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Let $B(n,p,r)$ denote the binomial distribution function (DF) with parameters $n \in \mathbb N$ and $p \in (0,1)$ evaluated at $r \in \{0,1,\ldots,n\}$: \begin{equation} B(n,p,r) = \sum_{i=0}^r \binom{n}{i} p^i (1-p)^{n-i}, \end{equation} and let $F(\nu,r)$ denote the Poisson DF with parameter $a \in \mathbb R^+$ evaluated at $r \in \{0,1,2,\ldots\}$: \begin{equation} F(a,r) = e^{-a} \sum_{i=0}^r \frac{a^i}{i!}. \end{equation}

Consider $p \rightarrow 0$, and let $n$ be defined as $\lceil a/p-d \rceil$, where $d$ is a constant of the order of $1$. Since $np \rightarrow a$, the function $B(n,p,r)$ converges to $F(a,r)$ for all $r$, as is well known.

With the above definition for $n$, I'm interested in determining the values of $a$ for which \begin{equation} B(n,p,r) > F(a,r) \quad \forall p \in (0,1), \end{equation} and similarly those for which \begin{equation} B(n,p,r) < F(a,r) \quad \forall p \in (0,1). \end{equation} I have been able to prove that the first inequality holds for $a$ sufficiently smaller than $r$; more specifically, for $a$ lower than a certain bound $g(r)$, with $g(r)<r$. Similarly, the second inequality holds for $a$ sufficiently larger than $r$, i.e. for $a$ greater than a certain bound $h(r)$, with $h(r)>r$. (The expressions of the bounds $g(r)$ and $h(r)$ are irrelevant here. I will provide the details to anyone interested.) However, numerical results suggest that those inequalities hold for less stringent bounds, that is, for $a$ closer to $r$ than I can prove.

So, I'd like to know if there is some theorem or result which establishes under which conditions each inequality holds (for all $p$); that is, when the binomial DF is guaranteed to be above/below its limiting Poisson DF. If such theorem doesn't exist, any idea or pointer in the right direction would be appreciated.

Please note that a similar question, phrased in terms of incomplete beta and gamma functions, was posted in math.stackexchange.com but got no answer.

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Luis Mendo
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    This is an interesting question, though I think it would help to clarify a few things, particularly which are the "moving parts" and which are not. It seems you want a bound that holds *uniformly* in $p$ for each *fixed* $r$. But, what is the role of $d$ here? It shouldn't matter much, but is it's introduction necessary? One approach might be to look at things in terms of waiting times of a Poisson process and couple them to associated geometric waiting times (via taking the ceiling of each) for your binomial random variable. But that might not yield the *uniform* bound you are seeking. – cardinal Sep 19 '13 at 14:39
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    @cardinal Thanks for taking the time. Yes, I want the bound to be uniform in p. All other parameters are fixed (but selectable). $d$ is just one such free parameter. For example, one hypothetic result could be as follows: "For any natural $r$ greater than $2$ and any $d \in (-1,1)$, the first inequality holds for all $a < r - \sqrt{r}$ and for all $p \in (0,1)$; and the second holds for all $a > r + \sqrt{r}$ and for all $p \in (0,1)$. – Luis Mendo Sep 19 '13 at 14:48
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    There is a stein chen theory which estimate errors when you use poisson rv to estimate sum of not necessary independent bernoulli variables. Not sure about your question thoufh. – Lost1 Jan 06 '14 at 05:05
  • For finite $n$, the Binomial distribution has closed support from above. Its size may be selectable (by choosing $n$) but it is closed. On the other hand, the Poisson distribution has unbounded support. Since we are looking at the CDF's, for any finite $n$ we will always have $$B(n,p,r=n) = 1 > F(a,n) $$ for any permissible values of $p,a$. So the conditions for the 2nd inequality the OP is after, will always include, at least, "for $r – Alecos Papadopoulos Sep 25 '14 at 18:44
  • See Did's answer here: http://math.stackexchange.com/questions/37018/poisson-cdf-as-lower-bound-to-binomial-cdf – Alex R. Aug 13 '15 at 03:24
  • @AlexR. Thanks. But that's a different (and easier) question. Here $n = \lceil a/p−d \rceil$; there $n = a/p$ – Luis Mendo Aug 13 '15 at 06:44

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With regard to the following:

  • the mean of a Binomial dist is $np$

  • the variance is $np(1-p)$

  • the mean of a Poisson dist is $\lambda$, which we can imagine as $n\times p$

  • the variance of a Poisson is the same as the mean

Now, if a Poisson is the limit to a Binomial with parameters $n$ and $p$, such that $n$ increases to infinity and $p$ decreases to zero while their product remains constant, then assuming that $n$ and $p$ are not converged to their respective limits, the expression $np$ is always greater than $np(1-p)$, therefore the variance of Binomial is less than that of Poisson. That would imply that the Binomial is below in the tails and above elsewhere.

Glen_b
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Germaniawerks
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  • Thank you for your contribution. It seems to me it fails to address the question, though, because (1) the O.P. is interested in the CDF, not the PDF. (2) He asks for a quantitative answer. – whuber Feb 07 '14 at 15:49