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Following set up:

We have $i.i.d.$ integer valued random variables $\tau_1,\dots,\tau_n$ with \begin{align*} \mathbb{P}(\tau_1=k)=2k^{k-2}\frac{e^{-k}}{k!} \end{align*} for $k\in\mathbb{N}$. It holds that \begin{align*} \mathbb{E}[\tau_1]=2,\quad \mathrm{Var}(\tau_1)=\infty. \end{align*} The aim is to investigate the asymptotic behaviour of \begin{align*} \mathbb{P}\left(S_{n/2}=n\right), \end{align*} where $S_{n/2}:=\sum\nolimits_{i\leq n/2}\tau_i$, as $n$ tends to infinity. Obviously $\mathbb{E}[S_{n/2}]=n$, but as the variance is not finite, there is no known version of the local central limit theorem, which is applicable in this case.

Question:

Do you know any version of the local central limit theorem, or any other method (I tried combinatorial counting of the several possibilities, but can't derive useful estimates) to determine the asymptotic behaviour of this probability?

Thank you very much in advance.

1 Answers 1

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Using Stirling approximation $$k ! \approx k^{k}e^{-k}/\sqrt{2\pi k}$$ I obtain $$\mathbb{P}(\tau_1=k) \approx \sqrt{2 \pi^{-1}} k^{-5/2}.$$ The rest is straightforward.

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    Using this approximation I obtain \begin{align*} \mathbb{P}(S_{n/2}=n)=\sum_{l_1+\dots+l_{n/2}=n}\prod_{i\leq n/2}\mathbb{P}(\tau_i=l_i)\sim\sum_{l_1+\dots+l_{n/2}=n}\prod_{i\leq n/2} \sqrt{2\pi^{-1}}l_i^{-5/2}. \end{align*} There are $\binom{n}{n/2}$ possibilities for $n/2$ integers to sum up to $n$. But how can I handle the product? I know that \begin{align*} n/2+1\leq\prod_{i\leq n/2}l_i\leq 2^{n/2} \end{align*} for any $l_1,\dots,l_{n/2}$ such that the sum equals $n$, but this estimate is way too rough. – 2017-01-15