Densities Of The Iid Sample Minimum And Maximum

Why the Sample Extremes Have Different Densities

Suppose a set of independent measurements comes from the same continuous distribution. The smallest and largest measurements are not distributed like a single observation from that parent distribution: their cdfs describe events involving all the observations. For example, for the minimum to exceed a threshold, every observation must exceed it. For the maximum to be at most a threshold, every observation must be at most it.

Those event descriptions, together with independence, give the cdfs directly. Differentiating them gives the densities. The results are useful for questions about operating times, losses, measurements, and other settings where a sample’s earliest, smallest, or largest value matters.

Definition: Let \(X_1,\ldots,X_n\) be independent and identically distributed continuous random variables with parent cdf \(F_X(x)\) and density \(f_X(x)\). Define the sample minimum and maximum by \(X_{\min}=\min(X_1,\ldots,X_n)\) and \(X_{\max}=\max(X_1,\ldots,X_n)\). The formulas in this tutorial assume \(n\geq 1\), a common parent distribution, independence, and a density for that distribution.

From the Cdfs to the Densities

Write \(F_{\min}(x)=P(X_{\min}\leq x)\) and \(F_{\max}(x)=P(X_{\max}\leq x)\). The minimum exceeds \(x\) precisely when every observation exceeds \(x\). The maximum is at most \(x\) precisely when every observation is at most \(x\). Independence lets us multiply the corresponding probabilities.

F_{\min}(x)=1-\bigl(1-F_X(x)\bigr)^n, \qquad F_{\max}(x)=\bigl(F_X(x)\bigr)^n.

A density is the derivative of the cdf where that derivative exists. Applying the chain rule to the two cdfs gives the central formulas:

Theorem: For an iid continuous sample with parent density \(f_X\) and cdf \(F_X\),

f_{\min}(x)=n\bigl(1-F_X(x)\bigr)^{n-1}f_X(x), \qquad f_{\max}(x)=n\bigl(F_X(x)\bigr)^{n-1}f_X(x).

These densities are zero outside the support of the parent distribution. Within the support, \(F_X(x)\) is the probability one observation is at most \(x\), while \(1-F_X(x)\) is the probability it is greater than \(x\). The powers \(n-1\) account for the other observations; the factor \(n\) accounts for which one observation lies in a small interval around \(x\).

Here is a direct justification by differentiation. For the minimum, differentiate \(1-(1-F_X(x))^n\). The derivative of \(1-F_X(x)\) is \(-f_X(x)\), and the outer negative sign cancels the resulting negative sign. For the maximum, differentiate \(F_X(x)^n\). In each case, the chain rule produces the factor \(n\).

\frac{d}{dx}\left[1-(1-F_X(x))^n\right] =n(1-F_X(x))^{n-1}f_X(x), \qquad \frac{d}{dx}\left[F_X(x)^n\right] =nF_X(x)^{n-1}f_X(x).

The formulas require the same parent distribution for every observation and independence across observations. If the observations are dependent, the probabilities of their joint events generally cannot be obtained by multiplying the individual probabilities. If the observations have different distributions, the iid formulas do not apply as written. Also, if the parent distribution has point masses, differentiating its cdf alone does not describe all the probability: the sample extremes may have point masses too. The formulas here are for continuous parent distributions.

Worked Examples

Worked Example: Minimum and Maximum of Uniform Measurements

A hypothetical laboratory records four independent calibration errors, each uniformly distributed between 2 and 10 units. Find the density of the minimum at 4 units, the density of the maximum at 8 units, and the probability that the minimum exceeds 4 units. State densities to four decimal places and the probability to four decimal places.

Model and method. Let each error have \(X_i\sim\operatorname{Uniform}(2,10)\), with \(n=4\). On \(2\leq x\leq10\), the parent cdf and density are \(F_X(x)=(x-2)/8\) and \(f_X(x)=1/8\). Apply the minimum and maximum density formulas, and use the minimum survival probability for the event in the question.

At \(x=4\), \(F_X(4)=(4-2)/8=1/4\), so \(1-F_X(4)=3/4\). Therefore,

f_{\min}(4) =4\left(\frac{3}{4}\right)^3\left(\frac{1}{8}\right) =\frac{4\cdot27}{64\cdot8} =\frac{27}{128} =0.2109375.

Thus the minimum density at 4 units is \(0.2109\) per unit. At \(x=8\), \(F_X(8)=(8-2)/8=3/4\), and the maximum density is

f_{\max}(8) =4\left(\frac{3}{4}\right)^3\left(\frac{1}{8}\right) =\frac{27}{128} =0.2109375.

Thus the maximum density at 8 units is also \(0.2109\) per unit. The two values happen to agree because the thresholds are symmetrically placed in this uniform interval; this is not a general rule that the two densities agree at the same threshold.

For the probability that the minimum exceeds 4, all four measurements must exceed 4. Each does so with probability \(3/4\), independently. Hence

P(X_{\min}>4) =\left(1-F_X(4)\right)^4 =\left(\frac{3}{4}\right)^4 =\frac{81}{256} =0.31640625.

The requested probability is \(0.3164\). As a check, the cdf of the minimum at 4 is \(1-(3/4)^4=175/256\), so the probability it exceeds 4 is the complement, \(81/256\). The method avoids treating the minimum as if it had the parent uniform density.

Worked Example: Operating Times with an Exponential Parent

A hypothetical system has two independent components with operating times that are exponentially distributed at rate 3 per hour. Find the probability that the system’s first component failure occurs after 0.2 hours, the density of that first-failure time at 0.2 hours, and the density of the later of the two component failure times at 0.2 hours. Give probabilities to four decimal places and densities to four decimal places per hour.

