I would like to prove that driftless Geometric Brownian Motion process has only one Equivalent Local Martingale Measure.
$dX_t=\sigma X_t\,dW_t,\: \: X_0=1$,
where $W$ is a Brownian motion process. Assume that $\mathbb{F}$ is the natural filtration of $X$ and $\mathcal{F}=\mathcal{F}_T$.
I would like to prove that there is only one Equivalent Local Martingale Measure for X.
Best Answer
For a general market model consisting of $n$ risky assets driven by $d$ Brownian motions given by $$ dX^i(t,\omega)= \mu^i(t,\omega)X^i(t,\omega)dt + \sum_{k=1}^d\sigma^{i,k}(t,\omega)X^i(t,\omega)dW^k(t,\omega) $$ for $i=1,\dots,n$ or in short $$ d\hat{X}_t=\mu_t\hat{X}_tdt+\sigma_t\hat{X}_tdW_t $$ for an $\mathbb{R}^n$-valued progressive process $\mu$ and an $\mathbb{R}^{n\times d}$-valued progressive process $\sigma$ such that $$\int_0^T\sqrt{\sum_{i=1}^n(\mu_s^i)^2}ds < \infty $$ and $$ \int_0^T\sum_{i=1}^n\sum_{k=1}^d(\sigma_s^{i,k})^2ds<\infty $$ it holds that the market is complete if and only if the volatility matrix $\sigma(t,\omega)$ has a left-inverse, i.e. there exists a progressive process $\lambda(t,\omega)$ with values in $\mathbb{R}^{d\times n}$ s.t. $$ \lambda(t,\omega)\sigma(t,\omega)=I_d\quad\text{for a.e. }(t,\omega)\in[0,T]\times\Omega $$
In the special case of the Black-Scholes model we have $n=d=1$ and $\sigma(t,\omega)\equiv\sigma$ is constant (and deterministic) and thus the market is obviously complete. Since completeness is defined by the existence of only one equivalent local martingale measure this is the answer to your question.