$X$ is a Poisson random variable and the probability mass function is given by: $$\Pr(X = k) = e^{-\lambda}\frac{{\lambda}^k}{k!}$$

I’ve got a probability function $f(\lambda)$ $$f(\lambda) = \sum\limits_{k=4}^{\infty}\Pr(X=k)[{\Pr(X\le k)}^6 - {\Pr(X\le k-4)}^6]$$

To date, I only find that ${\Pr(X\le k)}^6 - {\Pr(X\le k-4)}^6$ can be factorized as \begin{align*} &{\Pr(X\le k)}^6 - {\Pr(X\le k-4)}^6 \\&= [\Pr(X=k)+ \Pr(X=k-1) + \Pr(X=k-2) + \Pr(X=k-3)]\cdot[ {\Pr(X\le k)}^5+{\Pr(X\le k)}^4{\Pr(X\le k-4)}+…+ {\Pr(X\le k-4)}^5] \end{align*} But I have no idea what to do next… Can I assume that $\Pr(X=k) \approx \Pr(X=k-1)$ if $\lambda \to +\infty$? Are there any better approximation for $f(\lambda)$?

If there is a simple expression for $ f(\lambda) $ about $\lambda \to +\infty$ that would be best, but I’m open to whatever can be suggested. Thank you in advance!

  • $\begingroup$ You will probably end up with $f(\lambda)\to 0$ pretty quickly. How tight a bound would you need? $\endgroup$ – Tim Jun 20 '13 at 13:14
  • $\begingroup$ @Tim Thanks for your comment. The bound would not need to be very tight. Actually, I'm looking for a polynomial of $\lambda$ (i.e. ${\lambda }^{-a}$,$a>0$)to approximate the decaying speed of $f{(\lambda)}$ as $\lambda$ goes to $+\infty$. $\endgroup$ – Erdos Yi Jun 20 '13 at 15:16
  • $\begingroup$ There's an easy(ish) bound $f(\lambda) \leq\frac{28}{\sqrt{2\pi\lambda}}$ which I'd assume is not quite good enough. I'll see if I can find a way to improve it. $\endgroup$ – Tim Jun 20 '13 at 15:35
  • $\begingroup$ @Tim You're right, I think the decaying speed of $f(\lambda)$ is much faster than ${\lambda}^{-\frac{1}{2}}$ as well. I tried to bound $f(\lambda)$ as ${\lambda}^{-a}\le f(\lambda)\le {\lambda}^{-b}$, where $a>0$ and $b>0$, but I could not find appropriate ${a}$ and ${b}$ to approximate the decaying speed of $f(\lambda)$ accurately. $\endgroup$ – Erdos Yi Jun 21 '13 at 0:54

By definition, $\sum\limits_kP(X=k)u(k)=E(u(X))$ for every bounded function $u$, hence, considering a Poisson $\lambda$ random variable $Y$ independent of $X$, one gets $$ f(\lambda)=P(X\leqslant Y;Y\geqslant4)^6-P(X\leqslant Y-4)^6=(u(\lambda)-v(\lambda))^6-(u(\lambda)-w(\lambda))^6, $$ with $$ u(\lambda)=P(X\leqslant Y),\quad v(\lambda)=P(X\leqslant Y\leqslant3),\quad w(\lambda)=P(Y-3\leqslant X\leqslant Y). $$ Since $Z_\lambda=(Y-X)/\sqrt{2\lambda}$ is approximately standard normal and $$ u(\lambda)=P(Z_\lambda\geqslant0),\qquad w(\lambda)=P(0\leqslant Z_\lambda\leqslant3/\sqrt{2\lambda}), $$ one sees that $u(\lambda)\to\frac12$ and $w(\lambda)\to0$. Likewise, $v(\lambda)\leqslant P(Y\leqslant3)\to0$ hence $$ f(\lambda)=6\cdot\left(\tfrac12\right)^5(w(\lambda)-v(\lambda))+o(w(\lambda)+v(\lambda)). $$ Now, $v(\lambda)\ll1/\sqrt\lambda$ and, if $(Z_\lambda)$ satisfies a local central limit theorem, the integer interval $[0,3]$ having length $4$, one might have $w(\lambda)\sim4/\sqrt{2\lambda}\cdot1/\sqrt{2\pi}$, hence $$ f(\lambda)\sim6\cdot\left(\tfrac12\right)^5\cdot\frac2{\sqrt{\pi\lambda}}=\frac3{8\sqrt{\pi\lambda}}. $$ This result is conditional on the (plausible) assertion that, for every fixed integer $k$, when $\lambda\to\infty$, $$ P(Y-X=k)\sim\frac1{\sqrt{2\lambda}}\cdot\frac1{\sqrt{2\pi}}. $$ Fortunately, asymptotics of Bessel functions of the first kind indicate that, for every $k\geqslant0$, $$ P(Y-X=k)=\sum_{n\geqslant0}\mathrm e^{-2\lambda}\frac{\lambda^n}{n!}\frac{\lambda^{n+k}}{(n+k)!}=\mathrm e^{-2\lambda}\mathrm i^{-k}J_k(2\mathrm i\lambda)\sim\frac1{2\sqrt{\pi\lambda}}. $$ Thus, the heuristics above holds and the result is proved.

  • $\begingroup$ You are welcome. Nice question, by the way (where does it come from?). $\endgroup$ – Did Jun 23 '13 at 15:45
  • $\begingroup$ It comes from one of my former quesiton.Link $\endgroup$ – Erdos Yi Jun 23 '13 at 15:49
  • $\begingroup$ I see. It is good that you abstracted the present question from there... $\endgroup$ – Did Jun 23 '13 at 17:11

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