Drift Corrections And Common Geometric Brownian Motion Errors

The Drift Correction in Geometric Brownian Motion

For geometric Brownian motion, two drift rates appear in closely related formulas, and confusing them changes the model. The level process has arithmetic drift rate $\mu$, while the logarithm of the process has drift rate $\mu-\tfrac12\sigma^2$. The difference comes from the second-derivative term in Itô’s formula. In particular, the exponent in the explicit solution does not have drift $\mu$.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a Brownian motion $B_t$ with respect to this filtration: for $s<t$, $B_t-B_s$ is independent of $\mathcal{F}_s$ and has distribution $N(0,t-s)$. Fix constants $S_0>0$, $\mu\in\mathbb{R}$, and $\sigma\in\mathbb{R}$. Consider the Itô SDE

dS_t=\mu S_t\,dt+\sigma S_t\,dB_t.

Here $\mu S_t$ is the drift coefficient in the level equation, $\sigma S_t$ is its diffusion coefficient, and $\mu$ is the arithmetic drift rate. The stochastic integral is an Itô integral: on every finite time interval $[0,T]$, its integrand $\sigma S_t$ is progressively measurable and belongs to $\mathcal{L}^2$, the class of progressively measurable $H_t$ such that $\mathbb{E}\int_0^T H_t^2\,dt<\infty$. We verify this integrability below.

Definition: For this SDE, the logarithmic drift rate is the coefficient of $dt$ in the Itô equation for $\log S_t$. It is $\mu-\tfrac12\sigma^2$. It is a rate, not the level-process drift coefficient $\mu S_t$; the two rates $\mu$ and $\mu-\tfrac12\sigma^2$ differ by $\tfrac12\sigma^2$.

Deriving and Verifying the Explicit Solution

Theorem: The unique strong solution of the SDE with $S_{t=0}=S_0>0$ is $$ S_t=S_0\exp\!\left(\left(\mu-\tfrac12\sigma^2\right)t+\sigma B_t\right). $$ In particular, $S_t>0$ for every $t\ge0$.

Proof. The drift and diffusion coefficients $x\mapsto\mu x$ and $x\mapsto\sigma x$ are globally Lipschitz: their differences at $x,y$ are bounded by $|\mu||x-y|$ and $|\sigma||x-y|$, respectively. They also satisfy linear growth, since $|\mu x|^2+|\sigma x|^2=(\mu^2+\sigma^2)|x|^2$. The standard existence-and-uniqueness theorem for globally Lipschitz SDE coefficients therefore gives a unique strong solution. To identify it, define $G(t,b)=S_0\exp((\mu-\tfrac12\sigma^2)t+\sigma b)$. Its derivatives are $G_t=(\mu-\tfrac12\sigma^2)G$, $G_b=\sigma G$, and $G_{bb}=\sigma^2G$. Itô’s formula, with the second-derivative correction shown, gives

dG(t,B_t)=G_t\,dt+G_b\,dB_t+\tfrac12G_{bb}\,dt =\left(\mu-\tfrac12\sigma^2\right)G\,dt+\sigma G\,dB_t+\tfrac12\sigma^2G\,dt =\mu G\,dt+\sigma G\,dB_t.

Since $G(0,B_0)=S_0$, this candidate satisfies the SDE, and uniqueness identifies it with $S_t$. Its exponential form is strictly positive. Moreover, the normal moment-generating function gives $\mathbb{E}[S_t^2]=S_0^2e^{(2\mu+\sigma^2)t}$, so $\mathbb{E}\int_0^T(\sigma S_t)^2\,dt=\sigma^2S_0^2\int_0^T e^{(2\mu+\sigma^2)t}\,dt<\infty$; if $\sigma=0$, the left side is zero. Thus the Itô integral in the SDE has an integrand in $\mathcal{L}^2$. $\square$

The same correction can be derived by applying Itô’s formula to the logarithm. Because the solution is positive, $f(x)=\log x$ is defined along its paths. Its derivatives are $f'(x)=1/x$ and $f''(x)=-1/x^2$. Using $dS_t=\mu S_t\,dt+\sigma S_t\,dB_t$ and $dB_t\,dB_t=dt$, the formula gives

d\log S_t =f'(S_t)\,dS_t+\tfrac12 f''(S_t)(\sigma S_t)^2\,dt =\mu\,dt+\sigma\,dB_t-\tfrac12\sigma^2\,dt =\left(\mu-\tfrac12\sigma^2\right)dt+\sigma\,dB_t.

Equivalently, integrating this identity from $0$ to $t$ gives $\log S_t=\log S_0+(\mu-\tfrac12\sigma^2)t+\sigma B_t$. The correction is a consequence of quadratic variation, $[B]_t=t$, not of an ordinary chain rule. In the Itô multiplication table, $dB_t\,dB_t=dt$, $dB_t\,dt=0$, and $dt\,dt=0$.

