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I want to prove that the Binomial distributions $X_n$ with $p_n$ probability converges in total variation norm to the poisson distribution $X$ if we have that $p_n \rightarrow 0$ and $np_n \rightarrow \lambda$, where $\lambda $ is the parameter of the poisson distribution.

So far I showed that this is true pointwise, but I have troubles with the total variation norm.

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A key-result in this domain is that the distance in total variation between a Bernoulli distribution and a Poisson distribution with common parameter $p$ is $$p(1-\mathrm e^{-p}).$$ Since binomial distributions correspond to sums of independent Bernoulli random variables and Poisson distributions correspond to sums of independent Poisson random variables, the distance in total variation between a binomial $(n,p)$ distribution and the Poisson distribution with parameter $np$ is at most $$np(1-\mathrm e^{-p}).$$ Likewise, the distance in total variation between a binomial $(n,p)$ distribution and a binomial $(n,q)$ distribution is at most $$n\,|p-q|.$$ Thus, using the triangle inequality from a binomial $(n,p)$ distribution to a Poisson distribution with parameter $\lambda$ using a binomial $(n,\lambda/n)$ distribution as middle point yields that the distance in total variation between a binomial $(n,p)$ and a Poisson distribution with parameter $\lambda$ is at most $$|np-\lambda|+\lambda(1-\mathrm e^{-\lambda/n})\leqslant|np-\lambda|+\lambda^2/n.$$ This nonasymptotic upper bound holds for every fixed $(n,p,\lambda)$.

Using it for some parameters $(n,p_n,\lambda)$ such that $\lambda$ is fixed, $n\to\infty$ and $np_n\to\lambda$, one sees that the distance in total variation between the binomial $(n,p_n)$ and the Poisson distribution with parameter $\lambda$ goes to zero when $n\to\infty$.

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  • $\begingroup$ could you just show for instance how you got the first equation? $\endgroup$
    – user66906
    May 20, 2014 at 20:54
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    $\begingroup$ The distance in total variation is $\frac12\sum\limits_n|p_n-q_n|$. Here $p_0=1-p$, $p_1=p$, $p_n=0$ for $n\geqslant2$, $q_0=\mathrm e^{-p}$, $q_1=p\mathrm e^{-p}$, and so on, so one just computes. $\endgroup$
    – Did
    May 20, 2014 at 20:57
  • $\begingroup$ yes, it worked out...and how do you exactly make the factor n transition from bernoulli to binomial distribution? $\endgroup$
    – user66906
    May 20, 2014 at 21:23
  • $\begingroup$ If $X_i$ and $Y_i$ have distributions $\mu_i$ and $\nu_i$, $(X_i)$ independent, $(Y_i)$ independent, then the distance in TV between the distributions of $\sum\limits_iX_i$ and $\sum\limits_iY_i$ is at most the sum of the distances in TV between $\mu_i$ and $\nu_i$ (this is called the tensorization property of the distance in TV). Apply this to every $\mu_i$ Bernoulli $p$ and every $\nu_i$ Poisson $p$. $\endgroup$
    – Did
    May 20, 2014 at 21:38
  • $\begingroup$ okay, don't know how to prove it and especially don't know how to use independence there. unforuntately, the tensorization property is not that famous either; so, i don't know how to continue from here. $\endgroup$
    – user66906
    May 20, 2014 at 23:19

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