The Joint Law of Sample Minimum and Maximum

Why the Two Extremes Have a Joint Law

The sample minimum and maximum summarize opposite ends of a sample, but they are not separate summaries: both are computed from the same observations. Their joint distribution describes how the two extremes can occur together. For example, it can answer questions about the probability that the minimum falls below one threshold while the maximum exceeds another, or about the likely maximum when the minimum is known.

Let $X_1,\ldots,X_n$ be independent and identically distributed random variables with cumulative distribution function $F$ and density $f$, where $n\geq 2$. Write $X_{(1)}=\min_i X_i$ for the sample minimum and $X_{(n)}=\max_i X_i$ for the sample maximum. We assume the common distribution is absolutely continuous. This condition ensures that the sample has no ties with positive probability and that the extremes have a joint density. A continuous distribution need not be absolutely continuous, so continuity alone would not guarantee a density.

Definition: The joint law of the sample minimum and maximum is the probability distribution of the pair $(X_{(1)},X_{(n)})$. For an absolutely continuous iid sample of size $n\geq 2$, the pair is supported on $u<v$, apart from boundary events of probability zero.

Deriving the Joint Density

Suppose the observed minimum is near $u$ and the observed maximum is near $v$, with $u<v$. One observation must be the minimum and a different observation must be the maximum. There are $n$ choices for the observation at the minimum and $n-1$ choices for the one at the maximum. Once those two observations are specified, each of the remaining $n-2$ observations must lie between $u$ and $v$, which has probability $F(v)-F(u)$. This gives the form of the density; the argument below makes the calculation precise.

Theorem: If $X_1,\ldots,X_n$ are iid with an absolutely continuous distribution having density $f$ and CDF $F$, then the joint density of $X_{(1)}$ and $X_{(n)}$ is $$ f_{X_{(1)},X_{(n)}}(u,v) = \begin{cases} n(n-1)\,[F(v)-F(u)]^{n-2} f(u)f(v), & u<v,\\ 0, & u\geq v. \end{cases} $$ The density formula holds almost everywhere with respect to two-dimensional Lebesgue measure.

Proof. Ties have probability zero because the common distribution is absolutely continuous. On the event that the minimum and maximum are distinct, their observation labels are therefore a uniquely determined ordered pair $(i,j)$, where $i$ labels the minimum and $j$ labels the maximum. There are $n(n-1)$ such ordered pairs.

Fix distinct labels $i$ and $j$. For the minimum to lie in an infinitesimal interval around $u$, the observation $X_i$ contributes density $f(u)$; for the maximum to lie around $v$, $X_j$ contributes $f(v)$. Each other observation must lie in $(u,v)$, an event of probability $F(v)-F(u)$. Independence makes the probability contribution from those $n-2$ observations equal to $[F(v)-F(u)]^{n-2}$. Integrating these contributions over any measurable region of pairs $(u,v)$ with $u<v$, and summing over the $n(n-1)$ possible label pairs, gives the stated density. The restriction $u<v$ is necessary because a minimum cannot exceed a maximum. This proves the formula. $\square$

The same structure can be checked using the joint CDF. For $x<y$, the event $\{X_{(1)}\leq x,\ X_{(n)}\leq y\}$ is the event that all observations are at most $y$, except for the case in which they all lie in $(x,y]$. Thus

P(X_{(1)}\leq x,\ X_{(n)}\leq y) = F(y)^n-[F(y)-F(x)]^n,\qquad x<y.

For $x\geq y$, the event reduces to $\{X_{(n)}\leq y\}$, so its probability is $F(y)^n$. Differentiating the expression for $x<y$ with respect to $x$ and then $y$ gives the joint density above at points where the derivatives exist. The label-counting argument establishes the density directly and also explains why the formula applies to the entire joint law.

Marginals and Conditional Distributions

Integrating the joint density over one coordinate recovers the familiar marginal densities of the extremes. For the minimum, integrate over $v>u$. The chain rule for absolutely continuous $F$ gives

\begin{aligned} f_{X_{(1)}}(u) &=n(n-1)f(u)\int_u^\infty [F(v)-F(u)]^{n-2}f(v)\,dv\\ &=n f(u)[1-F(u)]^{n-1}. \end{aligned}

For the maximum, integrate over $u<v$ instead; the result is $f_{X_{(n)}}(v)=n f(v)F(v)^{n-1}$. These marginals also follow by asking that all observations exceed a proposed minimum, or that all are below a proposed maximum.

