Actuarial Statistics · Random sampling and sampling distributions

Order Statistics and Sample Quantiles: Distribution of Minimum and Maximum

Q: What is the distribution of the maximum of a random sample?

For an iid sample of size n with cdf F, the cdf of the maximum is F(x)ⁿ. The pdf is n F(x)ⁿ⁻¹ f(x) for a continuous distribution. It works because the maximum is at most x only when every value is at most x.

Q: How do I find the pdf of the kth order statistic?

Use f(k)(x) = n! ÷ [(k − 1)!(n − k)!] × F(x)ᵏ⁻¹ [1 − F(x)]ⁿ⁻ᵏ f(x). It counts k − 1 values below x, one at x, and n − k above. You can derive it from the binomial cdf by differentiation.

Q: Is the sample median the same as the population median?

No. The sample median is a random variable that changes from sample to sample. The population median is a fixed parameter. For many distributions, the sample median is used to estimate it.

Q: What is the distribution of the minimum of exponential variables?

If X₁, ..., Xₙ are iid Exp(λ), the minimum is Exp(nλ). This is because P(min > x) = (e^(−λx))ⁿ = e^(−nλx). Its mean is 1 ÷ (nλ).

Updated 11 October 2026 · Fact-checked

Order statistics are the sample values sorted from smallest to largest. For a random sample of size n with cdf F and pdf f, the maximum has cdf F(x)ⁿ and the minimum has cdf 1 − [1 − F(x)]ⁿ. Differentiate to get pdfs. The kth smallest has a binomial-based cdf.

Understand Order Statistics and Sample Quantiles

Take a random sample X₁, ..., Xₙ, which are independent and identically distributed (iid) with cdf F(x). Sort the values from smallest to largest. The sorted values are written X(1) ≤ X(2) ≤ ... ≤ X(n). X(k) is the kth order statistic. X(1) is the sample minimum and X(n) is the sample maximum.

The sorted values are random and not independent. Knowing X(1) is large tells you that all the others are large too. But you can still find the distribution of each one.

The key idea is to turn the event into a statement about how many observations fall below x. The maximum is at most x only if every observation is at most x. By independence, P(X(n) ≤ x) = F(x)ⁿ. The minimum exceeds x only if every observation exceeds x. So P(X(1) > x) = [1 − F(x)]ⁿ.

For the general case, X(k) ≤ x means at least k observations are at most x. The number of observations at most x is Binomial(n, F(x)). So the cdf of X(k) is a binomial tail sum. Differentiating gives a pdf with a clear meaning: k − 1 values below x, one value at x, and n − k values above x.

The sample median is X((n+1)/2) when n is odd, and the average of the two middle values when n is even. A sample quantile is read from the sorted data, so it is an order statistic or an interpolation between two of them. IAI questions usually use a simple distribution such as the uniform or exponential, and ask for a cdf, pdf, expectation or probability.

Key rules to remember

Cdf of the maximum: F(n)(x) = P(X(n) ≤ x) = [F(x)]ⁿ; Needs iid observations. Pdf: n [F(x)]ⁿ⁻¹ f(x).
Cdf of the minimum: F(1)(x) = 1 − [1 − F(x)]ⁿ; Pdf: n [1 − F(x)]ⁿ⁻¹ f(x).
Cdf of the kth order statistic: F(k)(x) = Σ (j = k to n) C(n, j) [F(x)]ʲ [1 − F(x)]ⁿ⁻ʲ; At least k of the n values are at most x.
Pdf of the kth order statistic: f(k)(x) = n! ÷ [(k − 1)! (n − k)!] × [F(x)]ᵏ⁻¹ [1 − F(x)]ⁿ⁻ᵏ f(x); For a continuous distribution. Valid for k = 1, ..., n.
Minimum of exponentials: If Xᵢ ~ Exp(λ) iid, then X(1) ~ Exp(nλ); Mean of the minimum is 1 ÷ (nλ).
Uniform order statistics: If Xᵢ ~ U(0,1) iid, X(k) ~ Beta(k, n − k + 1), E[X(k)] = k ÷ (n + 1); Variance = k(n − k + 1) ÷ [(n + 1)²(n + 2)].
Sample median: n odd: X((n+1)/2). n even: [X(n/2) + X(n/2 + 1)] ÷ 2; Definitions of sample quantiles differ for other p. Use the one given in the question.

How to solve Order Statistics and Sample Quantiles questions

Use this method for any question on the distribution of an order statistic from a continuous iid sample.

1Write down the parent distribution: its cdf F(x), pdf f(x), the support and the sample size n.
2Identify which order statistic is asked for: minimum, maximum, or the kth.
3For the minimum or maximum, write the event in terms of all n values. Use P(max ≤ x) = F(x)ⁿ or P(min > x) = [1 − F(x)]ⁿ.
4For the kth, either use the pdf formula directly, or use the binomial tail with parameters n and F(x).
5Differentiate the cdf to get the pdf if needed. Keep the support limits.
6Compute what is asked: a probability, a mean, a variance or a percentile. Integrate only over the support.
7Check: the pdf must be non-negative, the cdf must run from 0 to 1, and the answer must lie in a sensible range.

