Novikov and Kazamaki Conditions¶

In the unifying framework of this section, the Novikov and Kazamaki conditions are the technical guarantees that control is valid — they ensure the stochastic exponential is a true martingale, so that the Girsanov measure change produces a well-defined probability measure.

When does a stochastic exponential define a valid probability measure? The Novikov condition and Kazamaki condition provide sufficient conditions for the stochastic exponential to be a true martingale, ensuring that measure changes are well-defined.

Prerequisites

This section assumes familiarity with:

The Problem¶

Consider the stochastic exponential of a continuous local martingale \(M_t = \int_0^t \theta_s\,dW_s\):

\[ Z_t = \mathcal{E}(M)_t = \exp\left(M_t - \frac{1}{2}\langle M \rangle_t\right) = \exp\left(\int_0^t \theta_s\,dW_s - \frac{1}{2}\int_0^t \theta_s^2\,ds\right) \]

In short, this determines whether \(\mathbb{Q}\) exists as a valid equivalent measure. We want to use \(Z_T\) as a Radon–Nikodym derivative:

\[ \frac{d\mathbb{Q}}{d\mathbb{P}}\bigg|_{\mathcal{F}_T} = Z_T \]

For this to be valid, we need:

\(Z_t \geq 0\) (automatically satisfied by exponential)
\(\mathbb{E}^{\mathbb{P}}[Z_T] = 1\) (normalization)

The second condition is not automatic for local martingales!

When Stochastic Exponentials Fail¶

The Issue¶

By Itô's formula, \(Z_t\) satisfies \(dZ_t = Z_t \theta_t\,dW_t\), which shows \(Z_t\) is always a local martingale.

But local martingales can have \(\mathbb{E}[Z_T] < Z_0 = 1\) due to:

Mass escaping to infinity
The integrand \(\theta\) being too large too often

Example of Failure¶

Consider \(\theta_t = \frac{c}{\sqrt{T-t}}\) for \(t < T\), where \(c > 0\) is a constant. Then:

\[ \int_0^T \theta_s^2\,ds = \int_0^T \frac{c^2}{T-s}\,ds = +\infty \]

The quadratic variation explodes, and Novikov's condition fails. More precisely, for any \(\epsilon > 0\):

\[ \int_0^{T-\epsilon} \theta_s^2\,ds = c^2 \log\left(\frac{T}{\epsilon}\right) \to \infty \quad \text{as } \epsilon \to 0 \]

The stochastic exponential \(Z_t\) converges to 0 as \(t \to T\), and \(\mathbb{E}[Z_T] < 1\).

Intuition: The integrand \(\theta\) blows up near the terminal time, causing the local martingale to "leak mass to infinity."

Novikov's Condition¶

Theorem (Novikov, 1972): Let \(M_t = \int_0^t \theta_s\,dW_s\) be a continuous local martingale. If

\[ \boxed{ \mathbb{E}^{\mathbb{P}}\left[\exp\left(\frac{1}{2}\langle M \rangle_T\right)\right] = \mathbb{E}^{\mathbb{P}}\left[\exp\left(\frac{1}{2}\int_0^T \theta_s^2\,ds\right)\right] < \infty } \]

then \(Z_t = \mathcal{E}(M)_t\) is a true martingale on \([0,T]\), and in particular:

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

Interpretation¶

Novikov's condition requires that the "energy" \(\langle M \rangle_T = \int_0^T \theta_s^2\,ds\) has finite exponential moments. This prevents the integrand from being too large too often.

Key Properties¶

Sufficient but not necessary: There exist true martingale exponentials that violate Novikov.
Sign-independent: Only \(\theta^2\) appears, so the sign of \(\theta\) is irrelevant.
Deterministic case simplifies: If \(\theta_t = \theta(t)\) is deterministic, Novikov reduces to \(\int_0^T \theta(t)^2\,dt < \infty\).

Proof Sketch for Novikov's Condition¶

The proof proceeds in several steps. We follow the approach in Revuz–Yor (Chapter VIII) and Karatzas–Shreve (Section 3.5).

