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?

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 $$T_i=\alpha_{t_i}\,.$$

• 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 $\langle\int_0^t f(s,\omega)dB_s\rangle=\int_0^t f(s,\omega)^2ds$. Jan 19, 2022 at 21:59
• 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 $\langle M\rangle_t=\alpha_t\,.$ Jan 20, 2022 at 10:28
• 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? Jan 20, 2022 at 11:55
• 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. Jan 20, 2022 at 12:59
• 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. Jan 20, 2022 at 13:26