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Let $W_t$ be a Brownian motion under probability measure $\mathbb{P}$. Let $X_t$ be defined as follows.

$$\mathrm{d}X_t = a \mathrm{d}t + 2\sqrt{ X_t} \mathrm{d}W_t.$$

Also define: $$L_t = \exp\left(-\frac{k}{2}\int_0^t \sqrt{X_s}\mathrm{d}W_s-\frac{k^2}{8}\int_0^t X_s\mathrm{d}W_s\right).$$

If $L_t=\frac{\mathrm{d}\mathbb{Q}}{\mathrm{d}\mathbb{P}}$ is used to change measure from $\mathbb{P}$ to $\mathbb{Q}$, what is the dynamic of process $X_t$ under the new probability measure $\mathbb{Q}$?

In this question, the form of $L_t$ is different from what is used in the Girsanov theorem (i.e., the Doleans-Dade exponential). How do we use the theorem to change the measure in this case?

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    $\begingroup$ $L(t)$ is not a martingale under $P$ and doesn't form an admissible radon nikodym derivative. It is exponent raised to the power a (ito integral) martingale, which isn't a martingale. $\endgroup$
    – Arshdeep
    Jun 12, 2021 at 4:28
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    $\begingroup$ I suspect there is a typo, and that the $dW_s$ in the right should be $dt$, so that this forms the Doleans-Dade expotential. $\endgroup$
    – Arshdeep
    Jun 12, 2021 at 4:29

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