Step 1: Exponential martingale inequality

For any continuous local martingale \(M\) with \(M_0 = 0\) and any \(\lambda > 0\):

\[ \mathbb{P}\left(\sup_{t \leq T} M_t \geq \lambda\right) \leq e^{-\lambda} \mathbb{E}\left[\exp(M_T)\mathbf{1}_{\{\sup_{t \leq T} M_t \geq \lambda\}}\right] \]

Step 2: Control the maximum

Using the inequality from Step 1 and properties of the stochastic exponential, one can show:

\[ \mathbb{E}\left[\sup_{t \leq T} \mathcal{E}(M)_t\right] \leq 2\,\mathbb{E}\left[\mathcal{E}(M)_T^{1/2} \cdot \exp\left(\frac{1}{4}\langle M \rangle_T\right)\right] \]

Step 3: Apply Cauchy–Schwarz

By Cauchy–Schwarz:

\[ \mathbb{E}\left[\mathcal{E}(M)_T^{1/2} \cdot \exp\left(\frac{1}{4}\langle M \rangle_T\right)\right] \leq \mathbb{E}\left[\mathcal{E}(M)_T\right]^{1/2} \cdot \mathbb{E}\left[\exp\left(\frac{1}{2}\langle M \rangle_T\right)\right]^{1/2} \]

Step 4: Bootstrap

Under Novikov's condition, the second factor is finite. A careful stopping time argument shows that the local martingale \(\mathcal{E}(M)\) is dominated by an integrable random variable, hence is a true martingale.

Reference: For the complete proof, see Karatzas–Shreve, Brownian Motion and Stochastic Calculus, Proposition 3.5.12, or Revuz–Yor, Continuous Martingales and Brownian Motion, Proposition VIII.1.15.

Kazamaki's Condition¶

Kazamaki's condition is weaker than Novikov's and involves the local martingale \(M\) itself rather than its quadratic variation.

Theorem (Kazamaki, 1977): Let \(M_t = \int_0^t \theta_s\,dW_s\) be a continuous local martingale with \(M_0 = 0\). If

\[ \boxed{ \mathcal{E}\left(\frac{1}{2}M\right)_t = \exp\left(\frac{1}{2}M_t - \frac{1}{8}\langle M \rangle_t\right) \text{ is a submartingale} } \]

then \(\mathcal{E}(M)_t\) is a true martingale.

Equivalent Formulations¶

The submartingale condition is equivalent to:

\[ \sup_{t \leq T} \mathbb{E}\left[\exp\left(\frac{1}{2}M_t\right)\right] < \infty \]

A sufficient (but slightly stronger) condition often stated in practice:

\[ \mathbb{E}\left[\exp\left(\frac{1}{2}M_T\right)\right] < \infty \]

Why Kazamaki Is Weaker¶

Proposition: Novikov's condition implies Kazamaki's condition.

Proof: Assume Novikov holds: \(\mathbb{E}[\exp(\frac{1}{2}\langle M \rangle_T)] < \infty\).

The process \(\exp(\frac{1}{2}M_t)\) can be written as:

\[ \exp\left(\frac{1}{2}M_t\right) = \mathcal{E}\left(\frac{1}{2}M\right)_t \cdot \exp\left(\frac{1}{8}\langle M \rangle_t\right) \]

Taking expectations and using that \(\mathcal{E}(\frac{1}{2}M)\) is a non-negative local martingale (hence a supermartingale):

\[ \mathbb{E}\left[\exp\left(\frac{1}{2}M_t\right)\right] \leq \mathbb{E}\left[\exp\left(\frac{1}{8}\langle M \rangle_t\right)\right] \leq \mathbb{E}\left[\exp\left(\frac{1}{2}\langle M \rangle_T\right)\right]^{1/4} < \infty \]

where we used Jensen's inequality in the last step. Thus Kazamaki's condition is satisfied. \(\square\)

The converse is false: there exist processes satisfying Kazamaki but not Novikov.

