Plotting Positions from Uniform Order-Statistic Means

Why Order-Statistic Means Give Plotting Positions

A quantile plot pairs ordered observations with reference quantiles. The reference probabilities need to be chosen: should the smallest observation be paired with the theoretical 10th percentile, the 20th percentile, or some other value? A principled starting point is to ask where an ordered observation lies, on average, in a uniform sample. The answer gives a plotting position that is simple, has an exact order-statistic interpretation, and extends naturally to comparisons with continuous reference distributions.

The key result is that the expected value of the kth smallest observation in a sample of size $n$ from the uniform distribution on $(0,1)$ is $k/(n+1)$. This is an expectation of a percentile coordinate, not a claim that the observed rank is fixed or that every plotting-position convention is identical.

The Uniform Order Statistic

Definition: Let $U_1,\ldots,U_n$ be independent random variables, each uniformly distributed on $(0,1)$. Their order statistics are the sorted values $U_{(1)}\leq\cdots\leq U_{(n)}$. The variable $U_{(k)}$ is the kth order statistic: the kth smallest observation.

For $0<u<1$, the event $U_{(k)}\leq u$ occurs exactly when at least $k$ of the $n$ observations are no greater than $u$. Each observation has probability $u$ of being no greater than $u$, independently of the others. Thus the count of observations no greater than $u$ has a binomial distribution with parameters $n$ and $u$. Differentiating that binomial probability gives the density of the order statistic.

Density formula: For $k=1,\ldots,n$, the density of $U_{(k)}$ on $0<u<1$ is $$ f_{U_{(k)}}(u)=\frac{n!}{(k-1)!(n-k)!}u^{k-1}(1-u)^{n-k}. $$

One way to verify the density is to consider a small interval $(u,u+h]$. For $U_{(k)}$ to lie in this interval, to first order in $h$, one observation must fall in the interval, exactly $k-1$ must fall below $u$, and the rest must fall above $u+h$. There are $n!/((k-1)!(n-k)!)$ ways to assign these roles. The probability of each such arrangement is $u^{k-1}h(1-u)^{n-k}+O(h^2)$. Configurations with two or more observations in the interval have probability $O(h^2)$, since $n$ is fixed. Dividing by $h$ and taking $h$ down to zero yields the displayed density.

The density has the form of a beta density. The beta integral and its recurrence provide a direct way to calculate the required moments, so we establish the recurrence rather than relying on an unverified formula.

Definition: For $a,b>0$, the beta function is $$ B(a,b)=\int_0^1 u^{a-1}(1-u)^{b-1}\,du. $$

Beta Recurrence and the Expected Plotting Position

For $a,b>0$, integration by parts gives an intermediate identity. Take $u^a/a$ as an antiderivative of $u^{a-1}$, and differentiate $(1-u)^b$. The boundary term vanishes at both endpoints because $a,b>0$. Therefore

B(a,b+1)=\frac{b}{a}B(a+1,b).

To obtain the recurrence in the form needed for moments, also use $1=u+(1-u)$ inside the beta integral. This gives $B(a,b)=B(a+1,b)+B(a,b+1)$. Substituting the integration-by-parts identity and solving for $B(a+1,b)$ yields the following result.

Beta recurrence: For $a,b>0$, $$ B(a+1,b)=\frac{a}{a+b}B(a,b). $$

Indeed, $B(a,b)=B(a+1,b)+(b/a)B(a+1,b)=((a+b)/a)B(a+1,b)$, which rearranges to the stated identity. For positive integers, $B(a,b)=(a-1)!(b-1)!/(a+b-1)!$, as follows by iterating the recurrence and using $B(1,b)=1/b$.

The density of $U_{(k)}$ is the beta density with parameters $k$ and $n-k+1$. Integrating $u$ against that density changes the first beta parameter from $k$ to $k+1$. The recurrence then gives its mean.

