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Cumulative Problem Sets and Mixed Proofs · Tutorial 576 of 600

Exponential Martingales and Geometric Brownian Moments

Use Ito's correction and conditional expectations to test exponential martingales and compute geometric Brownian moments.

Advanced 12 min read

What You'll Learn

  • Apply Ito's formula to the stochastic exponential and identify its drift cancellation.
  • Verify the true martingale property for a constant-kernel exponential using conditional Gaussian expectations.
  • State Novikov's condition and distinguish its sufficient condition from a necessary one.
  • Derive geometric Brownian motion's first and second moments and check them independently.
  • Determine exactly when a geometric Brownian motion is a martingale, including degenerate cases.

From Ito's Correction to an Exponential Martingale

Exponentials turn Brownian motion into positive processes that often appear to be martingales. The decisive point is the correction in Ito's formula: the quadratic-variation term cancels the deterministic drift placed in the exponent. That cancellation produces a local martingale, but a separate argument is needed to establish a true martingale. For constant coefficients, conditional Gaussian expectations give that argument directly.

Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a Brownian motion \(B_t\) relative to the filtration. For a fixed finite horizon \(T\), an Ito integral \(\int_0^t H_s\,dB_s\) is used with a progressively measurable integrand \(H\) satisfying \(\mathbb{E}\int_0^T H_s^2\,ds<\infty\); this is the class \(\mathcal{L}^2\). An integral whose integrand is only locally square-integrable is understood after stopping at times that make it square-integrable.

Definition: For a progressively measurable process \(\theta_t\) with \(\int_0^T\theta_s^2\,ds<\infty\) almost surely, its stochastic exponential is \(Z_t=\exp\!\left(\int_0^t\theta_s\,dB_s-\tfrac12\int_0^t\theta_s^2\,ds\right)\). It is positive and satisfies \(dZ_t=\theta_tZ_t\,dB_t\) locally. A local martingale need only become a martingale after suitable stopping; a martingale, by contrast, is integrable at each time and satisfies \(\mathbb{E}[Z_t\mid\mathcal{F}_s]=Z_s\) for \(s\le t\).

For the general expression, the Ito integral is defined locally when its integrand is locally square-integrable; in the constant-kernel results below, the relevant integrands are in \(\mathcal{L}^2\) on every finite horizon. A useful sufficient condition for the general stochastic exponential to be a true martingale on \([0,T]\) is Novikov's condition: \(\mathbb{E}\exp\!\left(\tfrac12\int_0^T\theta_s^2\,ds\right)<\infty\). Novikov's theorem is a standard result; its proof belongs with the later material on Girsanov's theorem. The condition is sufficient, not necessary. We will verify the constant-kernel case directly instead.

The Constant-Kernel Exponential

Theorem: Let \(\theta\in\mathbb{R}\) be constant and set \(Z_t=\exp(\theta B_t-\tfrac12\theta^2t)\). On every finite time interval, \(Z\) is a true martingale and \(\mathbb{E}[Z_t]=1\), \(\mathbb{E}[Z_t^2]=e^{\theta^2t}\). For \(\theta=0\), these identities give \(Z_t=1\).

Proof. Take \(f(t,x)=\exp(\theta x-\tfrac12\theta^2t)\). Its derivatives are \(f_t=-\tfrac12\theta^2f\), \(f_x=\theta f\), and \(f_{xx}=\theta^2f\). Ito's formula, including its second-derivative correction, gives

$$ dZ_t=f_t\,dt+f_x\,dB_t+\tfrac12 f_{xx}\,dt =-\tfrac12\theta^2Z_t\,dt+\theta Z_t\,dB_t+\tfrac12\theta^2Z_t\,dt =\theta Z_t\,dB_t. $$

Thus Ito's formula gives a local martingale. For the true martingale property, use the independent Brownian increment \(B_t-B_s\), which is independent of \(\mathcal{F}_s\) and has distribution \(N(0,t-s)\). The moment-generating function of this normal variable yields

$$ \mathbb{E}[Z_t\mid\mathcal{F}_s] =Z_s\,\mathbb{E}\!\left[\exp\!\left(\theta(B_t-B_s)-\tfrac12\theta^2(t-s)\right)\middle|\mathcal{F}_s\right] =Z_s e^{\theta^2(t-s)/2}e^{-\theta^2(t-s)/2} =Z_s. $$

This conditional expectation is well-defined: the same Gaussian calculation gives \(\mathbb{E}[Z_t]=1\), so \(Z_t\) is integrable. For the second moment, \(Z_t^2=\exp(2\theta B_t-\theta^2t)\), and the normal moment-generating function gives \(\mathbb{E}[Z_t^2]=e^{2\theta^2t-\theta^2t}=e^{\theta^2t}\). When \(\theta=0\), the formula itself gives \(Z_t=1\), and all the claims still hold. \(\square\)

Worked Example: Checking a Constant Exponential

Set \(\theta=\tfrac32\) and \(t=2\). Compute the first and second moments of \(Z_t\), and verify the conditional martingale identity from time \(s=1\).

