\,dW_s)] < \infty

for all

t \le T

implies

Z_t

is a true martingale. This is strictly weaker than Novikov for some

\theta

. However, it can still fail for aggressive

\theta

's.

Qualitatively: Kazamaki recovers some cases on the boundary of Novikov's failure (the MGF of $\int\theta\,dW$ is finite but the MGF of $\int\theta^2\,ds$ is not — possible because Itô integrals have lower tails than their variance bounds). Both conditions are sufficient, not necessary — there are cases where both fail yet $Z_T$ still has $\mathbb{E}[Z_T] = 1$ . Standard workarounds include localisation by stopping times or direct computation of $\mathbb{E}[Z_T]$ .

Part 4

Heston and Novikov. In the Heston model, the market price of volatility risk

\theta^v_t

is often modelled as proportional to

\sqrt{v_t}

or to

v_t/\sqrt{v_t}

. Under certain parameter regimes (particularly high vol-of-vol

\xi

, low mean-reversion

\kappa

), the integral

\int_0^T (\theta^v_s)^2\,ds

can have infinite MGF — Novikov fails, and the risk-neutral measure is not uniquely well-defined on long horizons.

Practical consequences:

$\mathbb{Q}$ prices computed via Girsanov-based Monte Carlo may be biased; the measure is a sub-probability.
Long-dated options (swaptions, LTAM products) are particularly affected.
Numerical schemes that compute option prices by $\mathbb{E}^{\mathbb{Q}}[\text{payoff}]$ under the naive $\mathbb{Q}$ may converge to the wrong answer — a silent failure mode.
Production quant libraries (e.g. QuantLib, ORE) include horizon-dependent checks for martingale-defect issues, and some practitioners use alternative parameterisations or reduced horizons to avoid Novikov violation.

Bottom line: for long-dated derivatives in stochastic-vol models, always check that the chosen parameters keep you on the safe side of Novikov (or verify

\mathbb{E}^{\mathbb{P}}[Z_T] = 1

directly by simulation). Blind trust in

\mathbb{Q}

prices is dangerous.

Takeaways

Novikov is sufficient, not necessary. If it holds, you have a true martingale; if it fails, additional work is needed.
$Z_t$ can be a strict local martingale with $\mathbb{E}[Z_T] < 1$ . This corresponds to a sub-probability measure $\mathbb{Q}$ — pathological.
Stochastic-vol and heavy-drift models are where Novikov issues bite. In practice, check horizon-dependent criteria; don't assume Girsanov "just works."
Kazamaki and Beneš provide weaker sufficient conditions. Worth knowing but not panaceas. Always validate $\mathbb{E}[Z_T] = 1$ numerically when in doubt.