Why does $k \to np$?

57 Views Asked by At

I am trying to understand the DeMoivre-Laplace Theorem in this page.

There's a section called Proof, where k is defined as $k=np + \sqrt{npq}x$. Then it says that "From this definition we have the approximations $k \to np$" when $n \to \infty$ and I get stuck here. I don't understand why $k \to np$.

1

There are 1 best solutions below

6
On BEST ANSWER

Note that $np+(\sqrt{npq})x=np\left(1+\frac{(\sqrt{npq})x}{np}\right)=np(1+O(1/\sqrt n))$. Thus, as $n\to\infty$, we have $k\sim np$. In particular, all the author of the article means is that $k\to\infty$ at exactly the same rate as $np$ as $n\to\infty$. These types of arguments are particularly useful when needing to work with asymptotics rather than exact values; it gives us an easy way to bound quantities.