Exponential Transforms of Ito Processes

Positive quantities such as prices, intensities and likelihood ratios are modelled as exponentials \(Y_t=e^{X_t}\) of an Ito process \(X\). The first question to ask is whether \(Y\) is a martingale, because the answer decides whether \(\mathbb{E}[Y_t]\) stays equal to \(\mathbb{E}[Y_0]\), and whether \(Y\) can serve as a change-of-measure density. Throughout, \(X_t=X_0+\int_0^t b_s\,ds+\int_0^t\sigma_s\,dW_s\), where \(b\) and \(\sigma\) are progressively measurable with \(\int_0^T(|b_s|+\sigma_s^2)\,ds<\infty\) almost surely.

Ito's Formula for the Exponential

Proposition: The process \(Y_t=e^{X_t}\) satisfies \[ dY_t=Y_t\Big(b_t+\tfrac12\sigma_t^2\Big)dt+Y_t\sigma_t\,dW_t. \] Consequently \(Y\) is a local martingale if and only if \(b_t=-\tfrac12\sigma_t^2\) for \(dt\otimes\mathbb{P}\)-almost every \((t,\omega)\).

Proof. Apply Ito's formula to \(f(x)=e^x\), for which \(f'=f''=e^x\), and use \(d\langle X\rangle_t=\sigma_t^2\,dt\). This gives \(dY_t=Y_t\,dX_t+\tfrac12Y_t\sigma_t^2\,dt\), which is the displayed equation. The stochastic integral \(\int Y\sigma\,dW\) is a continuous local martingale, so \(Y\) is a local martingale exactly when the finite-variation part \(\int_0^t Y_s(b_s+\tfrac12\sigma_s^2)\,ds\) is a local martingale. A continuous local martingale of finite variation is constant, so that integral must vanish identically. Since \(Y_s>0\), this forces \(b_s+\tfrac12\sigma_s^2=0\) almost everywhere. The converse is immediate. \(\square\)

Worked Example: Constant Coefficients

Let \(X_t=\mu t+\sigma W_t\) with \(X_0=0\). Since \(\sigma W_t\sim N(0,\sigma^2t)\), the normal moment generating function gives \(\mathbb{E}[e^{X_t}]=e^{(\mu+\sigma^2/2)t}\). Take \(\mu=0.1\), \(\sigma=0.4\) and \(t=2\). Then \(\mu+\sigma^2/2=0.1+0.08=0.18\), and \[ \mathbb{E}[Y_2]=e^{0.18\cdot2}=e^{0.36}\approx1.4333\ne1=Y_0. \] So \(Y\) is not a martingale. Choosing \(\mu=-\sigma^2/2=-0.08\) makes the exponent zero, and then \(\mathbb{E}[Y_t]=1\) for every \(t\). The correction \(-\tfrac12\sigma^2\) exactly cancels the convexity gain of the exponential, in the sense of Jensen's inequality: \(\mathbb{E}[e^{\sigma W_t}]=e^{\sigma^2t/2}>1\).

A Positive Local Martingale Is a Supermartingale

Proposition: Suppose \(b=-\tfrac12\sigma^2\) and \(\mathbb{E}[e^{X_0}]<\infty\). Then \(Y\) is a nonnegative supermartingale, so \(\mathbb{E}[Y_t]\le\mathbb{E}[Y_0]\) for all \(t\). It is a martingale if and only if \(\mathbb{E}[Y_t]=\mathbb{E}[Y_0]\) for every \(t\).

Proof. Choose stopping times \(\tau_n\uparrow\infty\) such that each stopped process \(Y^{\tau_n}\) is a martingale. For \(s<t\), \(\mathbb{E}[Y_{t\wedge\tau_n}\mid\mathcal{F}_s]=Y_{s\wedge\tau_n}\). Let \(n\to\infty\). Since \(Y\ge0\) and \(Y_{t\wedge\tau_n}\to Y_t\) almost surely, the conditional Fatou lemma gives \(\mathbb{E}[Y_t\mid\mathcal{F}_s]\le\liminf_n Y_{s\wedge\tau_n}=Y_s\). Taking \(s=0\) and expectations shows that \(Y_t\) is integrable, so \(Y\) is a supermartingale. For the last statement, a supermartingale whose expectation is constant is a martingale: if \(\mathbb{E}[Y_t\mid\mathcal{F}_s]\le Y_s\) and both sides have the same expectation, then the nonnegative difference has zero mean and is zero almost surely. \(\square\)

This is why the question is never whether \(\mathbb{E}[Y_t]\le\mathbb{E}[Y_0]\), which is automatic, but whether the inequality is an equality. Mass can leak out of \(Y\) only through very large values of \(\sigma\).