The exponent is \(\tfrac32 B_t-\tfrac12(\tfrac94)t\). At \(t=2\), this is \(\tfrac32B_2-\tfrac94\). The theorem gives

$$ \mathbb{E}[Z_2]=1,\qquad \mathbb{E}[Z_2^2]=e^{(9/4)2}=e^{9/2}. $$

For \(s=1\), the increment \(B_2-B_1\) is independent of \(\mathcal{F}_1\) and has variance \(1\). Therefore its exponential factor has conditional expectation \(e^{(1/2)(9/4)}e^{-9/8}=e^{9/8}e^{-9/8}=1\), so \(\mathbb{E}[Z_2\mid\mathcal{F}_1]=Z_1\). Check: taking expectations of this conditional identity gives \(\mathbb{E}[Z_2]=\mathbb{E}[Z_1]=1\), in agreement with the direct first-moment calculation.

Geometric Brownian Motion and Its Moments

Let \(S_0\ge0\), and let \(\mu,\sigma\in\mathbb{R}\) be constants. The geometric Brownian motion (GBM) with drift \(\mu\) and diffusion coefficient \(\sigma\) is the process solving \(dS_t=\mu S_t\,dt+\sigma S_t\,dB_t\), with initial value \(S_0\). Here the drift is the coefficient of \(dt\), and the diffusion coefficient is the coefficient of \(dB_t\). Its explicit candidate is

$$ S_t=S_0\exp\!\left((\mu-\tfrac12\sigma^2)t+\sigma B_t\right). $$
Proposition: The displayed process solves the GBM equation. For every \(t\ge0\), \(\mathbb{E}[S_t]=S_0e^{\mu t}\) and \(\mathbb{E}[S_t^2]=S_0^2e^{(2\mu+\sigma^2)t}\). If \(S_0>0\), it is a martingale on a time interval precisely when \(\mu=0\). If \(S_0=0\), it is the zero process and hence a martingale for every \(\mu,\sigma\).

Proof. First suppose \(S_0>0\), and apply Ito's formula to \(f(t,x)=S_0\exp((\mu-\tfrac12\sigma^2)t+\sigma x)\). The derivatives are \(f_t=(\mu-\tfrac12\sigma^2)f\), \(f_x=\sigma f\), and \(f_{xx}=\sigma^2f\). Hence

$$ dS_t=(\mu-\tfrac12\sigma^2)S_t\,dt+\sigma S_t\,dB_t +\tfrac12\sigma^2S_t\,dt =\mu S_t\,dt+\sigma S_t\,dB_t. $$

The Gaussian moment-generating function gives, for each real \(p\), \(\mathbb{E}[S_t^p]=S_0^p\exp((p\mu+\tfrac12p(p-1)\sigma^2)t)\) when \(S_0>0\). In particular, \(p=1\) and \(p=2\) give the stated moments. For \(0\le s\le t\), independence of \(B_t-B_s\) from \(\mathcal{F}_s\) gives \(\mathbb{E}[S_t\mid\mathcal{F}_s]=S_s e^{\mu(t-s)}\). The conditional expectation exists because the first moment is finite. If \(\mu=0\), this equals \(S_s\), proving the martingale property. If \(\mu\ne0\) and \(S_0>0\), then for any \(t>s\), \(e^{\mu(t-s)}\ne1\), so \(S_s e^{\mu(t-s)}\ne S_s\) almost surely; the martingale condition fails. If \(S_0=0\), the explicit solution is identically zero, including when \(\sigma=0\), and its conditional expectations are zero. \(\square\)

The moment formulas can also be checked using Ito's formula on powers of the process. For \(f(x)=x^2\), the derivatives are \(f'(x)=2x\) and \(f''(x)=2\); the correction term is therefore \(\tfrac12 f''(S_t)(dS_t)^2=\sigma^2S_t^2\,dt\). The resulting identity is

$$ d(S_t^2)=(2\mu+\sigma^2)S_t^2\,dt+2\sigma S_t^2\,dB_t. $$

Its stochastic integral has zero mean: the integrand \(2\sigma S_t^2\) is progressively measurable and in \(\mathcal{L}^2\) on \([0,T]\), since \(\mathbb{E}\int_0^T4\sigma^2S_s^4\,ds<\infty\) by the Gaussian fourth-moment formula above. The Ito isometry then gives a square-integrable martingale with zero expectation. Taking expectations in the integral identity yields the moment equation corresponding to the displayed second-moment formula. This provides an Ito-based check independent of directly evaluating the second moment at the final time.

Worked Example: First and Second Moments at a Fixed Time

Take \(S_0=4\), \(\mu=\tfrac14\), \(\sigma=\tfrac12\), and \(t=2\). Find the mean, second moment, and variance.

