Why are -/+0.5 scale divisions used in continuity corrections for binomial approximation to normal distribution?

Question

Why are half-scale divisions used in continuity corrections for binomial approx to normal distribution?

Why specifically half and not another correction?

Answer 1

Let $X \sim \mathsf{Binom}(n=30,p=1/3),$ Then by computation in R statistical software $$P(9 \le X \le 11) = \sum_{i=9}^{11} P(X=i)\\ = P(X\le 11) - P(X\le 8) = 0.4378,$$ without using a normal approximation. (In R, 'dbinom' is a binomial PDF and 'pbinom' is a binomial CDF.) With some patience, you could also use the formula for the binomial PDF.

sum(dbinom(9:11, 30, 1/3))
[1] 0.4378488
diff(pbinom(c(8,11), 30, 1/3))
[1] 0.4378488

This probability can be written in the following three formats as follows: $$P(8 < X < 12) = P(9 \le X \le 11) = P(8.5 < X < 11.5).$$ It turns out that the last of these three ways gives the best normal approximation.

Recall that $\mu = E(X) = np = 30(1/3) = 10,$ $\sigma^2=Var(X) = np(1-p) = 20/3 \approx 6.6667,$ and $\sigma=SD(X) \approx 2.5820.$ Thus

$$P(8.5 < X < 11.5) = P\left(\frac{8.5-10}{2.5820} < \frac{X-\mu}{\sigma} < \frac{11.5-10}{2.5820}\right)$$

$$\approx P(-0.5809 < Z < 0.5809) \approx 0.4381.$$ where $Z \sim \mathsf{Norm}(0,1)$ and the probability can be found from printed tables of the standard normal CDF. Such normal approximations can usually be relied upon to give two places of accuracy. Here the exact binomial probability and its normal approximation both round to $0.438.$

In the plot below the histogram bars represent the discrete distribution $\mathsf{Binom}(30. 1/3)$ and curve is the density of the approximating continuous normal distribution with $\mu = 10, \sigma = 2.5820.$

In order to get the best normal approximation, we need to include the area under the normal curve between 8.5 and 11.5. This use of half-integers is sometimes called a 'continuity correction. The relevant histogram bars are the ones centered at 9, 10, and 11.

Answer 2

If I understand you correctly, your main question is why we use .5 as correction and not some other number between 0 and 1.

As a further illustration, consider a binomial distribution with parameters $p=.5$ and $n=20$. Now, we try to approximate that with a normal distribution with mean $np$ and variance $n(p)(1-p)$. We use continuity corrections 0 (no correction), 0.25, 0.5, and 0.75. In the below graph the coloured lines are the corresponding densities of the normal distribution.

As you can see, the one with continuity correction 0.5 seems to "fit best".

Now, let us look at an asymmetric example with $p=0.2$.

The one with continuity correction 0.5 again seems to "fit best".

Here is the R code, if you want to try out different examples.

n<-20  # number trials 
p<-.5  # success probability
barplot(dbinom(0:n,n,p), names.arg=0:n, space=0, ylim=c(0,.2))  # binomial distribution
cols<-rainbow(5)  # colors
xseq<-seq(0,n,.01)  # x-axis
for (i in 1:5)
{
  c<-0+(1-i)*.25  # continuity correction
  xcorr<-seq(0+c,n+c,.01)  # continuity corrected x's
  lines(xseq, dnorm(xcorr,n*p,sqrt(n*p*(1-p))), lwd = 1, col = cols[i])  # normal density
}
legend("topright", legend=paste(seq(0,.75,.25)), lwd=2, col=cols)