A Converse of the Skorokhod Embedding Theorem

Question

I am wondering whether the following "sort of converse" of Skorokhod's embedding theorem holds:

Suppose that $\{D_t\}_{t \geq 0}$ is a stochastic process with continuous paths, $D_0 = 0$, and suppose that the following is true:

For any mean zero, finite variance random variable $X$ there exists a stopping time $\tau$ with respect to the filtration $\mathcal{F}_t \stackrel{def}{=} \sigma\{D_s:0\leq s\leq t\}$, such that $X \stackrel{d}{=} D_\tau$ and $\mathbb{E}X^2 = \mathbb{E}\tau$.

My question is, does this imply that the process $\{D_t\}$ is a Brownian motion?

Even if not, under what condition(s) does it follow that $\{D_t\}$ is a Brownian motion? For example does strong Markov property or independent increment assumption of $\{D_t\}$ help?

Any help will be greatly appreciated.

Answer 1

Here is a thought: take a 50-50 mixture of Brownian motions with volatilities $\sigma_1^2, \sigma_2^2$ satisfying $\frac{1}{2}(\frac 1 {\sigma_1^2} + \frac 1 {\sigma_2^2}) = 1$. The mixer ought to be $F_0$ measurable since you can observe the volatility in any time interval $(0, \epsilon)$. Given $X$ use the stopping time $\tau_i$ you would use on either process by itself. Then as each satisfies $\sigma_i^2 E(\tau_i) = E(X^2)$ the package should satisfy $E(\tau) = .5*(E(\tau_1) + .5 E(\tau_2) = E(X^2)\frac{1}{2}(\frac 1 {\sigma_1^2} + \frac 1 {\sigma_2^2}) = E(X^2)$