Poisson versus Binomial and estimation of variance Let X denotes the number of successes in n trials and p is the probability of success.
Then, X ~ B(n, p) and EX = np, VarX = npq.
If n is really large (e.g. n > 20000) then it seems also reasonable to calculate the expected number of successes to be λ = np and model X as Poisson(λ).
Doing so we find EX = VarX = λ.
The two models provide the same expected value but different variances. I understand that the Binomial approach is the correct one.
The question is:
Is the Poisson model a false approach for this problem?
or:
What is the reason that the Poisson approach give "inflated" estimation for the variance?

The important part isn't whether $n$ is large, it's whether $p$ is small.
The Binomial approach is correct, but the Poisson is a good approximation when $p$ is small. Thinking about the variances, if $p$ is small, $q=1-p\approx 1$, and $\lambda=np\approx npq$.  The approximation gets better as $p$ gets smaller.  For example, if $p=0.001$ then $q=0.999$ and the Poisson and Binomial variances are equal to three significant digits.
The same general result is true for a lot of other properties apart from the variance, but the proof is more complicated.
