To build a deterministic time-changed Brownian motion, we have $$\int_0^t \zeta_s dW_s \sim \int_0^t dW_{\xi_s} = \int_0^{\xi_t} dW_s = W_{\xi_t},$$ where $${\xi_t}=\int_0^t \zeta_s^2 ds.$$ Is there a similar time change that can be built for $\int_0^t \zeta_s^2 dW_s^2=$$\int_0^t \zeta_s^2 (ds+2W_s dW_s)$ ?
Aside from the abusive notation, is it possible to reconcile the two following (it would work for $\zeta_t=$ constant)?
- $\zeta_t^2 dW_t^2=\zeta_t^2 (dt+2W_t dW_t)=\zeta_t^2dt+2\zeta_t^2 W_t dW_t =d\xi_t+2\zeta_t W_t dW_{\xi_t}$
- $dW_{\xi_t}^2=d\xi_t+2W_{\xi_t} dW_{\xi_t}$
Leaving aside the abuse of notations... With $\xi_{t}=\int_{0}^t\zeta_{s}^2ds$, $$W_{\xi_{t}}^{2}=\int_{0}^{\xi_{t}}dW_{s}^{2} \\=\int_{0}^{\xi_{t}}(ds+2W_{s}dW_{s}) \\=\int_{0}^{t}(d\xi_{s}+2W_{\xi_{s}}dW_{\xi_{s}}) \\\overset{\mathrm{law}}{=}\int_{0}^{t}\left(\zeta_{s}^{2}ds+2\sqrt{\frac{\xi_{s}}{\zeta_{s}^{2}s}}\zeta_{s}W_{s}\zeta_{s}dW_{s}\right) \\=\int_{0}^{t}\zeta_{s}^{2}\left(ds+2\sqrt{\frac{\xi_{s}}{\zeta_{s}^{2}s}}W_{s}dW_{s}\right) \\=\int_{0}^{t}\zeta_{s}^{2}\left(\left(1-\sqrt{\frac{\xi_{s}}{\zeta_{s}^{2}s}}\right)ds+\sqrt{\frac{\xi_{s}}{\zeta_{s}^{2}s}}(ds+2W_{s}dW_{s})\right) \\=\int_{0}^{t}\zeta_{s}\left((\zeta_{s}-\sqrt{\xi_{s}/s})ds+\sqrt{\xi_{s}/s}dW_{s}^{2}\right)$$