Exponential martingales and measure changes

Definition Fix a finite horizon $T > 0$ and a filtered probability space $(\Omega, \mathcal{F}, (\mathcal{F}_t)_{0 \le t \le T}, \mathbb{P})$ with $\mathcal{F}_T \subseteq \mathcal{F}$. A change of measure keeps $\Omega$, the events of $\mathcal{F}_T$, the filtration and every random variable as they are, and replaces $\mathbb{P}$ by another probability measure $\mathbb{Q}$ on $\mathcal{F}_T$. The new measure is absolutely continuous with respect to $\mathbb{P}$, written $\mathbb{Q} \ll \mathbb{P}$, when every $\mathbb{P}$-null event in $\mathcal{F}_T$ is also $\mathbb{Q}$-null. The two measures are equivalent, written $\mathbb{Q} \sim \mathbb{P}$, when they have the same null events,

for every $A \in \mathcal{F}_T$. If $\mathbb{Q} \ll \mathbb{P}$, every statement that holds $\mathbb{P}$-almost surely also holds $\mathbb{Q}$-almost surely, and under equivalence the two measures agree on which events are impossible and which are certain. Other probabilities, expectations and distributions can all change, so a process that is a Brownian motion under $\mathbb{P}$ need not be one under $\mathbb{Q}$. The pricing measures of later lessons are equivalent measures chosen so that discounted traded prices are martingales, and a change of measure alone does not select one.

Definition Let $\mathbb{Q} \ll \mathbb{P}$ on $\mathcal{F}_T$. The Radon-Nikodym theorem, stated here without proof, gives an $\mathcal{F}_T$-measurable random variable $Z_T \ge 0$, unique up to $\mathbb{P}$-null events, such that

for every $A \in \mathcal{F}_T$. This $Z_T$ is the Radon-Nikodym derivative, or density, of $\mathbb{Q}$ with respect to $\mathbb{P}$ on $\mathcal{F}_T$, written

The choice $A = \Omega$ gives $\mathbb{E}^{\mathbb{P}}[Z_T] = 1$, and approximation by simple functions gives $\mathbb{E}^{\mathbb{Q}}[X] = \mathbb{E}^{\mathbb{P}}[Z_T X]$ for every $\mathcal{F}_T$-measurable $X \ge 0$ and for every $X$ with $\mathbb{E}^{\mathbb{P}}[Z_T \lvert X \rvert] < \infty$. Conversely, every $\mathcal{F}_T$-measurable $Z_T \ge 0$ with $\mathbb{E}^{\mathbb{P}}[Z_T] = 1$ defines a probability measure $\mathbb{Q} \ll \mathbb{P}$ by the same formula. The measures are equivalent exactly when $Z_T > 0$ holds $\mathbb{P}$-almost surely, and then $1/Z_T$ is the density of $\mathbb{P}$ with respect to $\mathbb{Q}$. On a finite set of states that all have positive probability, $Z_T$ is the likelihood ratio of Risk neutral probabilities and physical probabilities. The density process

for $0 \le t \le T$ is a $\mathbb{P}$-martingale by the tower property, and it is the density of $\mathbb{Q}$ on $\mathcal{F}_t$, because for $A \in \mathcal{F}_t$ the indicator $\mathbf{1}_A$ is $\mathcal{F}_t$-measurable and

This course takes $\mathcal{F}_0$ trivial up to null events, so $Z_0 = \mathbb{E}^{\mathbb{P}}[Z_T] = 1$. With a nontrivial $\mathcal{F}_0$, a start at one also needs $\mathbb{E}^{\mathbb{P}}[Z_T \mid \mathcal{F}_0] = 1$. The weights at different dates must come from one terminal density in this way, since normalizing unrelated weights separately at each date does not give consistent probabilities. Under equivalence $Z_t > 0$ almost surely for every $t$.

Definition Let $\mathbb{Q} \ll \mathbb{P}$ on $\mathcal{F}_T$ with density $Z_T$ and density process $Z_t = \mathbb{E}^{\mathbb{P}}[Z_T \mid \mathcal{F}_t]$, let $0 \le t \le T$, and let $X$ be $\mathcal{F}_T$-measurable with $\mathbb{E}^{\mathbb{Q}}[\lvert X \rvert] < \infty$. Then

holds $\mathbb{Q}$-almost surely. The denominator is positive $\mathbb{Q}$-almost surely, since $\mathbb{Q}(Z_t = 0) = \mathbb{E}^{\mathbb{P}}[Z_t \mathbf{1}_{\{Z_t = 0\}}] = 0$, and under equivalence the identity also holds $\mathbb{P}$-almost surely. To prove it, call the right side $Y$, set to $0$ where $Z_t = 0$. It is $\mathcal{F}_t$-measurable, and for every $A \in \mathcal{F}_t$ the density on $\mathcal{F}_t$, the definition of conditional expectation and $\mathbb{Q}(Z_t = 0) = 0$ give

which is the defining property of $\mathbb{E}^{\mathbb{Q}}[X \mid \mathcal{F}_t]$. The formula moves martingale properties between measures. When $\mathbb{Q} \sim \mathbb{P}$ and $M$ is adapted with $\mathbb{E}^{\mathbb{Q}}[\lvert M_t \rvert] < \infty$ for every $t$, applying it to $X = M_t$ and using the tower property gives, for $s \le t$,

so $M$ is a $\mathbb{Q}$-martingale exactly when the product $ZM$ is a $\mathbb{P}$-martingale.

