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Find conditions on $\lambda, n, p$, so that the characteristic function of the Binomial converges to that of the Poisson

Binomial distribution is given as $P(X=k)=\binom{n}{k}p^k(1-p)^{n-k}$

Poisson distribution is given as $P(X=k)=\frac{\lambda^ke^{-\lambda}}{k!}$

So characteristic function of binomial: $\phi_X(t)=E(e^{itX})=\sum_1^n \binom{n}{k} e^{itk} p^k(1-p)^{n-k}=\sum \binom{n}{k} (e^{it}p)^k (1-p)^{n-k}=(pe^{it}+(1-p))^n$

and that of Poisson in a similar way:$\quad$$e^{\lambda(e^{it}-1)}$

Now my questions:

1) We defined Expectation with integral, why here with a sum ?

2) Can you give me a hint to solve the problem ?

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2 Answers 2

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Concerning your questions

  1. Expectation is defined as an integral in the continuous case. Now you have discrete random variables. The discrete analogon of the integral is the sum (actually vice versa, i.e. the integral denotes an infite sum with infinitesimal (instead of integer valued) increment). So, for random variables with discrete values you will have sums, and for random variables that take values in continuous intervals you will have integrals.
  2. Write the characteristic function of the binomial as follows $$(pe^{it}+(1-p))^n=\left(1+\frac{np(e^{it}-1)}{n}\right)^n$$ Denote $np$ with $λ$ and use the fact that $$\lim_{n \to \infty}\left(1+\frac xn\right)^n=e^x$$ to conclude as required.
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  • $\begingroup$ discrete values means that there are countably many ? $\endgroup$
    – derivative
    Apr 1, 2014 at 9:38
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    $\begingroup$ @derivative Yes. In the binomial case there are finitely many (from 0 to n) but in the Poisson case there are countably many (0,1,2,...). Since these are probability distributions the sum of their probabilities in the discrete case (or the integral of the pdf in the continuous case) should be equal to 1. $\endgroup$
    – Jimmy R.
    Apr 1, 2014 at 9:41
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Answer to (1): If $X$ is a discrete random variable then there exists a countable set $\{x_1,x_2,\ldots\}$ such that $P(X^{-1}(\Bbb{R}\setminus\{x_1,x_2,\ldots\}))=P(X\notin \{x_1,x_2,\ldots\})=0\implies P(X^{-1}(\{x_1,x_2,\ldots\}))=1$

Let $A=X^{-1}(\{x_1,x_2,\ldots\})=\bigsqcup\limits_{i=1}^\infty X^{-1}(x_i)$. Then for any event $E$, $P(E)=P(E\cap A)$.

So for any measurable map $f$,

$\int_\Omega f(X)\ dP=\int_E f(X)\ dP+\int_{E^c} f(X)\ dP=\int_E f(X)\ dP=\int_{\bigsqcup\limits_{i=1}^\infty X^{-1}(x_i)} f(X)\ dP=\sum\limits_{i=1}^\infty \int_{X^{-1}(x_i)}f(X)\ dP=\sum\limits_{i=1}^\infty f(x_i)P(X=x_i)$

Here $f(x)=e^{itx}$

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