Worked Example: Separate the Two Drift Rates

Take $S_0=100$, $\mu=\tfrac14$, and $\sigma=\tfrac12$. Find the level equation’s drift coefficient at time zero, the logarithmic drift rate, and the explicit solution.

The level drift coefficient is $\mu S_0=\tfrac14\cdot100=25$. The correction is $\tfrac12\sigma^2=\tfrac12\cdot(\tfrac12)^2=\tfrac18$. Thus the logarithmic drift rate is $\mu-\tfrac12\sigma^2=\tfrac14-\tfrac18=\tfrac18$, and

S_t=100\exp\!\left(\tfrac18t+\tfrac12B_t\right).

Check: Applying Itô’s formula to $G(t,b)=100e^{t/8+b/2}$ gives $G_t=G/8$, $G_b=G/2$, and $G_{bb}=G/4$. Therefore $dG=(G/8+G/8)dt+(G/2)dB_t=(G/4)dt+(G/2)dB_t$, which is exactly $dS_t=\tfrac14S_t\,dt+\tfrac12S_t\,dB_t$. The level drift coefficient at time zero is $25$, whereas the two rates being compared are $\tfrac14$ and $\tfrac18$; those are not interchangeable quantities.

Why the Exponential Martingale Does Not Remove the Correction

The factor $\exp(\sigma B_t-\tfrac12\sigma^2t)$ is a true martingale, not merely a local martingale. This fact is a useful check on drift calculations. A local martingale is a process that becomes a martingale after stopping at a suitable sequence of stopping times; without an integrability argument, that property alone does not guarantee the conditional-expectation identity required of a true martingale.

Proposition: For every finite $T$, $N_t=\exp(\sigma B_t-\tfrac12\sigma^2t)$, $0\le t\le T$, is an integrable martingale. Consequently $e^{-\mu t}S_t=S_0N_t$ is a martingale, and $\mathbb{E}[S_t\mid\mathcal{F}_s]=S_s e^{\mu(t-s)}$ for $0\le s\le t$.

Proof. The normal moment-generating function gives $\mathbb{E}[N_t]=e^{-\sigma^2t/2}\mathbb{E}[e^{\sigma B_t}]=1$, so $N_t$ is integrable. For $s\le t$, write $B_t=B_s+(B_t-B_s)$. The increment is independent of $\mathcal{F}_s$, and $B_t-B_s$ has distribution $N(0,t-s)$. Hence

\mathbb{E}[N_t\mid\mathcal{F}_s] =N_s e^{-\sigma^2(t-s)/2}\mathbb{E}[e^{\sigma(B_t-B_s)}] =N_s e^{-\sigma^2(t-s)/2}e^{\sigma^2(t-s)/2} =N_s.

The explicit solution gives $e^{-\mu t}S_t=S_0N_t$, so this discounted process is also an integrable martingale. Multiplying the conditional identity by $e^{\mu t}$ gives $\mathbb{E}[S_t\mid\mathcal{F}_s]=S_s e^{\mu(t-s)}$. In particular, $S$ itself is a martingale when $\mu=0$; for general $\mu$, its conditional mean has the factor $e^{\mu(t-s)}$. $\square$

This conditional-expectation calculation depends on Brownian increments being independent of the stated filtration. Adaptedness by itself would not justify the step. Also, $S$ is not generally a martingale just because its solution contains a Brownian exponential: the deterministic factor $e^{\mu t}$ remains.

Worked Example: A Wrong Exponent Fails Itô’s Formula

Set $\mu=0$, $\sigma=2$, and $S_0=1$. A common incorrect candidate is $\widetilde S_t=e^{2B_t}$, obtained by putting $\mu=0$ in the exponent and omitting the correction. Test it directly.

Use $F(t,b)=e^{2b}$. Its derivatives are $F_t=0$, $F_b=2F$, and $F_{bb}=4F$. Itô’s formula gives

d\widetilde S_t=0\,dt+2\widetilde S_t\,dB_t+\tfrac12(4\widetilde S_t)\,dt =2\widetilde S_t\,dt+2\widetilde S_t\,dB_t.

The required SDE is $dS_t=2S_t\,dB_t$, with zero drift, so the candidate does not solve it. The corrected solution is $S_t=e^{-2t+2B_t}$, since $\mu-\tfrac12\sigma^2=0-\tfrac12\cdot4=-2$.

Check: For the incorrect candidate, the normal moment-generating function gives $\mathbb{E}[\widetilde S_t]=\mathbb{E}[e^{2B_t}]=e^{(2^2)t/2}=e^{2t}$, not $1$. For the corrected candidate, $\mathbb{E}[S_t]=e^{-2t}\mathbb{E}[e^{2B_t}]=e^{-2t}e^{2t}=1$, consistent with the proposition because $\mu=0$. Both the SDE substitution and the expectation test expose the missing correction.

