I have a seemingly simple question that has me stumped.
Suppose $(B_t)_{t\geq0}$ is a Brownian motion, and consider its rescaled version $(B_{\alpha t})_{t\geq0}$ for some $\alpha>0$. It seems intuitive enough that the rescaled Brownian motion should be an Ito process, i.e., can be written as $$B_{\alpha t}=\int_0^tX_s~ds+\int_0^tY_s~dB_s$$ for some adapted and square integrable $X_s$ and $Y_s$.
If we could write $B_{\alpha t}$ as $f(t,B_t)$ for some deterministic $f$, then Ito's Lemma would provide a direct way of doing this, but this is apparently not the case. Am I missing something obvious?
This representation does not seem to exist whatsoever. Indeed, as Ian noticed, it only might make sense if $a<1$. Now, let's rewrite $$B_{\alpha t}=\int_0^tX_s~ds+\int_0^tY_s~dB_s$$ in the form $$\int_0^{t} (\mathbf 1_{[0; \alpha t]}(s) - Y_s)~dB_s = \int_0^tX_s~ds,$$ where $\mathbf 1_{[0; \alpha t]}$ is the indicator function of the set $[0; \alpha t] \subset [0; t]$. Taking quadratic variations of both sides of the expression, we will get to $$\int_0^{t} (\mathbf 1_{[0; \alpha t]}(s) - Y_s)^2~ds = 0,$$ or $$\int_0^{\alpha t} (1 - Y_s)^2~ds + \int_{\alpha t}^t Y_s^2~ds = 0,$$ which is contradictory.