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.


1 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 $E(\tau) = .5*(E(\tau_1) + .5 E(\tau_2) = E(X^2)\frac{1}{2}(\frac 1 {\sigma_1^2} + \frac 1 {\sigma_2^2}) = E(X^2)$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.