Solving Geometric Brownian Motion with Ito's Formula

From a Multiplicative SDE to an Additive One

Geometric Brownian motion models a quantity whose instantaneous changes are proportional to its current level. That proportionality makes the equation nonlinear in appearance, but taking a logarithm turns it into an additive process. The transformation is not the ordinary chain rule: Brownian motion has quadratic variation, so Ito’s formula contributes a drift correction. Keeping that correction is essential to obtaining the solution.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a standard Brownian motion $B_t$. Fix constants $\mu,\sigma\in\mathbb{R}$, a finite time horizon $T>0$, and a deterministic initial value $s_0>0$. The drift coefficient is $\mu S_t$, and the diffusion coefficient is $\sigma S_t$. A solution on $[0,T]$ means a continuous adapted process satisfying the integral equation below, with progressively measurable integrands and $\mathbb{E}\int_0^T S_t^2\,dt<\infty$.

S_t=s_0+\int_0^t\mu S_s\,ds+\int_0^t\sigma S_s\,dB_s,\qquad 0\le t\le T.

The stochastic integral here and throughout is the Ito integral. Its integrand $\sigma S_s$ is progressively measurable and square-integrable in expectation over each finite time horizon. The logarithm is initially defined only while $S_t>0$, so positivity must be justified rather than assumed for all times.

Key idea: For $f(x)=\log x$, the derivatives are $f'(x)=1/x$ and $f''(x)=-1/x^2$. Ito’s second-order term therefore changes the drift from $\mu$ to $\mu-\tfrac12\sigma^2$. The multiplication rules behind this calculation are $dB_t\,dB_t=dt$, $dB_t\,dt=0$, and $dt\,dt=0$.

The Logarithmic Transformation and Positivity

Proposition: If $S_t$ is a continuous solution with $S_0=s_0>0$, then $S_t$ does not hit zero on any finite time interval. Before a possible first hit of zero, it satisfies $$ \log S_t=\log s_0+\left(\mu-\tfrac12\sigma^2\right)t+\sigma B_t. $$

Proof. Let $\tau_0=\inf\{t\ge0:S_t=0\}$, with the infimum of the empty set defined as $+\infty$. Choose integers $n$ sufficiently large that $1/n<s_0$, and define $\tau_n=\inf\{t\ge0:S_t\le1/n\}$. Ito’s formula can be applied to $\log S$ up to $\tau_n$; if needed, first stop also when $S$ reaches an upper level and then let that level increase. This localization keeps the derivatives of the logarithm bounded on the stopped range. Since $f'(x)=1/x$ and $f''(x)=-1/x^2$, the formula, including its second-order term, is $$ d\log S_t=\frac{1}{S_t}(\mu S_t\,dt+\sigma S_t\,dB_t) +\frac12\left(-\frac{1}{S_t^2}\right)(\sigma S_t)^2\,dt =\left(\mu-\tfrac12\sigma^2\right)dt+\sigma\,dB_t. $$ The stopped stochastic integral has integrand $\sigma\mathbf{1}_{\{s\le\tau_n\}}$, which is progressively measurable and satisfies $\mathbb{E}\int_0^T\sigma^2\mathbf{1}_{\{s\le\tau_n\}}\,ds\le\sigma^2T<\infty$. Thus, up to $\tau_n$, integration gives $$ \log S_{t\wedge\tau_n}=\log s_0+\left(\mu-\tfrac12\sigma^2\right)(t\wedge\tau_n)+\sigma B_{t\wedge\tau_n}. $$ If $\tau_0\le T$, continuity implies that each sufficiently small positive level $1/n$ is reached before or at $\tau_0$, that $\tau_n\uparrow\tau_0$, and that $S_{\tau_n}=1/n$. The left side at $t=T$ is then $-\log n$, which tends to $-\infty$. The right side converges to the finite random variable $\log s_0+(\mu-\tfrac12\sigma^2)\tau_0+\sigma B_{\tau_0}$, since $\tau_0\le T$ and Brownian paths are continuous. This is a contradiction. Hence $S$ does not hit zero on $[0,T]$. Letting $n$ increase in the stopped identity now gives the asserted logarithmic formula for every $t\le T$. $\square$

The correction has a specific source: in Ito’s formula the second derivative is multiplied by one half of the squared diffusion coefficient. It is not an adjustment to the original drift coefficient in the SDE; it appears when expressing the dynamics of $\log S_t$.

The Explicit Solution

Theorem: For $s_0>0$, the unique continuous adapted solution of the stated SDE on each finite horizon is $$ S_t=s_0\exp\left(\left(\mu-\tfrac12\sigma^2\right)t+\sigma B_t\right). $$

Proof. The proposition shows that any solution must satisfy the displayed logarithmic identity. Exponentiating that identity gives the stated formula, so there can be at most one solution. To prove existence and verify the formula, set $c=\mu-\tfrac12\sigma^2$ and define $Y_t=s_0e^{ct+\sigma B_t}$. Apply Ito’s formula to $f(t,x)=s_0e^{ct+\sigma x}$. Its derivatives are $f_t=cf$, $f_x=\sigma f$, and $f_{xx}=\sigma^2f$. Consequently, $$ dY_t=f_t(t,B_t)\,dt+f_x(t,B_t)\,dB_t+\tfrac12 f_{xx}(t,B_t)\,dt =\left(c+\tfrac12\sigma^2\right)Y_t\,dt+\sigma Y_t\,dB_t =\mu Y_t\,dt+\sigma Y_t\,dB_t. $$ Also $Y_0=s_0e^0=s_0$. The integrands satisfy the required integrability condition: the Gaussian moment-generating function gives $$ \mathbb{E}[Y_t^2]=s_0^2e^{2ct}\mathbb{E}[e^{2\sigma B_t}] =s_0^2e^{(2\mu+\sigma^2)t}, \qquad \mathbb{E}\int_0^T(\sigma Y_t)^2\,dt =\sigma^2s_0^2\int_0^T e^{(2\mu+\sigma^2)t}\,dt<\infty. $$ The drift integrand is square-integrable as well, since $\mathbb{E}\int_0^T(\mu Y_t)^2\,dt<\infty$. Thus the Ito integral in the verification is defined in the specified class, and $Y$ is a solution. Since every solution was forced by the logarithmic identity to equal $Y$, uniqueness follows. $\square$

