# Derivation of an estimate for CDF given ordered sample

Given a sample $x_1,\ldots,x_n$ one may order them $x_{(1)},\ldots,x_{(n)}$ and an estimate for the CDF $P(X\leq x_{(i)})$ is $\frac{i}{n}$ which is valid given that the CDF of a sample is uniform. However, the text I'm reading claims the following estimate to be a valid (and better) estimate for the CDF

$$P(X\leq x_{(i)}) \approx \frac{i-\frac{3}{8}}{n+\frac{1}{4}}$$

but it doesn't explain how to derive this estimate, so I was wondering how this estimate can be found?

The text later assumes $x_i$ to be normally distributed, if that is of relevance.

I found the answer after searching the web: This is due to Blom (1954), who used a general estimate for the empirical distribution function given by

$$P(X\leq x_{(i)}) = F(x_{(i)}) \approx \hat{F}_n(x_{(i)}) = \frac{i-\alpha}{n+1-2\alpha}$$

where $\alpha \in [0,0.5]$. He attempted to find $\alpha$ by solving $$\operatorname{E}(x_{(i)}) = \Phi^{-1}\left(\frac{i-\alpha}{n+1-2\alpha}\right)$$

which yields

$$\alpha_{i,n} = \frac{1-(n+1)\Phi(E(x_{(i)})}{1-2\Phi(E(x_{(i)})}$$

Blom conjectured this value to lie in the interval $(0.33,0.5)$ which he found was true for small $n$, so a good compromise was $0.375$, or $3/8$. Inserting this value for $\alpha$ in the original estimate yields

$$\hat{F}_n(x) = \frac{i-3/8}{n+1/4}$$

Sources: