Let $W$ be a brownian motion and $p>0$.
For which $p$ does $S_t=W_t+t^p$ admit an equivalent martingale measure?
I recently saw at my lectures that:
NFLVR cond: There does not exist a sequence $\{H_n\}_{n \geq 1}$ of predictable processes, integrable wrt S, such that there exists $t_0 , b , \epsilon >0$ $$ \int_0^{t_0} H_n(s) S_s > -1/n$$ and $$ P(\int_0^{t_0} H_n(s) S_s >b)> \epsilon$$
Is equivalent to having such a measure and I think that is pretty much my only tool, so I guess it has to be that. On the other hand I can't get any ideas on how to use the condition.
Let me just give a short answer since probably nobody cares:
It can bee seen as a corollary to (at least the proof i have seen) of Girsanov that if we are looking for a equivalent martingale measure for a continuous semimartingale $S_t = M_t + A_t$ where $M_t$ is the local martingale part and $A_t$ is the bounded variation part, then $A_t$ must satisfy that there exists predictable $h_t$ such that $A_t=\int _0^t h_s d<M_s>$ and $\int_0^t h_s^2 d<M_s> < \infty $. ($<M>$ is quadratic variation). In our case this rules out $p\leq 1/2$ since $A_t=\int_0^t p\cdot s^{p-1} ds $ which is only square integrable if $p > 1/2$.
Assuming from now on $p>1/2$. We will see an Martingale measure does indeed exist. This is simply and application of Girsanov - e.g. by applying Novikovs condition which is here trivial - on $h_s = - p s^{p-1}$. This directly yields that $W_t - \int h_s d<W>_s=W_t + t^p$ is a brownian motion under $\mathbb{Q}$ (in particular a local martingale) and $\mathbb{P}$ and $\mathbb{Q}$ are equivalent.