Comparison of Conditions¶

Condition	Requires	Strength
Novikov	\(\mathbb{E}\left[\exp\left(\frac{1}{2}\langle M \rangle_T\right)\right] < \infty\)	Stronger
Kazamaki	\(\sup_{t \leq T}\mathbb{E}\left[\exp\left(\frac{1}{2}M_t\right)\right] < \infty\)	Weaker

\[ \text{Novikov satisfied} \Rightarrow \text{Kazamaki satisfied} \Rightarrow \mathcal{E}(M) \text{ is a true martingale} \]

When to use which:

Novikov: Easier to verify when \(\theta\) has known bounds or growth conditions
Kazamaki: Useful when direct control of \(\langle M \rangle\) is difficult but \(M\) itself is well-behaved

Practical Verification¶

Case 1: Deterministic θ¶

If \(\theta_t = \theta(t)\) is deterministic:

\[ \int_0^T \theta(t)^2\,dt < \infty \Rightarrow \text{Novikov satisfied} \]

The exponential of a constant is finite, so the condition reduces to \(\theta \in L^2[0,T]\).

Case 2: Bounded θ¶

If \(|\theta_t| \leq M\) almost surely for all \(t \in [0,T]\):

\[ \mathbb{E}\left[\exp\left(\frac{1}{2}\int_0^T \theta_s^2\,ds\right)\right] \leq \exp\left(\frac{M^2 T}{2}\right) < \infty \]

Novikov is satisfied.

Case 3: Black–Scholes Model¶

The market price of risk \(\theta = (\mu - r)/\sigma\) is constant.

\[ \mathbb{E}\left[\exp\left(\frac{1}{2}\theta^2 T\right)\right] = \exp\left(\frac{\theta^2 T}{2}\right) < \infty \quad \checkmark \]

Novikov is satisfied for any finite \(T\).

Case 4: Heston Stochastic Volatility Model¶

In the Heston model, the variance process \(V_t\) follows:

\[ dV_t = \kappa(\bar{V} - V_t)\,dt + \xi\sqrt{V_t}\,dW_t^V \]

The market price of risk for the stock is \(\theta_t = (\mu - r)/\sqrt{V_t}\), so:

\[ \int_0^T \theta_s^2\,ds = (\mu - r)^2 \int_0^T \frac{1}{V_s}\,ds \]

Novikov verification requires: \(\mathbb{E}\left[\exp\left(\frac{(\mu-r)^2}{2}\int_0^T V_s^{-1}\,ds\right)\right] < \infty\)

This holds when the Feller condition is satisfied:

\[ 2\kappa\bar{V} \geq \xi^2 \]

The Feller condition ensures \(V_t > 0\) for all \(t\) (the process never hits zero), which prevents \(\int_0^T V_s^{-1}\,ds\) from exploding. When Feller is violated, \(V_t\) can hit zero, and verification becomes more delicate—one must check whether the time spent near zero is short enough.

Reference: For detailed analysis, see Andersen and Piterbarg, Interest Rate Modeling, Vol. 1, Chapter 8.

Applications in Finance¶

Girsanov's Theorem¶

For Girsanov's theorem to produce a valid measure change, we need the Radon–Nikodym derivative:

\[ Z_T = \exp\left(\int_0^T \theta_s\,dW_s - \frac{1}{2}\int_0^T \theta_s^2\,ds\right) \]

to satisfy \(\mathbb{E}[Z_T] = 1\). Novikov or Kazamaki provides this guarantee.

Risk-Neutral Measure Construction¶

The risk-neutral measure \(\mathbb{Q}\) exists and is equivalent to \(\mathbb{P}\) if:

A market price of risk \(\theta\) exists (completeness/no-arbitrage)
The stochastic exponential \(\mathcal{E}(\int \theta\,dW)\) is a true martingale (Novikov/Kazamaki)

When condition 2 fails, \(\mathbb{Q}\) may still exist but could fail to be equivalent to \(\mathbb{P}\), or the market may admit arbitrage of the first kind.

Bubbles and Strict Local Martingales¶

When both Novikov and Kazamaki fail, \(Z_t\) may be a strict local martingale with \(\mathbb{E}[Z_T] < 1\).

Consequence of failure

Failure of both conditions means: density \(Z_t\) is not a true martingale \(\Rightarrow\) \(\mathbb{Q}\) is not a valid equivalent probability measure \(\Rightarrow\) the entire risk-neutral pricing framework collapses.