Arithmetic Drift, Logarithmic Drift, and Stratonovich Form

“Drift” must be interpreted in context. In the Itô level equation, the drift coefficient is $\mu S_t$; its rate relative to the level is $\mu$. In the Itô equation for $\log S_t$, the drift coefficient is the constant rate $\mu-\tfrac12\sigma^2$. These are statements about different processes. In particular, the logarithmic drift rate is not the expected rate of change of $S_t$ itself.

The Stratonovich form provides another place where a correction appears. For a scalar diffusion coefficient $b(x)=\sigma x$, conversion from Stratonovich drift $a(x)$ to Itô drift adds $\tfrac12 b(x)b'(x)=\tfrac12\sigma^2x$. Thus the same Itô SDE can be written

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

This Stratonovich drift rate happens to equal the logarithmic drift rate for this model. They arise for different reasons: one is the drift rate in the Stratonovich equation for $S_t$, and the other is the drift rate in the Itô equation for $\log S_t$. The Stratonovich chain rule is the ordinary chain rule for Stratonovich integrals; it does not permit replacing the Itô equation’s correction by zero while keeping its Itô interpretation.

Worked Example: Check a Logarithmic Mean and a Level Mean

Let $\mu=\tfrac12$, $\sigma=1$, and $S_0=1$. Find the solution at time $t=2$, its expected logarithm, and its expected level.

The logarithmic drift rate is $\tfrac12-\tfrac12\cdot1^2=0$, so $S_2=e^{B_2}$. Since $B_2\sim N(0,2)$, $\mathbb{E}[B_2]=0$, and therefore $\mathbb{E}[\log S_2]=\mathbb{E}[B_2]=0$. In contrast, the normal moment-generating function gives $\mathbb{E}[S_2]=\mathbb{E}[e^{B_2}]=e^{1^2\cdot2/2}=e$. More generally, for these parameters the explicit solution gives $S_t=e^{B_t}$, and hence $\mathbb{E}[S_t]=e^{t/2}=S_0e^{\mu t}$.

Check: At $t=2$, the mean of $\log S_2$ is $0$, while the logarithm of the mean level is $\log(e)=1$. These are not equal: in general, $\mathbb{E}[\log S_t]$ is not $\log\mathbb{E}[S_t]$. The level mean $e$ also agrees with $S_0e^{\mu t}=1\cdot e^{(1/2)2}=e$.

A Practical Error-Checking Checklist

When a proposed geometric Brownian motion formula is written down, check the object and the convention before interpreting its drift. The following comparisons keep the distinct rates separate.

Quantity	For the Itô SDE	Interpretation
Level drift coefficient	$\mu S_t$	Coefficient of $dt$ in the equation for $S_t$
Arithmetic drift rate	$\mu$	Rate in $\mathbb{E}[S_t]=S_0e^{\mu t}$
Logarithmic drift rate	$\mu-\tfrac12\sigma^2$	Coefficient of $dt$ in the Itô equation for $\log S_t$
Exponent’s Brownian coefficient	$\sigma$	Coefficient of $B_t$ in the explicit solution

The recurring incorrect calculation is to assert $S_t=S_0e^{\mu t+\sigma B_t}$ for an Itô SDE with drift $\mu S_t$. Applying Itô’s formula to that candidate adds $\tfrac12\sigma^2S_t\,dt$, so its actual Itô drift is $(\mu+\tfrac12\sigma^2)S_t$, not $\mu S_t$. The correction belongs in the exponent with a minus sign: $-\tfrac12\sigma^2t$. This diagnosis is specific to the Itô interpretation; switching to a Stratonovich equation changes the drift coefficient and must be done by the conversion rule.

Check Your Understanding

For each question, distinguish rates from coefficients and state which stochastic-integral interpretation is being used.

For $dS_t=\tfrac13S_t\,dt+2S_t\,dB_t$, what is the logarithmic drift rate, and what is the exponential solution?
Apply Itô’s formula to $e^{\mu t+\sigma B_t}$. What drift coefficient does this candidate actually have?
For which value of $\mu$ is $S_t$ itself a martingale, and which process is a martingale for arbitrary $\mu$?
Explain why independence of $B_t-B_s$ from $\mathcal{F}_s$ is needed to verify the conditional expectation of the exponential martingale.
Write the Stratonovich drift rate corresponding to the Itô level equation, and explain why it should not be confused with the level drift coefficient $\mu S_t$.

Quantity	For the Itô SDE	Interpretation
Level drift coefficient	\(\mu S_t\)	Coefficient of \(dt\) in the equation for \(S_t\)
Arithmetic drift rate	\(\mu\)	Rate in \(\mathbb{E}[S_t]=S_0e^{\mu t}\)
Logarithmic drift rate	\(\mu-\tfrac12\sigma^2\)	Coefficient of \(dt\) in the Itô equation for \(\log S_t\)
Exponent’s Brownian coefficient	\(\sigma\)	Coefficient of \(B_t\) in the explicit solution