**6. Conclusions**

The asymptotic distributions of the sample correlation coefficient of vectors with random dimensions are normal scale mixtures. From (43) and (52), one can conclude that random dimension and corresponding scaling have significant influence on limit distributions A scale mixture of a normal distribution change the tail behavior of the distribution. Students *t*-Distributions have polynomial tails, as one class of heavy-tailed distributions, they can be used to model heavy-tail returns data in finance. The Laplace distributions have heavier tails than normal distributions.

**Author Contributions:** Conceptualization, G.C. and V.V.U.; methodology, V.V.U. and G.C.; writing–original draft, G.C. and V.V.U.; writing–review and editing, V.V.U. and G.C. All authors have read and agreed to the published version of the manuscript.

**Funding:** This research received no external funding. It was done within the framework of the Moscow Center for Fundamental and Applied Mathematics, Lomonosov Moscow State University, and HSE University Basic Research Programs.

**Acknowledgments:** The authors would like to thank the Managing Editor and the Reviewers for the careful reading of the manuscript and pertinent comments. Their constructive feedback helped to improve the quality of this work and shape its final form.

**Conflicts of Interest:** The authors declare no conflict of interest.

#### **Appendix A. Proofs of the Theorems and Lemmas**

#### *Appendix A.1. Proofs of Theorems 1 and 2*

**Proof of Theorem 1.** The proof follows along the similar arguments of the more general transfer Theorem 3.1 in Bening et al. [24]. Since in Theorem 3.1 in Bening et al. [24] the constant *γ* has to be non-negative and in our Theorem 1, we also need *γ* = −1/2, therefore we repeat the proof. Conditioning on *Nn*, we have

$$\mathbb{P}\left(\mathbf{g}\_n^\gamma T\_{\mathcal{N}\_n} \le \mathbf{x}\right) = \mathbb{P}\left(\mathbf{N}\_n^\gamma T\_{\mathcal{N}\_n} \le \mathbf{x} \left(\mathcal{N}\_n / \mathcal{g}\_n\right)^\gamma\right) \\ = \sum\_{m=1}^\infty \mathbb{P}\left(m^\gamma T\_m \le \mathbf{x} (m/\mathcal{g}\_n)^\gamma\right) \mathbb{P}(\mathcal{N}\_n = m).$$

Using now (18) with <sup>Φ</sup>*m*(*z*) := <sup>Φ</sup>(*z*) + *m*<sup>−</sup><sup>1</sup> *f*2(*z*), we find

$$\sum\_{m=1}^{\infty} \left| \mathbb{P} \left( m^{\gamma} T\_m \le \mathbf{x} (m / \mathcal{g}\_n)^{\gamma} \right) - \Phi\_m (\mathbf{x} (m / \mathcal{g}\_n)^{\gamma}) \right| \: \mathbb{P} (\mathcal{N}\_n = m) $$

$$\stackrel{(18)}{\leq} \mathbb{C}\_1 \sum\_{m=1}^{\infty} m^{-a} \: \mathbb{P} (\mathcal{N}\_n = m) = \mathbb{C}\_1 \, \mathbb{E} (\mathcal{N}\_n^{-a}). \tag{A1}$$

Taking into account P *Nn*/*gn* < 1/*gn* = P *Nn* < 1 = 0, we obtain

$$\begin{aligned} \sum\_{m=1}^{\infty} \Phi\_{\mathfrak{m}}(\mathbf{x} \,(m/\mathfrak{g}\_{\mathfrak{n}})^{\gamma}) \mathbb{P}(N\_{\mathfrak{n}} = m) &= \, \, \mathbb{E} \left( \Phi\_{\mathbf{N}\_{\mathfrak{n}}} (\mathbf{x} \,(N\_{\mathfrak{n}}/\mathfrak{g}\_{\mathfrak{n}})^{\gamma}) \right) \\ &= \int\_{1/\mathfrak{g}\_{\mathfrak{n}}}^{\infty} \Lambda\_{\mathfrak{n}}(\mathbf{x}, \mathfrak{y}) d\mathbb{P} \left( \frac{N\_{\mathfrak{n}}}{\mathfrak{g}\_{\mathfrak{n}}} \le \mathfrak{y} \right) = \mathsf{G}\_{\mathfrak{n}}(\mathbf{x}, 1/\mathfrak{g}\_{\mathfrak{n}}) + I\_{1/2} \end{aligned}$$

where <sup>Δ</sup>*n*(*<sup>x</sup>*, *y*) := <sup>Φ</sup>(*xy<sup>γ</sup>*) + *f*2(*xy<sup>γ</sup>*)/(*gny*), *Gn*(*<sup>x</sup>*, 1/*gn*) is defined in (21) and

$$I\_1 = \int\_{1/\mathcal{g}\_n}^{\infty} \Lambda\_{\mathcal{U}}(x, y) d\left(\mathbb{P}\left(\frac{\mathcal{N}\_n}{\mathcal{g}\_n} \le y\right) - H(y) - \frac{h\_2(y)\,\mathbb{I}\_{\{b > 1\}}(b)}{n}\right).$$

Estimating integral *I*1, we use the integration by parts for Lebesgue–Stieltjes integrals, the boundedness of *f*2(*z*), say sup*z* | *f*2(*z*)| ≤ *<sup>c</sup>*<sup>∗</sup>1, and estimates (19)

$$\begin{split} |I\_{1}| &\leq \sup\_{\boldsymbol{\lambda} \in \mathbb{T}^{n}} \lim\_{\boldsymbol{l} \to \infty} |\Delta\_{\boldsymbol{n}}(\boldsymbol{x}, \boldsymbol{y})| \left| \mathbb{P} \{ \mathcal{N}\_{\boldsymbol{n}} / \mathcal{g}\_{\boldsymbol{n}} \leq \boldsymbol{y} \} - H(\boldsymbol{y}) - n^{-1} h\_{2}(\boldsymbol{y}) \mathbb{I} \{ b > 1 \} (b) \right| \Big|\_{\boldsymbol{y} = 1/\mathcal{g}\_{\boldsymbol{n}}}^{\boldsymbol{y} = L} \\ &+ \sup\_{\boldsymbol{x}} \int\_{1/\mathcal{g}\_{\boldsymbol{n}}}^{\infty} \left| \frac{\partial}{\partial \boldsymbol{y}} \Delta\_{\boldsymbol{n}}(\boldsymbol{x}, \boldsymbol{y}) \right| \left| \mathbb{P} \{ \mathcal{N}\_{\boldsymbol{n}} / \mathcal{g}\_{\boldsymbol{n}} \leq \boldsymbol{y} \} - H(\boldsymbol{y}) - n^{-1} h\_{2}(\boldsymbol{y}) \mathbb{I}\_{\{b > 1\}}(b) \right| dy \\ &\leq \quad \left( 1 + c\_{1}^{\*} \right) \operatorname{C}\_{2} n^{-b} + \operatorname{C}\_{2} D\_{\boldsymbol{n}} n^{-b} . \end{split}$$

where *Dn* is defined in (22). Together with (A1), we obtain (20) and Theorem 1 is proved. **Proof of Theorem 2.** Using (23), we find for *b* > 0

$$\int\_0^{1/\xi\_n} \Phi(\mathbf{x} \, y^\gamma) dH(y) \le \int\_0^{1/\xi\_n} dH(y) = H(1/\sqrt{\xi\_n}) - H(0) \stackrel{(23)}{\le} c\_1 \varrho\_n^{-b}$$

.

.

Let now *b* > 1. Since *f*2(*z*) is supposed to be bounded, it follows from | *f*2(*z*)| ≤ *c*∗1 < ∞ and (24i) that

$$\int\_0^{1/\xi\_n} |f\_2(\mathbf{x}\,y^\gamma)| \, y^{-1} dH(y) \le c\_1^\* \int\_0^{1/\xi\_n} y^{-1} dH(y) \stackrel{(24)}{\le} c\_1^\* c\_2 \mathbf{g}\_n^{-b+1}.$$

Integration by parts, |*z*|*ϕ*(*z*) ≤ *c*<sup>∗</sup> = (2 *π e*)−1/2, (24ii) and (24iii) lead to

$$\left| \left| \int\_0^{1/\xi\_n} \Phi(\mathbf{x} \, y^\gamma) dh\_2(y) \right| \le |h\_2(1/\xi\_n)| + \gamma c^\* \int\_0^{1/\xi\_n} y^{-1} |h\_2(y)| dy \le (c\_3 + \gamma c^\* c\_4) n \, \mathcal{g}\_n^{-b} \dots$$

Theorem 2 is proved.

