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$\text { Give a measure change so that } X_{t}=e^{B_{t}}\left(B_{t}-t / 2\right) \text { is a martingale, } 0 \leq t \leq T$

My attempt

Using Ito's lemma on $X_{t}$ we get:

$-\frac{e^{B t}}{2} d t+\left(X_{t}+e^{B_{t}}\right) d B_{t}+\left(\left(2 B_{t}-t+4\right)e^{B_{t}} d t\right. \\ =(\left.2 B_{t}-t+4-\frac{1}{2}\right) e^{B_{t}} d t+\left(X_{t}+e^{B_{t}}\right) d B_{t}$

Then I used Girsanov's theorem:

$d \hat{B}_{t}=d B_{t}+\int_{0}^{+} H_{S} ds$

$H_{t}=\frac{\left(2 B_{t}-t+4-\frac{1}{2}\right) e^{B_{t}}}{X_{t}+e^{B_{t}}}$

I am some what scared to go any further because it seems like I am heading in the wrong direction.

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Let $Y_t= e^{B_t}$ and $Z_t = B_{t}-t / 2$. Then, \begin{align*} dX_t &= Z_t dY_t + Y_t dZ_t + d\langle Y, Z\rangle_t\\ &=(B_{t}-t / 2)e^{B_t}\big( dB_t + 1/2\,dt \big) + e^{B_t}\big(dB_t -1/2\, dt\big) + e^{B_t} dt\\ &=e^{B_t}(B_t-t / 2+1)dB_t + e^{B_t}(B_t/2-t / 4 -1/2+1)dt\\ &=e^{B_t}(B_t-t / 2+1)d\big(B_t+1/2t\big). \end{align*} We define the probability measure $Q$ on $\mathscr{F}_T$ by \begin{align*} \frac{dQ}{dP}=e^{-\frac{1}{8}t-\frac{1}{2}B_t}. \end{align*} Then $\{W_t, \, t\ge 0\}$, where $W_t = B_t+1/2\,t$, is a standard Brownian motion under $Q$. Moreover, \begin{align*} dX_t = e^{W_t-\frac{1}{2}t}(W_t-t+1)dW_t. \end{align*} That is, $\{X_t, \, 0 \le t \le T\}$ is a martingale.

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    $\begingroup$ Thank you! I am confused about this part: $\frac{d Q}{d P}=e^{-\frac{1}{2} t-B_{t}}$. I used Girsanov's theorem where: $W_{t}=B_{t}+\int_{0}^{t} H_{s} d s$ and $\frac{d Q}{d P}$ = $e^{-\int_{0}^{T} H_{s} d B_{s}-\frac{1}{2} \int_{0}^{T} H_{s}^{2} d s}$ I got $H_t = 1/2$ Hence: $\frac{d Q}{d P}$ = $e^{-\frac{1}{2} B_t -\frac{1}{8}T}$ $\endgroup$ – codelearner May 16 at 3:12
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    $\begingroup$ Thanks. revised. $\endgroup$ – Gordon May 16 at 12:57
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(Now that I saw Gordon's solution, I can finish my attempt; I had noticed $dB_t +1/2dt$ immediately from product rule for $V_t$, zero quadratic covariation between $t/2$ and $e^{B_t}$, but hours later :) I was still perplexed by $U_t$.)

$$ X_t = U_t - V_t $$ $$ V_t = e^{B_t}t/2$$ $$ U_t = e^{B_t}B_t $$

$$ dV_t = \boxed{1/2e^{B_t} dt} + 1/2V_t(dB_t + 1/2dt) $$

$$ dU_t = (U_t + e^{B_t}) dB_t + 1/2(U_t + 2e^{B_t})dt $$

$$ = (U_t + e^{B_t})(dB_t +1/2dt) + \boxed{1/2e^{B_t} dt} $$

So, by subtraction (and lucky cancellation of the boxed terms):

$$ dX_t = (U_t + e^{B_t} - 1/2V_t)(dB_t + 1/2dt) $$

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