0
$\begingroup$

I am looking for an efficient lower bound of the expression

$$ E(p)=\sum_{i=0}^{\left \lceil \frac{l}{2}\right \rceil-1}\sum_{j=0}^{i}\binom{l}{i}\binom{i}{j}\left ( -1 \right )^{j}p^{l-i+j}, $$ where $\frac{1}{2}<p<1$.

EDIT: As I suggested in comments, the expression should converge to $1$. I am very much interested in a lower-bound approximation, in terms of $p$ (you may assume that $l$ is large, and of any parity).

Ultimately, I would be interested in a rate of convergence of the sequence $p=p_0,p_1,\dots $ , where $ p_{i+1}=E(p_i)$.

$\endgroup$
4
  • $\begingroup$ If it helps, you may assume any parity of $l$. $\endgroup$ Oct 20, 2016 at 12:44
  • $\begingroup$ It's easier to understand if you write your sum shorter $\sum\limits_{i=0}^{\left \lceil \frac{l}{2}\right \rceil-1}p^{l-i}(1-p)^i$ . $\endgroup$
    – user90369
    Oct 20, 2016 at 13:28
  • 1
    $\begingroup$ At first glance, the whole summation would probably lead to some $\, _2F_1\left(a,b;c;\frac{p-1}{p}\right)$ hypergeometric function. $\endgroup$ Oct 20, 2016 at 13:35
  • $\begingroup$ @ClaudeLeibovici: I´m not familiar with these, but thanks! If you define a sequence recursively as $p_{i+1}=E(p_i)$, it should converge to 1. I´m actually just interested in the speed of convergence and I believe your comment may be helpful. Could you give some hint how could I prove an equivalence? $\endgroup$ Oct 20, 2016 at 14:04

1 Answer 1

1
$\begingroup$

I think there is no general answer. The speed of convergence depends on $l$ and $p$.

Be $\enspace \frac{1}{2}<p<E(p)<1$ .

Speed of convergence:

$E(1-p)=1-E(p)\leq c(1-p)^m$ with $0<c<1$ and $m\geq 1$

For odd $l$ we get $E(p)|_{p\to \frac{1}{2}+0}\to \frac{1}{2}+0$ which means $c\to 1^-$ and $m\to 1^+$ or in words: Very weak convergence for $p$ near by $\frac{1}{2}$ . Much better is it with e.g. $p>\frac{2}{3}$ .

$E(1-p)=\sum\limits_{i=\lceil l/2\rceil}^l \binom{l}{i}p^{l-i}(1-p)^i \leq c(1-p)^m$

It remains the question what we can choose for $m$ so that

$\sum\limits_{i=\lceil l/2\rceil}^l \binom{l}{i}p^{l-i}(1-p)^{i-m} \leq c<1$

If $p$ is near by $1$ then we can choose $m$ with $m<\lceil l/2\rceil$ because the left side is very small and we can find $c\in(0;1)$ .

The condition is e.g. $\sum\limits_{i=\lceil l/2\rceil}^l \binom{l}{i}p^{l-i}(1-p)^{i-\lceil l/2\rceil} \leq \frac{c}{1-p}<\frac{1}{1-p}$ . In this case we have convergence of order $\lceil l/2\rceil-1$.

$\endgroup$
2
  • $\begingroup$ Thank you for a response. I am not sure I understand your definition of the speed of convergence. In my opinion, the value of $l$ should not affect much the rate of convergence, as defined here, especially when $l$ is large (please note the updated question). In addition, could you clarify the step where you introduce $c$ and $m$? $\endgroup$ Oct 27, 2016 at 13:20
  • $\begingroup$ I am sorry to answer too late. To define a speed of convergence is a bit complicate because there are different ideas about that. But I have used the case in de.wikipedia.org/wiki/Konvergenzgeschwindigkeit (German Wikipedia) where is written "Konvergenz der Ordnung p" (I have used $m$ instead of $p$ here). The translation is *convergence of order p". The higher the order, the higher the speed. $\endgroup$
    – user90369
    Nov 11, 2016 at 15:50

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .