Geometric Brownian Motion for Stock Prices

From a Stock SDE to a Price Distribution

In the Black–Scholes model, the stock’s random movement is proportional to its current price. This multiplicative structure means that the stock price itself is not normally distributed; its logarithm is. The explicit solution makes that fact precise and provides direct formulas for conditional distributions, moments and risk-neutral discounted expectations.

Let $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ be a filtered probability space, and let $\mathbb{Q}$ be a probability measure equivalent to $\mathbb{P}$ on the time horizon under study. Assume $B_t$ is a standard Brownian motion with respect to $(\mathcal{F}_t)$ under $\mathbb{Q}$: it is adapted, has continuous paths and independent normal increments, and $B_t-B_s$ is independent of $\mathcal{F}_s$ under $\mathbb{Q}$ for $s<t$. In particular, adaptedness alone is not enough for the Brownian and Itô properties used below. Expectations under the two measures are denoted by $\mathbb{E}$ and $\mathbb{E}^{\mathbb{Q}}$, respectively.

We study the risk-neutral Black–Scholes stock equation $dS_t=rS_t\,dt+\sigma S_t\,dB_t$, with deterministic initial price $S_0>0$, constant interest rate $r\in\mathbb{R}$, and constant volatility $\sigma\ge0$. Thus $rS_t$ is the drift coefficient and $\sigma S_t$ is the diffusion coefficient. Any Itô integral below is taken under $\mathbb{Q}$. For a finite horizon $[0,T]$, its integrand is progressively measurable and belongs to $\mathcal{L}^2(\mathbb{Q})$, meaning $\mathbb{E}^{\mathbb{Q}}\int_0^T H_s^2\,ds<\infty$; where only local square-integrability is available, the integral is understood after stopping.

Definition: A solution of this stock SDE is a continuous adapted process satisfying, for each $t\ge0$, $$ S_t=S_0+\int_0^t rS_s\,ds+\int_0^t \sigma S_s\,dB_s. $$ The stochastic integral is the Itô integral. Under $\mathbb{Q}$, the symbol $dB_t\,dB_t=dt$ records Brownian quadratic variation; the accompanying multiplication rules are $dB_t\,dt=0$ and $dt\,dt=0$.

The Explicit Solution

Theorem: For $S_0>0$, $r\in\mathbb{R}$, and $\sigma\ge0$, the stock SDE has the unique continuous adapted solution $$ S_t=S_0\exp\!\left(\left(r-\frac{\sigma^2}{2}\right)t+\sigma B_t\right). $$ For every finite $T$, the integrand $\sigma S_t$ is in $\mathcal{L}^2(\mathbb{Q})$. The solution is strictly positive at every finite time.

Proof. Define $f(t,x)=S_0\exp((r-\sigma^2/2)t+\sigma x)$. Its derivatives are $f_t=(r-\sigma^2/2)f$, $f_x=\sigma f$, and $f_{xx}=\sigma^2 f$. Itô’s formula, including its second-derivative correction, gives $$ df(t,B_t)=f_t\,dt+f_x\,dB_t+\frac12 f_{xx}\,dt =r f(t,B_t)\,dt+\sigma f(t,B_t)\,dB_t. $$ Also $f(0,B_0)=S_0$, so this candidate satisfies the integral equation once its stochastic integral is square-integrable. The normal moment-generating function gives $\mathbb{E}^{\mathbb{Q}}[S_t^2]=S_0^2e^{(2r+\sigma^2)t}$. This is finite and bounded as a function of $t$ on any finite interval, so $\mathbb{E}^{\mathbb{Q}}\int_0^T(\sigma S_t)^2\,dt =\sigma^2 S_0^2\int_0^T e^{(2r+\sigma^2)t}\,dt<\infty$. The candidate’s stochastic integral is therefore in the stated Itô class. It is positive because it is a positive constant times an exponential.

For uniqueness, let $\widetilde S_t$ be any other continuous adapted solution with the same initial value. Set $G_t=\exp(-(r-\sigma^2/2)t-\sigma B_t)$. For $g(t,x)=\exp(-(r-\sigma^2/2)t-\sigma x)$, the derivatives are $g_t=-(r-\sigma^2/2)g$, $g_x=-\sigma g$, and $g_{xx}=\sigma^2g$. Itô’s formula, with the correction $\frac12g_{xx}\,dt$, yields $dG_t=(-r+\sigma^2)G_t\,dt-\sigma G_t\,dB_t$. The Itô product rule for $G_t\widetilde S_t$ has cross term $(-\sigma G_t)(\sigma\widetilde S_t)dB_t\,dB_t=-\sigma^2G_t\widetilde S_t\,dt$. Its total drift is $(-r+\sigma^2+r-\sigma^2)G_t\widetilde S_t\,dt=0$, and its diffusion coefficient is $-\sigma G_t\widetilde S_t+\sigma G_t\widetilde S_t=0$. Hence $G_t\widetilde S_t=G_0\widetilde S_0=S_0$, first locally and therefore on every finite interval. It follows that $\widetilde S_t=S_0/G_t$, the displayed solution. This proves pathwise uniqueness as well as the formula. $\square$

