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For $1 \leq s' < s \leq n$ and $t,t' \in \mathbb{R}$, one has $$ n(1-L_n) > t, (S(n)-1)(1-\frac{L_{S(n)-1}}{L_n}) > t',S(n)=s, Z(n) = s' $$ precisely when $$ U_1,\dots,U_{s'-1} < U_{s'},\\ U_{s'+1},\ …

As pointed out by fedja, one can not expect the inequality yo hold with the same epsilon. However, the following holds : if $\mathbf{P}_{z}( |f(z)| \leq \epsilon ) \geq 1 - \delta$, with $\delta \ll n …

One should assume that $X$ is not a.s. $0$, in which case $\tau$ is a.s. integer-valued and non-constant, hence a.s. non-continuous, hence a.s. non-Lipschitz.

The simplest idea is to estimate the variance. One has $$ \left(\prod^{d}_{j=1}\hat{\mu}_{j} \right)^2=\frac{1}{n^{2d}}\sum^{n}_{i_1,...,i_{2d}=1}\prod^{d}_{j=1}x^{(i_j)}_j x^{(i_{j+d})}_j $$ When $i_ …

Yes. Indeed, if $s = \sum_{i \geq 1} t_i^2 <1$, then $$ \mathbb{P}[ \ \ \forall n, \sum_{i=1}^n X_i \in [-1,1] \ \ ] \geq 1-s > 0. $$ To see this, note that $M_n = |\sum_{i=1}^n X_i|$ is a nonnegativ …

For fixed $d$, one can actually achieve a bound independent of $n$. More precisely, $K=K(d)=O(\sqrt{d})$ is fine, uniformly in $n$. Proof : the unit ball of $\mathbb{R}^d$ can be covered by $2C^d$ ba …