Model and method. Let \(X_1,X_2\) be independent exponential lifetimes with rate \(\lambda=3\) per hour, so \(n=2\), \(F_X(t)=1-e^{-3t}\), and \(f_X(t)=3e^{-3t}\) for \(t\geq0\). The first failure time is the minimum; the later failure time is the maximum.

The first failure occurs after \(0.2\) hours only if both lifetimes exceed \(0.2\). Therefore,

P(X_{\min}>0.2) =\left(e^{-3(0.2)}\right)^2 =e^{-1.2} =0.301194\ldots \approx0.3012.

For the minimum density, \(f_{\min}(t)=2(1-F_X(t))f_X(t)\). At \(t=0.2\), \(e^{-3t}=e^{-0.6}\), and \(e^{-1.2}=0.301194\ldots\). Thus

f_{\min}(0.2) =2e^{-0.6}\cdot3e^{-0.6} =6e^{-1.2} =1.807165\ldots \approx1.8072\text{ per hour}.

For the maximum density, \(f_{\max}(t)=2F_X(t)f_X(t)\). Since \(F_X(0.2)=1-e^{-0.6}\),

f_{\max}(0.2) =2(1-e^{-0.6})3e^{-0.6} =6(e^{-0.6}-e^{-1.2}) =1.485705\ldots \approx1.4857\text{ per hour}.

The results have the expected units: probabilities are unitless, while densities are per hour. As a check, the minimum density is larger than the maximum density at this early time, where a first failure is more concentrated than a last failure. The common mistake is to confuse the density \(1.8072\) per hour with the probability \(0.3012\) of surviving past 0.2 hours; a density at a point is not a probability at that point.

Worked Example: Extremes of a Normal Sample

Three independent measurements have distribution \(X_i\sim N(0,1)\), where the second parameter is the variance. Find the density of the sample maximum at zero and the probability that the maximum is at most one. Give both answers to four decimal places.

Model and method. The parent standard normal cdf is denoted by \(\Phi\), and its density is \(\phi(x)=(1/\sqrt{2\pi})e^{-x^2/2}\). The maximum density formula gives \(f_{\max}(x)=3\Phi(x)^2\phi(x)\), and its cdf is \(F_{\max}(x)=\Phi(x)^3\).

At zero, symmetry of the standard normal distribution gives \(\Phi(0)=0.5\), and \(\phi(0)=1/\sqrt{2\pi}=0.398942\ldots\). Therefore,

f_{\max}(0) =3(0.5)^2(0.398942\ldots) =0.299206\ldots \approx0.2992.

For the probability that the maximum is at most one, use the maximum cdf. The standard normal value is \(\Phi(1)=0.8413447\ldots\), so

P(X_{\max}\leq1) =\Phi(1)^3 =(0.8413447\ldots)^3 =0.5956.

The maximum density is \(0.2992\) per unit, and the probability is \(0.5956\). As a check on the event, all three observations must be at most one, so the result is the cube of the single-observation probability. The method avoids substituting a normal value into a uniform or exponential formula; the parent \(F_X\) and \(f_X\) must match the stated distribution.

Common Exam Trap: Using the Wrong Extreme

The most important distinction is the power of the parent cdf or survival probability. A maximum is controlled by the event that every observation is at most \(x\), so its density contains \(F_X(x)^{n-1}\). A minimum is controlled by the event that the other observations exceed \(x\), so its density contains \((1-F_X(x))^{n-1}\).

For example, in the four-observation uniform model above, the incorrect use of the maximum formula for the minimum at 4 would give

4F_X(4)^3f_X(4) =4\left(\frac14\right)^3\left(\frac18\right) =\frac{1}{128} =0.0078125.

That is not the minimum density at 4. The correct calculation uses the probability that the other observations exceed 4:

f_{\min}(4) =4\left(1-\frac14\right)^3\left(\frac18\right) =\frac{27}{128} =0.2109375.

A second trap is to interpret a density value as a probability. For a continuous random variable, \(P(X_{\min}=x)=0\) at any single point \(x\). A density describes probability per unit near that point; probabilities require an interval or an event such as \(X_{\min}>x\).

Key takeaway: For an iid continuous sample, \(F_{\min}(x)=1-(1-F_X(x))^n\) and \(F_{\max}(x)=F_X(x)^n\). Differentiate the appropriate cdf, retain the parent density \(f_X(x)\), and check the support and units before interpreting the result.

Check Your Understanding

For each problem, identify whether the event or density concerns the minimum or maximum before applying a formula.

Three independent measurements are uniform on the interval from 0 to 5. Find the density of the minimum at 1 and the density of the maximum at 4. (Answer: both densities are \(0.3840\) per unit.)
Four independent lifetimes are exponential with rate 2 per day. Find the probability that the maximum lifetime is at most 0.5 days. (Answer: \((1-e^{-1})^4\approx0.1597\).)
Two independent measurements have distribution \(N(10,4)\), with variance 4. Find the density of the maximum at 10 and the probability that the minimum exceeds 10. (Answer: density \(0.1995\) per unit; probability \(0.2500\).)
Five independent measurements are uniform from 0 to 12. Find the density of the maximum at 9 and the probability that the minimum exceeds 3. (Answer: density \(0.1318\) per unit; probability \(0.2373\).)
Three independent operating times are exponential with rate 4 per hour. Find the density of the maximum at 0.25 hours and the probability that the minimum exceeds 0.25 hours. (Answer: density \(1.7640\) per hour; probability \(0.0498\).)

What You'll Learn