Definition Let $W$ be a standard Brownian motion under $\mathbb{P}$ with respect to $(\mathcal{F}_t)$, and let $\lambda$ be a predictable process with

Then the Itô integral $\int_0^t \lambda_s \, dW_s$ is defined as a continuous local martingale, and the stochastic exponential used in this course is

for $0 \le t \le T$, written $Z = \mathcal{E}\bigl(-\int \lambda \, dW\bigr)$. More generally, the stochastic exponential of a continuous local martingale $M$ with $M_0 = 0$ is $\mathcal{E}(M)_t = \exp(M_t - \tfrac12 [M]_t)$, which is the unique solution of $dZ_t = Z_t \, dM_t$ with $Z_0 = 1$, a uniqueness result stated without proof. For the course’s case write $Y_t$ for the exponent, so that $dY_t = -\lambda_t \, dW_t - \tfrac12 \lambda_t^2 \, dt$ and $d[Y]_t = \lambda_t^2 \, dt$. Itô’s lemma for $Z = e^{Y}$ gives

because the time terms $-\tfrac12 Z_t \lambda_t^2 \, dt$ and $+\tfrac12 Z_t \lambda_t^2 \, dt$ cancel. So $Z$ is a strictly positive continuous local martingale with $Z_0 = 1$. Without the half term, the exponential of $-\int \lambda \, dW$ alone would carry the drift $\tfrac12 Z_t \lambda_t^2 \, dt$. The minus sign in front of the stochastic integral is this course’s convention, and with it the theorem of Girsanov theorem and risk neutral dynamics adds $\int_0^t \lambda_s \, ds$ to $W$.

Definition A stochastic exponential that is a true martingale on $[0, T]$, not only a local martingale, is an exponential martingale. A nonnegative local martingale is a supermartingale by Fatou’s lemma, so for $Z = \mathcal{E}(-\int \lambda \, dW)$ the map $t \mapsto \mathbb{E}^{\mathbb{P}}[Z_t]$ is nonincreasing and at most $1$, and it can drop below $1$. The process is a true martingale on $[0, T]$ exactly when $\mathbb{E}^{\mathbb{P}}[Z_T] = 1$. Indeed, if $\mathbb{E}^{\mathbb{P}}[Z_T] = 1$ then $1 = \mathbb{E}^{\mathbb{P}}[Z_T] \le \mathbb{E}^{\mathbb{P}}[Z_t] \le 1$, so the nonnegative variable $Z_t - \mathbb{E}^{\mathbb{P}}[Z_T \mid \mathcal{F}_t]$ has mean zero and vanishes almost surely. Only then does $Z_T$ define a probability measure, and that measure is equivalent to $\mathbb{P}$ because $Z_T > 0$. Which positive local martingales are strict local martingales is beyond this course. For a constant $\lambda$ the property can be checked directly. Since $W_T \sim N(0, T)$ has $\mathbb{E}^{\mathbb{P}}[e^{a W_T}] = e^{a^2 T/2}$, the choice $a = -\lambda$ gives a factor $e^{\lambda^2 T/2}$ that cancels the deterministic factor $e^{-\lambda^2 T/2}$ of $Z_T$. Squaring gives $Z_T^2 = e^{-2\lambda W_T - \lambda^2 T}$, and the choice $a = -2\lambda$ gives $e^{2\lambda^2 T} e^{-\lambda^2 T}$. Hence

so $\operatorname{Var}^{\mathbb{P}}(Z_T) = e^{\lambda^2 T} - 1$, which grows with the size of the tilt and with the horizon. Completing the square in the weighted normal density,

shows that $W_T \sim N(-\lambda T, T)$ under $\mathbb{Q}$. The same computations hold for a deterministic $\lambda$ with $\lambda^2 T$ replaced by $\int_0^T \lambda_s^2 \, ds$, because $\int_0^T \lambda_s \, dW_s$ is then normal.

Definition Novikov’s theorem, stated here without proof, says the following. Let $M$ be a continuous local martingale on $[0, T]$ with $M_0 = 0$. If $\mathbb{E}^{\mathbb{P}}[\exp(\tfrac12 [M]_T)] < \infty$, then $\mathcal{E}(M)$ is a true martingale on $[0, T]$ and $\mathbb{E}^{\mathbb{P}}[\mathcal{E}(M)_T] = 1$. For $M = -\int \lambda \, dW$ with $\lambda$ predictable and $\int_0^T \lambda_s^2 \, ds < \infty$ almost surely, the condition reads

and then $d\mathbb{Q}/d\mathbb{P} = Z_T$ defines a probability measure $\mathbb{Q}$ on $\mathcal{F}_T$ equivalent to $\mathbb{P}$. The condition is sufficient and not necessary, so its failure does not show that a density is invalid. It holds whenever $\lvert \lambda_t \rvert \le K$ for a constant $K$, because the expectation is then at most $e^{K^2 T/2}$. It concerns the integrated square of $\lambda$, not a moment of $Z$ itself.