*Appendix A.2. Proofs of Lemmas 1 to 5*

**Proof of Lemma 1.** To estimate *Dn*, we consider three cases:

> *Dn* = sup*x* |*Dn*(*x*)| = max{sup*x*><sup>0</sup> |*Dn*(*x*)|, sup*x*<<sup>0</sup> |*Dn*(*x*)|, |*Dn*(0)|}.

Let *x* > 0. Since *∂∂y*<sup>Φ</sup>(*<sup>x</sup> y<sup>γ</sup>*) = *γ x yγ*−<sup>1</sup>*ϕ*(*x y<sup>γ</sup>*) ≥ 0, we find

$$\int\_{1/\varrho\_{\mathbb{R}}}^{\infty} \left| \frac{\partial}{\partial y} \Phi(\mathbf{x} \, y^{\gamma}) \right| dy = \int\_{1/\varrho\_{\mathbb{R}}}^{\infty} \gamma \mathbf{x} \, y^{\gamma - 1} \varphi(\mathbf{x} \, y^{\gamma}) dy = \int\_{\mathbf{x} \, \mathbb{S}\_{\mathbb{R}}}^{\infty} \varphi(u) du = \Phi(\infty) - \Phi(\mathbf{x} \, \mathbb{S}\_{\mathbb{R}}^{-\gamma}) \le 1/2.$$

Consider now *f*2(*<sup>x</sup> yγ*; *a*)=(*a*(*x y<sup>γ</sup>*)<sup>3</sup> − 5 *x y<sup>γ</sup>*) *ϕ*(*x y<sup>γ</sup>*)/4 with *a* = 1 or *a* = 1/3. Then,

$$\frac{\partial}{\partial y}\left(\frac{f\_2(\mathbf{x};y^\gamma;a)}{y}\right) = \frac{Q\mathbf{\tilde{z}}(\mathbf{x};y^\gamma;a)}{4y^2}, \quad Q\_5(\mathbf{z};a) = -\left(\gamma a \, z^5 - \left(\left(3a+5\right)\gamma - a\right) \, z^3 + \dots \left(\gamma - 1\right) \mathbf{z}\right) q(\mathbf{z}). \tag{A2}$$

Since sup*z* |*Q*5(*<sup>z</sup>*; *a*)| ≤ *c*(*γ*; *a*) < ∞ and *g*<sup>−</sup><sup>1</sup> *n* ∞1/*gn y*<sup>−</sup><sup>2</sup>*dy* = 1, inequality (29) holds for *x* > 0. Taking into account |*Dn*(*x*)| = |*Dn*(−*<sup>x</sup>*)| and *Dn*(0) = 0, Lemma 1 is proved.

**Proof of Lemma 2.** Using (30), we find *Gr*,*<sup>r</sup>*(1/*gn*) ≤ *<sup>c</sup>*1*g*<sup>−</sup>*<sup>r</sup> n* with *c*1 = *<sup>r</sup>r*−1/Γ(*r*). For *r* > 1, then 1/*gn* 0 *<sup>y</sup>*<sup>−</sup><sup>1</sup>*dGr*,*<sup>r</sup>*(*y*) ≤ *<sup>c</sup>*2*g*<sup>−</sup>*r*+<sup>1</sup> *n* with *c*2 = *r<sup>r</sup>*/-(*r* − <sup>1</sup>)Γ(*r*). Since *gr*,*<sup>r</sup>*(0) = 0, *<sup>h</sup>*2;*r*(0) = 0 and *gn* ≤ *r n* for *r* > 1, then (24ii) and (24iii) hold with *c*3 = *c*∗*r* and *c*4 = *c*∗*r* /(*r* − <sup>1</sup>), where *c*∗*r* = *rr* 2*r*<sup>Γ</sup>(*r*)sup*y*{*e*<sup>−</sup>*<sup>r</sup> y* (|*y* − 1||2 − *r*| + 1)} < ∞.

It remains to prove the bounds in (34) and (35). Let first *r* < 1. With *c*∗1= sup*z*| *f*2(*<sup>z</sup>*; *<sup>a</sup>*)|, we find

$$|I\_1(x,n)| \le \frac{c\_1^\* r^r}{g\_n \Gamma(r)} \int\_{1/\mathcal{S}\_n}^{\infty} y^{r-2} dy \le \frac{c\_1^\* r^r}{(r-1)\Gamma(r)} \, g\_n^{-r} \quad \text{with} \quad c\_5 = \frac{c\_1^\* r^r}{(r-1)\Gamma(r)}.$$

If *r* = 1 with *<sup>c</sup>*∗∗1 = sup*z*{|*<sup>a</sup> z*2 − <sup>5</sup>|*ϕ*(*z*/√2)}, we find | *f*2(*<sup>z</sup>*; *a*)| ≤ *<sup>c</sup>*∗∗1 |*z*| *ϕ*(*z*/√2) and

$$|I\_1(\mathbf{x}, n)| \le \frac{c\_1^{\ast \ast} |\mathbf{x}|}{\sqrt{2\pi}n} \int\_{1/n}^{\infty} y^{\gamma - 1} e^{-\left(y + x^2 y^{2\gamma}/4\right)} dy \quad \text{with} \quad \gamma \in \{-1/2, 0, 1/2\}.$$

For *γ* = 1/2 using |*x*|(<sup>1</sup> + *x*2/4)−1/2 ≤ 2, we obtain

$$|I\_1(\mathbf{x}, n)| \le \frac{c\_1^{\*\*} |\mathbf{x}|}{\sqrt{2\pi}n} \int\_{1/n}^{\infty} y^{1/2 - 1} e^{-(1 + \mathbf{x}^2/4)y} \, dy \le \frac{c\_1^{\*\*} |\mathbf{x}| \Gamma(1/2)}{\sqrt{2\pi} \left(1 + \mathbf{x}^2/4\right)^{1/2}} n^{-1} \le c\_6 n^{-1}, \quad c\_6 = \sqrt{2} \, c\_1^{\*\*}.$$

If *γ* = −1/2, then Prudnikov et al. [37], formula 2.3.16.3, for *x* -= 0 leads to

$$I\_1(x,n) \le \frac{c\_1^{\*\*} \, |\mathbf{x}|}{\sqrt{2\pi}n} \int\_{1/n}^{\infty} y^{-1-1/2} e^{-(2y+x^2/(4y))} dy \le \frac{c\_1^{\*\*} \, |\mathbf{x}|}{\sqrt{2\pi}n} \frac{2\sqrt{\pi}}{|\mathbf{x}|} e^{-\sqrt{2}|\mathbf{x}|} \le \frac{\sqrt{2}c\_1^{\*\*}}{n}, \quad c\_6 = \sqrt{2} \, c\_1^{\*\*}.$$

Finally, if *γ* = 0, then *f*2(*<sup>x</sup> yγ*; *a*) = *f*2(*<sup>x</sup>*; *a*) does not depend on *y*. Using now

$$0 \le \ln n - \int\_{1/n}^{1} y^{-1} dG\_{1,1}(y) = \int\_{1/n}^{1} \frac{1 - e^{-y}}{y} dy \le 1 \quad \text{and} \quad \int\_{1}^{\infty} y^{-1} dG\_{1,1}(y) \le e^{-1},$$

then (34) for *r* = 1 holds with *c*6 = *c*∗1(1 + *<sup>e</sup>*<sup>−</sup><sup>1</sup>).

