the above identity yields (\[Gmotion\]). Finally, the rest assertions follow from $\phi_{\mathbb{B}_{p,1/p^*}}=E_{\mu^{\uplus(p-1)}}=(p-1)\sigma^2G_\nu$ and (\[Gmotion\]).

The identity (\[motion\]) indicates that $\mu^{\boxplus p}$ has mean zero and finite variance $p\sigma^2$ if $p\geq1$. The next result shows that this is also true for the measure $\mu^{\boxplus
p}$ whenever it is defined.

\[4.1\] Suppose that $\mu\in\mathcal{M}$ has mean zero and variance $\sigma^2$. If the measure $\mu^{\boxplus p}$ is defined for some $p>0$ then it has mean zero and variance $p$, in which case $$\begin{aligned}
&\Phi\left(\left(\mu^{\boxplus
p}\right)^{\uplus1/(p\sigma^2)}\right)=\Phi\left(\mu^{\uplus1/\sigma^2}\right)
\boxplus\gamma_{(p-1)\sigma^2},\;\;\;\;\;p\geq1, \\
&\Phi\left(\left(\mu^{\boxplus
p}\right)^{\uplus1/(p\sigma^2)}\right)\boxplus\gamma_{(1-p)\sigma^2}=\Phi
\left(\mu^{\uplus1/\sigma^2}\right),\;\;\;\;\;p<1.\end{aligned}$$

By (\[motion\]), it suffices to show the lemma for the case $1-\mathrm{Ind}(\mu)\leq p<1$. Let $\nu=\Phi(\mu^{\uplus1/\sigma^2})$, $\mu_p=\mu^{\boxplus p}$, and $H(z)=z/p+(1-1/p)F_{\mu_p}(z)$. Then it follows from (\[p&lt;1\]) that $F_{\mu_p}(iy)=F_\mu(H(iy))$ or, equivalently, $E_{\mu_p}(iy)=iy-H(iy)+\sigma^2G_\nu(H(iy))$ for sufficiently large $y>0$. Since $z-H(z)=(1-1/p)E_\mu(z)$, we see that $E_{\mu_p}(iy)=p\sigma^2G_\nu(H(iy))$ for sufficiently large $y>0$. Next, we claim that $E_{\mu_p}/(p\sigma^2)\in\mathcal{G}$. Indeed, since $\lim_{y\to\infty}H(iy)/(iy)=1$, for any $\alpha>0$ there exists a number $\beta>0$ such that $$\left|\frac{H(iy)}{iy}-1\right|<c_\alpha:=\frac{\alpha}{\sqrt{1+\alpha^2}},\;\;\;\;\;y>\beta,$$ from which we deduce that $H(iy)\in\Gamma_{\alpha,\beta'}$ for $y>\beta$, where $\beta'=(1-c_\alpha)\beta$. By \[\[HV2\], Proposition 5.1\], we obtain $$\lim_{y\to\infty}iyE_{\mu_p}(iy)=p\sigma^2\left(\lim_{y\to\infty}\frac{iy}{H(iy)}\right)
\left(\lim_{y\to\infty}H(iy)G_\nu(H(iy))\right)=p\sigma^2,$$ which yields that $E_{\mu_p}/(p\sigma^2)\in\mathcal{G}$, as desired. Finally, let $\nu_p=\Phi((\mu^{\boxplus p})^{\uplus1/(p\sigma^2)})$. Then $E_{\mu^{\boxplus p}}=p\sigma^2G_{\nu_p}$ and (\[motion\]) show that $$p\sigma^2G_{\Phi(\mu^{\uplus1/\sigma^2})}=E_{\mu^{\uplus
p}}=E_{\mathbb{B}_{1/p-1}(\mu^{\boxplus
p})}=p\sigma^2G_{\nu_p\boxplus\gamma_{(1-p)\sigma^2}},$$ which gives the last assertion. If $p>1$ then $E_\mu=\sigma^2G_{\Phi(\mu^{\uplus1/\sigma^2})}$ and (\[motion\]) yield $$E_{\mathbb{B}_{p-1}(\mu)}=\sigma^2G_{\Phi(\mu^{\uplus1/\sigma^2})\boxplus\gamma_{(p-1)\sigma^2}},$$ 