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Please note: this question was posted first (September 4) in math.stackeschange.com and then (September 16) in stats.stackeschange.com. It got no answers in neither of those sites.


Let the regularized incomplete beta and gamma functions be defined as usual: \begin{equation} I_p(z,w) = \frac a {B(z,w)} \int_0^p t^{z-1} (1-t)^{w-1} \,\mathrm dt, \end{equation} \begin{equation} \gamma(r,t) = \frac 1 {\Gamma(r)} \int_0^t s^{r-1} e^{-s} \,\mathrm ds. \end{equation}

Let $r \in \mathbb N$, and consider $a, b > 0$. As is well known, $I_p(r,a/p-b) \rightarrow \gamma(r,a)$ as $p \rightarrow 0$.

I have been able to prove that the difference between the incomplete beta function and its limit, \begin{equation} I_p(r,a/p-b) - \gamma(r,a), \end{equation} is positive for all $p \in (0,1)$ if $a$ is sufficiently larger than $r$, and negative if $a$ is sufficiently smaller than $r$. For example, if you have access to Matlab you can check that (note that Matlab's gammainc uses reverse order for its arguments)

p = 1e-3;  r = 5; a = 10; b = 5; betainc(p,r,a/p-b) - gammainc(a,r)

is positive, whereas

p = 1e-3;  r = 5; a = 2; b = 5; betainc(p,r,a/p-b) - gammainc(a,r)

is negative.

I'd like to know the widest range of values of $a$ for which that difference is assured to be positive/negative.

So, my question is: Is there a known theorem that gives sufficient conditions on $a, b, r$ that assure that $I_p(r,a/p-b) - \gamma(r,a)$ is positive/negative? Even if you don't know any such theorem, any pointer in the right direction would be appreciated.


Alternate formulation

Since the incomplete gamma is related to the distribution function of a Poisson random variable, and the incomplete beta is related to that of a binomial random variable, the question can also be posed from a statistical point of view, as follows.

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 $n$, it is interesting to know 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 (in either of the two formulations) holds for $a$ sufficiently larger than $r$; more specifically, for $a$ greater than a certain bound $g(r)$, with $g(r)>r$. Similarly, the second inequality holds for $a$ sufficiently smaller than $r$, i.e. for $a$ lower 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.

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  • $\begingroup$ You may be interested in contributing to a proposal Spanish language version of math stackexchange; it could use some input from fluent professors: area51.stackexchange.com/proposals/64529/… $\endgroup$ – Brian Rushton Feb 1 '14 at 3:12
  • $\begingroup$ @BrianRushton Thanks for letting me know about this. On one hand I like the idea. On the other hand, I think it's better to stick to a single language (English) so that everything is "centralized". Anyway, I've just joined it; we'll see :-) $\endgroup$ – Luis Mendo Feb 1 '14 at 15:12

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