Theorem: If $U_1,\ldots,U_n$ are independent Uniform$(0,1)$ random variables, then for $k=1,\ldots,n$, $$ E[U_{(k)}]=\frac{k}{n+1}. $$

Proof. By the density formula and the definition of $B$, $$ E[U_{(k)}]=\frac{B(k+1,n-k+1)}{B(k,n-k+1)}. $$ Apply the beta recurrence with $a=k$ and $b=n-k+1$. The ratio is $k/(k+n-k+1)=k/(n+1)$. This proves the result. $\square$

Thus $k/(n+1)$ is the expected value of the kth ordered percentile in a uniform sample. This is often called the Weibull plotting position. In a quantile plot, it supplies a natural probability to pair with the kth ordered observation.

Worked Example: The Middle Order Statistic for Four Observations

Let $n=4$, and find the mean and variance of $U_{(2)}$, the second-smallest observation. Its density is $$ f_{U_{(2)}}(u)=\frac{4!}{1!2!}u(1-u)^2=12u(1-u)^2,\qquad 0<u<1. $$

The expected value follows from the theorem: $$ E[U_{(2)}]=\frac{2}{4+1}=0.4. $$ For the variance, first calculate the second moment using the beta integral: $$ E[U_{(2)}^2]=\frac{B(4,3)}{B(2,3)} =\frac{3!2!/6!}{1!2!/4!} =\frac{1/60}{1/12}=\frac{1}{5}. $$

Consequently, $$ \operatorname{Var}(U_{(2)})=E[U_{(2)}^2]-\{E[U_{(2)}]\}^2 =\frac{1}{5}-0.4^2=0.2-0.16=0.04. $$ The standard deviation is $0.2$. The expected percentile is $0.4$, but the observed second order statistic varies from sample to sample.

How Much Does an Order Statistic Vary?

A plotting position is a useful reference point, but it does not say that the kth order statistic is concentrated exactly there. Its variance quantifies the spread around its mean. The same beta recurrence derives this variance without requiring a separate distributional result.

Result: Under the same independent Uniform$(0,1)$ assumptions, $$ \operatorname{Var}(U_{(k)})= \frac{k(n-k+1)}{(n+1)^2(n+2)}. $$

Derivation. Integrating $u^2$ against the order-statistic density gives $$ E[U_{(k)}^2]=\frac{B(k+2,n-k+1)}{B(k,n-k+1)} =\frac{k}{n+1}\frac{k+1}{n+2} =\frac{k(k+1)}{(n+1)(n+2)}, $$ where the two ratios follow by applying the beta recurrence twice. Subtract the square of the mean: $$ \operatorname{Var}(U_{(k)})= \frac{k(k+1)}{(n+1)(n+2)}-\frac{k^2}{(n+1)^2} =\frac{k\{(k+1)(n+1)-k(n+2)\}}{(n+1)^2(n+2)} =\frac{k(n-k+1)}{(n+1)^2(n+2)}. $$ This proves the formula. $\square$

The variance is positive for every $k=1,\ldots,n$, and it is symmetric in the sense that the variances at ranks $k$ and $n-k+1$ are equal. The spread is not the same at all ranks: the formula itself shows how it depends on the rank and sample size.

Worked Example: Plotting Positions at Three Ranks

Suppose $n=9$. The expected uniform values of the first, fifth, and ninth order statistics are $$ E[U_{(1)}]=\frac{1}{10}=0.1,\qquad E[U_{(5)}]=\frac{5}{10}=0.5,\qquad E[U_{(9)}]=\frac{9}{10}=0.9. $$

Thus the nine plotting positions are $0.1,0.2,\ldots,0.9$. They are equally spaced by $1/(n+1)=0.1$, and the middle rank is centered at $0.5$. The smallest and largest observations are not assigned probabilities 0 and 1; their expected percentile coordinates are $0.1$ and $0.9$. This keeps the reference quantiles finite for distributions whose quantiles diverge at probability 0 or 1.

