Pome

Proposition 18.41.132E997A26D38A

Proposition 18.41.13.2E997A26D38A Fix a real $\tau $. Choose $z_ m\ne 0$ with

\[ m|z_ m|^{2r}=\frac{q_ m}{r}+\tau \sqrt{\frac{q_ m}{r}}, \]

which is positive for all sufficiently large $m$. Then

\[ \frac{\pi B_{m,r}^{(q_ m)}(z_ m)}{mr^2|z_ m|^{2r-2}} \longrightarrow \Phi (\tau ). \]

In particular, this applies both when $q_ m/m\to \lambda >0$ and when $q_ m/m\to 0$.

Proof. We first recall the central limit theorem for Poisson variables in the form used here. For a fixed real $M>0$, Stirling's formula for $\ell !$ gives, uniformly over integers $\ell $ with $|\ell -T|\leq M\sqrt T$,

\[ \mathbb P(N_ T=\ell )=\frac1{\sqrt{2\pi T}} \exp \left(-\frac{(\ell -T)^2}{2T}\right) (1+O_ M(T^{-1/2}))\quad (T\to \infty ). \]

Indeed, insert the expansion of the factorial in $\mathrm{e}^{-T}T^\ell /\ell !$ and expand the logarithm in powers of $(\ell -T)/T$; the cubic remainder is $O_ M(T^{-1/2})$. Riemann sums over bounded intervals therefore converge to Gaussian integrals, and Chebyshev's bound $\mathbb P(|N_ T-T|>M\sqrt T)\leq M^{-2}$ then allows $M\to \infty $. Consequently, if real thresholds $a_ T$ satisfy $(a_ T-T)/\sqrt T\to a\in \mathbb R$, then $\mathbb P(N_ T\geq a_ T)\to 1-\Phi (a)$; rounding changes the normalized threshold by at most $T^{-1/2}$, and $\Phi $ is continuous.

Set $T_ m=m|z_ m|^{2r}$. Then $T_ m\to \infty $ and, for every integer $0\leq s<r$,

\[ \frac{L_ s(q_ m)-T_ m}{\sqrt{T_ m}}\longrightarrow -\tau . \]

Apply the preceding Poisson limit in (18.41.7). The uniform error there tends to zero, and each of the finitely many probabilities tends to $1-\Phi (-\tau )=\Phi (\tau )$. $\square$

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