In finance, this corresponds to asset price bubbles: the discounted price process is a local martingale but not a true martingale under the pricing measure. See Local Martingales for the connection between strict local martingales and bubbles.

Extension to General Continuous Local Martingales¶

For a general continuous local martingale \(M\) (not necessarily an Itô integral), the Doléans-Dade exponential is:

\[ \mathcal{E}(M)_t = \exp\left(M_t - \frac{1}{2}\langle M \rangle_t\right) \]

General Novikov Condition: If \(\mathbb{E}[\exp(\frac{1}{2}\langle M \rangle_T)] < \infty\), then \(\mathcal{E}(M)\) is a true martingale on \([0,T]\).

General Kazamaki Condition: If \(\mathcal{E}(\frac{1}{2}M)\) is a submartingale, then \(\mathcal{E}(M)\) is a true martingale.

For discontinuous local martingales (with jumps), the Doléans-Dade exponential has the more complex form:

\[ \mathcal{E}(M)_t = \exp\left(M_t - \frac{1}{2}\langle M^c \rangle_t\right) \prod_{0 < s \leq t}(1 + \Delta M_s)e^{-\Delta M_s} \]

where \(M^c\) is the continuous part and \(\Delta M_s = M_s - M_{s-}\) are the jumps. Conditions for this to be a true martingale are more involved.

Summary¶

Condition	Statement	Verifies
Novikov	\(\mathbb{E}\left[\exp\left(\frac{1}{2}\langle M \rangle_T\right)\right] < \infty\)	\(\mathcal{E}(M)\) is a true martingale
Kazamaki	\(\mathcal{E}\left(\frac{1}{2}M\right)\) is a submartingale	\(\mathcal{E}(M)\) is a true martingale

\[ \boxed{ \text{Novikov} \Rightarrow \text{Kazamaki} \Rightarrow \mathcal{E}(M) \text{ is a true martingale} \Rightarrow \mathbb{E}[\mathcal{E}(M)_T] = 1 } \]

Key Takeaways

Both conditions are sufficient but not necessary for the stochastic exponential to be a true martingale.
Novikov is easier to verify when the quadratic variation \(\langle M \rangle\) is controlled.
Kazamaki is weaker and can apply in cases where Novikov fails.
In finance, these conditions ensure Girsanov's theorem produces valid measure changes, which is essential for risk-neutral pricing.
Failure of both conditions may indicate the presence of asset price bubbles (strict local martingales).

References¶

Novikov, A.A. (1972). "On an identity for stochastic integrals." Theory of Probability and Its Applications, 17(4), 717–720.
Kazamaki, N. (1977). "On a problem of Girsanov." Tôhoku Mathematical Journal, 29(4), 597–600.
Karatzas, I. and Shreve, S.E. (1991). Brownian Motion and Stochastic Calculus, 2nd ed. Springer. Section 3.5.
Revuz, D. and Yor, M. (1999). Continuous Martingales and Brownian Motion, 3rd ed. Springer. Chapter VIII.

Exercises¶

Exercise 1. In the Black-Scholes model, the market price of risk is \(\theta = (\mu - r)/\sigma\) with \(\mu = 0.10\), \(\sigma = 0.25\), and \(r = 0.03\). Verify Novikov's condition explicitly for \(T = 10\) and conclude that the Girsanov measure change is valid.

Solution to Exercise 1

With \(\mu = 0.10\), \(\sigma = 0.25\), and \(r = 0.03\), the market price of risk is:

\[ \theta = \frac{\mu - r}{\sigma} = \frac{0.10 - 0.03}{0.25} = 0.28 \]

Since \(\theta\) is constant, the Novikov condition becomes:

\[ \mathbb{E}\left[\exp\left(\frac{1}{2}\int_0^{10} \theta^2\,ds\right)\right] = \exp\left(\frac{\theta^2 \cdot 10}{2}\right) = \exp\left(\frac{0.0784 \cdot 10}{2}\right) = \exp(0.392) \approx 1.480 < \infty \]

The quantity inside the expectation is deterministic and finite, so Novikov's condition is trivially satisfied. Therefore \(\mathcal{E}(-\theta W^{\mathbb{P}})_T\) is a true martingale with \(\mathbb{E}[Z_T] = 1\), and the Girsanov measure change from \(\mathbb{P}\) to \(\mathbb{Q}\) is valid for \(T = 10\).

