Exercise: Girsanov Failure — Novikov Condition Violation

Prerequisites: Girsanov's Theorem

Problem

Novikov's condition —

\mathbb{E}^{\mathbb{P}}[\exp(\tfrac{1}{2}\int_0^T\theta_s^2\,ds)] < \infty

— is sufficient for the Doléans-Dade exponential

Z_t

to be a true martingale (not merely a local martingale). When Novikov fails,

Z_t

can still be a strict local martingale with

\mathbb{E}[Z_T] < 1

, and then the proposed measure

\mathbb{Q}

from Girsanov is a sub-probability measure — not a valid probability measure.

Consider a fixed horizon $T$ and suppose $\theta_t = \alpha\cdot W_t$ for some $\alpha > 0$ , where $W_t$ is a $\mathbb{P}$ -BM. Check whether Novikov's condition holds. (Hint: compute $\int_0^T W_s^2\,ds$ 's distribution; you will find that $\mathbb{E}[\exp(\tfrac{\alpha^2}{2}\int_0^T W_s^2\,ds)]$ is an integral of an unbounded exponential against a gaussian — check whether it is finite.)
For what values of $\alpha T$ does $\mathbb{E}^{\mathbb{P}}[\exp(\tfrac{\alpha^2}{2}\int_0^T W_s^2\,ds)] = \infty$ ? Use the fact that $\int_0^T W_s^2\,ds$ has a known distribution related to chi-squared.
When Novikov fails, can we use a weaker sufficient condition (Kazamaki: $\mathbb{E}[\exp(\tfrac{1}{2}\int \theta\,dW)] < \infty$ ) to rescue the argument? Explain qualitatively.
Practical implication. In the Heston stochastic-volatility model $dv_t = \kappa(\theta_v - v_t)\,dt + \xi\sqrt{v_t}\,dW^v_t$ , certain parameter combinations can violate Novikov — the martingale measure is not well-defined on all horizons. What does this mean for using Heston for long-dated-options pricing? Would you blindly trust the $\mathbb{Q}$ prices?

Hint

Part 1's integral $\int_0^T W_s^2\,ds$ has mean $T^2/2$ and a distribution that, via the Cameron-Martin-like expansion, relates to the quadratic functional of Brownian motion. The key question is whether the MGF of this integral is finite at $\alpha^2/2$ .

Jump to the solution when you're ready.