Using Plotting Positions in Quantile Comparisons

Let $X_1,\ldots,X_n$ be a sample from a continuous, strictly increasing distribution with cumulative distribution function $F$, and let $X_{(k)}$ denote its kth order statistic. The probability integral transform says that $F(X_1),\ldots,F(X_n)$ are independent Uniform$(0,1)$ random variables. Since $F$ is increasing, their kth order statistic is $F(X_{(k)})$. Therefore $$ E[F(X_{(k)})]=\frac{k}{n+1}. $$ This gives an interpretation of the plotting position under a continuous reference model: it is the expected percentile rank of the kth ordered observation on the probability scale.

To compare a sample with a specified continuous reference distribution $F$, one common construction pairs $X_{(k)}$ with the theoretical quantile $F^{-1}(k/(n+1))$, for $k=1,\ldots,n$. Here $F^{-1}$ is the usual inverse of the strictly increasing $F$. The resulting points can be inspected for departures from the reference distribution. The calculation determines the probability coordinate; the quantile transform then puts that coordinate on the reference distribution’s measurement scale.

Worked Example: A Normal Quantile Comparison

Consider four ordered measurements, $8.1, 9.4, 10.8, 12.2$, and compare them with the reference distribution $N(10,4)$, whose standard deviation is $2$. Use the uniform-order-statistic plotting positions $k/(n+1)$. For $n=4$, these are $0.2,0.4,0.6,0.8$.

The corresponding standard normal quantiles are approximately $-0.8416,-0.2533,0.2533,0.8416$. A quantile of $N(10,4)$ is $10+2z$, where $z$ is the corresponding standard normal quantile. The reference values are therefore $8.3168,9.4934,10.5066,11.6832$, respectively.

Rank $k$	Plotting position	Observed order statistic	$N(10,4)$ reference quantile
1	0.2	8.1	8.3168
2	0.4	9.4	9.4934
3	0.6	10.8	10.5066
4	0.8	12.2	11.6832

Each row pairs the observation at rank $k$ with the reference quantile at its expected percentile coordinate. The observations are slightly below the reference at the two lower ranks and above it at the two upper ranks. The calculation provides coordinates for a plot; four points alone do not establish that the reference distribution is or is not appropriate.

Other Conventions and a Common Pitfall

The expected uniform order-statistic value gives the Weibull convention $k/(n+1)$, but it is not the only plotting-position rule. For example, the Hazen convention uses $(k-\tfrac12)/n$, while the Blom convention uses $(k-\tfrac38)/(n+\tfrac14)$. These rules assign different probabilities, especially near the ends of a small sample. Blom’s formula is often used as an approximation for expected normal order statistics; it is not the exact mean of a uniform order statistic. A plotting-position rule should therefore be identified when exact reproducibility matters.

Common mistake: treating the expected percentile as a percentile of the expected observation. From $E[F(X_{(k)})]=k/(n+1)$, it does not follow in general that $F(E[X_{(k)}])=k/(n+1)$. Nor does it follow that $E[X_{(k)}]=F^{-1}(k/(n+1))$. The expectation is taken after applying $F$, and nonlinear transformations do not generally commute with expectation. The plotting-position result specifies an expected probability-scale location; it does not generally calculate the expected observation on the original measurement scale.

Key takeaway: In an independent Uniform$(0,1)$ sample, the kth order statistic has mean $k/(n+1)$. This exact mean motivates the Weibull plotting position for quantile comparisons, while other conventions may be chosen for different approximation goals.

Check Your Understanding

Work through these questions to check the interpretation and calculations.

For a sample of size $n=6$, what plotting position does the uniform-order-statistic mean assign to the fourth order statistic?
For $n=5$, compare the Weibull and Hazen plotting positions for $k=1$. Show both calculations.
Why does $E[F(X_{(k)})]=k/(n+1)$ not generally imply $E[X_{(k)}]=F^{-1}(k/(n+1))$?
For a uniform sample of size $n=4$, calculate the variance of $U_{(1)}$ using the variance formula.
In a quantile comparison, why are the first and last Weibull plotting positions strictly between zero and one?