Originally Posted by

**akbar** Using this result, in the case of the 3d random walk, the correlation matrix of the increment vector is diagonal (this is immediate from the probability mass function). E.g. the vector components are uncorrelated (yet not independent).

This ensures that the component of the limiting Gaussian vector are independent (2 by 2), since in the case of multivariate Gaussian variables it is implied by the fact they are uncorrelated. This enables to move to the independence case and apply your solution as it is with the limiting distribution.

I may be wrong, but I doubt that this argument (i.e. independence at the limit) will be enough to turn the sketch into a rigorous proof (i.e. get independence between components).

On the other hand, I could find excerpts from the book on GoogleBooks and I noticed that the result on p138 can be used for a simpler proof. This result tells that $\displaystyle P(W_{2n}=0)\sim_n c n^{-d/2}$.

Since the walk is symmetric, it is very tempting to say that $\displaystyle P(W_{2n}=x)$ is maximum when $\displaystyle x=0$, i.e. there are more paths of length $\displaystyle 2n$ from 0 back to 0 than from 0 to $\displaystyle x$ if $\displaystyle x\neq 0$. There's probably a simple proof for that. More about this later.

Then (assuming this), we have $\displaystyle P(|W_{2n}|<n^{\frac{1}{6}-\varepsilon})\leq \#B(0,n^{\frac{1}{6}-\varepsilon}) P(W_{2n}=0)$ (where $\displaystyle \#B(0,n^{\frac{1}{6}-\varepsilon})$ is the number of vertices inside the ball of radius $\displaystyle n^{\frac{1}{6}-\varepsilon}$ centered at 0). In dimension 3, $\displaystyle \#B(0,n^{\frac{1}{6}-\varepsilon})\sim_n C (n^{\frac{1}{6}-\varepsilon})^3 = C n^{\frac{1}{2}-3\varepsilon}$, hence (using the result p138):

$\displaystyle P(|W_{2n}|<n^{\frac{1}{6}-\varepsilon})\leq C' n^{\frac{1}{2}-3\varepsilon}n^{-3/2}=\frac{C'}{n^{1+3\varepsilon}}$

and you can apply Borel-Cantelli. That's it.

Coming back to the inequality $\displaystyle P(W_{2n}=x)\leq P(W_{2n}=0)$, one way to prove it is again based on p138. You have (cf. the book): (set $\displaystyle d=3$)

$\displaystyle P(W_{2n}=0)=\frac{1}{d^{2n}}\int_{-\pi}^\pi\cdots\int_{-\pi}^\pi (\cos\lambda_1+\cdots+\cos\lambda_d)^{2n}d\lambda_ 1\cdots d\lambda_d$.

Using the same proof but multiplying by $\displaystyle e^{-i(x,\lambda)}$ before integrating, we get (the result is real, so we can take the real part):

$\displaystyle P(W_{2n}=x)=\frac{1}{d^{2n}}\int_{-\pi}^\pi\cdots\int_{-\pi}^\pi (\cos\lambda_1+\cdots+\cos\lambda_d)^{2n}\cos(\lam bda_1 x_1+\cdots+\lambda_d x_d)d\lambda_1\cdots d\lambda_d$.

On this formula, it is clear (bounding the right-end cos by 1, since the other factor is positive) that indeed $\displaystyle P(W_{2n}=x)\leq P(W_{2n}=0)$.