Proving a Density-Weighted Process Is a Martingale with Girsanov
Summary
The document considers a process formed by multiplying an Ito process with a stochastic exponential built from its drift. The question observes that applying Ito’s rule yields a differential with no drift, then asks whether that alone proves a true martingale. The answer uses Novikov’s condition to establish that the stochastic exponential is a martingale and defines a new probability measure using it as the density.
Under that measure, Girsanov’s theorem makes the drift-adjusted process Brownian and therefore a martingale. The conditional expectation identity is then transferred back to the original measure with Bayes’ formula, showing that the product process is a martingale there. This gives a measure-change argument rather than relying only on a drift-free stochastic differential. The proof depends on Novikov’s condition and the stated integrability framework; a zero drift term by itself generally establishes only a local martingale without further conditions.
Key ideas
- A stochastic differential without a drift term establishes a local martingale, not automatically a true martingale.
- Novikov’s condition ensures the stochastic exponential can define a valid probability density.
- Girsanov’s theorem removes the drift from the process under the changed measure.
- Bayes’ formula transfers the martingale conditional expectation back to the original measure.
- The argument depends on the required integrability and measure-change conditions.
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Full text
# Prove uniqueness, and prove $Y_t$ is a martingale by considering $dZ_t$ and $dL_t$
# Prove uniqueness, and prove $Y_t$ is a martingale by considering $dZ_t$ and $dL_t$
Suppose we are given a filtered probability space $(\Omega, \mathscr{F}, \{\mathscr{F}_t\}_{t \in [0,T]}, \mathbb{P})$, where $\{\mathscr{F}_t\}_{t \in [0,T]}$ is the filtration generated by standard $\mathbb P$-Brownian motion.
Let $dX_t = \theta_tdt +dW_t$ be an Ito process where $(\theta_t)_{t \in [0,T]}$ is $\mathscr{F}_t$-adapated and $E[\int_0^T \theta_s^2 ds] < \infty$ and
$$Y_t := X_tL_t, \ \ L_t = \exp Z_t, \ \ Z_t = -\int_0^t \theta_s dW_s - \frac{1}{2}\int_0^t\theta_s^2ds$$
Suppose Novikov's condition holds.
Prove $Y_t$ is a $(\mathscr{F}_t, \mathbb{P})$-martingale.
I was able to show that $dY_t = (L_t - \theta_tY_t)dW_t$ from deriving that $dZ_t = -\theta_tdW_t -\frac 1 2 \theta_t^2 dt$ and $dL_t = e^{Z_t}(-\theta_tdW_t)$.
Assuming that this is right, does the fact that there is no drift term in $dY_t$ already establish that $Y_t$ is a $(\mathscr{F}_t, \mathbb{P})$-martingale and not merely that it is a local martingale or merely that $E[Y_t | \mathscr{F}_u] = Y_u$?
Edit: It seems that according to this, a solution of an SDE is a martingale if it is unique.
- $E[Y_0^2] = E[X_0^2]< \infty$, I guess? No initial condition is given for $X_t$
- Show $\exists K \in \mathbb R$ s.t.
$|L_t - \theta_tx| \le K(1+|x|)$
$|(L_t - \theta_tx) - (L_t - \theta_ty)| \le K|x-y|$
We have:
$$|L_t - \theta_tx| \le |L_t| + |\theta_t||x| \le |\theta_t|(1+|x|)$$
$$|(L_t - \theta_tx) - (L_t - \theta_ty)| \le |\theta_t||x-y|$$
I don't suppose $E[\int_0^T \theta_s^2 ds] < \infty$ means that $\theta_t$ is bounded, does it?
## Answer by Gordon (score 4)
https://quant.stackexchange.com/a/22270
We consider the case where the Novikov condition is satisfied, that is, \begin{align*} E\left[\exp\left(\frac{1}{2}\int_0^T \theta^2_s ds \right)\right] < \infty. \end{align*} Then $\{L_t \mid t \ge 0\}$ is a $(\mathscr{F}_t, \mathbb{P})$-martingale. On $\mathscr{F}_T$, we define the probability measure $Q$ by \begin{align*} \frac{dQ}{dP}\big|_{\mathscr{F}_T} &= L_T\\ &= \exp\left(-\frac{1}{2}\int_0^T \theta_s^2 ds - \int_0^T \theta_s dW_s \right). \end{align*} By Girsanov theorem, $X=\{X_t \mid t \ge 0\}$, where \begin{align*} X_t = \int_0^t \theta_s ds + W_t, \end{align*} is standard Brownian motion with respect to the measure $Q$, and then it is a $(\mathscr{F}_t, \mathbb{Q})$-martingale. That is, for $0\leq u \leq t \leq T$, \begin{align*} E_Q(X_t \mid \mathscr{F}_u) = X_u.\tag{1} \end{align*} On the other hand, by the abstract Bayes formula, \begin{align*} E_Q(X_t \mid \mathscr{F}_u) &= E_P\left(\frac{L_t}{L_u} X_t \mid \mathscr{F}_u\right)\\ &= \frac{1}{L_u}E_P\left(L_t X_t \mid \mathscr{F}_u\right).\tag{2} \end{align*} Combining (1) and (2), we have that \begin{align*} E_P\left(L_t X_t \mid \mathscr{F}_u\right) = L_u X_u. \end{align*} That is, $Y=LX$ is a martingale with respect to the measure $P$.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.