Stochastic integral change of variable. 
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?
 A: 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 $T_i=\alpha_{t_i}\,.$
