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We are given $X_n\sim B(n,p_n)$ where $np_n\rightarrow\lambda$, and $\lambda>0$.

The goal is to prove $X_n$ converges in distribution to Poisson($\lambda$) by use of characteristic functions.

My issue is proving this with $p_n$ being arbitrary as it is. Here are my current steps:

$\phi_{X_n}(t)=(1-p_n+p_ne^{it})^n=(1+p_ne^{it}-p_n)^n=(1+\frac{np_n(e^{it}-1)}{n})^n=e^{n\ln(1+\frac{np_n(e^{it}-1)}{n})}=e^\frac{\ln(1+\frac{np_n(e^{it}-1)}{n})}{1/n}$

Then evaluating the limit means examining the exponent:

$\displaystyle\lim_{n\rightarrow\infty}\frac{\ln(1+\frac{np_n(e^{it}-1)}{n})}{1/n}=\frac{\ln(1+\frac{\lambda(e^{it}-1)}{\displaystyle\lim_{n\rightarrow\infty} n})}{\displaystyle\lim_{n\rightarrow\infty}1/n}=\frac{\ln(1)}{0}=\frac{0}{0}$.

At this point I look to applying L'Hoptials rule, but I need to tweak something because when doing a u-substitution of $u = 1+\frac{np_n(e^{it}-1)}{n}$ I'm not sure how to evaluate the derivative to get the result.

I would imagine it is just zero, treating $p_n$ as a constant, but then this falls apart. As for treating the derivative with $p_n$, I don't see the trick. My first thought was the difference quotient manually, but that did not lead anywhere obvious.

The result should be $\displaystyle\lim_{n\rightarrow\infty}\phi_{X_n}(t)=e^{\lambda(e^{it}-1)}$, which is the $Poisson(\lambda)$ characteristic function.

Any feedback is much appreciated.

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  • $\begingroup$ Your mistake is distributing the limit operations too early, in the equations following the sentence with "examining the exponent". You should expand the $\ln$ a bit, and simplify, before taking limits. $\endgroup$ Commented Feb 4, 2019 at 23:34
  • $\begingroup$ Could you elaborate more on which expansion? Do you mean to go back to $\ln(1+p_ne^{it}-p_n)$ or to move to $\ln(n+np_n(e^{it}-1))-\ln(n)$? $\endgroup$
    – OGV
    Commented Feb 4, 2019 at 23:49
  • $\begingroup$ In the first expression after "examining the exponent", expand your $\ln$ more agressively in powers of $1/n$, and then divide by $1/n$ and only then take the limit. $\endgroup$ Commented Feb 4, 2019 at 23:53
  • $\begingroup$ Do you mean to make the power $\frac{n}{n}$ to expand and cancel the $\frac{1}{n}$ leaving me with the expression $\ln((1+\frac{np_n(e^{it}-1)}{n})^n)$? Just to be certain my intent is to prove $\displaystyle\lim_{n\rightarrow\infty}(1+\frac{np_n(e^{it}-1)}{n})^n=e^{\lambda(e^{it}-1)}$, and not to jump to the identity without verifying having $np_n$ in the numerator holds. $\endgroup$
    – OGV
    Commented Feb 5, 2019 at 0:01

1 Answer 1

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At the crucial step you have (writing $\lambda_n=np_n$) $$\ln \phi_n(t)= n \ln \left(1 + \frac {\lambda_n (e^{it}-1)}n\right)$$ which you should expand as $$\ln \phi_n(t)= n \left(\frac {\lambda_n (e^{it}-1)}n + o( \frac {\lambda_n (e^{it}-1)}n) \right)$$ which is $$\ln \phi_n(t)= \lambda_n (e^{it}-1)+ o(\lambda_n (e^{it}-1)) $$ which converges to $\lambda(e^{it}-1)$ as $n\to\infty$ and $\lambda_n\to\lambda$.

If you like you can write the expansion $\ln (1+x) = x + \theta x^2/2$, where $|\theta|\le 1$ when you plug in $x=\lambda_n (e^{it}-1)$ here, for added comfort.

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