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Let $W$ be a Brownian motion and $\alpha$ be a progressively measurable process taking values in $\mathbb R_+$. Set $\beta_t:=\max(\alpha_t, 1)$ for all $t\ge 0$. Define respectively $X$, $Y$ by

$$X_t:=\int_0^t \alpha_s \,dW_s,\quad Y_t:=\int_0^t \beta_s\, dW_s,\quad \forall t\ge 0.$$

It is known, e.g. by Dubins–Schwarz's theorem, that there exist some Brownian motion $B$ (with respect to some suitable filtration) and a non-negative stochastic process $h$ s.t.

$$\DeclareMathOperator\Law{Law}\Law(X)=\Law\bigl((B_{H_t})_{t\ge 0} \bigr),\quad\mbox{with}\quad H_t:=\int_0^t h_s^2\,ds.$$

Define further $g_t:=\max(h_t, 1)$ for all $t\ge 0$. Can we prove

$$\quad\Law(Y)=\Law\bigl((B_{G_t})_{t\ge 0}\bigr), \quad\mbox{with}\quad G_t:=\int_0^t g_s^2 \,ds?$$

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  • $\begingroup$ To be clear the two Brownian motions will be different because Dubins-Schwarz is only a statement on distribution not almost surely i.e. $Law(X)=B_{H}$ and $Law(Y)=\tilde{B}_{G}$. $\endgroup$ Apr 11, 2023 at 12:07
  • $\begingroup$ @ThomasKojar For two different Brownian motions the statement is clearly true. Here the desired identification is just in law, so I wish to known whether we may take the same Brownian motion $\endgroup$
    – Fawen90
    Apr 11, 2023 at 12:18
  • $\begingroup$ But that doesn't make sense really. The distribution statement is about probability measures i.e. comparing integrals over all omega. Whereas you are asking for having the same realization omega i.e. you are asking for almost sure statement. If it was true, then Dubin's Schwarz would be true almost surely. $\endgroup$ Apr 11, 2023 at 12:20
  • $\begingroup$ @ThomasKojar I don's really see what you mean. For the deterministic case, my desired equality is clearly true $\endgroup$
    – Fawen90
    Apr 11, 2023 at 12:56
  • $\begingroup$ @ThomasKojar What I look for is just a equality in law $\endgroup$
    – Fawen90
    Apr 11, 2023 at 12:57

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