A stochastic process $X$ is a martingale $\iff$ $X$ is driftless.

A Collector's Guide to Martingales: If $X$ is a stochastic process with volatility $\sigma _t$ (that is, $dX_t = \sigma _t dW_t + \mu _t dt$) which satisfies $\mathbb{E}[(\int_{0}^{T} \sigma^2 _s \, ds)^{\frac{1}{2}}] < \infty$, then:

$X$ is a martingale $\iff$ $X$ is driftless ($\mu _t \equiv 0$).

Remark: If the technical condition fails, a driftless process may not be a martingale. Such processes are called local martingales.

"Proof:"

"$\implies$: " If $X_t$ is a $\mathbb{P}$-martingale, then with $W_t$ a $\mathbb{P}$-Brownian motion, (by the Martingale Representation Theorem) we have an $\mathcal{F}$-previsible process $\phi _t$ such that: $X_t = X_0 + \int_{0}^{t} \phi _s \, dW_s$. This is just the integral form of the increment, $dX_t = \phi _t \, dW_t$, which has no drift term.

"$\Longleftarrow$: " Suppose a stochastic process $X_t = X_0 + \int_{0}^{t} \sigma _s \, dW_s + \int_{0}^{t} \mu _t \, dt$ has no drift term so that $X_t = X_0 + \int_{0}^{t} \sigma _s \, dW_s$. Since $W_t$ is a $\mathbb{P}$-Brownian motion ($\implies$ $\mathbb{P}$-martingale), then $X_t$ is also a $\mathbb{P}$-martingale since it has form $N_t = N_0 + \int_{0}^{t} \phi _s dM_s$ by the Martingale Representation Theorem. QED.

I am wondering if my proof for the reverse direction (if $X$ driftless, then $X$ is a martingale) works. In general, I would also like to know if the proof works overall. Also, can anyone explain the purpose/reason behind the technical condition ($\mathbb{E}[(\int_{0}^{T} \sigma^2 _s \, ds)^{\frac{1}{2}}] < \infty$)? How does one check that the technical condition is satisfied??? Thanks!

Edit: For reference, Martingale Representation Theorem:

Suppose $M_t$ is a $\mathbb{Q}$-martingale process whose volatility $\sigma _t$ satifies that it is (with probability one) always non-zero. Then if $N_t$ is any other $\mathbb{Q}$-martingale, there is an $\mathcal{F}$-previsible process $\phi$ such that $\int_{0}^{T} \phi^2 _t \sigma^2 _t \, dt < \infty$ with probability 1, and $N$ can be written as: $N_t = N_0 + \int_{0}^{t} \phi _s dM_s$.

• Hi unfortunately $\Longleftarrow$ is not working what is true in general is that a process of the form $X_t = X_0 + \int_{0}^{t} \sigma _s \, dW_s$ is a local martingale but not a martingale in most general setting, the technical condition ( which is a sufficient condition) is here to ensure that the local martingale is indeed a real martingale, more on this here : math.stackexchange.com/questions/38908/… . Best regards May 26 '16 at 6:46
• Hey, thank you so much for your help. I have been confused about the technical condition for some time now. May 26 '16 at 18:45