> Let *r* > 1. Integration by parts for Lebesgue–Stieltjes integrals in *<sup>I</sup>*2(*<sup>x</sup>*, *n*) in (35) and (A2) lead to

$$\mathcal{I}\_2(\mathbf{x}, n) \le \frac{1}{n \lg n} \left( c\_1^\* \, g\_{\mathbb{H}} \left| h\_{2\mathcal{F}}(1/\mathcal{g}\_n) \right| + \int\_{1/\mathcal{g}\_n}^{\infty} \frac{|Q\_5(\mathbf{x} y^\gamma; a)|}{4 \, y^2} \, |h\_{2\mathcal{F}}(y)| dy \right). \tag{A3}$$

Since *c*(*γ*; *a*) = sup*z* |*Q*5(*<sup>z</sup>*; *a*)| < ∞ and with above defined *c*∗*r* , we find

$$\int\_{1/\mathcal{g}\_n}^{\infty} \frac{|h\_{2;r}(y)|}{y^2} dy \le c\_r^\* \int\_{1/\mathcal{g}\_n}^{\infty} y^{r-3} dy = \frac{c\_r^\*}{(2-r)} \mathcal{g}\_n^{-r+2} \quad \text{for} \quad 1 < r < 2$$

and with *<sup>c</sup>*∗∗*r* = *rr*−<sup>1</sup> 2 <sup>Γ</sup>(*r*) sup*y*{(*e*<sup>−</sup>*<sup>r</sup> y*/2 (|*y* − 1| |2 − *r*| + 1)} < <sup>∞</sup>, we obtain

$$\int\_{1/\mathcal{g}\_n}^{\infty} \frac{|h\_{2;r}(y)|}{y^2} dy \le c\_r^{\*\*} \int\_{1/\mathcal{g}\_n}^{\infty} y^{r-3} e^{-ry/2} dy \le \frac{c\_r^{\*\*} \Gamma(r-2)}{(r/2)^{r-2}} \quad \text{for} \quad r > 2.$$

Hence, we obtain (35) for *r* > 1, *r* -= 2 with some constant 0 < *c*7 < ∞.

For *r* = 2, the second integral in line above is an exponential integral. Therefore, we estimate the integral *<sup>I</sup>*2(*<sup>x</sup>*, *n*) in (35) more precisely like in estimating *<sup>I</sup>*1(*<sup>x</sup>*, *n*) above, taking into account the given function *f*2(*<sup>z</sup>*; *<sup>a</sup>*).

Using |*h*2;2(*y*)| ≤ 4 *y e*<sup>−</sup>2*<sup>y</sup>* and consider (A2), define *<sup>P</sup>*4(*<sup>z</sup>*; *a*) by *Q*5(*<sup>z</sup>*; *a*) = −*z <sup>P</sup>*4(*<sup>z</sup>*; *a*)*ϕ*(*z*/√2) with *c*∗2 = sup*z* |*<sup>P</sup>*4(*<sup>z</sup>*; *a*)|*ϕ*(*z*/√2) < <sup>∞</sup>, we obtain in (A3)

$$\int\_{1/\mathcal{G}^n}^{\infty} \frac{|Q\_{\overline{\pi}}(\mathbf{x}\,y^{\gamma})|}{4\,y^2} |h\_{2,2}(y)| dy \le \frac{c\_2^\* |\mathbf{x}|}{\sqrt{2\pi}} \int\_{1/\mathcal{G}^n}^{\infty} y^{\gamma-1} e^{-(2y+x^2y^{2\gamma}/4)} dy.$$

We estimate the latter integral in the same way as *<sup>I</sup>*1(*<sup>x</sup>*, *n*) for the two cases *γ* = 1/2 *γ* = −1/2 and find (35) for *r* = 2 with some constants 0 < *c*7 < ∞.

In order to prove (35) for *r* = 2 and *γ* = 0, we consider for *α* > 0 the following inequalities:

$$\int\_{1/\mathcal{g}\_n}^{\infty} y^{-1} e^{-ay} dy \left\{ \begin{array}{ll} \leq \int\_{1/\mathcal{g}\_n}^1 y^{-1} dy + \int\_1^{\infty} e^{-ay} dy \leq \ln \mathcal{g}\_n + a^{-1} e^{-a} \\\\ \geq \int\_{1/\mathcal{g}\_n}^1 y^{-1} e^{-ay} dy \geq e^{-a} \int\_{1/\mathcal{g}\_n}^1 y^{-1} dy \geq e^{-a} \ln \mathcal{g}\_n \end{array} \right. \tag{A4}$$

The upper bound in (A4) leads to (35) for *r* = 2, *γ* = 0, too. The lower bound in (A4) shows that the ln *n*-term cannot be improved.

$$\text{Bound (36) for } n \ge 2, \ r > 1 \text{ results from } 0 \le \frac{1}{\mathcal{S}^n} - \frac{1}{r!!} = \frac{r - 1}{r^2 n^2 (1 - (r - 1)/(rn))} \le \frac{2(r - 1)}{r^2 n^2}. \quad \square$$

**Proof of Lemma 3.** Let *r* > 0. If *n* = 1, then <sup>P</sup>(*<sup>N</sup>*1(*r*) = 1) = 1 and (37) holds with *<sup>C</sup>*(*r*) = 1. Let *n* ≥ 2 and *α* > 0 

$$\mathbb{E}\left(N\_n(r)\right)^{-\alpha} = \frac{1}{n^r} \left(1 + \sum\_{k=2}^{\infty} \frac{\Gamma(k+r-1)}{k^\alpha \Gamma(r) \Gamma(k)} \left(1 - \frac{1}{n}\right)^{k-1}\right).$$

It follows from the relations (49) and (50) with their corresponding bounds in the proof of Theorem 1 in Christoph et al. [12] that

$$\frac{\Gamma(k+r-1)}{\Gamma(r)\Gamma(k)} = \frac{1}{(k+r-1)\,\,B(r\,k)} = \frac{k^{r-1}}{\Gamma(r)} \Big(1 + R\_1(k)\Big), \quad |R\_1(k)| \le \frac{c\_1(r)}{k}.\tag{A5}$$

.

For *x* ≥ *k* ≥ 2 using (1 − 1/*n*)*<sup>x</sup>* ≤ *<sup>e</sup>*<sup>−</sup>*<sup>x</sup>*/*<sup>n</sup>*, we find

$$\frac{k^{r-1}(1-1/n)^{k-1}}{k^a} \le \int\_k^{k+1} \frac{\mathbf{x}^r (1 - 1/n)^{\mathbf{x}-2}}{(\mathbf{x}-1)^{1+a}} d\mathbf{x} \le 2^{3+a} \int\_k^{k+1} \mathbf{x}^{r-3} e^{-\mathbf{x}/n} d\mathbf{x}.$$

Then, with *<sup>c</sup>*2(*r*) = 23+*α*(1 + *<sup>c</sup>*1(*r*))/Γ(*r*), we obtain

$$\mathbb{E}\left(\mathbf{N}\_{\mathbf{n}}(\mathbf{r})\right)^{-\mathbf{a}} \le c\_2(\mathbf{r})\mathbf{n}^{-r}\mathbf{J}\_{\mathbf{f}}(\mathbf{n}), \quad \text{where} \quad \mathbf{J}\_{\mathbf{f}}(\mathbf{n}) = \int\_1^\infty \mathbf{x}^{r-\mathbf{a}-1} \, e^{-\mathbf{x}/\mathbf{n}} d\mathbf{x} = n^{r-\mathbf{a}} \int\_{1/\mathbf{n}}^\infty \mathbf{y}^{r-\mathbf{a}-1} \, e^{-\mathbf{y}} d\mathbf{y}.$$

Since *Jr*(*n*) ≤ (*α* − *r*)−<sup>1</sup> for 0 < *r* < *α*, *Jr*(*n*) ≤ *nr*−*<sup>α</sup>* <sup>Γ</sup>(*r* − *α*) for *r* > *α* and using (A4) with *r* = *α Jr*(*n*) ≤ ln *n* + *e*<sup>−</sup>1, the upper bound (37) is proved.

