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Suppose $X\sim \mathsf{Poi}(c\cdot n)$. Consider the expectation \begin{align} F(n,c) = \mathbb{E}\left[ \prod_{r = 0}^{X-1} \left ( \frac{n+r}{2n+r} \right ) \right]. \end{align}

Clearly, $e^{-cn/2}=\mathbb{E}[1/2^X] \leq F(n,c)\leq 1$.

My question is, how can we compute the quantities \begin{align} \liminf_{n\to \infty} \frac{1}{n} \ln F(c,n), \limsup_{n\to \infty} \frac{1}{n} \ln F(c,n) \end{align}

We already know that the quantity $\frac{1}{n}\ln F(c,n)$ is bounded between $-c/2$ and $0$. Can we get a precise limit result?

user401582
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1 Answers1

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With a bit of computation, one can check that

$$ F(c,n) = \frac{\Gamma(2n)}{\Gamma(n)^2} \int_{0}^{1}\left[x(1-x)e^{-cx}\right]^{n-1}e^{-cx}\,dx. $$

(I will add a proof later; I have to go to bed now.) Now

  • By various means, one can check that $\frac{1}{n}\log\left(\Gamma(2n)/\Gamma(n)^2\right)$ converges to $\log 4$ as $n\to\infty$. Perhaps the easiest argument is to show that $ \frac{2^{2n}}{n+1} \leq \binom{2n}{n} \leq 2^{2n} $, which easily follows by noticing that $\binom{2n}{n}$ is the largest among the binomial coefficients $\binom{2n}{k}$ for $0 \leq k \leq 2n$. Of course, Stirling's approximation works perfectly here.

  • We know that $\| f \|_{L^p(\mu)} \to \| f\|_{L^{\infty}(\mu)}$ as $p\to\infty$ for a measurable function $f$ on a finite measure space. (See this answer, for instance.) Applying this to the function $f(x) = x(1-x)e^{-cx}$ and the measure $\mu(dx) = e^{-cx}\mathbf{1}_{[0,1]}(x)\,dx$, we get

    $$ \frac{1}{p}\log\left( \int_{0}^{1}\left[x(1-x)e^{-cx}\right]^p e^{-cx}\,dx \right) \xrightarrow[p\to\infty]{} \max_{0\leq x\leq 1} x(1-x)e^{-cx}.$$

Combining altogether, we obtain

$$ \frac{1}{n}\log F(c, n) = \log \left( 4 \max_{0\leq x\leq 1} x(1-x)e^{-cx} \right), $$

which can be explicitly computed as

$$ = \frac{\sqrt{c^2+4}-c-2}{2}-\log\left(\frac{\sqrt{c^2+4}+2}{4}\right). $$

Sangchul Lee
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