The first-moment exponent is \(\mu t=(\tfrac14)2=\tfrac12\), so \(\mathbb{E}[S_2]=4e^{1/2}\). The second-moment exponent is \((2\mu+\sigma^2)t=(\tfrac12+\tfrac14)2=\tfrac32\), so \(\mathbb{E}[S_2^2]=4^2e^{3/2}=16e^{3/2}\). Subtracting the squared mean gives

$$ \operatorname{Var}(S_2)=16e^{3/2}-(4e^{1/2})^2 =16e^{3/2}-16e =16e(e^{1/2}-1). $$

Check: the general variance formula is \(S_0^2e^{2\mu t}(e^{\sigma^2t}-1)\). Substitution gives \(16e^{(1/2)2}(e^{(1/4)2}-1)=16e(e^{1/2}-1)\), exactly the same result.

When Is the Process a Martingale?

The condition \(\mu=0\) concerns the process under the stated probability measure \(\mathbb{P}\). It is not a statement about a risk-neutral measure \(\mathbb{Q}\), nor does it assert that a discounted asset price is a martingale. If a constant interest rate \(r\) is introduced, then the conditional expectation of the discounted process is \(\mathbb{E}[e^{-rt}S_t\mid\mathcal{F}_s]=e^{-rs}S_s e^{(\mu-r)(t-s)}\). For \(S_0>0\), this is a martingale under \(\mathbb{P}\) exactly when \(\mu=r\). A change from \(\mathbb{P}\) to \(\mathbb{Q}\) is a separate operation and is not needed for these moment calculations.

Worked Example: A GBM That Is Not a Martingale

Let \(S_0=3\), \(\mu=\tfrac13\), and \(\sigma=\tfrac23\). Compare the conditional expectation from time \(s=1\) to \(t=2\) with \(S_1\).

The conditional-moment formula gives \(\mathbb{E}[S_2\mid\mathcal{F}_1]=S_1e^{(1/3)(2-1)}=S_1e^{1/3}\). Since \(S_1=3\exp((\tfrac13-\tfrac12\cdot\tfrac49)+\tfrac23B_1)=3\exp(\tfrac19+\tfrac23B_1)>0\), the factor \(e^{1/3}\) is strictly greater than \(1\), and the conditional expectation is not \(S_1\). Thus the process is not a martingale. Its mean at time \(2\), computed independently from the first-moment formula, is \(3e^{(1/3)2}=3e^{2/3}\). Check: \(\mathbb{E}[S_1e^{1/3}]=e^{1/3}(3e^{1/3})=3e^{2/3}\), agreeing with that unconditional mean.

Common Pitfall: Omitting the Ito Correction

A tempting but incorrect calculation is to differentiate \(S_0e^{\mu t+\sigma B_t}\) and conclude that it solves \(dS_t=\mu S_t\,dt+\sigma S_t\,dB_t\). The missing term is the second derivative correction: for \(f(t,x)=S_0e^{\mu t+\sigma x}\), \(f_t=\mu f\), \(f_x=\sigma f\), and \(f_{xx}=\sigma^2f\), so Ito's formula gives drift \((\mu+\tfrac12\sigma^2)f\), not \(\mu f\). The correct exponent must subtract \(\tfrac12\sigma^2t\).

A second error is to infer that every positive local martingale is a true martingale. The local-martingale conclusion from Ito's formula alone does not establish integrability and the conditional-expectation identity. In the constant-kernel exponential, those properties were verified explicitly using the normal moment-generating function. For more general kernels, conditions such as Novikov's condition provide a sufficient route, but they are not automatic.

Process or conditionConclusionReason
\(\exp(\theta B_t-\tfrac12\theta^2t)\), constant \(\theta\)True martingaleConditional Gaussian expectation equals its current value
GBM with \(S_0>0\) under \(\mathbb{P}\)Martingale exactly when \(\mu=0\)Conditional mean is \(S_se^{\mu(t-s)}\)
GBM with \(S_0=0\)Zero martingale for all coefficientsThe explicit solution is identically zero
General stochastic exponential satisfying NovikovTrue martingale on the stated finite horizonNovikov's condition is sufficient

Check Your Understanding

Use the conditional-expectation identities and moment formulas above to justify each answer.

  1. For \(\theta=-2\), compute \(\mathbb{E}[Z_3]\) and \(\mathbb{E}[Z_3^2]\).
  2. For GBM with \(S_0=2\), \(\mu=0\), and \(\sigma=1\), find the first and second moments at time \(t\).
  3. Explain why the Ito formula for \(S_t^2\) contains the term \(\sigma^2S_t^2\,dt\).
  4. For \(S_0>0\), use the conditional expectation to decide whether GBM with \(\mu=-1\) is a martingale on a nontrivial time interval.
  5. State Novikov's condition and explain why it is sufficient rather than necessary for the conclusion it guarantees.