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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?

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    $\begingroup$ In fact $B_{ct} $ is $c^{1/2} B'_t$ where $B'_t $ is another Brownian motion. This follows by checking that it is Gaussian and has the correct mean and covariance functions. $\endgroup$ – Ian Mar 15 '16 at 14:21
  • $\begingroup$ This is a very simple solution, I did not think of it. However, is it possible to express $B_{\alpha t}$ as an Ito process with respect to the same Brownian motion $B_t$, and not some other $B_t'$ (which may be annoying to work with since $B_t'$ is not exactly the same as $B_t$, yet it is not independent of if either). $\endgroup$ – user78270 Mar 15 '16 at 16:57
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    $\begingroup$ For $c>1$ definitely no, by measurability concerns. For $c<1$ those concerns go away so you might have a chance. $\endgroup$ – Ian Mar 15 '16 at 17:10
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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.

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