Let *r* = *α* > 0. Considering the formula (A5), 0 ≤ ∑∞*<sup>k</sup>*=<sup>2</sup> *<sup>k</sup>*−<sup>1</sup>|*<sup>R</sup>*1(*k*)| ≤ *<sup>c</sup>*1(*r*)*π*2/(<sup>6</sup> <sup>Γ</sup>(*r*)) < <sup>∞</sup>, ∑(*n*) := ∑*<sup>n</sup>*−<sup>1</sup> *k*=2 *k*−<sup>1</sup> ≥ ln *n* − ln 2 and ∑*<sup>n</sup>*−<sup>1</sup> *k*=2 1 − (1 − 1/*n*)*<sup>k</sup>*−<sup>1</sup> *k* ≤ ∑*<sup>n</sup>*−<sup>1</sup> *k*=2 *k* − 1 *k n* ≤ 1, we find:

$$\mathbb{E}(N\_{\mathbf{n}}(r))^{-r} \geq \frac{1}{n^{r}\Gamma(r)} \left(\sum\_{k=2}^{n-1} \frac{1}{k} \left(1 - \frac{1}{n}\right)^{k-1} - c\_{3}\right) \geq \frac{1}{n^{r}\Gamma(r)} \left(\sum\_{k=2}^{n-1} \frac{1}{k} - c\_{4}\right) \geq \frac{1}{n^{r}\Gamma(r)} \left(\ln n - c\_{5}\right) > \frac{1}{n^{r}\Gamma(r)}$$

where *c*3 = *<sup>c</sup>*1(*r*)*π*2/6, *c*4 = 1 + *c*3 and *c*5 = *c*4 − ln 2. Hence, the ln *n*-term cannot be dropped.

**Proof of Lemma 4.** The upper bounds in the estimates (23) and (24) with *Hs*(*y*), *<sup>h</sup>*2;*s*(*y*) and *<sup>I</sup>*2(*<sup>x</sup>*, *n*) given in (40) are *<sup>C</sup>*(*s*)*e*<sup>−</sup>*<sup>s</sup> n*/2. For example, (24ii): 1/*n* 0 *<sup>y</sup>*<sup>−</sup><sup>1</sup>|*h*2;*s*(*y*)|*dy* ≤ *s*(*s* + 1)/2 1/*n* 0 *y*<sup>−</sup><sup>3</sup> *e*<sup>−</sup>*<sup>s</sup>*/*ydy* ≤ (*s* + <sup>1</sup>)/(<sup>2</sup>*s*) ∞*s n z e*<sup>−</sup>*zdz* ≤ (*s* + <sup>1</sup>)/(<sup>2</sup>*s*)*e*<sup>−</sup>*sn*/2.

**Proof of Lemma 5.** Proceeding as in Bening et al. [24] using

$$\mathbb{P}(N\_n(s) = k) = \left(\frac{k}{s+k}\right)^n - \left(\frac{k-1}{s+k-1}\right)^n = s \ln \int\_{k-1}^k \frac{x^{n-1}}{(s+x)^{n+1}} dx$$

and Formula 2.2.4.24 in Prudnikov et al. [37], p. 298, then

$$\mathbb{E}(N\_n^{-n}) = \operatorname{s} \operatorname{n} \sum\_{k=1}^{\infty} \frac{1}{k^n} \int\_{k-1}^k \frac{\mathbf{x}^{n-1}}{(\mathbf{s}+\mathbf{x})^{n+1}} d\mathbf{x} \le \operatorname{s} \operatorname{n} \int\_0^{\infty} \frac{\mathbf{x}^{n-1-n}}{(\mathbf{s}+\mathbf{x})^{n+1}} d\mathbf{x} = \operatorname{s} \operatorname{n} \operatorname{B}(n-n, 1+n).$$

$$\text{Using } B(n-a, 1+a) = \Gamma(1+a) \left(n+1\right)^{-1+a} \left(1 + R\_1/n\right) \text{ with } |R\_1| \le c < \infty \text{, we obtain (41).}\quad \square$$

#### *Appendix A.3. Proofs of Theorems 3 to 8*

**Proof of Theorem 3.** Since the additional assumptions (23) and (24) in the transfer Theorem 2 for the limit Gamma distribution *<sup>H</sup>*(*x*) = *Gr*,*<sup>r</sup>*(*x*) of the normalized sample size *Nn*(*r*) are satisfied by Lemma 2 with *b* = *r* > 0 and by Lemma 3 for *α* = 2, it remains to calculate the integrals in (26). Define

$$\begin{array}{rcl}f\_1^\*(\mathbf{x}) &=& \int\_0^\infty \Phi(\mathbf{x}\sqrt{y}) d\mathbf{G}\_{rJ}(y), & f\_2^\*(\mathbf{x}) = \int\_0^\infty \frac{a\left(\mathbf{x}\sqrt{y}\right)^3 - 5\mathbf{x}\sqrt{y}\ \mathbf{e}(\mathbf{x}\sqrt{y})}{4y} d\mathbf{G}\_{rJ}(y), & \text{and} \\ f\_3^\*(\mathbf{x}) &=& \int\_0^\infty \Phi(\mathbf{x}\sqrt{y}) d\mathbf{h}\_{2r}(y) \quad \text{with} \quad h\_{2r}(y) = \left( \left(y - 1\right)\left(2 - r\right) + 2Q\_1\left(\left(r(n-1) + 1\right)y\right) \right) \frac{\mathcal{G}\_{lr}(y)}{2r}, & \end{array}$$

and *Q*1(*y*) = 1/2 − (*y* − [*y*]). Then,

$$G\_{2;n}(\mathbf{x};0) = f\_1^\*(\mathbf{x}) + \frac{f\_2^\*(\mathbf{x})}{g\_n} + \frac{f\_3^\*(\mathbf{x})}{n} \quad \text{with} \quad g\_n = \mathbb{E}N\_{\mathbb{H}}(r) = r(n-1) + 1. \tag{A6}$$

Using formula 2.3.3.1 in Prudnikov et al. [37], p. 322, with *α* = *r* − 1/2, *r* + 1/2, *p* = 1 + *x*2/(<sup>2</sup> *r*) and *q* = 1:

$$M\_{\mathfrak{a}}(\mathbf{x}) = \frac{r^r}{\Gamma(r)\sqrt{2\pi}} \int\_0^\infty y^{a-1} e^{-(r+x^2/2)y} dy = \frac{\Gamma(a)}{\Gamma(r)\sqrt{2\pi}} \left(1 + \mathbf{x}^2/(2r)\right)^{-a} \tag{A7}$$

we calculate the integrals occurring in (A6). Consider

$$\begin{split} \frac{\partial}{\partial \mathbf{x}} J\_1^\*(\mathbf{x}) &= \int\_0^\infty y^{1/2} \varrho(\mathbf{x}\sqrt{y}) g\_{r,r}(y) dy = \frac{r^r}{\Gamma(r)\sqrt{2\pi}} \int\_0^\infty y^{r-1/2} e^{-(r+x^2/2)y} dy \\ &= \quad M\_{r+1/2}(\mathbf{x}) = \mathbf{s}\_{2\mathbf{r}}(\mathbf{x}) \quad \text{and} \qquad J\_1^\*(\mathbf{x}) = \mathbf{S}\_{2\mathbf{r}}(\mathbf{x}) \,. \end{split}$$

The integral *J*∗2 (*x*) in (A6) we calculate again with (A7) using *Mr*−1/2(*x*) = *<sup>s</sup>*2*r*(*x*) (<sup>2</sup>*r* + *x*<sup>2</sup>)/(<sup>2</sup>*r* − 1) and *Mr*+1/2(*x*) = *<sup>s</sup>*2*r*(*x*)

$$\begin{split} J\_2^\*(\mathbf{x}) &:= \quad \frac{r^r}{\sqrt{2\pi}\Gamma(r)} \int\_0^\infty \frac{1}{y} \left( a \, x^3 y^{3/2} - 5 \, x \, y^{1/2} \right) y^{r-1} \, e^{-(r+x^2/2)y} dy \\ &= \quad \left( a \, x^3 \, M\_{r+1/2}(\mathbf{x}) - 5 \, x \, M\_{r-1/2}(\mathbf{x}) \right) = \left( a \, x^3 - \frac{10 \, r \, x + 5 \, x^3}{2r - 1} \right) s\_{2r}(\mathbf{x}). \end{split}$$

The integral *J*∗3 (*x*) in (A6) is the same as the integral *J*4(*x*) in the proof of Theorem 2 in Christoph et al. [12] with the estimate

$$\sup\_{x} \left| J\_3^\*(x) - \frac{(2 - r)x(x^2 + 1)}{4r(2r - 1)} s\_{2r}(x) \right| \le c(r) n^{-r + 1}.$$

With (36), we proved (44).