This result concerns a strong solution: it is constructed on the given probability space from the Brownian motion $B_t$. The uniqueness proved here is pathwise uniqueness: any two solutions driven by that same Brownian motion and having the same initial value agree. The proof does not require a separate assertion about uniqueness in law.

Worked Solutions

Worked Example: A Positive Drift and a Large Volatility

Solve the equation with $s_0=3$, $\mu=1$, and $\sigma=2$. First compute the drift in the logarithmic equation: $$ \mu-\tfrac12\sigma^2=1-\tfrac12(2^2)=1-2=-1. $$ The logarithmic identity is therefore $\log S_t=\log 3-t+2B_t$. Exponentiating gives $$ S_t=3e^{-t+2B_t}. $$

Check: For $f(t,x)=3e^{-t+2x}$, the derivatives are $f_t=-f$, $f_x=2f$, and $f_{xx}=4f$. Ito’s formula gives $dS_t=(-1+\tfrac12\cdot4)S_t\,dt+2S_t\,dB_t=S_t\,dt+2S_t\,dB_t$, which has the required coefficients. At $t=0$, $B_0=0$, so $S_0=3e^0=3$.

Worked Example: A Vanishing Logarithmic Drift

Take $s_0=4$, $\mu=\tfrac12$, and $\sigma=1$. The correction is $$ \mu-\tfrac12\sigma^2=\tfrac12-\tfrac12(1^2)=0. $$ Thus $\log S_t=\log4+B_t$, and the explicit solution is $S_t=4e^{B_t}$.

Check: For $f(t,x)=4e^x$, the partial derivatives are $f_t=0$, $f_x=f$, and $f_{xx}=f$. Ito’s formula gives $dS_t=\tfrac12S_t\,dt+S_t\,dB_t$, matching $\mu=\tfrac12$ and $\sigma=1$. Its initial value is $4e^{B_0}=4$. The zero drift of $\log S$ does not mean the original process has zero drift: its Ito differential has drift $\tfrac12S_t$.

Worked Example: Detecting a Missing Ito Correction

Let $s_0=2$, $\mu=-1$, and $\sigma=2$. The corrected logarithmic drift is $$ \mu-\tfrac12\sigma^2=-1-\tfrac12(2^2)=-1-2=-3, $$ so $S_t=2e^{-3t+2B_t}$. At a time $t=1$ when evaluating a path with $B_1=\tfrac12$, the exponent is $-3+2(\tfrac12)=-2$, and the formula gives $S_1=2e^{-2}$.

Check: For $f(t,x)=2e^{-3t+2x}$, $f_t=-3f$, $f_x=2f$, and $f_{xx}=4f$. The Ito drift is $(-3+\tfrac12\cdot4)f=-f$, and the diffusion is $2f\,dB_t$, exactly the coefficients in the SDE. The initial value is $2e^0=2$. By contrast, omitting the correction would give the incorrect candidate $2e^{-t+2B_t}$; its Ito drift would be $(-1+2)S_t=S_t$, not $-S_t$.

Common Pitfalls and Scope

Using the ordinary chain rule. The incorrect calculation $d\log S_t=(1/S_t)dS_t$ drops the term $\tfrac12 f''(S_t)(\sigma S_t)^2dt=-\tfrac12\sigma^2dt$. The correct logarithmic drift is $\mu-\tfrac12\sigma^2$, not $\mu$. This is an Ito calculation; it is not a Stratonovich integral or a Stratonovich chain rule.

Taking a logarithm without checking positivity. The logarithmic transform is valid only when $S_t>0$. The stopping argument proves that a solution starting at $s_0>0$ cannot reach zero on a finite horizon. If $s_0=0$, the logarithm is unavailable; the formula with $s_0=0$ gives the identically zero process and can be directly verified, but it is not obtained by taking $\log S_t$.

Confusing the transformed drift with the original drift. The coefficient $\mu-\tfrac12\sigma^2$ describes the drift of $\log S_t$. The original SDE still has drift coefficient $\mu$ multiplying $S_t$. The second example makes this distinction explicit.

The explicit formula is useful because it turns the multiplicative equation into a direct expression in terms of the driving Brownian path. The derivation also shows exactly where quadratic variation enters: the correction is what makes the exponential process satisfy the original Ito SDE rather than an equation with a different drift.

Check Your Understanding

Use Ito’s formula and the solution formula to answer each question.

For $s_0=5$, $\mu=2$, and $\sigma=2$, what is the drift coefficient in the equation for $\log S_t$, and what is the explicit solution?
For $s_0=1$, $\mu=0$, and $\sigma=1$, compute the Ito drift of the candidate $e^{-t/2+B_t}$ by displaying the second-derivative correction.
Why is it necessary to stop before applying Ito’s formula to $\log S_t$ in the positivity proof?
For which value of $\mu$, in terms of $\sigma$, does the logarithmic process have zero drift?
Explain why the expression $s_0e^{\mu t+\sigma B_t}$ generally fails to solve the Ito SDE when $\sigma\ne0$.