$\log(K/\beta+1)$, $K\sim\operatorname{Poisson}(\lambda)$, is asymptotically normal as $\lambda\to\infty$?

Question

Previously posted on the MSE here.

For $K\sim\operatorname{Poisson}(\lambda)$ and $\beta>0$ let $$ Y=\log(K/\beta+1). $$

Using first-order Taylor expansion on $Y$ we obtain $$ \tilde Y=\log(\lambda/\beta+1)+\frac{K-\lambda}{\lambda+\beta} $$ so that $$ \mathsf E\tilde Y\approx\log(\lambda/\beta+1) $$ and $$ \mathsf{Var}\tilde Y\approx\frac{\lambda}{(\lambda+\beta)^2}. $$ Defining $$ Z=\frac{Y-\mathsf E\tilde Y}{\sqrt{\mathsf{Var}\tilde Y}}=\frac{\lambda+\beta}{\sqrt{\lambda}}\log\left(\frac{K/\beta+1}{\lambda/\beta+1}\right), $$ is it true that $$ Z\overset{d}{\to}\mathcal N(0,1) $$ as $\lambda\to\infty$?

Answer 1

$\newcommand\la\lambda\newcommand\be\beta$Yes, this is true. Indeed, on the event $A_\la:=\{|K-\la|\le\la^{5/8}\}$, $$Z=Z_\la:=\frac{\la+\be}{\sqrt\la}\, \ln\Big(1+\frac{K-\la}{\la+\be}\Big) \\ =\frac{\la+\be}{\sqrt\la}\, \Big(\frac{K-\la}{\la+\be}+O\Big(\frac{(K-\la)^2}{\la^2}\Big)\Big)\\ =\frac{K-\la}{\sqrt\la}+O\Big(\frac{(K-\la)^2}{\la^{3/2}}\Big) \\ =\frac{K-\la}{\sqrt\la}+O\Big(\frac1{\la^{1/4}}\Big) \\ =\frac{K-\la}{\sqrt\la}+o(1),$$ and $P(A_\la^c)\le E(K-\la)^2/\la^{5/4}=\la^{-1/4}\to0$, where $A_\la^c$ is the complement of $A_\la$, so that $Z_\la\,1_{A_\la^c}\to0$ in probability and hence in distribution (as $\la\to\infty$). Also, $\frac{K-\la}{\sqrt\la}\to V$ in distribution, where $V$ is a standard normal random variable. So, $$Z_\la =\Big(\frac{K-\la}{\sqrt\la}+o(1)\Big)\,1_{A_\la} +Z_\la\,1_{A_\la^c}.$$ Thus, $$Z_\la\to (V+0)(1-0)+0=V$$ in distribution. $\quad\Box$