**Proof of Theorem 4.** By Lemma 2, the additional assumptions (23) and (24) in the transfer Theorem 2 are satisfied with the limit Gamma distribution *<sup>H</sup>*(*x*) = *Gr*,*<sup>r</sup>*(*x*) of the normalized sample size *Nn*(*r*) with *b* = *r* > 0. In Transfer Theorem 1 for *TNn* = √*Nn RNn* , the right-hand side of (20) is estimated by Lemma 1 and Lemma 3 for *α* = 2 for the case *γ* = 0. Then, we have by (21) with (35)

$$\mathcal{G}\_{\mathbb{H}}(\mathbf{x}, 1/\mathcal{g}\_{\mathbb{H}}) = \Phi(\mathbf{x}) \left( \mathbf{1} - \mathcal{G}\_{\mathbb{H}}(1/\mathcal{g}\_{\mathbb{H}}) - n^{-1} h\_{2\mathcal{r}}(1/\mathcal{g}\_{\mathbb{H}}) \mathbb{I}\_{\{r > 1\}}(r) \right) + \frac{f\_{\mathbb{2}}(\mathbf{x}; a)}{\mathcal{g}\_{\mathbb{n}}} \int\_{1/\mathcal{g}\_{\mathbb{m}}}^{\infty} \frac{1}{y} d\mathcal{G}\_{\mathbb{r}, \mathcal{r}}(y) \mathbb{I}\_{\{r > 1\}}(r) .$$

The estimates (23), (24i), (24ii), (34) and ∞0 *<sup>y</sup>*<sup>−</sup><sup>1</sup>*dGr*,*<sup>r</sup>*(*y*) = *r* <sup>Γ</sup>(*r* − <sup>1</sup>)/Γ(*r*) for *r* > 1 lead to (46) with <sup>Φ</sup>*n*;2(*<sup>x</sup>*; 1) defined in (47). Thus, Theorem 4 is proved.

**Proof of Theorem 5.** By Lemma 2, the additional assumptions (23) and (24) in the transfer Theorem 2 are satisfied with the limit Gamma distribution *<sup>H</sup>*(*x*) = *Gr*,*<sup>r</sup>*(*x*) of the normalized sample size *Nn*(*r*) with *b* = *r* > 0. In Transfer Theorem 1 for *TNn* = *g*<sup>−</sup>1/2 *n Nn RNn* , the right-hand side of (20) is estimated by Lemma 1 and Lemma 3 for *α* = 2 for the case *γ* = −1/2. Then, we have in (25)

$$\mathbb{E}\left[G\_{2\mathbb{Z}}(\mathbf{x};0)\right] = l\_{1\mathcal{I}}^{\*}\left(\mathbf{x}\right) + \left(\frac{l\_{2\mathcal{I}}^{\*}\left(\mathbf{x}\right)}{\mathcal{g}\_{\mathcal{I}}} + \frac{l\_{3\mathcal{I}}^{\*}\left(\mathbf{x}\right)}{n}\right)\mathbb{I}\_{\{r>1\}}\left(r\right) \quad \text{with} \quad \mathbb{g}\_{\mathcal{U}} = \mathbb{EN}\_{\mathbb{n}}(r) = r(n-1) + 1 \tag{A8}$$

$$\begin{aligned} \mathcal{J}\_{1\mathcal{T}}^{\*}(\mathbf{x}) &= \int\_{0}^{\infty} \Phi(\mathbf{x}/\sqrt{y}) d\mathcal{G}\_{r\mathcal{T}}(y), \ \mathcal{J}\_{2\mathcal{T}}^{\*}(\mathbf{x}) = \int\_{0}^{\infty} \frac{(\mathbf{a}\,\mathbf{x}^{3}y^{-3/2} - 5\mathbf{x}\,y^{-1/2})\varphi(\mathbf{x}/\sqrt{y})}{4y} d\mathcal{G}\_{r\mathcal{T}}(y), \text{ and} \\\mathcal{J}\_{3\mathcal{T}}^{\*}(\mathbf{x}) &= \int\_{0}^{\infty} \Phi(\mathbf{x}/\sqrt{y}) d\mathcal{h}\_{2\mathcal{T}}(y) \ \text{ with } \mathcal{h}\_{2\mathcal{T}}(y) = \left( (y-1)\left(2-r\right) + 2Q\_{1}((r(n-1)+1)y) \right) \frac{\mathcal{g}\_{Tr}(y)}{2r}. \end{aligned}$$

Consider formula 2.3.16.1 in Prudnikov et al. [37], p. 444:

$$I\_a := \int\_0^\infty y^{a-1} \, e^{-py-q/y} dy = 2 \left(\frac{q}{p}\right)^{a/2} K\_a(2\sqrt{p\,q}) \quad p>0, \quad q>0,$$

where *<sup>K</sup>α*(*u*) is the *α*-order Macdonald function (or *α*-order modified Bessel function of the second kind), see, e.g., Oldham et al. [30], Chapter 51, for properties of these functions.

Let us calculate the integral *J*<sup>∗</sup>1;*r*(*x*) occurring in (A8). Consider

$$\begin{split} \frac{d}{dx} l\_{1,r}^\*(\mathbf{x}) &= \quad \frac{r^r}{\sqrt{2\pi}\,\Gamma(r)} \int\_0^\infty y^{r-3/2} e^{-ry-(\mathbf{x}^2/(2y))} dy \\ &= \quad \frac{2r^r}{\sqrt{2\pi}\,\Gamma(r)} \left(\frac{|\mathbf{x}|}{2r}\right)^{r-1/2} K\_{r-1/2}(\sqrt{2r}|\mathbf{x}|) =: l\_r(\mathbf{x}). \end{split} \tag{A9}$$

If *α* = ±1/2, ±3/2, ±5/2, ... the integral *Iα* and consequently *<sup>K</sup>α*(*x*) are computable in closed-form expressions with formula 2.3.16.2 in Prudnikov et al. [37], p. 444:

$$I\_m^\* = \int\_0^\infty y^{m-1/2} e^{-py-q/y} dy = (-1)^m \sqrt{\pi} \frac{\partial^m}{\partial p^m} \left( p^{-1/2} e^{-2\sqrt{pq}} \right)\_{\prime\prime} p > 0, \neq 0, \neq 0, m = 0, 1, 2, \dots \tag{A10}$$

and with formula 2.3.16.3 in Prudnikov et al. [37], p. 444:

$$I\_{-m}^{\*} = \int\_{0}^{\infty} y^{-m-1/2} e^{-py-q/y} dy = (-1)^{m} \sqrt{\frac{\pi}{p}} \frac{\partial^{m}}{\partial q^{m}} e^{-2\sqrt{p\cdot q}}, \ p > 0, \ q > 0, \ m = 0, 1, 2, \ldots$$

