= Solution
Put $\lambda_i=z|V_i|$. Then
$$
N(V_i)\sim\operatorname{Pois}(\lambda_i),
\qquad
\mathbb E N(V_i)=\lambda_i,
$$
and $\lambda_i\to\infty$. The <Poisson central limit theorem> gives
$$
\frac{N(V_i)-\lambda_i}{\sqrt{\lambda_i}}
\xrightarrow{\mathrm d}N(0,1).
$$
Since $\sqrt{\lambda_i}=\sqrt z\,\sqrt{|V_i|}$,
$$
\frac{N(V_i)-\mathbb E N(V_i)}{\sqrt{|V_i|}}
=\sqrt z\,
\frac{N(V_i)-\lambda_i}{\sqrt{\lambda_i}}
\xrightarrow{\mathrm d}\boxed{N(0,z)}.
$$
Equivalently, the <characteristic function> of the left-hand side is
$$
\exp\left\{
\lambda_i\left(
e^{it/\sqrt{|V_i|}}-1-\frac{it}{\sqrt{|V_i|}}
\right)\right\}
\longrightarrow e^{-zt^2/2},
$$
which is the characteristic function of that centered <normal distribution>.
Back to article page
