Consider the following claim:

Claim: Let $x$ be a real random variable distributed according to the uniform distribution on the unit interval $U(0,1)$. Then for any quadratic irrational number $\alpha$, and real $\epsilon>0$, there exist finite real numbers $c,\gamma,\Delta>0$ such that $$ \mathrm{Prob}\left( \min_{\substack{(p,q) \in \mathbb{Z}^2 \\ q \neq 0}}|\alpha q - p - x|q^{1+\epsilon} < \delta\right) \leq c \delta^\gamma $$ is true for all $\delta < \Delta$.

My Question: Does the claim hold?

It may be helpful to note that one can always find a $p,q$ independent constant $C(\alpha,\epsilon)$ such that the bound $$ \left|\alpha - \frac{p}{q} \right| \geq \frac{C(\alpha,\epsilon)}{q^{2+\epsilon}} $$ is tight. This inequality holds for any algebraic $\alpha$ (Roth's theorem) and holds with probability one for $\alpha$ drawn from a piecewise continuous distribution (Gauss Kuzmin statistics). Thus for $x = 0$, we have that $$ \min_{(p,q) \in \mathbb{Z}^2, \,q \neq 0}|\alpha q - p|q^{1+\epsilon} = C(\alpha,\epsilon), $$ and the question is essentially about how frequently the constant on the RHS becomes small when $x$ is varied away from zero.

My suspicion is that the claim holds, and that the bound is tight for $\gamma = 1$. Thus I would also be grateful for any insight regarding the follow up questions

  • Can we choose $\gamma = 1$ always?
  • Can $c$ be bounded?
  • Does the claim hold with probability one for $\alpha$ itself drawn randomly from the unit interval (i.e. for $\alpha$ with Gauss-Kuzmin statistics)?
  • $\begingroup$ are you sure it's $\alpha p + q$? I would have expected $\alpha q + p$ (occurring as the numerator of $\alpha + p/q$) $\endgroup$ Commented Mar 11, 2021 at 20:39
  • $\begingroup$ @AndreaMarino why not $\alpha q - p$, since $\alpha-\frac{p}{q}$ is of interest? $\endgroup$ Commented Mar 11, 2021 at 21:53
  • 1
    $\begingroup$ @AndreaMarino I think it doesn't change the answer, but in response to your comment I have changed it, as you are right, it is off-putting, and potentially needlessly confusing. $\endgroup$ Commented Mar 11, 2021 at 21:58
  • $\begingroup$ @mathworker21 sure why not, I changed it, though I hope the audience is aware that $p \in \mathbb{Z}$ implies $(-p) \in \mathbb{Z}$. :) $\endgroup$ Commented Mar 11, 2021 at 21:59
  • $\begingroup$ @ComptonScattering you need to rule out $q=0$ from the minimum. $\endgroup$ Commented Mar 11, 2021 at 22:05


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