For *r* = 1, 2, 3 using (A10) with *m* = *r* − 1, we find

$$l\_r(\mathbf{x}) = \frac{d}{d\mathbf{x}} l\_{1,r}^\*(\mathbf{x}) = \frac{r^r}{\Gamma(r)\sqrt{2\pi}} \int\_0^\infty y^{r-3/2} e^{-ry-\mathbf{x}^2/(2y)} dy = \frac{r^r}{\Gamma(r)\sqrt{2\pi}} I\_{r-1}^\*$$

and we obtain the densities *lr*(*x*) in (49) with

$$I\_m^+(\mathbf{x}) = \begin{cases} \sqrt{2\pi} \frac{1}{|\mathbf{x}|} \int\_{\mathbf{x}} e^{-\sqrt{2\pi} |\mathbf{x}|} \, & \mathbf{x} \neq \mathbf{0} & m = -1, \\ \sqrt{\pi} \, e^{-\sqrt{2\pi} |\mathbf{x}|} \, & m = 0, \\ \sqrt{\pi} \left( \frac{1}{2r^{3/2}} + \frac{|\mathbf{x}| \sqrt{2}}{2r} \right) e^{-\sqrt{2\pi} |\mathbf{x}|} \, & m = 1, \\ \sqrt{\pi} \left( \frac{3}{4r^{5/2}} + \frac{3 \, |\mathbf{x}| \sqrt{2}}{4r^2} + \frac{|\mathbf{x}|^2}{2r^{3/2}} \right) e^{-\sqrt{2\pi} |\mathbf{x}|} \, & m = 2. \end{cases}$$

Consider now *J*2;*r*(*x*) for *r* = 2 and *r* = 3:

$$J\_{2r}^\*(\mathbf{x}) = \int\_0^\infty \frac{\left(a \, \mathbf{x}^3 \, y^{-3/2} - 5 \, \mathbf{x} \, y^{-1/2}\right) r^r \, y^{r-1} \, e^{-ry-\mathbf{x}^2/(2y)}}{4 \, y \sqrt{2\pi} \, \Gamma(r)} dy = \frac{r^r}{4\sqrt{2\pi} \, \Gamma(r)} \left(a \, \mathbf{x}^3 \, I\_{r-3}^\*(\mathbf{x}) - 5 \, \mathbf{x} \, I\_{r-2}^\*(\mathbf{x})\right).$$

Hence,

$$f\_{2,2}^\*(\mathbf{x}) = \left(a \ge |\mathbf{x}| - 5 \ge \sqrt{2}\right)e^{-2\left|\mathbf{x}\right|} \quad \text{and} \quad f\_{2,3}^\*(\mathbf{x}) = \frac{27}{8}\left(\frac{a \ge^3}{\sqrt{2}} - 5 \ge \left(\frac{1}{6\sqrt{6}} + \frac{|\mathbf{x}|}{6}\right)\right)e^{-\sqrt{6}|\mathbf{x}|}.$$

Integration by parts in the integral *J*<sup>∗</sup>3;*r*in (A8) leads to

$$\begin{split} f\_{3\mathcal{T}}^{\*}(\mathbf{x}) &:= \int\_{0}^{\infty} \Phi(\mathbf{x}\,y^{-1/2})d(\hbar\_{2\mathcal{T}}(y)) = \frac{\mathbf{x}}{2} \int\_{0}^{\infty} y^{-5/2} \varphi(\mathbf{x}\,y^{-1/2}) \, \mathrm{h}\_{2\mathcal{T}}(y) dy \\ &= \frac{r^{r}\mathbf{x}}{2r\,\Gamma(r)\sqrt{2\pi}} \int\_{0}^{\infty} y^{r-5/2} e^{-ry-\mathbf{x}^{2}/(2y)} \left( (y-1)\left(2-r\right) + 2Q\_{1}(g\_{\mathcal{B}}y) \right) dy \\ &= \frac{r^{r-1}\mathbf{x}}{2\,\Gamma(r)\sqrt{2\pi}} \int\_{0}^{\infty} y^{r-5/2} (y-1)\left(2-r\right) e^{-ry-\mathbf{x}^{2}/(2y)} dy \\ &+ \frac{r^{r-1}\mathbf{x}}{\Gamma(r)\sqrt{2\pi}} \int\_{0}^{\infty} y^{r-5/2} Q\_{1}(g\_{\mathcal{B}}y) e^{-ry-\mathbf{x}^{2}/(2y)} dy = f\_{3\mathcal{T},1}(\mathbf{x}) + f\_{3\mathcal{T},2}(\mathbf{x}). \end{split}$$

Since *J*3;2,1(*x*) vanishes, we calculate *J*3;3,1(*x*):

$$J\_{3;3,1}(\mathbf{x}) = \frac{9\,\mathrm{x}}{2\sqrt{2\pi}}(I\_1^\*(\mathbf{x}) - I\_2^\*(\mathbf{x})) = \frac{9\,\mathrm{x}}{2}\left(\frac{1}{12\sqrt{6}} + \frac{|\mathbf{x}|}{12} - \frac{|\mathbf{x}|^2}{6\sqrt{6}}\right)e^{-\sqrt{6}|\mathbf{x}|}.$$

It remains to estimate *J*3;2,2(*x*) and *J*3;3,2(*x*). The function *Q*1(*y*) is periodic with period 1:

$$Q\_1(y) = Q\_1(y+1) \text{ for all } y \in \mathbb{R} \text{ and } \\ Q\_1(y) := 1/2 - y \text{ for } 0 \le y < 1$$

It is right-continuous and has the jump 1 at every integer point *y*. The Fourier series expansion of *Q*1(*y*) at all non-integer points *y* is

$$Q\_1(y) = 1/2 - (y - [y]) = \sum\_{k=1}^{\infty} \frac{\sin(2\pi k y)}{k \,\pi} \quad y \neq [y]. \tag{A11}$$

see formula 5.4.2.9 in Prudnikov et al. [37], p. 726, with *a* = 0.

Using the Fourier series expansion (A11) of the periodic function *Q*1(*y*) and interchange sum and integral, we find

$$J\_{3;r,2}^{\*} = \frac{\mathbf{x}}{\sqrt{2\pi}} \sum\_{k=1}^{\infty} \frac{1}{k} \int\_{0}^{\infty} y^{r-5/2} e^{-r\,y - \mathbf{x}^{2}/(2y)} \sin(2\pi k \, g\_{n} \, y) dy. \tag{A12}$$

First, we consider *r* = 2. Let *p* > 0, *q* > 0 and *b* > 0 be some real constants. Formula 2.5.37.4 in Prudnikov et al. [37], p. 453 reads

$$\int\_0^\infty y^{-1/2} \, e^{-y\cdot y - q/y} \cdot \sin(b \, y) dy = \sqrt{\frac{\pi}{p^2 + b^2}} \, e^{-2\sqrt{q}z\_+} \left( z\_+ \sin(2\sqrt{q}z\_-) + z\_- \cos(2\sqrt{q}z\_-) \right) . \tag{A13}$$

with 2 *z*2± = 2*p*<sup>2</sup> + *b*2 ± *p*. Consider *z*± with *p* = *r*, *q* = *x*2/2, *b* = <sup>2</sup>*πkgn*, *k* ≥ 1 and *n* ≥ 1: Then,

$$
\sqrt{\frac{\pi}{p^2 + b^2}} = \sqrt{\frac{\pi}{r^2 + (2\pi \text{kg}\_n)^2}} \le \frac{\sqrt{\pi}}{2\pi \text{kg}\_n}, \quad \sqrt{2}|\mathbf{x}| z\_+ e^{-\sqrt{2}|\mathbf{x}|z\_+} \le e^{-1} \quad \text{and} \quad 0 < z\_- \le z\_+ 
$$

leads to

$$\begin{split} |f\_{3,2,2}^{\*}(\mathbf{x})| &\leq \ \ \ \frac{2\left|\mathbf{x}\right|}{\sqrt{2\pi}} \sum\_{k=1}^{\infty} \frac{1}{k} \sqrt{\frac{\pi}{p^{2}+b^{2}}} \ e^{-2\sqrt{q}z\_{+}} \left(z\_{+} \sin(2\sqrt{q}z\_{-}) + z\_{-} \cos(2\sqrt{q}z\_{-})\right) \\ &\leq \ \ \ \frac{2}{\sqrt{2\pi}} \sum\_{k=1}^{\infty} \frac{1}{k} \frac{\sqrt{\pi}\sqrt{2}e^{-1}}{2\pi k g\_{n}} = \frac{1}{2\pi e \, g\_{n}} \sum\_{k=1}^{\infty} \frac{1}{k^{2}} = \frac{\pi}{12\varepsilon \, g\_{n}}. \end{split}$$

Together with *gn* ≥ *n*, we find *<sup>n</sup>*<sup>−</sup><sup>1</sup>|*J*<sup>∗</sup>3;2,2(*x*)| ≤ *C n*<sup>−</sup>2.