Exercise 2. Consider a deterministic but time-varying market price of risk \(\theta(t) = \alpha e^{\beta t}\) with \(\alpha > 0\) and \(\beta > 0\). Compute \(\int_0^T \theta(t)^2\,dt\) and determine for which values of \(T\) the Novikov condition holds. What happens as \(T \to \infty\)?

Solution to Exercise 2

With \(\theta(t) = \alpha e^{\beta t}\):

\[ \int_0^T \theta(t)^2\,dt = \int_0^T \alpha^2 e^{2\beta t}\,dt = \frac{\alpha^2}{2\beta}\left(e^{2\beta T} - 1\right) \]

Since \(\theta(t)\) is deterministic, the Novikov condition reduces to:

\[ \exp\left(\frac{1}{2}\int_0^T \theta(t)^2\,dt\right) = \exp\left(\frac{\alpha^2}{4\beta}(e^{2\beta T} - 1)\right) < \infty \]

This is finite for every finite \(T\), since the exponential of any finite number is finite. Thus Novikov's condition holds for all \(T < \infty\).

As \(T \to \infty\): \(\int_0^T \theta(t)^2\,dt \to \infty\) (exponential growth), so \(\exp(\frac{1}{2}\int_0^\infty \theta(t)^2\,dt) = +\infty\). On any finite horizon, the Girsanov measure change is valid, but the condition does not extend to infinite horizon. The exponential growth of \(\theta(t)\) means the market price of risk becomes arbitrarily large, which would require increasingly extreme probability reweighting.

Exercise 3. Prove that Novikov's condition implies Kazamaki's condition. Specifically, show that if \(\mathbb{E}[\exp(\frac{1}{2}\langle M \rangle_T)] < \infty\), then \(\sup_{t \leq T}\mathbb{E}[\exp(\frac{1}{2}M_t)] < \infty\). (Hint: write \(\exp(\frac{1}{2}M_t)\) in terms of \(\mathcal{E}(\frac{1}{2}M)_t\) and \(\exp(\frac{1}{8}\langle M \rangle_t)\), then use the supermartingale property.)

Solution to Exercise 3

Assume Novikov holds: \(\mathbb{E}[\exp(\frac{1}{2}\langle M \rangle_T)] < \infty\). We want to show \(\sup_{t \leq T}\mathbb{E}[\exp(\frac{1}{2}M_t)] < \infty\).

Write:

\[ \exp\left(\frac{1}{2}M_t\right) = \mathcal{E}\left(\frac{1}{2}M\right)_t \cdot \exp\left(\frac{1}{8}\langle M \rangle_t\right) \]

This follows from the definition of the stochastic exponential:

\[ \mathcal{E}\left(\frac{1}{2}M\right)_t = \exp\left(\frac{1}{2}M_t - \frac{1}{2}\cdot\frac{1}{4}\langle M \rangle_t\right) = \exp\left(\frac{1}{2}M_t - \frac{1}{8}\langle M \rangle_t\right) \]

Now \(\mathcal{E}(\frac{1}{2}M)\) is a non-negative local martingale, hence a supermartingale: \(\mathbb{E}[\mathcal{E}(\frac{1}{2}M)_t] \leq \mathcal{E}(\frac{1}{2}M)_0 = 1\).

Taking expectations using Cauchy–Schwarz:

\[ \mathbb{E}\left[\exp\left(\frac{1}{2}M_t\right)\right] = \mathbb{E}\left[\mathcal{E}\left(\frac{1}{2}M\right)_t \cdot \exp\left(\frac{1}{8}\langle M \rangle_t\right)\right] \]

\[ \leq \left(\mathbb{E}\left[\mathcal{E}\left(\frac{1}{2}M\right)_t^2\right]\right)^{1/2} \cdot \left(\mathbb{E}\left[\exp\left(\frac{1}{4}\langle M \rangle_t\right)\right]\right)^{1/2} \]

