# Central limit theorem and application to binomial distribution

It is said that if the product of $$n$$, the number of trials, and $$p$$, the probability of a success is large, then a binomial distribution can be accurately approximated by a normal distribution.

Is the theory supporting this the Central Limit Theorem? When I think of central limit theorems, I usually think of the sum or mean of a series of IID random variables, where the sum or mean approaches a normal distribution as the number of variables approaches infinity. However, in the current case, I don't see sums or means, so is the idea that the binomial distribution can be approximately as normal because of CLT or some other condition?

• A binomial r.v. is automatically the sum of independent $0-1$ variables. – kimchi lover Jul 17 '20 at 17:17
• @kimchilover Ah yes, but, why is the condition for binomial also dependent on $p$ instead of just $n$? CLT doesn't care about the underlying distribution of the original variable (in this case bernoulli), just as long as $n$ is large and the variables are IID. – user5965026 Jul 17 '20 at 17:22
• It would help if you supplied at least one concrete example where "it is said..." – kimchi lover Jul 17 '20 at 17:53
• @kimchilover Here is one where they recommend that the product of $np$ and $n(1-p)$ is above some threshold sphweb.bumc.bu.edu/otlt/MPH-Modules/BS/BS704_Probability/…. – user5965026 Jul 17 '20 at 17:56

Elaborating on the document cited in the OP's comment, the claim is, that the hypotheses $$0\le p_n,q_n\le 1$$, $$np_n\to\infty$$, and $$nq_n\to\infty$$ (where $$q_n=1-p_n$$) together imply that the CLT applies to $$X_n\sim \operatorname{Bin}(n,p_n)$$, in the sense that $$\lim_{n\to\infty}P\left(\frac{X_n-np_n}{\sqrt{np_nq_n}} for all $$x$$. (Where $$F_n$$ is the cdf of the standardized version of $$X_n$$, and $$\Phi$$ is the standard normal cdf.) In the same vein, one could ask if the same conclusion followed under the simpler-looking hypothesis that $$np_nq_n\to\infty$$.
This version differs from the original statement of the problem by imposing a condition on $$n q_n$$, as does the web page cited in the OP's comment.
By the Berry-Essen theorem (a sharpening of the usual central limit theorem) we know that there is a constant $$C$$ such that for all $$n$$, and real $$x$$, $$|F_n(x)-\Phi(x)|\le \frac {C\rho_n}{\sigma_n^3\sqrt n}=B_n\text{ say},$$ where $$C$$ is a constant, $$\sigma_n=\sqrt{p_nq_n}$$, and $$\rho_n=p_nq_n(p_n^2+q_n^2)$$. Note that if $$Z_n=-p_n$$ with probabilty $$q_n$$ and $$Z_n=1-p_n$$ with probability $$p_n$$, then $$E[Z_n]=0$$, $$\sigma_n^2=E[Z_n^2]$$, and $$\rho_n=E[|Z_n|^3]$$.
So now one just notices that $$B_n=\frac{C(p_n^2+q_n^2)}{\sqrt{np_nq_n}}=\frac{Cp_n^{3/2}}{\sqrt{n q_n}} + \frac{Cq_n^{3/2}}{\sqrt{n p_n}} = O\left(\frac 1{\sqrt{np_n}} + \frac 1{\sqrt {nq_n}}\right) = o(1),$$ under the hypotheses first set of hypoptheses. If one assumes $$np_nq_n\to\infty$$ the result follows similarly: $$B_n\le C/\sqrt{np_nq_n}=o(1)$$. In either case, if you want $$B_n$$ to be less than (say) $$1/10$$, this tells you what your thresholds on $$np_n$$ and $$nq_n$$ must be, and so on.