Simple question on interpreting Geometric Brownian Motion SDE

Question

I'm writing an overview that is more economic than mathematical and I want to explain shortly the stochastic differential equation of Geometric Brownian Motion as simple and clear as possible

$$dS_t = S_t \mu dt + S_t \sigma dW_t $$

(I have already described the idea of Efficient market and why the Brownian motion is used).

So I wrote: In finance time series are often modeled as combination of drift (non-random trend) and volatility. This stochastic differential equation has two addends, the first controls the trend and is not random and the second is responsible for random fluctuations around the trend curve. Namely, after the very short time interval $dt$, the value of the stock price process $S_t$ will change by the small amount that is normally distributed with mean $ S_t \mu dt$ and variance $(S_t \sigma)^2dt$.

The last words were taken from Wikipedia article about SDEs and the problem is that I'm actually unsure in what is said there:

A heuristic (but very helpful) interpretation of the stochastic differential equation is that in a small time interval of length $\delta$ the stochastic process $X_t$ changes its value by an amount that is normally distributed with expectation $\mu(X_t, t)\delta$ and variance $\sigma(X_t, t)^2 \delta$ and is independent of the past behavior of the process. This is so because the increments of a Wiener process are independent and normally distributed.

How can this value be independent from the history of $X_t$ if both mean $\mu(X_t, t) \delta$ and variance $\sigma(X_t, t)^2 \delta$ depend on $X_t$? However, I understand that if this is not so, it wouldn't be Brownian motion.

Answer 1

This part which you quoted:

This is so because the increments of a Wiener process are indipendent an normally distributed

almost answers your question.

The next step is to recognize that a Wiener process, that is, an $\mathscr{F}_t$-adapted Brownian Motion $W_t$, is a Markov Process and as such possesses the so called Markov Property. This states that, given the usual sigma-algebra generated by the process $\mathscr{F}_t^W=\sigma(W_s |\;0 \le s \le t)$, for any $0\le s \le t$ the increment $W_t - W_s$ is indipendent of $\mathscr{F}_s^W$.

Intuitively you can think of the sigma-algebra generated by the process $W_t$ as a mathematical structure which encodes the information on the process up to time $t$. Then the meaning of the Markov Property is that at any point in time $t$, the next increment of the Brownian Motion is indipendent of its past history up to (and including) $t$.

Now, to explain the heuristic reasoning you were quoting, if you consider your SDE

$$ \frac{dS_t}{S_t}=\mu dt+ \sigma dW_t\\ d\log(S_t)=\mu dt + \sigma dW_t $$

you will notice that up to a deterministic drift part, the log-return of the stock movement is only determined by the increment of the Wiener process, and as such indipendent of its past history. To be precise, the log returns assume the form of an infinitesimal random walk with drift, which also is one of the starting assumptions of the famous Black&Scholes model in Quantitative Finance.

As a last note you are correct when saying that "if so, it wouldn't be a Brownian Motion". In fact the indipendence of a filtration-adapted stochastic process from its past history, together with the normality $\mathcal{N}(0,dt)$ property of the increments and the continuity of paths, is one of the ways to actually characterize a Wiener Process.