Under the Wiener measure $\Bbb{W}$ the process $x(t)$ is a brownian motion.
This means that $\Bbb{E}[{x(t)-x(s)\mid \mathcal{F}_s}]=0$.
Let $P$ be a measure in $C([0,\infty),\Bbb{R}^d)$ such that for every $\theta \in \Bbb{R}^d$
$$X_\theta(t):=\exp\bigg\{ \langle\theta,x(t)\rangle - \frac{1}{2} t |\theta|^2\bigg\}$$
is a martingale. Then $P = \Bbb{W}$ is the Wiener measure.
Let $\sigma$ be a simmetric matrix:
Assume now we define $$\tilde{P}(x(t_1)\in A_1, \ldots x(t_k) \in A_k) := \Bbb{W}(\sigma(x(t_1)-x) \in A_1,\ldots ,\sigma(x(t_k)-x) \in A_k)$$
Let $a = \sigma \sigma^T$
Is it true that $$Y_\theta(t):=\exp\bigg\{ \langle\theta,x(t)\rangle - \frac{d}{2} t \langle\theta,a\theta \rangle^2\bigg\} $$
Is a martingale?
Attempt: $\langle\theta,\sigma(x_t-x)\rangle = \sum_{i,j}\theta_i \sigma_{i,j}(x(t)-x)_j$
We could use Ito calculus to conclude that
\begin{align*} \exp\{\langle\theta,\sigma(x_t-x\rangle)\} &= \exp\{\langle\theta,\sigma(x_0-x)\rangle\} + \sum_{j=1}^d\int_0^t \partial_i\langle\theta\sigma(x(s)-x)\rangle \exp\{\langle\theta,\sigma(x_s-x)\rangle\}\, dx_i(s) \\ &+ \sum_{j = 1}^d \frac{1}{2}\int_0^t \partial_{i}\exp\{\langle\theta,\sigma(x_s-x)\rangle\} \partial_i \exp\{\langle\theta,\sigma(x_s-x)\rangle\} \, ds \end{align*} Since $\partial_j \sum_{i,j}\theta_i \sigma_{i,j}(x(t)-x)_j = \sum_{i}\theta_i \sigma_{i,j}$ \begin{align*} \exp\{\langle\theta,\sigma(x_t-x)\rangle\} &= \exp\{\langle\theta,\sigma(x_0-x)\rangle\} + \sum_{j=1}^d\int_0^t \sum_{i}\theta_i \sigma_{i,j}\exp\{\langle\theta,\sigma(x_s-x)\rangle\}\, dx_i(s) \\ &+ \sum_{j = 1}^d \frac{1}{2}\int_0^t (\sum_{i}\theta_i \sigma_{i,j})(\sum_{i}\theta_i \sigma_{i,j})\exp\{\langle\theta,\sigma(x_s-x)\rangle\} \partial_i \exp\{\langle\theta,\sigma(x_s-x)\rangle\} \, ds \end{align*} Now note that $(\sum_{i}\theta_i \sigma_{i,j})(\sum_{i}\theta_i \sigma_{i,j}) = \sum_{i,j}\theta_i a_{i,j}\theta_j= \langle \theta, a\theta \rangle$
So we conclude
\begin{align*} \exp\{\langle\theta,\sigma(x_t-x)\rangle\} &= \exp\{\langle\theta,\sigma(x_0-x)\rangle\} + \sum_{j=1}^d\int_0^t \sum_{i}\theta_i \sigma_{i,j}\exp\{\langle\theta,\sigma(x_s-x)\rangle\}\, dx_i(s) \\ &+ \frac{d}{2}\int_0^t \langle \theta, a\theta \rangle \exp\{\langle\theta,\sigma(x_s-x)\rangle\} \, ds \end{align*}
Now it suffices to note that once $\phi(t)$ is a continuous bounded (on intervals) martingale and $\psi(t)$ is a continuous function of bounded variation such that $\Bbb{E}[\|\psi \|_[0,t]]<\infty$ (the total variation on the interval $[0,t]$ has finite expectation) then $$\phi(t)\psi(t)- \int_0^t \phi(s)\,d\psi(s) $$ is also a martingale.
Take \begin{align*}\phi(t)&=\exp\{\langle\theta,\sigma(x_t)-x\rangle\} - \frac{d}{2}\int_0^t \langle \theta, a\theta \rangle \exp\{\langle\theta,\sigma(x_s)-x\rangle\} \, ds \\ \psi(t) &= \exp{-\frac{d}{2} t \langle \theta, a\theta \rangle } \end{align*}
So in a sense, the result is already proved.
Question: is there a different approach to this question? I was thinking something along the following lines:
$$\tilde{P}(x(t_1)\in A_1, \ldots x(t_k) \in A_k) := \Bbb{W}(\sigma(x(t_1)-x) \in A_1,\ldots ,\sigma(x(t_k)-x) \in A_k)$$
Implies that $$\Bbb{E}_{\tilde{P}}[f(x_t)] = \Bbb{E}_{\Bbb{W}}[f(\sigma(x_t-x))] $$
then the condition $$\Bbb{E}_{\Bbb{W}}\bigg[\exp\bigg\{ \langle\theta,x(t)\rangle - \frac{d}{2} t |\theta|^2\bigg\}\bigg]=1$$
Should imply that $$\Bbb{E}_{\tilde{P}}\bigg[\exp\bigg\{ \langle\theta,\sigma(x(t)-x)\rangle - \frac{1}{2} t \langle\theta, a\theta\rangle\bigg\}\bigg]=1$$ But I don't see it