Approximating Distributions

thm. A Normal Distribution with $\mu =np,\sigma =\sqrt{np(1-p)}$ will approximate a Binomial distribution with $n$ trials and success probability $p$ when $np>1$

$$\mathbb{P}(a\le X\le b)=\sum _{k=a}^{b}P(X=k)\simeq {\int }_a^b{f}_{\text{normal}}(\mu =np,\sigma =\sqrt{np(1-p)})dx$$

REMARK. A Normal Distribution is equivalent to the following standard normal distribution:

$$\begin{gathered} {\int }_{a-0.5}^{b+0.5}\text{Normal}(\mu ,\sigma )dx={\int }_{a+0.5}^{b+0.5}\frac{e^{-\frac{1}{2}(\frac{x-\mu }{\sigma }{)}^2}}{\sigma \sqrt{2\pi }}dx \\ ={\int }_{\frac{a-0.5-\mu }{\sigma }}^{\frac{b+0.5-\mu }{\sigma }}\text{Normal}(0,1)du={\int }_{\frac{a-0.5-\mu }{\sigma }}^{\frac{b+0.5-\mu }{\sigma }}\frac{e^{-x^2/2}}{\sqrt{2\pi }}dx \end{gathered}$$

where the factor of $+0.5$ to the integrand’s range is the continuity correction.

thm. A Poisson Distribution with parameters $\lambda =np$ will approximate a Binomial distribution with $n$ trials and success probability $p$ when $np\le 1$.

$$X\sim \text{B}(n,p)\Rightarrow \mathbb{P}(X=k)=\frac{{\lambda }^k{e}^{-\lambda }}{k!}=\frac{(np{)}^k{e}^{-(np)}}{k!}$$

thm. A Binomial Distrubition with parameters $p=\frac{K}{N}$ will approxmiate a Hypergeometric Distribution when $N>>n$

$$\begin{array}{r}X\sim \text{Binomial}(n,p)\\ \Downarrow \\ \mathbb{P}(X=k)\simeq \left(\genfrac{}{}{0ex}{}{n}{k}\right)(\frac{K}{N}{)}^k(\frac{N-K}{N}{)}^{n-k}\end{array}$$ $$x_1\times x_2\int f(x_1,x_2)dx$$