When is binomial distribution function above/below its limiting Poisson distribution function?

Question

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.

Answer 1

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.