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Define $\alpha_t=\frac12\ln(1+\frac23 t^3)$. If $B_t$ is a Brownian motion, prove that there exists another Brownian motion $\tilde{B}_r$ such that $$\int_0^{\alpha_t}e^sdB_s=\int_0^t rd\tilde{B}_r.$$

I would like to use this specific Lemma 4 which I found here

Lemma 4 Let X be a semimartingale and ${\xi}$ be a predictable, X-integrable process. Suppose that ${\{\tau_t\}_{t\ge 0}}$ are finite stopping times such that ${t\mapsto\tau_t}$ is continuous and increasing. Define the time-changes ${\mathcal{\tilde F}_t=\mathcal{F}_{\tau_t}}, {\tilde X_t=X_{\tau_t}}$ and ${\tilde \xi_t=\xi_{\tau_t}}$. With respect to the filtration ${\mathcal{\tilde F}_t}, {\tilde X}$ is a semimartingale, ${\tilde\xi}$ is predictable and ${\tilde X}$-integrable, and $\displaystyle \int_0^t\tilde\xi\,d\tilde X=\int_{\tau_0}^{\tau_t}\xi\,dX$.

By the lemma we get $$\int_0^{\alpha_t}e^sdB_s=\int_0^te^{\alpha_s}dB_{\alpha_s}=\int_0^t \sqrt{1+\frac23s^3}dB_{\alpha_s}$$

Now I want to somehow take care of $dB_{\alpha_s}$. Maybe I can find a martingale $M_s$ with quadratic variation $\langle M\rangle_s=\alpha_s$ and then I would have $B_{\alpha_s}=B_{\langle M \rangle_s}=M_s$ but how do I find this martingale?

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1 Answer 1

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The Lemma is the right one and we need to show the second equals sign in $$\tag{1} \int_0^te^{\alpha_s}\,dB_{\alpha_s}=\int_0^{\alpha_t}e^s\,dB_s=\int_0^tr\,d\bar{B}_r\,. $$

With $f(t):=\alpha'_t$ we define the continuous martingale $$\tag{2} \bar{B}_t:=\int_0^t\frac{1}{\sqrt{f(r)}}\,dB_{\alpha_r} $$ which has quadratic variation $$\tag{3} \langle \bar B\rangle_t=\int_0^t\frac{1}{f(r)}\, d\alpha_r=\int_0^t\frac{1}{\alpha'_r}\alpha'_r\,dr=t $$ and is therefore a Brownian motion. Clearly by (2), $$\tag{4} \int_0^t\frac{r}{\sqrt{f(r)}}\,dB_{\alpha_r}=\int_0^tr\,d{\bar B}_r\,. $$ Finally, $$ \alpha'_t=\frac{t^2}{1+\frac{2}{3}t^3}\,,\quad\frac{1}{\sqrt{f(t)}}=\frac{\sqrt{1+\frac{2}{3}t^3}}{t}=\frac{e^{\alpha_t}}{t}\,,\quad\frac{t}{\sqrt{f(t)}}=e^{\alpha_t}\,. $$ Putting the last expression into (4) shows (1).

To elaborate on (3): This follows from two facts:

  • The quadratic variation of $\int_0^tY_s\,dM_s$ ($M$ a local martingale) is $\int_0^tY^2_s\,d\langle M\rangle_s\,.$

  • In the present case $M_t=B_{\alpha_t}$ and the second fact is $\langle B_{\alpha_.}\rangle_t=\alpha_t\,.$ To see why this is true observe that for any continuous increasing function $t\mapsto\alpha_t$

$$ \langle B_{\alpha_.}\rangle_t=\lim_{n\to\infty}\sum_{i=1}^n\big(B_{\alpha_{t_i}}-B_{\alpha_{t_{i-1}}}\big)^2=\alpha_t $$ because, as we know, $$ \langle B\rangle_T=\lim_{n\to\infty}\sum_{i=1}^n\big(B_{T_i}-B_{T_{i-1}}\big)^2=T $$ and $\alpha_{t_0},...,\alpha_{t_n}$ is a sequence of finer and finer partitions of the interval $[0,\alpha_t]\,.$ Just take $T=\alpha_t$ and $T_i=\alpha_{t_i}\,.$

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  • $\begingroup$ How did you compute $\langle \bar B\rangle_t=\int_0^t\frac{1}{f(r)}\, d\alpha_r$? Is there a formula for this or even for the quadratic variation $\langle B_{\alpha_r}\rangle$? For the moment, I know just the formula $\langle\int_0^t f(s,\omega)dB_s\rangle=\int_0^t f(s,\omega)^2ds$. $\endgroup$
    – roi_saumon
    Jan 19, 2022 at 21:59
  • $\begingroup$ This is just the general formula that for a local martingale $M_t$ the quadratic variation of $\int_0^tY_s\,dM_s$ is $\int_0^tY^2_s\,d\langle M\rangle_s\,.$ In your case $M_t=B_{\alpha_t}$ and $\langle M\rangle_t=\alpha_t\,.$ $\endgroup$
    – Kurt G.
    Jan 20, 2022 at 10:28
  • $\begingroup$ Thank you @Kurt.G I found a proof that $\int_0^t Y_s dM_s=\int_0^t Y_s^2 d\langle M\rangle_s$. For the second remark, I have a result saying that for a martingale $M_t$ we have $M_t=B_{\langle M\rangle_t}$ but I didn't find anything saying that for some process $\alpha_t, \langle B_{\alpha_t}\rangle =\alpha_t$. Does one follow from the other? $\endgroup$
    – roi_saumon
    Jan 20, 2022 at 11:55
  • $\begingroup$ I have added a proof. Since $\alpha_t$ is a deterministic continuous increasing function the fact $\langle B_{\alpha_t}\rangle=\alpha_t$ is probably easier than you think. $\endgroup$
    – Kurt G.
    Jan 20, 2022 at 12:59
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    $\begingroup$ What matters I think is just the continuity of $\alpha$ not its monotonicity. It was a long time a go when I did this at university. Best if you take some time and think about it. $\endgroup$
    – Kurt G.
    Jan 20, 2022 at 13:26

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