Consider finally *J*<sup>∗</sup>3;3,2 given in (A12). In order to estimate *J*<sup>∗</sup>3;3,2(*x*), we apply Leibniz's rule for differentiation under the integral sign with respect to *p* in (A13) and obtain

$$\int\_0^\infty y^{1/2} e^{-y\cdot y - q/y} \sin(b\,y) dy \quad = \quad \frac{\partial}{\partial p} \left\{ \sqrt{\frac{\pi}{p^2 + b^2}} e^{-2\sqrt{q}z\_+} \left( z\_+ \sin(2\sqrt{q}z\_-) + z\_- \cos(2\sqrt{q}z\_-) \right) \right\}.$$

Simple calculation considering √*q* = |*x*|/√<sup>2</sup> and |*x*|*<sup>m</sup> e*−√<sup>2</sup> |*x*| *z*+ ≤ *m* 2*<sup>m</sup>*/2 *z*+ ≤ *m* 2*<sup>m</sup>*/2 *bm* for *m* = 1, 2, leads to

$$\frac{1}{2} \left| \mathbf{x} \right| \frac{\partial}{\partial p} \left\{ \sqrt{\frac{\pi}{p^2 + b^2}} e^{-2\sqrt{q}z\_+} \left( z\_+ \sin(2\sqrt{q}z\_-) + z\_- \cos(2\sqrt{q}z\_-) \right) \right\} \le \frac{\mathcal{C}}{b^2} = \frac{\mathcal{C}}{(2\pi \mathcal{S} \, \mathcal{g}\_2)^2}$$

and we find equation (A12) with *r* = 3 that *n*<sup>−</sup><sup>1</sup> |*J*<sup>∗</sup>3;3,2| ≤ *C n*<sup>−</sup><sup>3</sup> and (50) is proved. The approximation (52) holds since Lemmas 1, 2, and 3 are valid for arbitrary *r* > 0. Theorem 5 is proved.

**Proof of Theorem 6.** By Lemma 4, the additional assumptions (23) and (24) in the transfer Theorem 2 are satisfied with the limit inverse exponential distribution *Hs*(*y*) and *<sup>h</sup>*2;*s*(*y*) given in (40), *gn* = *n* and *b* = 2. In Transfer Theorem 1, the right-hand side of (20) is estimated by Lemma 1 and Lemma 5 for *α* = 2 for the case *γ* = 1/2. Then, we have in (25) with (35)

$$G\_{2; \mathfrak{u}}(\mathfrak{x}; 0) = f\_{1; \mathfrak{s}}^{\*}(\mathfrak{x}) + \mathfrak{n}^{-1} f\_{2; \mathfrak{s}}^{\*}(\mathfrak{x}) + \mathfrak{n}^{-1} f\_{3; \mathfrak{s}}^{\*}(\mathfrak{x}),$$

$$\begin{aligned} \text{where} \quad &f\_{1; \boldsymbol{\uprho}}^{\*}(\mathbf{x}) = \int\_{0}^{\infty} \Phi(\mathbf{x}\sqrt{\boldsymbol{y}}) d\boldsymbol{e}^{-s/y} \, &\quad &f\_{2; \boldsymbol{\uprho}}^{\*}(\mathbf{x}) = \int\_{0}^{\infty} \frac{(\mathbf{x}^{3}\mathbf{y}^{3/2} - 5\mathbf{x}\sqrt{\boldsymbol{y}})\boldsymbol{q}(\mathbf{x}\sqrt{\boldsymbol{y}})}{4y} d\boldsymbol{e}^{-s/y} \, &\quad \\ \text{and} \quad &f\_{3; \boldsymbol{\uprho}}^{\*}(\mathbf{x}) = \int\_{0}^{\infty} \Phi(\mathbf{x}\sqrt{\boldsymbol{y}}) d\boldsymbol{h}\_{2; \boldsymbol{\uprho}}(\boldsymbol{y}) \quad &\quad \text{with} \quad &h\_{2; \boldsymbol{\uprho}}(\boldsymbol{y}) = \mathrm{s} \, \mathrm{e}^{-s/y} \left(\mathrm{s} - 1 + 2Q\_{1}(\boldsymbol{n}\boldsymbol{y})\right) / \left(2\boldsymbol{y}^{2}\right). \end{aligned}$$

To obtain (53), we calculate the above integrals as in the proof of Theorem 5 in Christoph et al. [12]. Here, we use Formula 2.3.16.3 in Prudnikov et al. [37], p. 344 with *p* = *x*2/2 > 0, *s* > 0, *m* = 1, 2:

$$\int\_0^\infty \frac{e^{-\mathbf{x}^2 y/2}}{\sqrt{2\pi} \, y^{m-3/2}} dH\_s(y) = \int\_0^\infty \frac{\mathbf{s} \, e^{-\mathbf{x}^2 y/2 - s/y}}{\sqrt{2\pi} \, y^{m+1/2}} dy = (-1)^m \frac{\mathbf{s}}{|\mathbf{x}|} \frac{\partial^m}{\partial \mathbf{s}^m} e^{-\sqrt{2\pi}|\mathbf{x}|}.\tag{A14}$$

In the mentioned proof we obtained with (A14) for *m* = 1

$$\int\_0^\infty \Phi(x\sqrt{y}) dH\_s(y) = L\_{1/\sqrt{s}}(x)$$

and with (A14) for *m* = 2

$$n^{-1} \sup\_{\mathbf{x}} \left| l\_{3;\mathbf{x}}^{\*} (\mathbf{x}) - \frac{(1 - s)\mathbf{x}(1 + \sqrt{2s}|\mathbf{x}|)}{8s} l\_{1/\sqrt{s}}(\mathbf{x}) \right| \le n^{-1} c(s) e^{-\sqrt{\pi s}n/2} \le \mathcal{C}(s) n^{-2}.$$

Again using (A14) with *p* = *x*2/2 > 0, *s* > 0, *m* = 1, 2 we find

$$\begin{split} l\_{2\mathbb{A}}(\mathbf{x}) &= \ \frac{\mathbf{s}}{\sqrt{2\pi}} \int\_0^\infty (a\mathbf{x}^3 y^{-1-1/2} - \mathbf{5x} y^{-2-1/2}) e^{-(\mathbf{x}^2 y/2 + \mathbf{s}/y)} d\mathbf{y} \\ &= \ \frac{2s a \mathbf{x}^3 - \mathbf{5x} (\sqrt{2\mathbf{s}} \, |\mathbf{x}| + 1)}{8s} l\_{1/\sqrt{s}}(\mathbf{x}). \end{split}$$

**Proof of Theorem 7.** By Lemma 4, the additional assumptions (23) and (24) in Transfer Theorem 2 are satisfied with the limit inverse exponential distribution *Hs*(*y*) and *<sup>h</sup>*2;*s*(*y*) given in (40), *gn* = *n* and *b* = 2. In Transfer Theorem 1, the right-hand side of (20) is estimated by Lemma 1 and Lemma 5 for *α* = 2 in the case *γ* = 0. Then, we have in (21) with (35)

$$G\_{\mathfrak{n}}(\mathbf{x}, 1/n) = \Phi(\mathbf{x}) \left( 1 - e^{-s\mathfrak{n}} - n^{-1} h\_{2\mathfrak{z}}(1/n) \right) + \frac{f\_2(\mathbf{x}; a)}{n} \int\_{1/n}^{\infty} \frac{1}{y} d e^{-s/y} \dots$$

The estimates (24i), (24ii) for *b* = 2 and ∞0 *y*<sup>−</sup>1*de*−*<sup>s</sup>*/*<sup>y</sup>* = *s* ∞0 *y*<sup>−</sup>3*e*<sup>−</sup>*<sup>s</sup>*/*ydy* = *s*2 ∞0 *ze*<sup>−</sup>*zdz* = *s*2 lead to

$$\left| G\_{\mathfrak{n}}(\mathfrak{x}, 1/g\_{\mathfrak{n}}) - \Phi(\mathfrak{x}) - n^{-1} s^2 f\_2(\mathfrak{x}; a) \right| \le \mathbb{C}\_{\mathfrak{s}} n^{-2}$$

and Theorem 7 is proved.

