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.

  1. $E[Y_0^2] = E[X_0^2]< \infty$, I guess? No initial condition is given for $X_t$

  2. 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?


1 Answer 1


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 $P$.

  • $\begingroup$ Thanks @Gordon. However the problem specifies $Y_t$ must be shown to be a martingale by considering $dL_t$ and $dZ_t$. How about transferring this answer to my other question? $\endgroup$
    – BCLC
    Dec 11, 2015 at 20:56
  • 1
    $\begingroup$ If they are square integrable , that is $E\left(\int_0^T(L_s-\theta_sY_s)^2 ds\right)$ is finite, then driftless implies the martingale property. For your other question, you can proceed as we show in (1). $\endgroup$
    – Gordon
    Dec 11, 2015 at 21:14
  • $\begingroup$ Thanks Gordon. Might you have a reference for that? I can't find that in my stocal notes, but I guess there's some integrability condition that allows us to say no drift $\to$ martingale $\endgroup$
    – BCLC
    Dec 11, 2015 at 21:29
  • $\begingroup$ Found it. $X_t$ is an Ito process that can apparently be rewritten as $$X_t = \int_0^t \frac{(L_s-\theta_sY_s)}{L_t} dW_s$$ According to my stocal notes for an Ito process we actually have $$E[\int_0^T (\frac{(L_s-\theta_sY_s)}{L_t})^2 ds] < \infty$$ I think $\endgroup$
    – BCLC
    Dec 11, 2015 at 21:43
  • $\begingroup$ Am I interpreting my notes right? i.stack.imgur.com/KgPLX.png There is the matter of the inner term having at in it... $\endgroup$
    – BCLC
    Dec 11, 2015 at 21:48

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