# show that the solution is a local martingale iff it has zero drift

Most financial maths textbook state the following:

Given an $n$-dimensional Ito-process defined by $$X_t = X_0 + \int_0^{t} \alpha_s \,d W_s + \int_0^{t} \beta_s \,d s,$$ where $(\alpha_t)_{t \geq0}$ is a predictable process that is valued in the space of $n \times d$ matrices and $(W_t)_{t \geq 0}$ is a $d$-dimensional Brownian motion, $$(X_t) \text{ is a local martingale } \quad \Longleftrightarrow \text{ It has zero drift.}$$ Can anyone show me a reference for the proof of this statement or at least give a hint of how to construct this proof? (I know how to prove the ($\Leftarrow$) direction, but I am not so sure about the other one.)

Suppose that $(X_t)_{t \geq 0}$ is a local martingale. Since

$$X_0 + \int_0^t \alpha_s dW_s$$

is also a local martingale, this means that

$$M_t := X_t - \left( X_0 + \int_0^t \alpha_s \, dW_s \right) = \int_0^t \beta_s \, ds$$

is a local martingale. Moreover, $(M_t)_{t \geq 0}$ is of bounded variation and has continuous sample paths. It is widely known that any continuous local martingale of bounded variation is constant, see e.g. Brownian Motion - An Introduction to Stochastic Processes by René Schilling and Lothar Partzsch, Proposition A.22.

• Do we need to assume that $(\beta_t)_{t \geq 0}$ is a continuous process in order to conclude that $(M_t)_{t \geq 0}$ is continuous? Nov 23, 2014 at 22:43
• @Richard No, Lebesgue-integrability of $(\beta_t)$ is enough. If it is continuous, then $(M_t)_t$ is even differentiable.
– saz
Nov 24, 2014 at 7:12
• Is it trivial to check that $X_0+\int_0^t \alpha_s dW_s$ is a martingale?
– asdf
Apr 4, 2018 at 8:07
• @asdf In general it's just a local martingale; I'll edit my answer accordingly.
– saz
Apr 4, 2018 at 9:01