Worked Example: One Simulated Log-Price Step

Take $S_0=100$, $r=\tfrac1{20}$, $\sigma=\tfrac15$, and a time step $h=\tfrac14$. Suppose a simulation draws a standard normal value $Z=1$. Compute the corresponding stock price after one step using the exact log-price update.

For a Brownian increment over a step of length $h$, write $\Delta B=\sqrt{h}\,Z$. Here $\sqrt{h}=\tfrac12$, so $\Delta B=\tfrac12$. The log-price increment is $$ \Delta\log S=\left(r-\frac{\sigma^2}{2}\right)h+\sigma\Delta B. $$ The drift adjustment is $r-\sigma^2/2=\tfrac1{20}-\tfrac1{50}=\tfrac3{100}$. Thus its time-step contribution is $\tfrac3{100}\cdot\tfrac14=\tfrac3{400}$; the random contribution is $\tfrac15\cdot\tfrac12=\tfrac1{10}=\tfrac{40}{400}$. Consequently $\Delta\log S=\tfrac{43}{400}$, and $S_h=100e^{43/400}$, approximately $111.4$.

Check: Substituting the simulated increment directly into the explicit solution gives $100\exp((\tfrac3{100})(\tfrac14)+(\tfrac15)(\tfrac12)) =100e^{(3+40)/400}=100e^{43/400}$, the same answer. The simulated normal value is not itself the Brownian increment: the factor $\sqrt h$ is essential.

Lognormal Distributions and Stock Moments

Proposition: For $0\le s<t$, conditionally on $\mathcal{F}_s$, $$ \log S_t=\log S_s+\left(r-\frac{\sigma^2}{2}\right)(t-s)+\sigma(B_t-B_s). $$ If $\sigma>0$, this is conditionally normal with mean $\log S_s+(r-\sigma^2/2)(t-s)$ and variance $\sigma^2(t-s)$, so $S_t$ is conditionally lognormal. If $\sigma=0$, the conditional distribution is a point mass at $S_se^{r(t-s)}$.

Proof. The explicit solution gives $S_t/S_s=\exp((r-\sigma^2/2)(t-s)+\sigma(B_t-B_s))$. The ratio is valid because the solution is positive. Taking logarithms gives the claimed identity. By the Brownian assumption, $B_t-B_s$ is independent of $\mathcal{F}_s$ and has distribution $N(0,t-s)$ under $\mathbb{Q}$. Multiplication by $\sigma$ and addition of the stated deterministic term give the conditional normal law when $\sigma>0$. When $\sigma=0$, the random term vanishes and the conditional price is $S_se^{r(t-s)}$. $\square$

Corollary: For every real $p$ and $t\ge0$, $$ \mathbb{E}^{\mathbb{Q}}[S_t^p] =S_0^p\exp\!\left(\left(pr+\frac{p(p-1)}2\sigma^2\right)t\right). $$ In particular, $\mathbb{E}^{\mathbb{Q}}[S_t]=S_0e^{rt}$ and $\operatorname{Var}_{\mathbb{Q}}(S_t)=S_0^2e^{2rt}(e^{\sigma^2t}-1)$.

Proof. Raising the explicit solution to the power $p$ gives $S_t^p=S_0^p\exp(p(r-\sigma^2/2)t+p\sigma B_t)$. Since $B_t\sim N(0,t)$, its moment-generating function gives $\mathbb{E}^{\mathbb{Q}}[e^{p\sigma B_t}]=e^{p^2\sigma^2t/2}$. Multiplication of the exponential factors proves the moment formula, including $p=0$ and $\sigma=0$. Taking $p=1$ gives the mean, and $p=2$ gives $\mathbb{E}^{\mathbb{Q}}[S_t^2]=S_0^2e^{(2r+\sigma^2)t}$. Subtracting the squared mean, $S_0^2e^{2rt}$, yields the stated variance. $\square$

Worked Example: Mean, Second Moment and Variance

Let $S_0=50$, $r=\tfrac1{10}$, $\sigma=\tfrac12$, and $t=2$. Find the mean and variance under $\mathbb{Q}$, and check the discounted mean.

