球面投影随机游走收敛于布朗运动的严格证明及步骤详解请求

Hey there! As someone who’s worked through similar convergence problems in stochastic processes, let me break this down step by step for you—since you’re a beginner, I’ll try to avoid overloading you with jargon where possible, but we’ll stay rigorous as requested.

This proof follows the exact blueprint of Donsker’s theorem, which requires two key ingredients: convergence of finite-dimensional distributions and tightness of the process family. Let’s dive in.

Background: Formalizing the Problem #

First, let’s define our processes clearly:

• Let ${X_i^\Delta t}_{i=1}^\infty$ be i.i.d. 3-dimensional random vectors with $\mathbb{E}[X_i^\Delta t] = 0$, $\mathbb{E}[X_i^\Delta t (X_i^\Delta t)^T] = \Delta t \cdot I_3$ (where $I_3$ is the 3x3 identity matrix), and bounded components (e.g., $|(X_i^\Delta t)_j| \leq C\sqrt{\Delta t}$ for some constant $C$).
• The $\mathbb{R}^3$ random walk at time $t$ is $S_t^\Delta t = X_1^\Delta t + X_2^\Delta t + ... + X_{\lfloor t/\Delta t \rfloor}^\Delta t$.
• The projected spherical random walk is $Y_t^\Delta t = \frac{S_t^\Delta t}{|S_t^\Delta t|}$, which is well-defined almost surely (a.s.) as $\Delta t \to 0$ (since 3D Brownian motion—our limit—never hits 0 a.s. at fixed times).

Our goal is to show $Y_t^\Delta t \xrightarrow{d} B_t^{S2}$ as $\Delta t \to 0$, where $B_t^{S2}$ is Brownian motion on the unit sphere $S^2$ (equivalently, the projection of 3D Brownian motion onto $S^2$, or the diffusion process with generator $\frac{1}{2}\Delta_{S^2}$, where $\Delta_{S^2}$ is the Laplace-Beltrami operator on $S^2$).

Step 1: Convergence of Finite-Dimensional Distributions #

To prove this, we use the continuous mapping theorem and the multi-dimensional version of Donsker’s theorem:

1. Multi-dimensional Donsker: The $\mathbb{R}^3$ walk $S_t^\Delta t$ converges in distribution to 3D Brownian motion $B_t$ as $\Delta t \to 0$. This means for any finite set of times $0 \leq t_1 < t_2 < ... < t_k$, the joint distribution $(S_{t_1}^\Delta t, ..., S_{t_k}^\Delta t)$ converges to $(B_{t_1}, ..., B_{t_k})$.
2. Continuous Mapping: The projection map $f(x) = \frac{x}{|x|}$ is continuous on $\mathbb{R}^3 \setminus {0}$. Since $B_t$ never hits 0 a.s. at fixed times, the set $\mathbb{R}^3 \setminus {0}$ has full measure under the finite-dimensional distributions of $B_t$.
3. Joint Distribution Convergence: Applying the continuous mapping theorem to the joint distribution, we get:
   $$(Y_{t_1}^\Delta t, ..., Y_{t_k}^\Delta t) = (f(S_{t_1}^\Delta t), ..., f(S_{t_k}^\Delta t)) \xrightarrow{d} (f(B_{t_1}), ..., f(B_{t_k}))$$
   which is exactly the finite-dimensional distribution of $B_t^{S2}$.

Step 2: Tightness of the Process Family #

Tightness ensures our sequence of processes doesn’t "escape to infinity" or develop pathological discontinuities. For processes taking values in $C([0,T], S^2)$ (continuous functions from $[0,T]$ to $S^2$), we use the Arzelà-Ascoli theorem: a set of continuous functions is compact iff it’s uniformly bounded and equicontinuous.

Uniform Boundedness #

This is trivial: every path of $Y_t^\Delta t$ lies on the unit sphere, so $|Y_t^\Delta t| = 1$ for all $t, \Delta t$.

Equicontinuity #

We need to show: for any $\epsilon > 0, \eta > 0$, there exists $\delta > 0$ such that for all $\Delta t > 0$,
$$\mathbb{P}\left(\sup_{|s-t| < \delta} |Y_s^\Delta t - Y_t^\Delta t| > \epsilon\right) < \eta$$

To prove this, use vector norm inequalities to bound the difference between projected points:
Let $s = n\Delta t$, $t = m\Delta t$ with $|m-n| \leq \lfloor \delta/\Delta t \rfloor$, and let $Z = S_t^\Delta t - S_s^\Delta t$. Then:
$$|Y_t^\Delta t - Y_s^\Delta t| = \left| \frac{S_s + Z}{|S_s + Z|} - \frac{S_s}{|S_s|} \right| \leq \frac{2|Z|}{|S_s|}$$
(This follows from the triangle inequality and reverse triangle inequality for norms.)

Now, split the probability into two parts:

1. Probability $S_s$ is close to 0: By convergence of $S_s^\Delta t$ to $B_s$, for any $M > 0$, we can choose $\Delta t$ small enough so $\mathbb{P}(|S_s^\Delta t| < M) < \eta/2$.
2. Probability $Z$ is large: $Z$ is the sum of $\lfloor \delta/\Delta t \rfloor$ i.i.d. vectors with variance $\Delta t I_3$, so $\mathbb{E}[|Z|^2] = 3\delta$. By Chebyshev’s inequality:
   $$\mathbb{P}(|Z| > \frac{\epsilon M}{2}) \leq \frac{3\delta}{(\epsilon M/2)^2}$$
   Choose $\delta$ small enough so this probability is $< \eta/2$.

Combining these gives:
$$\mathbb{P}(|Y_t^\Delta t - Y_s^\Delta t| > \epsilon) \leq \mathbb{P}(|S_s^\Delta t| < M) + \mathbb{P}(|Z| > \frac{\epsilon M}{2}) < \eta$$
This proves equicontinuity, so the process family is tight.

Step 3: Final Convergence Conclusion #

By Prokhorov’s theorem, a tight family of probability measures on a Polish space (like $C([0,T], S^2)$) is relatively compact—every subsequence has a weakly convergent subsubsequence. Since we already showed all such limits have the same finite-dimensional distributions as $B_t^{S2}$, the entire family converges weakly to the measure induced by $B_t^{S2}$. In short:
$$Y_t^\Delta t \xrightarrow{d} B_t^{S2} \quad \text{as } \Delta t \to 0$$

Extra Tips for Beginners #

• Why Donsker’s Blueprint?: It’s the standard way to prove discrete processes converge to continuous diffusions—finite-dimensional distributions capture the "marginal" behavior, while tightness ensures the limit is a continuous process (no jumps or wild oscillations).
• Well-Defined Projection: As $\Delta t$ shrinks, $S_t^\Delta t$ becomes more spread out (via CLT), so the probability it’s near 0 goes to 0—we can safely ignore the rare cases where projection is undefined.
• Generator Check: For deeper understanding, you can verify the limit process has generator $\frac{1}{2}\Delta_{S^2}$ by computing the generator of the spherical random walk and taking the limit as $\Delta t \to 0$. This is more advanced but connects convergence to the infinitesimal behavior of the process.

备注：内容来源于stack exchange，提问作者K252