This question is regarding the Girsanov theorem. Define on some filtered probability space $(\Omega,F,({\scr F}_t)_t,P)$ $$Z_t=\exp\Bigg[\sum_{i=1}^{n}\int_0^t \lambda_s^{(i)}dB^{(i)_s}-\frac{1}{2}\int_0^t \|\lambda_s\|^2 ds\Bigg]. $$ where $\lambda_i$ is such that $$\int_0^t(\lambda_s^{i})^2ds<\infty\text{ a.s. }$$ and $B$ is an $n$-dimensional Brownian motion. Define an equivalent probability measure $Q$ such that $$\frac{dQ}{dP}=Z_T\text{ on }{\scr F}_T. $$ What the Girsanov theorem (as stated in Karatzas and Shreve) says is if $Z_t$ is a martingale ($E[Z_t]=1)$ then the process $\tilde{W}$ defined by $$\tilde{W}_t=W_t- \int_0^t\lambda_s^ids\text{ for all }i=1,...,n,$$ is a Brownian motion on $(\Omega,F_T,Q)$. My question is does any probability measure $Q$ equivalent to $P$ take the form $$\frac{dQ}{dP}=\exp\Bigg[\sum_{i=1}^{n}\int_0^T \lambda_s^{(i)}dB^{(i)_s}-\frac{1}{2}\int_0^T \|\lambda_s\|^2 ds\Bigg]?$$ Or ( the converse?) if we define a $\tilde{W}$ as above for a given process $\lambda,$ does there exist an equivalent probability measure such that $$\frac{dQ}{dP}=Z_T\text{ on }F_T$$ such that $\tilde{W} $is a BM w.r.t $Q?$
I have not assumed here that the filtration is the one generated by the BM but if it helps that can be assumed. Any ideas?
In the following answer integrals are multivariate. Assume that martingale representation with respect to the Brownian motion holds, i.e., for any ${\scr F}_T$-measurable integrable random variable $X_T$ and the corresponding $P$-martingale $X_t=E_t[X_T]$ there is an $B$-integrable predictable process $\eta$ such that $$ X = E[X_T] + \int_0^\cdot \eta_s dB_s.$$
Now take $Z$ to be the unique $P$-martingale closed by $dQ/dP > 0$. From the martingale representation theorem one obtains $$ Z = 1 + \int_0^\cdot \eta_s dB_s > 0. $$ Set $\lambda = \eta/Z$ and perform change of variables to $\ln Z$ to conclude that $Z_T=dQ/dP$ is of the desired form. The 'converse' will only hold if $\lambda$ is well behaved so that ${\scr E}(\lambda\cdot B)$ stopped at $T$ is a martingale. There are plenty of $\lambda$-s that integrate $B$ but do not have this property.