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Consider the SDE $dX_t = \sqrt{1 -X_t^2}\;1_{|X_t| \leq 1}\;dW_t$ with $X_0 =0$. Lemma 6 of "Noise stability on the Boolean hypercube via a renormalized Brownian motion (Eldan, Mikulincer, Raghavendra. 2022)" shows that $\text{var}(X_t) = (1 - e^{-t})$. I don't understand the proof given in the paper, and that's probably because I'm only learning stochastic calculus.

In particular, I don't understand how one gets $dX_t^2 = \sqrt{1 -X_t^2}\;1_{|X_t| \leq 1}\;dW_t + (1-X_t^2)dt$ from the definition given above and Ito's lemma. It would be very helpful if someone can give more details.

Mathews Boban
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    I don't think it's true, but nevertheless their argument is correct. The point is that $$X^2 = 2X\cdot X+\langle X,X\rangle$$ and $2X\cdot X$ is a martingale, thus $E X_t^2 = E \langle X,X\rangle_t$ – Andrew Dec 25 '23 at 04:28

1 Answers1

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We have

$$d(X^{2}_{t})=2X_{t}dX_{t}+d[X]_{t}$$

$$=2X_{t}\sqrt{1-X^{2}_{t}}1_{|X_{t}|\leq 1}dW_{t}+(\sqrt{1-X^{2}_{t}}1_{|X_{t}|\leq 1})^{2}dt$$

or

$$X_{t}^{2}=\int^t_{0} 2X_{s}\sqrt{1-X^{2}_{s}}1_{|X_{s}|\leq 1}dW_{s}+\int^t_{0}(\sqrt{1-X^{2}_{s}}1_{|X_{s}|\leq 1})^{2}ds.$$

Here by uniqueness we can write the LHS as $X_{t}^{2}=X_{t}^{2}1_{|X_{t}|\leq 1}$ because once $X_{t}$ hits the boundary points $\pm 1$, it gets trapped there. The first term has zero expectation (Itō Integral has expectation zero) and so indeed as mentioned in the article we are left with

$$v_{t}=E[X_{t}^{2}1_{|X_{t}|\leq 1}]=0+E\int_{0}^{t}(1-X^{2}_{s})1_{|X_{s}|\leq 1}ds$$

or in derivative form

$$\frac{d}{dt}E[X_{t}^{2}1_{|X_{t}|\leq 1}]=E[(1-X_{t}^{2})1_{|X_{t}|\leq 1}]=P(|X_{t}|\leq 1)-v_{t},$$

which has the solution

$$v_{t}=e^{-t}\int_{0}^{t}e^{s}P(|X_{s}|\leq 1)ds.$$

Thomas Kojar
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