In the paper [*], the following time-change for a Brownian motion $B$ is performed in eq. (2.3): $$ \int_0^t \mathrm e^s\, \mathrm d B_{1-\mathrm e^{-2s}} = \int_0^{1-\mathrm e^{-2t}} \frac 1{\sqrt{1-s}}\, \mathrm d B_s. $$
I'm new to the field and I do not follow. It seems that they have "straightforwardly" performed the change of variable $\phi(s)=1-\mathrm e^{-2s}\mapsto s$. However, I understood that the time-changed Brownian motion carries a factor of $\sqrt{\phi'(s)}$, that is, $\mathrm dB_{\phi(s)}$ has the same distribution as (and therefore is the appropriate change-of-variables formula for) $\sqrt{\phi'(s)}\,\mathrm dB_s$.
I do not understand where the square-root factor is on the right-hand side. So my specific question is how do I understand this time change? And my more general question is what the general SDE time-change formula looks like? That is, for an Ito process satisfying $\mathrm d X_t=\mu(X,t)\,\mathrm d t + \sigma(X,t)\,\mathrm d B_t$, what the isdifferential form of the time-transformed process $Y(t)=X(\phi(t))$? My understanding was that it was $$\mathrm d Y_t = \mu(Y,\phi(t))\phi'(t)\,\mathrm d t + \sigma(Y,\phi(t))\sqrt{\phi'(t)}\,\mathrm d B_t.$$
[*] S. E. Graversen, G. Peskir, and A. N. Shiryaev. Stopping brownian motion without anticipation as close as possible to its ultimate maximum. Theory of Probability & Its Applications , 45(1):41–50, 2001.
UPDATE
I think the answer is that the integral is not really time-parametrised, and so a "straightforward" change of variables is justified. You can see this simply by observing that the Riemann sums of each side are identical.
I believe that the differential form $\mathrm d Y_t$ is also correct. To see this, we can split it into drift and diffusion. The drift part comes from the deterministic change-of-variables formula, and the diffusion part comes from the equi-distributed forms I wrote. The crucial difference here is that the differential form is time-parametrised. To get a feel for this, take the example of $\mu=0$, $\sigma=1$ and $\phi(s)=2s$. Then $X$ is the Brownian motion $B$ (assuming $X_0=0$) and the form simplifies to $\mathrm dX_{2t}=\sqrt 2\,\mathrm dB_t$, i.e. $ B_{2t} \overset d= \sqrt2 B_t$.
These final equalities seem true in distribution but not almost surely. I think that is an artefact of swapping the two differential forms that are equal only in distribution. It feels that a comment about the filtration is necessary here, perhaps the time-changed variable is adapted to the time-changed filtration or something similar?