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As is well known, the $\mathsf{Binomial}(n,p)$ distribution converges to the $\mathsf{Poisson}(a)$ distribution as $n\rightarrow \infty$, $p\rightarrow 0$ with $np=a$.

I'm pretty sure that the moments of $\mathsf{Binomial}(n,p)$ also converge to those of $\mathsf{Poisson}(a)$, but I don't know how to prove it. Convergence in distribution doesn't imply convergence of moments, in general. How can I prove that the moments converge?

I've found that the binomial probability (mass) function converges uniformly to the Poisson one. This is stronger than convergence in distribution, so perhaps it can be exploited (but if so I don't know how).

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Because the characteristic function (cf) of a Bernoulli$(p)$ variate is

$$\psi_p(t) = 1 + p(e^{it}-1),$$

the cf of a sum of $n$ independent such variates (which is a Binomial$(p,n)$ variable) is

$$\psi_p(t)^n = \left(1 + p(e^{it}-1)\right)^n = \left(1 + \frac{np(e^{it}-1)}{n}\right)^n.$$

It is well known (and easy to show, even for Complex numbers) that

$$\left(1 + \frac{x}{n}\right)^n$$

converges to $\exp(x)$ as $n\to \infty$. Keeping $np=a$ constant as $n$ increases allows us to write

$$x = np(e^{it}-1) = a(e^{it}-1).$$

Therefore

$$\psi_p(t)^n \to \exp(x) = \exp(a(e^{it}-1)).$$

Because this is the characteristic function of a Poisson$(a)$ distribution, and all the characteristic functions we have considered are analytic in a neighborhood of $t=0$ with power series whose coefficients give the moments, the moments of the Binomial distributions must have converged to the moments of this Poisson distribution, QED.

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  • $\begingroup$ Thanks! This the type of general method I was looking for. $\endgroup$ – Luis Mendo Aug 14 '15 at 22:25
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I think I found an answer using factorial moments. Still, I will accept someone else's answer if they can shed some light into the more general case, such as giving sufficient conditions that assure convergence.

The $r$-th factorial moment of the binomial distribution is easily computed as $$ \mathrm E[(X)_r] = (n)_r\, p^r, $$ where $(a)_r = a(a-1)\cdots(a-r+1)$ denotes the falling factorial; and that of the Poisson distribution is $$ \mathrm E[(X)_r] = a^r. $$ It is clear that $(n)_r\, p^r$ tends to $a^r$ as $n\rightarrow \infty$, $p \rightarrow 0$ with $np=a$. Thus the binomial factorial moments converge to the Poisson ones.

The $r$-th moment is a linear combination of the $0$-th, ..., $r$-th factorial moments: $$ \mathrm E[X^r] = \sum_{k=0}^r \left\lbrace\ r\atop k \right\rbrace \mathrm E[(X)_r], $$ where $\left\lbrace\ r\atop k \right\rbrace$ denotes the Stirling numbers of the second kind. This expression, together with the convergence of the factorial moments, implies the convergence of the moments.

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