Difference of Geometric Brownian Motions

Worked Solution

How to Think About It: A GBM is always positive -- that is one of its defining features. Since $S_1(t) > 0$ and $S_2(t) > 0$ almost surely, their difference $D(t) = S_1 - S_2$ can be positive, negative, or zero. That alone rules out GBM for the difference. But even beyond the sign issue, for a process to be GBM its drift and diffusion must be proportional to its own level. The difference process has drift and volatility that depend on $S_1$ and $S_2$ separately, not on $D$ alone.

Quick Sanity Checks: - GBM requires $D(t) > 0$ a.s. -- violated since $D$ can be negative. - GBM requires $dD = \mu D \, dt + \sigma D \, dW$ for some constants $\mu, \sigma$ -- but the dynamics of $D$ cannot be written in terms of $D$ alone. - The ratio, on the other hand, is always positive and has autonomous dynamics -- a good candidate.

Derivation:

Part 1: The difference is not GBM.

By linearity of the stochastic differential: $dD = dS_1 - dS_2 = (\mu_1 S_1 - \mu_2 S_2) \, dt + \sigma_1 S_1 \, dW_1 - \sigma_2 S_2 \, dW_2$

The drift term $\mu_1 S_1 - \mu_2 S_2$ and the diffusion terms $\sigma_1 S_1$ and $\sigma_2 S_2$ depend on the individual levels $S_1$ and $S_2$, not just on $D = S_1 - S_2$. Knowing $D$ does not tell you $S_1$ and $S_2$ individually (there are infinitely many decompositions), so the process is not Markov in $D$ alone and certainly not GBM.

Additionally, GBM processes are strictly positive, but $D(t)$ can be negative, zero, or positive.

Part 2: Special cases.

There is no non-trivial special case where $D(t)$ follows a GBM. Even if $\mu_1 = \mu_2$ and $\sigma_1 = \sigma_2$ and $W_1 = W_2$ (perfectly correlated), we get $D(t) = S_1(0) - S_2(0)$ times a common GBM factor only if $S_1(t) = c \cdot S_2(t)$ for a constant $c$ -- but then $D(t) = (c-1) S_2(t)$, which is a scaled GBM only if $c > 1$ (so $D > 0$). This is the degenerate case where the two assets are perfectly proportional.

In general, the difference of two GBMs is not even a diffusion in $D$ alone.

Part 3: The ratio follows a GBM.

Apply Ito's lemma to $\log R = \log S_1 - \log S_2$. Since each $\log S_i$ is a Brownian motion with drift: $d(\log S_i) = \left(\mu_i - \frac{\sigma_i^2}{2}\right) dt + \sigma_i \, dW_i$

we get: $d(\log R) = \left(\mu_1 - \mu_2 - \frac{\sigma_1^2 - \sigma_2^2}{2}\right) dt + \sigma_1 \, dW_1 - \sigma_2 \, dW_2$

This is a Brownian motion with drift, so $R(t) = S_1(t)/S_2(t)$ is a GBM. Specifically, $R(t)$ follows: $dR = \left(\mu_1 - \mu_2 + \sigma_2^2 - \rho \sigma_1 \sigma_2\right) R \, dt + R \left(\sigma_1 \, dW_1 - \sigma_2 \, dW_2\right)$ where $\rho = \text{Corr}(dW_1, dW_2)$.

The ratio is always positive (since both $S_1, S_2 > 0$), and its dynamics depend only on $R$ -- making it a proper GBM.

Practical Interpretation: This is why pairs trading works on log-price spreads, not price differences. The log-spread $\log(S_1/S_2)$ is a Brownian motion with drift, which is mean-reverting if you pick co-integrated pairs. The raw price difference has non-stationary, path-dependent dynamics that are much harder to model.

Answer: The difference $D(t) = S_1(t) - S_2(t)$ is not a GBM because (i) it can be negative, and (ii) its dynamics are not autonomous in $D$. The ratio $R(t) = S_1(t)/S_2(t)$ is a GBM, which is why practitioners work with log-price spreads rather than price differences.