Uniqueness and non-uniqueness of a SDE involving a random time change The other day, when I was reading the book of Ethier and Kurtz, I found the following problem (Chapter 6, p.332).
Let $W=\{W(s)\}_{s \ge 0}$ be standard Brownian motion (the book doesn't give any more details, but I think $W$ is a one-dimensional Brownian motion).
(a) Show that for $\alpha \in (0,1)$,
\begin{align*}
P\left[\int_0^t|W(s)|^{-\alpha}\,ds<\infty,\,t\ge0\right]=1,
\end{align*}
and for $\alpha\ge 1$,
\begin{align*}
P\left[\int_0^t|W(s)|^{-\alpha}\,ds=\infty,\,t>0\right]=1.
\end{align*}
(b) Show that for $\alpha \ge1$ the solution of
\begin{align*}
X(t)=X(0)+W\left(\int_{0}^t |X(s)|^\alpha\,ds\right),\quad t \ge 0
\end{align*}
is unique but it is not unique if $\alpha \in (0,1)$.
Problem (c) comes up after this, but for now I'm only interested in this problem (b).
I'm not sure if the problem (a) can be solved easily, but this indeed follows from the Engelbert--Schmidt Zero-One law (integrability near the origin of the map $t \mapsto |t|^{-\alpha}$ is important). However, I don't know how to solve (b) (in the first place, the meaning of the  uniqueness is not written in the problem, though).
If you know how to solve (b), please let me know.
 A: I am not sure if $X(0)$ is supposed to be part of the data of the problem or if we are free to choose it, but I will confine myself in this answer to the case $X(0) \equiv 0$.
Note that if $X$ solves the given equation, it must be necessarily continuous.
Set \begin{align*} \tau_0 &:= \inf\{t\geq 0~|~ W(t) = 0\}, \\ \tau_1&:=\inf\left\{t \geq 0 ~|~ \int_{0}^{t}|W(s)|^{-\alpha} ~ds = \infty \right\}.\end{align*}
Let's deal with the case $\alpha \geq 1$ first. Then, by a), $\tau_0 = \tau_1$ and hence, by Theorem 1.1 of Chapter 6 in the book of Ethier and Kurtz, the solution to the equation is unique. As $X \equiv 0$ solves the equation, this is the unique solution.
Suppose now that $\alpha \in (0,1)$. As before, $X \equiv 0$ solves the equation, so we're done if we can construct a solution that does not vanish identically. By a), we may define the time-change $\tau(t)$ through the equation
$$t = \int_{0}^{\tau(t)} |W(s)|^{-\alpha}~ds, \quad t\geq 0.$$
Note that $t \mapsto \tau(t)$ is almost surely strictly increasing and continuous (locally Lipschitz) with $$ \dot \tau(t) = |W(\tau(t))|^{\alpha}, \quad \text{almost every t > 0}.$$ We may in particular define the inverse time-change $T(t)$ such that $\tau(T(t)) = t$.  Now let $X(t) := W(\tau(t)).$   Then,
$$\int_{0}^{t} |X(s)|^\alpha~ds = \int_{0}^{t} |W(\tau(s))|^\alpha~ds =  \int_{0}^{\tau(t)}\frac{|W(s)|^{\alpha}}{ \dot \tau(T(s))}~ds = \tau(t).$$
It follows that $X$ solves the desired equation and clearly $X$ does not vanish identically, so the equation does not have a unique solution.