A Sufficient Condition: Bounded Volatility

Theorem: Let \(b=-\tfrac12\sigma^2\), suppose \(|\sigma_t|\le C\) for a constant \(C\), and suppose \(\mathbb{E}[e^{X_0}]<\infty\). Then \(Y_t=e^{X_t}\) is a true martingale on \([0,T]\) for every \(T\).

Proof. Write \(M_t=\int_0^t\sigma_s\,dW_s\), so that \(Y_t=e^{X_0}Z_t\) with \(Z_t=\exp(M_t-\tfrac12\langle M\rangle_t)\) and \(Z_0=1\).

Step 1: a second exponential. Because \(2M_t-\tfrac12\langle 2M\rangle_t=2M_t-2\langle M\rangle_t\), we have the identity \(Z_t^2=Z^{(2)}_t\,e^{\langle M\rangle_t}\), where \(Z^{(2)}_t=\exp(2M_t-2\langle M\rangle_t)\) is the same construction built from the bounded volatility \(2\sigma\). By the previous proposition \(Z^{(2)}\) is a nonnegative supermartingale with \(\mathbb{E}[Z^{(2)}_{t\wedge\tau}]\le1\) for every stopping time \(\tau\).

Step 2: an \(L^2\) bound. Since \(\langle M\rangle_t\le C^2t\), Step 1 gives \(\mathbb{E}[Z_{t\wedge\tau}^2]\le e^{C^2t}\) for every stopping time \(\tau\).

Step 3: \(Z\) is a martingale. Let \(\tau_n\uparrow\infty\) localize \(Z\). For fixed \(t\), the family \(\{Z_{t\wedge\tau_n}\}_n\) is bounded in \(L^2\), hence uniformly integrable, and it converges to \(Z_t\) almost surely. Therefore it converges in \(L^1\), and passing to the limit in \(\mathbb{E}[Z_{t\wedge\tau_n}\mid\mathcal{F}_s]=Z_{s\wedge\tau_n}\) gives \(\mathbb{E}[Z_t\mid\mathcal{F}_s]=Z_s\).

Step 4: restore the initial value. The variable \(e^{X_0}\) is nonnegative and \(\mathcal{F}_0\)-measurable, so it can be pulled out of the conditional expectation: \(\mathbb{E}[Y_t\mid\mathcal{F}_s]=e^{X_0}\mathbb{E}[Z_t\mid\mathcal{F}_s]=e^{X_0}Z_s=Y_s\). Taking expectations, \(\mathbb{E}[Y_t]=\mathbb{E}[e^{X_0}]<\infty\), so \(Y_t\) is integrable. \(\square\)

The hypothesis on \(X_0\) is not decoration. If \(\mathbb{E}[e^{X_0}]=\infty\), then \(\mathbb{E}[Y_t]=\mathbb{E}\big[e^{X_0}\,\mathbb{E}[Z_t\mid\mathcal{F}_0]\big]=\mathbb{E}[e^{X_0}]=\infty\), so \(Y_t\) is not even integrable and cannot be a martingale, however well behaved \(\sigma\) is.

Unbounded Volatility

Worked Example: Volatility Equal to the Brownian Path

Take \(X_0=0\), \(\sigma_s=W_s\) and \(b_s=-\tfrac12W_s^2\), so that \[ X_t=\int_0^tW_s\,dW_s-\tfrac12\int_0^tW_s^2\,ds=\tfrac12(W_t^2-t)-\tfrac12\int_0^tW_s^2\,ds. \] By the proposition, \(Y=e^X\) is a positive local martingale, but \(\sigma\) is unbounded, so the theorem does not apply. Novikov's criterion, proved later in this course, requires \(\mathbb{E}\big[\exp(\tfrac12\int_0^T W_s^2\,ds)\big]<\infty\). The Cameron-Martin formula gives the value \((\cos T)^{-1/2}\) for \(T<\pi/2\); at \(T=1\) it equals \(1/\sqrt{\cos 1}\approx1.3604\), which is finite. A Monte Carlo simulation with \(2\times10^5\) paths returns about \(1.362\). Hence \(Y\) is a martingale on \([0,T]\) for every \(T<\pi/2\).

