I would like to have an estimate for the series $$P(t) = \sum\limits_{k = 0}^\infty (e^{-t}\frac{t^k}{k!})^m,$$ where $e$ is the base of natural logarithm, $k!$ is the factorial of the integer $k$, $t$ represents the time and $t>0$, $m$ is a positive integer and $m>1$ (Obviously, $P(t)=1$ when $m=1$ since $P(t)$ is exactly the cdf of Poisson distribution with associated parameter $t$ in this case.).

I am interested in showing that the answer could look something like $$P(t) = O(t^{-\alpha m}),\alpha > 0.$$

It has come to my attention that in the paper “Rumors in a Network: Who's the Culprit?” D Shah and T Zaman proved that (Page 24-27) $$P(t) = O(\frac{1}{\sqrt{t}}),\text{if }m=2.$$

However, I find it difficult for me to extend their method to the cases when $m>2$.
I also tried to calculate $P(t)$ in Mathematica and it gave me the result like $$P(t) = \sum\limits_{k = 0}^\infty (e^{-t}\frac{t^k}{k!})^m = e^{-mt}\text{HypergeometricPFQ}[...,t^m]$$ It seems that the decreasing speed of $P(t)$ is much slower than that of $e^{-mt}$ as $t$ increases. But I could not find the closed-form bound when $m>2$ just like $O(1/\sqrt t)$ when $m=2$.

Does anyone know of a scale or a bound of $P(t)$ in the literature? Any comments and answers would be highly appreciated. Many thanks!

  • $\begingroup$ Multiply the $\ell^1$ norm ($1$) by the $\ell^\infty$-norm (about $t^{-1/2}$) to the power $m-1$. If you want a better estimate just look up the "Laplace asymptotic formula" on google. $\endgroup$ – fedja May 28 '13 at 14:19
  • $\begingroup$ @fedja Thanks for your comment, but I cannot follow the part "Multiply the...". I know what ${\ell}^1$ and ${\ell}^{\infty}$ mean but I do not know how this muplication comes from. Could you explain it for me? $\endgroup$ – Erdos Yi May 28 '13 at 15:28
  • $\begingroup$ Sure: $\sum a_j^m\le (\max a_j)^{m-1}\sum a_j$ (provided $a_j\ge 0$, etc.) $\endgroup$ – fedja May 29 '13 at 2:20

Actually this can be solved by standard asymtotic analysis methods (with help from Mathematica):

Expand the factorial in the sum with Sterling. Then do as if $k$ is continuous and find $k=k_{0}$ where the term in the sum is maximal (the result contains Lambert's $W$ function). Next expand quadratically around $k$ with expansion parameter, say, $\delta$. The quadratic expansion is then re-interpreted as quadratic expansion of a Gaussian function, which one can integrate over $-\infty<\delta<\infty$. The integral converges and one can expand its value (with big help from Mathematica) for large $t$. The result is $$ P(t)\sim (2\pi t)^{-(m-1)/2}m^{-1/2}, $$ and, indeed, direct tests with Mathematica confirm the result.


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