Quickest way: Event-counting shortcut

When to use it: Use for probability questions about the minimum, maximum or median where you do not need the full pdf.

Convert the statement to counts. 'Max ≤ x' means all n values ≤ x. 'Min > x' means all n values > x.
Compute p = F(x) once.
For the maximum use pⁿ. For the minimum use (1 − p)ⁿ in the survival form.
For the median or kth value, treat the count of values ≤ x as Binomial(n, p) and read the needed tail.
For exponential or uniform parents, use the known results (Exp(nλ), Beta) and skip integration.

Common mistakes in Order Statistics and Sample Quantiles

Writing the cdf of the minimum as [F(x)]ⁿ, or the maximum as 1 − [1 − F(x)]ⁿ.
The two formulas look alike and are easily swapped.
Fix: Ask: max ≤ x needs all values ≤ x, so use F(x)ⁿ. Min > x needs all values > x, so use survival to the power n.
Forgetting the factor n in the pdf of the minimum or maximum.
Students differentiate only the inside part and drop the chain rule factor.
Fix: Differentiate [F(x)]ⁿ properly to get n [F(x)]ⁿ⁻¹ f(x). Check that the pdf integrates to 1.
Treating the order statistics X(1), ..., X(n) as independent.
The original Xᵢ are independent, so students assume the sorted ones are too.
Fix: Sorting creates dependence. Only the marginal distributions follow from simple formulas.
Using E[X(n)] = n × E[X] or assuming the median of the sample equals the population median.
Confusing sums with maxima, and a sample statistic with a population parameter.
Fix: Find E[X(n)] from its own pdf. The sample median is random and has its own distribution.
Ignoring the support when integrating.
The pdf is written correctly but integrated over the wrong limits.
Fix: State the support first, for example 0 < x < 1, and use those limits.
Using the wrong index for the median with an even n.
The odd-n rule is applied automatically.
Fix: For even n average X(n/2) and X(n/2 + 1).

Worked examples

Example 1

A random sample of n = 5 is taken from a distribution with pdf f(x) = 2x for 0 < x < 1. Find (a) the pdf of the maximum, and (b) P(X(1) > 0.5).

Show the solution

The cdf is F(x) = x² for 0 < x < 1.
(a) The cdf of the maximum is F(x)⁵ = x¹⁰.
Differentiate: the pdf is 10x⁹ for 0 < x < 1.
(b) P(X(1) > 0.5) = [1 − F(0.5)]⁵.
F(0.5) = 0.25, so 1 − F(0.5) = 0.75.
0.75⁵ = 0.2373046875, so about 0.2373.

Answer: (a) f(x) = 10x⁹ for 0 < x < 1. (b) P(X(1) > 0.5) = 0.75⁵ ≈ 0.2373.

Example 2

A random sample of size 3 is taken from U(0, 1). Find the pdf, mean and variance of the sample median.

Show the solution

For n = 3 the median is X(2), so k = 2.
Here F(x) = x and f(x) = 1.
Use f(k)(x) = n! ÷ [(k − 1)!(n − k)!] × F^(k−1) (1 − F)^(n−k) f.
The constant is 3! ÷ (1! × 1!) = 6.
The pdf is 6x(1 − x) for 0 < x < 1. This is Beta(2, 2).
Mean = k ÷ (n + 1) = 2 ÷ 4 = 0.5.
Variance = k(n − k + 1) ÷ [(n + 1)²(n + 2)] = 2 × 2 ÷ (16 × 5) = 4 ÷ 80 = 0.05.

Answer: The pdf is 6x(1 − x) on (0, 1). The mean is 0.5 and the variance is 0.05.

Exam tips

Write the cdf of the maximum or minimum first. Marks are given for the setup even if the algebra slips.
In MCQs, test the answer with simple checks: the cdf must be 0 at the lower end of the support and 1 at the upper end.
Learn the exponential minimum result. It appears in survival and claims contexts, where the first of several lives or claims is needed.
In written answers, state the assumption that the observations are iid and from a continuous distribution.
In computer-based papers, you can check your result by simulating many samples and sorting each one. Use sort() or max() in R.

Practice questions from Random sampling and sampling distributions

Order Statistics and Sample Quantiles: frequently asked questions

What is the distribution of the maximum of a random sample?

For an iid sample of size n with cdf F, the cdf of the maximum is F(x)ⁿ. The pdf is n F(x)ⁿ⁻¹ f(x) for a continuous distribution. It works because the maximum is at most x only when every value is at most x.

How do I find the pdf of the kth order statistic?

Use f(k)(x) = n! ÷ [(k − 1)!(n − k)!] × F(x)ᵏ⁻¹ [1 − F(x)]ⁿ⁻ᵏ f(x). It counts k − 1 values below x, one at x, and n − k above. You can derive it from the binomial cdf by differentiation.

Is the sample median the same as the population median?

No. The sample median is a random variable that changes from sample to sample. The population median is a fixed parameter. For many distributions, the sample median is used to estimate it.

What is the distribution of the minimum of exponential variables?

If X₁, ..., Xₙ are iid Exp(λ), the minimum is Exp(nλ). This is because P(min > x) = (e^(−λx))ⁿ = e^(−nλx). Its mean is 1 ÷ (nλ).