Let
- $(\Omega,\mathcal A,\operatorname P)$ be a probability space
- $(\mathcal F_t)_{t\ge0}$ be complete and right-continuous filtration on $(\Omega,\mathcal A,\operatorname P)$
- $B$ be a $\mathcal F$-Brownian motion on $(\Omega,\mathcal A,\operatorname P)$
- $b:\mathbb R\to\mathbb R$ be Lipschitz continuous
- $\alpha>0$
We say that $X$ is a strong solution of $${\rm d}X_t=\frac\alpha2b(X_t){\rm d}t+\sqrt\alpha{\rm d}B_t\;\;\;\text{for all }t\ge0,\tag1$$ if $X$ is a $\mathcal F$-adapted process on $(\Omega,\mathcal A,\operatorname P)$ almost surely statisfying $(1)$. We say that strong existence holds for $(1)$, if there is a strong solution for every $(\mathcal F,B)$.
Now consider $${\rm d}Y_t=\frac12b(Y_t){\rm d}t+{\rm d}B_t\;\;\;\text{for all }t\ge0.\tag2$$
How are strong solutions of $(1)$ and $(2)$ related?
It's clear that $(1)$ and $(2)$ both admit strong existence (since $b$ is Lipschitz continuous). Now, if $Y$ is a strong solution of $(2)$, let $$X_t:=Y_{\alpha t}\;\;\;\text{for }t\ge0.$$ By the substitution rule, $$X_t=X_0+\frac12\int_0^tb(X_s){\rm d}s+B_{\alpha t}\;\;\;\text{for all }t\ge0\tag3.$$
What do we do with the $B_{\alpha t}$?
Clearly, we know that $(\alpha^{-1/2}B_{\alpha t})_{t\ge0}$ is a $(\mathcal F_{\alpha t})_{t\ge0}$-Brownian motion and hence $B_{\alpha t}\sim\sqrt\alpha B_t$ for all $t\ge0$.
But what can we conclude now? Only that $X$ is a strong solution for $((\mathcal F_{\alpha t})_{t\ge0},(\alpha^{-1/2}B_{\alpha t})_{t\ge0})$?