**Proof of Theorem 8.** By Lemma 4, the additional assumptions (23) and (24) in Transfer Theorem 2 are satisfied with the limit inverse exponential distribution *Hs*(*y*) and *<sup>h</sup>*2;*s*(*y*) given in (40), *gn* = *n* and *b* = 2. In Transfer Theorem 1, the right-hand side of (20) is estimated by Lemma 1 and Lemma 5 for *α* = 2 in the case *γ* = −1/2. Then, we have in (21) with (35)

$$G\_{2;n}(\mathbf{x};0) = f\_{1;\*}^\*(\mathbf{x}) + n^{-1} \, f\_{2;\*}^\*(\mathbf{x}) + n^{-1} \, f\_{3;\*}^\*(\mathbf{x}),$$

$$\begin{aligned} \text{where} \quad &f\_{1; \mathbf{x}}^{\*}(\mathbf{x}) = \int\_{0}^{\infty} \Phi(\mathbf{x} \mathbf{y}^{-1/2}) d\mathbf{e}^{-s/y}, \qquad &f\_{2; \mathbf{y}}^{\*}(\mathbf{x}) = \int\_{0}^{\infty} \frac{(\mathbf{a} \mathbf{x}^{3} \mathbf{y}^{-3/2} - 5 \mathbf{x} \mathbf{y}^{-1/2}) \varrho(\mathbf{x} \mathbf{y}^{-1/2})}{4y} d\mathbf{e}^{-s/y}, \\ \text{and} \quad &f\_{3; \mathbf{x}}^{\*}(\mathbf{x}) = \int\_{0}^{\infty} \Phi(\mathbf{x} \mathbf{y}^{-1/2}) d\mathbf{h}\_{2; \mathbf{x}}(\mathbf{y}) \qquad &\text{with} \quad h\_{2; \mathbf{x}}(\mathbf{y}) = \mathbf{s} \, \varepsilon^{-z/y} \left( \mathbf{s} - 1 + 2 \mathbf{Q}\_{1}(\mathbf{n} \mathbf{y}) \right) / \left( 2 \mathbf{y}^{2} \right). \end{aligned}$$

To obtain (54), we calculate the above integrals:

$$\frac{\partial}{\partial \mathbf{x}} \int\_0^\infty \Phi(\mathbf{x}\sqrt{y}) d e^{-s/y} = \quad \frac{\mathbf{s}}{\sqrt{2\pi}\pi} \int\_0^\infty y^{-3/2} e^{-(\mathbf{x}^2/2 + s)/y} dy = \frac{\mathbf{s}}{\sqrt{2\pi}\pi} \int\_0^\infty z^{1/2 - 1} e^{-(\mathbf{x}^2/2 + s)z} dz$$

$$= \frac{1}{2\sqrt{2\pi}} \left( 1 + \frac{\mathbf{x}^2}{2s} \right)^{-3/2} = \quad s\_2^\*(\mathbf{x}; \sqrt{s}), \qquad \text{and} \qquad \int\_0^\infty \Phi(\mathbf{x}\sqrt{y}) d e^{-s/y} = S\_2^\*(\mathbf{x}; \sqrt{s}). \tag{A15}$$

Define *K* = (*s* + *<sup>x</sup>*2/2). With *z* = *<sup>K</sup>*/*y* and <sup>Γ</sup>(*α*) = ∞0 *<sup>z</sup>α*−1*e*<sup>−</sup>*zdz*, *α* > 0, we obtain

$$\begin{split} \mathcal{J}^{\*}\_{2;s}(\mathbf{x}) &= \ \frac{s}{4\sqrt{2\pi}} \int\_{0}^{\infty} (a\mathbf{x}^{3}y^{-9/2} - 5\mathbf{x}y^{-7/2})e^{-K/y} dy = \frac{s}{4\sqrt{2\pi}} \int\_{0}^{\infty} (a\mathbf{x}^{3}z^{5/2} - 5\mathbf{x}z^{3/2})e^{-z} dz \\ &= \ \frac{s\,\mathrm{K}^{-7/2}}{4\sqrt{2\pi}} \left(a\mathbf{x}^{3}\Gamma(7/2) - 5\mathbf{x}\,K\Gamma(5/2)\right) = \frac{1}{4\left(\mathbf{x}^{2} + 2\mathbf{s}\right)^{2}} \left(15(a-1)\mathbf{x}^{3} - 30\mathbf{x}\right) \,s\_{2}^{\*}(\mathbf{x};\sqrt{s}). \end{split}$$

Integration by parts in *J*<sup>∗</sup>3;*s*(*x*) leads to

$$I\_{3\varphi}^\*(\mathbf{x}) = \frac{\mathbf{x}}{2\sqrt{2\pi}} \int\_0^\infty y^{-3/2} e^{-\mathbf{x}^2/(2y)} \, \_3y^{-2} e^{-s/y} \left( (s-1)/2 + Q\_1(\mathbf{n} \, y) \right) dy = I\_{4\varphi}^\*(\mathbf{x}) + I\_{5\varphi}^\*(\mathbf{x}),$$

where

$$f\_{4;3}^{\*}(\mathbf{x}) = \frac{s\left(\mathbf{s} - 1\right)\mathbf{x}}{4\sqrt{2\pi}} \int\_{0}^{\infty} y^{-7/2} e^{-\mathbf{K}/y} dy = \frac{s\left(\mathbf{s} - 1\right)\mathbf{x}\,\Gamma(5/2)}{4\sqrt{2\pi}\,\mathbf{K}^{5/2}} = \frac{\mathbf{3}\left(\mathbf{s} - 1\right)\mathbf{x}}{4\left(\mathbf{x}^{2} + 2\mathbf{s}\right)} s\_{2}^{\*}(\mathbf{x}; \sqrt{s})$$

and using the Fourier series expansion (A11) of the periodic function *Q*1(*y*) and interchange sum and integral, we find

$$\begin{split} J\_{5\mu}^{\*}(\mathbf{x}) &= \, \frac{\mathbf{s}\cdot\mathbf{x}}{2\sqrt{2\pi}} \int\_{0}^{\infty} y^{-7/2} e^{-K/y} Q\_{1}(\mathbf{n}\,y) dy = \frac{\mathbf{s}\cdot\mathbf{x}}{2\sqrt{2\pi}} \sum\_{k=1}^{\infty} \frac{1}{k} \int\_{0}^{\infty} y^{-7/2} e^{-K/y} \sin(2\pi k \, n\, y) dy \\ &= \, \frac{\mathbf{s}\cdot\mathbf{x}}{2\sqrt{2\pi}} \sum\_{k=1}^{\infty} \frac{1}{k} \int\_{0}^{\infty} y^{-7/2} e^{-K/y} \sin(2\pi k \, n\, y) dy. \end{split}$$

Integration by parts in the latter integral and |*x*|/√*<sup>K</sup>* ≤ √2 leads now to

$$\sup\_{x} |f\_{5,3}^\*(x)| \le \sup\_{x} \frac{s \, |x|}{(2\pi)^{3/2} n} \sum\_{k=1}^{\infty} \frac{1}{k^2} \int\_0^{\infty} \left(\frac{7}{2} y^{-9/2} + Ky^{11/2}\right) e^{-K/y} dy \le c\_s n^{-1/2}$$
 
$$\text{with } c\_s = \frac{s\sqrt{2}}{(2\pi)^{3/2} n} \left(\frac{7\Gamma(11/2)}{2\, s^3} + \frac{\Gamma(13/2)}{s^4}\right) \frac{\pi^2}{6} \text{ and Theorem 8 is proved.} \quad \Box$$
