# Change of measure for a stochastic process to be a martingale

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

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.

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.

• 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: $\frac{d Q}{d P}$ = $e^{-\frac{1}{2} B_t -\frac{1}{8}T}$ – codelearner May 16 at 3:12
• Thanks. revised. – Gordon May 16 at 12:57

(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):

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