Using $p=1$, the mean is $\mathbb{E}^{\mathbb{Q}}[S_2]=50e^{(1/10)2}=50e^{1/5}$. For $p=2$, the exponent coefficient is $2r+\sigma^2=\tfrac2{10}+\tfrac14=\tfrac9{20}$, so $\mathbb{E}^{\mathbb{Q}}[S_2^2]=50^2e^{(9/20)2}=2500e^{9/10}$. The variance is therefore $$ \operatorname{Var}_{\mathbb{Q}}(S_2) =2500e^{9/10}-(50e^{1/5})^2 =2500(e^{9/10}-e^{2/5}). $$ Discounting the mean gives $e^{-r\cdot2}\mathbb{E}^{\mathbb{Q}}[S_2] =e^{-1/5}50e^{1/5}=50$.

Check: The variance is also $S_0^2e^{2rt}(e^{\sigma^2t}-1) =2500e^{2/5}(e^{1/2}-1)$. Expanding gives $2500(e^{9/10}-e^{2/5})$, since $2/5+1/2=9/10$, which matches the subtraction calculation.

Why the Discounted Stock Is a Martingale

The discounted stock is $M_t=e^{-rt}S_t$. A martingale must be integrable at each time and satisfy the conditional-expectation identity, not merely have zero drift in a differential expression. Both requirements can be checked directly here.

Theorem: Under $\mathbb{Q}$, $M_t=e^{-rt}S_t$ is a true martingale: for $0\le s\le t$, it is integrable and $\mathbb{E}^{\mathbb{Q}}[M_t\mid\mathcal{F}_s]=M_s$.

Proof. The first-moment formula gives $\mathbb{E}^{\mathbb{Q}}[|M_t|]=e^{-rt}\mathbb{E}^{\mathbb{Q}}[S_t]=S_0<\infty$; positivity of $S_t$ was established in the explicit-solution theorem. From the conditional solution, $$ M_t=M_s\exp\!\left(-\frac{\sigma^2}{2}(t-s)+\sigma(B_t-B_s)\right). $$ The increment $B_t-B_s$ is independent of $\mathcal{F}_s$ and is normal with variance $t-s$. Its exponential factor has conditional expectation $e^{-\sigma^2(t-s)/2}e^{\sigma^2(t-s)/2}=1$, by the normal moment-generating function. Since $M_s$ is $\mathcal{F}_s$-measurable, conditional expectation gives $\mathbb{E}^{\mathbb{Q}}[M_t\mid\mathcal{F}_s]=M_s$. Thus the process is a true martingale, not just a local martingale. $\square$

This statement concerns the risk-neutral measure $\mathbb{Q}$, not an automatic identity under the physical measure $\mathbb{P}$. If the physical drift is different, the stock’s physical expected growth need not equal $r$. The risk-neutral discounted-martingale property is a model assumption used in pricing; it does not assert that investors’ physical expected return is the interest rate.

A Common Log-Drift Error

The stock drift $r$ is not the drift of $\log S_t$. Omitting the Itô correction $-\sigma^2/2$ gives the wrong distribution and wrong simulated log-price steps. In the simulated example, the incorrect step would use $rh+\sigma\Delta B$, giving $(\tfrac1{20})(\tfrac14)+(\tfrac15)(\tfrac12)=\tfrac1{80}+\tfrac1{10}=\tfrac9{80}$. The correct step is $\tfrac{43}{400}$; these differ because the omitted term is $(\sigma^2/2)h=(1/50)(1/4)=1/200$, and $\tfrac9{80}-\tfrac1{200}=\tfrac{43}{400}$.

The correction arises because Brownian quadratic variation is nonzero. Applying Itô’s formula to the logarithm uses $f(x)=\log x$, with $f'(x)=1/x$ and $f''(x)=-1/x^2$; for positive $S_t$, the second-derivative correction is $\tfrac12 f''(S_t)(\sigma S_t)^2dt=-\tfrac12\sigma^2dt$. The ordinary chain rule would miss this term. A further distinction is useful: $S_t$ is lognormal when $\sigma>0$, but its logarithm is normal; when $\sigma=0$, the stock is deterministic rather than lognormal.

Check Your Understanding

Use the explicit solution and the assumptions stated above. Treat conditional laws as statements under $\mathbb{Q}$.

For $S_0=40$, $r=0$, and $\sigma=\tfrac12$, write the distribution of $\log S_1$, including its mean and variance.
Compute $\mathbb{E}^{\mathbb{Q}}[S_t^3]$ for general $S_0,r,\sigma,t$.
For $\sigma=0$, state the stock price at time $t$ and explain why the lognormal-distribution description does not apply.
Explain why adaptedness of $B_t$ alone does not justify the conditional-expectation calculation for the discounted stock.
In the simulated step example, identify the contribution of the Itô correction to the log-price increment.

Geometric Brownian Motion for Stock Prices

What You'll Learn