# Geometric Brownian motion and semi-martingality

I am trying to learn by myself stochastic calculus, and I have a question regarding semi-martingale stochastic process and geometric Brownian motion (GBM).

We know that a stochastic process $$S_t$$ is said to follow a GBM if it satisfies the following stochastic differential equation (SDE): $$\textrm{d}S_t = \mu_t S_t \textrm{d}t+\sigma S_t \textrm{d}W_t$$ where $$W_{t}$$ is a Brownian motion

I was reading a paper where the authors assumed that under the null hypothesis, the real-valued logarithmic efficient price process denoted by $$p_t^e$$ is given by the Brownian semi-martingale: $$\textrm{d}p_t^e = \mu_t \textrm{d}t+\sigma \textrm{d}W_t$$

My question is, what is the difference between a GBM and a semi-martingale stochastic process? Are they the same stochastic process?

A local martingale $$(M_t)_t$$ (mart) is a stochastic process whose expected increments are zero that is $$E(\mathrm{d}M_t)=0$$. Under suitable technical conditions, a local martingale is also a (true) martingale that is the current value is the best estimate for its future value: $$E(M_t)=M_0$$. The difference between the two cases is roughly that a local martingale might have paths under which it experiences explosive behavior.
A process $$(A_t)_t$$ with locally bounded variation (BV) is roughly speaking a process which is "relatively smooth" in the sense that it never experiences an arbitrarily large jump over a very small time interval. This process is also required to be right-continuous with left limits and adapted.
A semimartingale $$(S_t)_t$$ is a stochastic process allowing the decomposition: $$S\equiv M+A$$ i.e. it can be written as the sum of a martingale and a bounded variation process. Geometric Brownian Motion (GBM) is a semimartingale: $$\textrm{d}S_t=\underbrace{\mu S_t\textrm{d}t}_{\text{BV}}+\underbrace{\sigma S_t\textrm{d}W_t}_{\text{mart}}$$ Indeed on every infinitesimal interval $$\textrm{d}t$$ the GBM (1) increases/decreases by a proportional rate $$\mu$$, the amount being smaller the smaller the interval or the value $$S_t$$, nothing jumpy going on here, and (2) a random amount which depends on the Brownian Motion $$W$$ whose expected increment is zero: $$E(\textrm{d}W_t)=0$$.