The criterion fails to be informative from \(T=\pi/2\) onward, since the expectation is then infinite, but that says nothing about the martingale property itself. Linear growth of the drift in the Girsanov-transformed equation, a criterion due to Benes, shows that \(Y\) is in fact a martingale on every finite horizon. Sufficient conditions are not necessary conditions.

Unbounded volatility can genuinely destroy the martingale property. A standard example, cited here rather than proved, is \(Y=1/R\) where \(R\) is a three-dimensional Bessel process started at \(R_0=1\). It is a positive local martingale satisfying \(dY=-Y^2\,dB\), so it has the exponential form above with volatility \(-Y\), which is unbounded near zero of \(R\). Its expectation is strictly decreasing in time, so it is a strict local martingale.

Deterministic Volatility and Geometric Brownian Motion

When \(\sigma\) is a deterministic function of time, every ingredient is explicit. Then \(M_t=\int_0^t\sigma_s\,dW_s\) is Gaussian with mean zero and variance \(\int_0^t\sigma_s^2\,ds\), so \(\mathbb{E}[e^{M_t}]=\exp\big(\tfrac12\int_0^t\sigma_s^2\,ds\big)\). The compensator \(-\tfrac12\langle M\rangle_t\) in \(Z_t\) cancels this factor exactly, and \(\mathbb{E}[Z_t]=1\). No localization is needed, because the moment generating function of a normal variable is finite.

Worked Example: Volatility Growing Linearly in Time

Let \(\sigma_s=s\) and \(t=2\). Then \(\langle M\rangle_2=\int_0^2s^2\,ds=8/3\), so \(M_2\sim N(0,8/3)\) and \[ \mathbb{E}[e^{M_2}]=e^{4/3}\approx3.7937,\qquad \mathbb{E}[Z_2]=e^{4/3}\cdot e^{-4/3}=1. \] Without the compensator the exponential would have mean about \(3.79\), nearly four times too large. This shows how quickly the missing \(-\tfrac12\sigma^2\) term distorts expectations when volatility grows.

The most familiar case is geometric Brownian motion. Take \(S_t=S_0\exp\big((\mu-\tfrac12\sigma^2)t+\sigma W_t\big)\) with constants \(\mu\) and \(\sigma\). Here \(X_t=\log S_t\) has drift \(b=\mu-\tfrac12\sigma^2\), so the process \(e^{-\mu t}S_t\) has exponent drift \(-\tfrac12\sigma^2\) and is a martingale, by the theorem with bounded \(\sigma\). Therefore \(\mathbb{E}[S_t]=S_0e^{\mu t}\).

Worked Example: Mean Versus Median of a Price

Let \(S_0=50\), \(\mu=0.08\), \(\sigma=0.3\) and \(t=1\). The mean is \(\mathbb{E}[S_1]=50e^{0.08}\approx54.16\). The median of \(S_1\) is obtained by setting the Gaussian part to zero: \(50\exp(0.08-\tfrac12\cdot0.09)=50e^{0.035}\approx51.78\). The mean exceeds the median because the exponential of a symmetric variable is right-skewed. A martingale statement concerns the mean, so it says nothing about the typical outcome, which is the median.

Exponential Martingales as Change-of-Measure Densities

The main use of a martingale \(Y\) with \(Y_0=1\) is as a density process. Define a new measure by \(d\mathbb{Q}=Y_T\,d\mathbb{P}\). This is a probability measure exactly when \(\mathbb{E}[Y_T]=1\), that is, when the supermartingale inequality is an equality at time \(T\). If \(\mathbb{E}[Y_T]<1\), then \(\mathbb{Q}(\Omega)<1\) and the construction fails, which is why the martingale question has practical weight.

Worked Example: A Shifted Brownian Motion

Let \(\sigma\equiv0.4\) and \(T=2\), so \(Y_T=\exp(0.4W_2-0.16)\) is a martingale by the theorem. Under \(\mathbb{Q}\), Girsanov's theorem says \(W_t-0.4t\) is a Brownian motion, so \(W_2\) has mean \(0.8\). We can verify this directly with the Gaussian identity \(\mathbb{E}[W_2e^{0.4W_2}]=0.4\cdot2\cdot e^{0.16}\): then \(\mathbb{E}_{\mathbb{Q}}[W_2]=e^{-0.16}\cdot0.8\,e^{0.16}=0.8\). The exponential martingale tilts the Brownian paths upward by the drift \(\sigma\).