Alternatively, a simpler bound: since \(\mathcal{E}(\frac{1}{2}M)_t \leq 1\) is not generally true, we use Jensen's inequality more carefully. Since \(\langle M \rangle_t \leq \langle M \rangle_T\) for \(t \leq T\):

\[ \mathbb{E}\left[\exp\left(\frac{1}{2}M_t\right)\right] \leq \mathbb{E}\left[\exp\left(\frac{1}{8}\langle M \rangle_T\right)\right] \leq \mathbb{E}\left[\exp\left(\frac{1}{2}\langle M \rangle_T\right)\right]^{1/4} < \infty \]

where the last step uses Jensen's inequality with the convex function \(x \mapsto x^4\) (or equivalently, \(\frac{1}{8} \leq \frac{1}{2}\) combined with monotonicity of the exponential). The bound is uniform in \(t \leq T\), so \(\sup_{t \leq T}\mathbb{E}[\exp(\frac{1}{2}M_t)] < \infty\), which is Kazamaki's condition.

Exercise 4. For the Heston model with \(\theta_t = (\mu - r)/\sqrt{V_t}\), explain why Novikov's condition involves \(\mathbb{E}[\exp(\frac{(\mu-r)^2}{2}\int_0^T V_s^{-1}\,ds)]\). State the Feller condition \(2\kappa\bar{V} \geq \xi^2\) and explain its role in ensuring \(V_t > 0\). Why does \(V_t\) hitting zero cause the Novikov condition to potentially fail?

Solution to Exercise 4

In the Heston model, \(\theta_t = (\mu - r)/\sqrt{V_t}\), so:

\[ \int_0^T \theta_s^2\,ds = \int_0^T \frac{(\mu-r)^2}{V_s}\,ds = (\mu-r)^2 \int_0^T \frac{ds}{V_s} \]

Novikov's condition requires:

\[ \mathbb{E}\left[\exp\left(\frac{(\mu-r)^2}{2}\int_0^T V_s^{-1}\,ds\right)\right] < \infty \]

The Feller condition \(2\kappa\bar{V} \geq \xi^2\) ensures that the CIR process \(V_t\) never reaches zero. Specifically, when Feller holds, \(V_t > 0\) for all \(t > 0\) a.s. (the boundary at zero is entrance, not accessible). This keeps \(1/V_s\) bounded in a neighborhood of zero, preventing \(\int_0^T V_s^{-1}\,ds\) from exploding.

When Feller is violated (\(2\kappa\bar{V} < \xi^2\)), \(V_t\) can hit zero. Near zero, \(1/V_s\) diverges, and \(\int_0^T V_s^{-1}\,ds\) may become infinite. If the process spends too much time near zero, the exponential moment in Novikov's condition blows up, and the stochastic exponential may fail to be a true martingale. This would invalidate the Girsanov measure change and potentially signal the presence of arbitrage or bubbles in the model.

Exercise 5. Consider \(\theta_t = c / \sqrt{T - t}\) for \(t < T\) with \(c > 0\). Show that \(\int_0^T \theta_s^2\,ds\) diverges logarithmically. Despite this divergence, explain why the stochastic exponential \(Z_t\) may still be well-defined as a local martingale, and identify the defect \(\delta = 1 - \mathbb{E}[Z_T]\).

Solution to Exercise 5

With \(\theta_t = c/\sqrt{T-t}\) for \(t < T\):

\[ \int_0^T \theta_s^2\,ds = \int_0^T \frac{c^2}{T-s}\,ds = c^2\left[-\ln(T-s)\right]_0^T = c^2 \lim_{\epsilon \to 0^+}(\ln T - \ln \epsilon) = +\infty \]

The integral diverges logarithmically: \(\int_0^{T-\epsilon} \theta_s^2\,ds = c^2 \ln(T/\epsilon)\).

Despite this, the stochastic exponential \(Z_t = \exp(\int_0^t \theta_s\,dW_s - \frac{1}{2}\int_0^t \theta_s^2\,ds)\) is still well-defined as a local martingale for \(t < T\), because \(\int_0^t \theta_s^2\,ds < \infty\) for each \(t < T\). The process \(Z_t\) satisfies \(dZ_t = Z_t \theta_t\,dW_t\) and is a non-negative local martingale, hence a supermartingale.

