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By central limit theorem, random walk in $R^1$ converges in distribution to the Brownian motion in $R^1$.

  1. For defining a 2D random walk, is there any difference between :

a) If we decompose a 2D random walk to two one dimensional random walks,

b) If we define the random walk by a unit vector like $a=(x,y)'X =(\cos(\theta), \sin(\theta))'X$? $\theta$ and $X$ have uniform distribution between $[0,2\pi]$ and $[-1,1]$ and they are independent.

  1. In the light of the first question, how can we prove that the 2D random walk (by 1a and/or 1b), can converge in distribution to the 2D Brownian motion?

  2. In 1a, we have two variances $\sigma_x$ and $\sigma_y$ and in 1b, the variance is a matrix with zero off-diagonal. How can we have two forms of expressing variance, whereas the variance of a 2d Brownian motion is only a scalar. I think $\sigma_{2dBM}^2=4t$.

Edit: I have just found that the variance of Brownian motion in higher dimensions is in matrix form. I am still confused, why in many sources the variance is only expressed as a scalar. Only having zero off-diagonals in the matrix? More interested to see if we can decompose safely the Brownian motion/random walk to one dimensional movements, and then use central limit theorem...

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    $\begingroup$ How do you define 2D Brownian motion? What sort of random walk do you consider in 1a? $\endgroup$ – Sextus Empiricus Aug 22 at 23:16
  • $\begingroup$ Terminology and definitions are not correct. A random walk converges weakly to the Brownian motion, not just in (finite dimensional) distribution. The 2D version of Brownian motion is called Brownian sheet. The invariance principle extends to the multidimensional setting. In particular, yes, a 2D version random walk converges weakly to the Brownian sheet, after normalization. What you define as $a$ is not a 2D random walk. $\endgroup$ – Michael Aug 23 at 7:42
  • $\begingroup$ @Michael the case 1b is indeed not a simple random walk. But it is some sort of random walk, not? $\endgroup$ – Sextus Empiricus Aug 23 at 8:02
  • $\begingroup$ @SextusEmpiricus just two simple 1d random walks . not sure if it is a right way... $\sum X_i$ $\endgroup$ – Rob Aug 23 at 9:00
  • $\begingroup$ @Michael i edited the definition of $a$. Also i added something to the post. Appreciate to have your explanation. $\endgroup$ – Rob Aug 23 at 9:28

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