이항 분포 함수가 제한 포아송 분포 함수보다 위 / 아래에 있는가?

하자 파라미터를 이항 분포 함수 (DF)를 나타내고 및 에서 평가 : $B(n,p,r)$ $n \in \mathbb N$ $p \in (0,1)$ $r \in \{0,1,\ldots,n\}$ 및하자매개 변수 포아송 DF를 나타낸다에서 평가:

B (n, p, r) = \sum_{i = 0}^{r} (\binom{n}{i}) p^{i} (1 - p)^{n - i},

$\begin{equation} B(n,p,r) = \sum_{i=0}^r \binom{n}{i} p^i (1-p)^{n-i}, \end{equation}$

F (ν, r)

$F(\nu,r)$

a \in R^{+}

$a \in \mathbb R^+$

r \in {0, 1, 2, \dots}

$r \in \{0,1,2,\ldots\}$

F (a, r) = e^{- a} \sum_{i = 0}^{r} \frac{a^{i}}{i!} .

$\begin{equation} F(a,r) = e^{-a} \sum_{i=0}^r \frac{a^i}{i!}. \end{equation}$

고려 및하자 으로 정의 될 , 정도의 일정 . 이후 , 함수 으로 수렴 모두 ,도 알려져있다. $p \rightarrow 0$ $n$ $\lceil a/p-d \rceil$ $d$ $1$ $np \rightarrow a$ $B(n,p,r)$ $F(a,r)$ $r$

With the above definition for $n$ , I'm interested in determining the values of $a$ for which

B (n, p, r) > F (a, r) \forall p \in (0, 1),

$\begin{equation} B(n,p,r) > F(a,r) \quad \forall p \in (0,1), \end{equation}$ and similarly those for which

B (n, p, r) < F (a, r) \forall p \in (0, 1) .

$\begin{equation} B(n,p,r) < F(a,r) \quad \forall p \in (0,1). \end{equation}$ I have been able to prove that the first inequality holds for

a

$a$ sufficiently smaller than

r

$r$ ; more specifically, for

a

$a$ lower than a certain bound

g (r)

$g(r)$ , with

g (r) < r

$g(r)<r$ . Similarly, the second inequality holds for

a

$a$ sufficiently larger than

r

$r$ , i.e. for

a

$a$ greater than a certain bound

h (r)

$h(r)$ , with

h (r) > r

$h(r)>r$

g (r)

$g(r)$

h (r)

$h(r)$

a

$a$

r

$r$

$p$ ); 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.

— Luis Mendo
소스

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

$p$ for each fixed

r

$r$ . But, what is the role of

d

$d$ 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

@cardinal Thanks for taking the time. Yes, I want the bound to be uniform in p. All other parameters are fixed (but selectable).

d

$d$ is just one such free parameter. For example, one hypothetic result could be as follows: "For any natural

r

$r$ greater than

2

$2$ and any

d \in (- 1, 1)

$d \in (-1,1)$ , the first inequality holds for all

a < r - \sqrt{r}

$a < r - \sqrt{r}$ and for all

p \in (0, 1)

$p \in (0,1)$ ; and the second holds for all

a > r + \sqrt{r}

$a > r + \sqrt{r}$ and for all

p \in (0, 1)

$p \in (0,1)$ .

— Luis Mendo

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

For finite

n

$n$ , the Binomial distribution has closed support from above. Its size may be selectable (by choosing

n

$n$ ) 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

$n$ we will always have

B (n, p, r = n) = 1 > F (a, n)

$B(n,p,r=n) = 1 > F(a,n)$ for any permissible values of

p, a

$p,a$ . So the conditions for the 2nd inequality the OP is after, will always include, at least, "for

r < n

$r<n$ ..."

— Alecos Papadopoulos

답변이 여기에 있습니다 : math.stackexchange.com/questions/37018/…

— Alex R.

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.

— Germaniawerks
소스

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