# Dispersion of random walk with scaled step sizes

Let $$Y_j$$ be a sequence of independent Gaussian random variables with mean zero and unit variance ($$\mathbb{E} Y_j = 0$$ and $$\mathbb{E} Y_j^2 = 1$$) and let $$\sigma:\mathbb{R}\to [1,2]$$.

We define the random variable $$X_k$$ so that $$X_0=0$$ and $$X_N = \sum_{j=1}^N \sigma(X_{j-1}) Y_j.$$ It is not so hard to show that $$N \leq \mathbb{E} X_N^2 \leq 4N$$. I would like to know whether one can show the anticoncentration bound $$\mathbb{P}( X_N \in [-1,1]) \leq C N^{-1/2},$$ with a constant $$C$$ independent of the function $$\sigma$$. More generally, I'd like to know if the probability density function of $$X_N$$ is bounded by $$CN^{-1/2}$$ (or $$CN^{-d/2}$$ in a $$d$$-dimensional generalization).

As an aside, this seems related to the question of $$L^1\to L^\infty$$ bounds for the variable coefficient parabolic PDE $$\partial_t u = \nabla \cdot (a(x) \nabla u),$$ which can be done elegantly with Nash's argument. In this case however (a naive translation of) Nash's argument fails because (1) there isn't an integration by parts formula and (2) the operator mapping the density of $$X_{N-1}$$ to the density of $$X_N$$ is not self-adjoint.

• Maybe I am missing somehitng basic, but since $\sigma$ only ever takes values in $[1,2]$, it seems easy to show that $\mathbb{P}( X_N \in [-1,1]) \leq \mathbb{P}( S_N \in [-1,1])$ where $S_N = \sum_{j=1}^N Y_j$. And since $S_N \sim N(0, \sqrt{N})$, the inequality seems trivial. Or did I misunderstand the question? Commented Aug 1 at 9:56
• The comparison you claim is what I am unable to prove. For example if you choose an interval other than $[-1,1]$ (perhaps $[x-1,x+1]$ where $x = 10\sqrt{N}$) your proposed inequality fails if $\sigma=2$. Commented Aug 1 at 10:32
• You are right, I forgot to take into account the correlations induced by using $X_{j-1}$ as an argument to $\sigma$. Indeed, I can choose $\sigma$ such that the inequality does not hold. Commented Aug 1 at 10:57
• $$\mathbb{E} X_N^2 \sim N$$ What does that mean? Are you using the letter $N$ for two different things in this expression? I'm tempted to construed${}\sim N$ as "is normally distributed" (i.e. Gaussian) but, to say the least, that doesn't seem right. Commented Aug 1 at 21:57
• If you allow σ to also depend on j , then the claim is false; See (for continuous time) arxiv.org/pdf/0903.3068 and for discrete time (but with bounded increments, which should not change much) arxiv.org/pdf/1402.2402. This shows that time-change ideas and embedding into Brownian motion are ill-suited for this problem. I am not sure yet what happens if σ is not allowed to depend on j . Commented Aug 3 at 14:42

• Does the arcsine law imply that the Brownian motion hits zero (and thereby any interval containing zero) between $N$ and $4N$ with constant probability? If so, does that imply that the worst-case stopping time you describe would stop in the interval with constant probability?
• I agree with Ziv. The probability that $B_N > 0$ and $B_{4N} <0$ is bounded below from a constant. For example there is a constant probability that $0<B_N<\sqrt{N}$, and then a constant probability that $B_{4N}-B_N < -\sqrt{N}$ ($B_{4N}-B_N$ being just another independent Gaussian). Commented Aug 2 at 5:34
• If you allow $\sigma$ to also depend on $j$, then the claim is false; See (for continuous time) arxiv.org/pdf/0903.3068 and for discrete time (but with bounded increments, which should not change much) arxiv.org/pdf/1402.2402. This shows that time-change ideas and embedding into Brownian motion are ill-suited for this problem. I am not sure yet what happens if $\sigma$ is not allowed to depend on $j$. Commented Aug 3 at 8:27