0
$\begingroup$

I want to show that if $Z_n$ has the binomial distribution with parameters $n$ and $\lambda/n$ with $\lambda$ fixed, then $Z_n $ converges in distribution to the Poisson distribution, parameter $\lambda$ as $n\rightarrow \infty$. How do I do this using characteristic functions?

Edit: i think the characteristic function of the binomial distribution is $(pe^{it}+(1-p))^n$ and that of the Poisson is $e^{\lambda(e^{it}-1)}$, but i dont know which limit to take.

$\endgroup$
2
  • $\begingroup$ You could at least show us that you know what the characteristic functions of the binomial and Poisson distributions are, and some indication that you know what limit you need to compute and what you know about characteristic functions that is relevant. $\endgroup$
    – Chappers
    Apr 27, 2015 at 18:31
  • $\begingroup$ @Chappers see edit $\endgroup$
    – JimmyP
    Apr 27, 2015 at 18:45

1 Answer 1

Reset to default
2
$\begingroup$

So the characteristic function of $\text{B}(n,\lambda/n)$ is $$ ((1-\lambda/n)+\lambda/n e^{it})^{n} = \left( 1 + \frac{1}{n} \lambda\left( e^{it}-1 \right) \right)^n. $$ Now use that $$ \lim_{n \to \infty} \left( 1 + \frac{x}{n} \right)^n = e^x. $$ Then the convergence and uniqueness theorems for characteristic functions imply that the distribution is $\text{Po}(\lambda)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.