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.

  • $\begingroup$ What are your thoughts on this problem? $\endgroup$
    – Math1000
    Nov 4 '19 at 4:44

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.

  • $\begingroup$ Thank you Leguan3000 I need to study this more to grasp your proof. Is this for general market model , where $\mu$ is not equal to 0? If yes does the same proof still holds , when $\mu$ is zero. $\endgroup$ Nov 4 '19 at 13:42
  • $\begingroup$ Yes, this is for a general market model where $\mu$ and $\sigma$ can be any progressive processes with the stated integrability conditions. Thus it still holds for $\mu=0$. I think you can find the proof in any book on mathematical finance as it a very fundamental statement. $\endgroup$
    – Leguan3000
    Nov 4 '19 at 14:11
  • $\begingroup$ Hi, you can also mark this answer as correct (if you think it's correct) $\endgroup$
    – Leguan3000
    Nov 19 '19 at 10:15
  • $\begingroup$ Thank you once again Leguan3000 , I marked answer as correct. $\endgroup$ Nov 23 '19 at 21:54

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