For a bounded stopping time \(\tau\le T\), optional stopping gives \(\mathbb{E}[Y_\tau]=\mathbb{E}[Y_0]\) when \(Y\) is a martingale, but only \(\mathbb{E}[Y_\tau]\le\mathbb{E}[Y_0]\) when \(Y\) is a mere supermartingale. This one-sided inequality is the version that survives in every case.

Why Localization Is Needed

It is tempting to skip the supermartingale step and argue directly that \(\mathbb{E}[Z_t\mid\mathcal{F}_s]=Z_s\). The difficulty is that the stochastic integral \(\int\sigma\,dW\) is only a local martingale in general, and the exponential of a local martingale can have larger tails than the martingale itself. The stopping times \(\tau_n\) cut the process off before it grows too large, where all expectations are finite and the martingale identity holds. The price of this device is the limit \(n\to\infty\), and Step 3 of the proof is precisely the argument that the limit can be taken inside the conditional expectation. Uniform integrability is what makes the interchange legitimate; without it, Fatou's lemma yields only an inequality.

Boundedness of \(\sigma\) is the cleanest way to obtain that uniform integrability, but any condition producing an \(L^p\) bound with \(p>1\) on the stopped exponentials will do. For instance, if \(\mathbb{E}\big[\exp\big(c\,\langle M\rangle_T\big)\big]<\infty\) for a large enough constant \(c\), the same computation as in Step 1 shows that \(\{Z_{t\wedge\tau}\}\) is bounded in \(L^p\) for some \(p>1\). Novikov's condition is the sharp version of this idea with \(p\) taken to its limit, which is why it features in the study of the stochastic exponential that follows.

How to Decide in Practice

A short checklist covers most problems. First, apply Ito's formula and read off the drift of \(e^{X}\); if it is not zero, the process is not even a local martingale. Second, if it is zero, the process is a supermartingale and \(\mathbb{E}[Y_t]\le\mathbb{E}[Y_0]\). Third, test whether \(\sigma\) is bounded and \(\mathbb{E}[e^{X_0}]\) is finite; if both hold, the process is a true martingale. Fourth, if \(\sigma\) is unbounded, turn to Novikov's or Kazamaki's criterion, or compute \(\mathbb{E}[Y_t]\) directly and test whether it equals \(\mathbb{E}[Y_0]\), using the equality criterion proved above.

Common Mistakes

The most frequent error is forgetting the Ito correction and declaring \(e^{X}\) a martingale because \(X\) has zero drift. A driftless \(X\) gives \(\mathbb{E}[e^{X_t}]=e^{\sigma^2t/2}\) in the constant case, which grows. A second error is to conclude from "local martingale" that \(\mathbb{E}[Y_t]=\mathbb{E}[Y_0]\); the supermartingale inequality is all that holds without extra conditions. A third is to check boundedness of \(\sigma\) but ignore \(\mathbb{E}[e^{X_0}]\) when the starting point is random, for example lognormal with a very large variance.

Takeaway: Ito's formula gives \(d e^{X}=e^{X}\big((b+\tfrac12\sigma^2)dt+\sigma\,dW\big)\), so \(e^{X}\) is a local martingale exactly when \(b=-\tfrac12\sigma^2\). Such a process is always a supermartingale. It is a true martingale when \(\sigma\) is bounded and \(\mathbb{E}[e^{X_0}]<\infty\), and for unbounded \(\sigma\) one needs a further criterion such as Novikov's.

Check Your Understanding

Work each question before looking back at the arguments above.

For \(X_t=\mu t+\sigma W_t\), find the value of \(\mu\) that makes \(e^{X_t}\) a martingale when \(\sigma=0.5\).
Where does the proof of the theorem use the bound \(|\sigma|\le C\)?
Why is a supermartingale with constant expectation a martingale?
Suppose \(X_0\) is independent of \(W\) and \(e^{X_0}\) has infinite mean. What happens to \(\mathbb{E}[Y_t]\), even when \(\sigma\equiv1\)?
Why does the example with \(\sigma_s=W_s\) fall outside the theorem, and what replaces it?