As \(t \to T\), the explosive growth of \(\theta_t\) causes \(Z_t \to 0\) a.s., leading to:

\[ \mathbb{E}[Z_T] < 1 \]

The defect \(\delta = 1 - \mathbb{E}[Z_T] > 0\) represents the "mass" lost to infinity. The exact value depends on \(c\) and the specific path structure, but \(\delta > 0\) confirms that \(Z\) is a strict local martingale and cannot serve as a valid Radon–Nikodym derivative for an equivalent measure.

Exercise 6. Suppose \(|\theta_t| \leq M\) almost surely for some constant \(M > 0\) and all \(t \in [0, T]\). Show that both the Novikov condition and the Kazamaki condition are satisfied. What is the upper bound on \(\mathbb{E}[\exp(\frac{1}{2}\int_0^T \theta_s^2\,ds)]\) in terms of \(M\) and \(T\)?

Solution to Exercise 6

If \(|\theta_t| \leq M\) a.s. for all \(t \in [0,T]\), then:

\[ \int_0^T \theta_s^2\,ds \leq M^2 T \quad \text{a.s.} \]

Novikov's condition:

\[ \mathbb{E}\left[\exp\left(\frac{1}{2}\int_0^T \theta_s^2\,ds\right)\right] \leq \exp\left(\frac{M^2 T}{2}\right) < \infty \]

So Novikov is satisfied with upper bound \(\exp(M^2 T / 2)\).

Kazamaki's condition: Since Novikov implies Kazamaki (proved in Exercise 3), Kazamaki is automatically satisfied. Alternatively, direct verification: \(M_t = \int_0^t \theta_s\,dW_s\) is a continuous martingale with \(\langle M \rangle_t \leq M^2 t\), and:

\[ \sup_{t \leq T}\mathbb{E}\left[\exp\left(\frac{1}{2}M_t\right)\right] \leq \sup_{t \leq T}\exp\left(\frac{1}{8}M^2 t\right) \cdot 1 \leq \exp\left(\frac{M^2 T}{8}\right) < \infty \]

(using the supermartingale bound from the proof that Novikov implies Kazamaki).

Exercise 7. Construct a process \(M_t\) for which Kazamaki's condition is satisfied but Novikov's condition fails. (Hint: consider a process where \(M_t\) has controlled moments but \(\langle M \rangle_T\) has heavy tails. You may describe the construction conceptually rather than giving an explicit formula.)

Solution to Exercise 7

Construction (conceptual): Let \(\tau\) be a random time with \(\mathbb{P}(\tau \leq T) = 1\) and define \(\theta_t\) to be a process that is bounded for \(t < \tau\) but has a carefully chosen blow-up at \(t = \tau\).

More concretely, consider \(M_t = \int_0^t \theta_s\,dW_s\) where \(\theta_s\) is chosen so that:

\(\langle M \rangle_T = \int_0^T \theta_s^2\,ds\) has a heavy-tailed distribution (e.g., comparable to an exponential or Pareto random variable), so \(\mathbb{E}[\exp(\frac{1}{2}\langle M \rangle_T)] = \infty\) — Novikov fails.
\(M_T\) itself remains "controlled" enough that \(\mathbb{E}[\exp(\frac{1}{2}M_T)] < \infty\) — Kazamaki holds.

This is possible because \(M_T\) has both positive and negative fluctuations (it is a martingale centered at zero), while \(\langle M \rangle_T\) is always non-negative and monotone increasing. The quadratic variation accumulates without cancellation, while the martingale \(M_T\) benefits from cancellations between positive and negative increments.

A classical example from Kazamaki (1977): take \(\theta_s\) to depend on \(W\) itself in such a way that \(\int_0^T \theta_s^2\,ds\) has moments growing faster than exponential, but \(M_T = \int_0^T \theta_s\,dW_s\) has controlled exponential moments due to the "centering" effect of the stochastic integral. The separation arises precisely because \(\langle M \rangle\) measures accumulated volatility (always additive), while \(M\) itself benefits from martingale cancellations.