Dividing the joint density by the marginal density of the minimum gives a conditional density for the maximum given the minimum. For almost every $u$ with $f_{X_{(1)}}(u)>0$, it is

f_{X_{(n)}\mid X_{(1)}}(v\mid u) = \frac{(n-1)f(v)[F(v)-F(u)]^{n-2}}{[1-F(u)]^{n-1}}, \qquad v>u,\quad F(u)<1.

Integrating this expression from $u$ to $y$, for $y\geq u$, gives the conditional CDF

P(X_{(n)}\leq y\mid X_{(1)}=u) = \left(\frac{F(y)-F(u)}{1-F(u)}\right)^{n-1}.

This is a conditional distribution in the regular-conditional sense: conditioning on the exact value of a continuous random variable is expressed through a density and is specified for almost every conditioning value. It describes the maximum among the $n-1$ remaining observations, which, given the minimum $u$, must all exceed $u$.

Worked Example: A Joint Probability for a Uniform Sample

Let $X_1,\ldots,X_4$ be iid Uniform$(0,1)$, and write $U=X_{(1)}$ and $V=X_{(4)}$. Find the joint density and calculate $P(U\leq 1/4,\ V\geq 3/4)$.

For this distribution $F(t)=t$ and $f(t)=1$ on $0<t<1$. The joint density is therefore

f_{U,V}(u,v)=12(v-u)^2,\qquad 0<u<v<1.

To calculate the probability, use the complement of the event that the minimum is at most $1/4$ and the maximum is at least $3/4$. The event $U>1/4$ has probability $(3/4)^4$, and $V<3/4$ also has probability $(3/4)^4$. Their intersection requires all four observations to lie in $(1/4,3/4)$, so its probability is $(1/2)^4$. Inclusion-exclusion gives

\begin{aligned} P(U\leq 1/4,\ V\geq 3/4) &=1-P(U>1/4)-P(V<3/4)+P(U>1/4,\ V<3/4)\\ &=1-\left(\frac34\right)^4-\left(\frac34\right)^4+\left(\frac12\right)^4\\ &=\frac{256-81-81+16}{256} =\frac{110}{256} =\frac{55}{128}. \end{aligned}

The intersection correction is essential: the events $U>1/4$ and $V<3/4$ can happen together when every observation lies between the thresholds.

Worked Example: Conditional Maximum Given the Minimum

For the same four-observation Uniform$(0,1)$ sample, find the conditional probability $P(V\leq 0.6\mid U=0.2)$, using the conditional distribution derived above.

Substitute $n=4$, $F(t)=t$, $u=0.2$, and $y=0.6$:

\begin{aligned} P(V\leq 0.6\mid U=0.2) &=\left(\frac{F(0.6)-F(0.2)}{1-F(0.2)}\right)^3\\ &=\left(\frac{0.6-0.2}{1-0.2}\right)^3 =\left(\frac{0.4}{0.8}\right)^3 =\frac18. \end{aligned}

Given that the minimum is $0.2$, the other three observations must exceed $0.2$. Each has conditional probability $(0.6-0.2)/(1-0.2)=1/2$ of being at most $0.6$, and all three must meet that condition for the maximum to be at most $0.6$. This independently checks the result.

Worked Example: Conditional Maximum in an Exponential Sample

Suppose $X_1,X_2,X_3$ are iid exponential random variables with rate $1$, so $F(t)=1-e^{-t}$ for $t>0$. Find $P(X_{(3)}-X_{(1)}>\log 2\mid X_{(1)}=u)$, for $u>0$.

Here $F(v)-F(u)=e^{-u}-e^{-v}$ and $f(v)=e^{-v}$. The conditional density formula with $n=3$ becomes, for $v>u$,

\begin{aligned} f_{X_{(3)}\mid X_{(1)}}(v\mid u) &=\frac{2e^{-v}(e^{-u}-e^{-v})}{(e^{-u})^2}\\ &=2\left(e^{-(v-u)}-e^{-2(v-u)}\right). \end{aligned}

Write $d=v-u$. Its conditional CDF at a nonnegative value $d$ is

\begin{aligned} P(X_{(3)}-u\leq d\mid X_{(1)}=u) &=\int_0^d 2(e^{-s}-e^{-2s})\,ds\\ &=2\left[(1-e^{-d})-\frac{1-e^{-2d}}{2}\right]\\ &=(1-e^{-d})^2. \end{aligned}

At $d=\log 2$, this CDF is $(1-1/2)^2=1/4$. Therefore the requested upper-tail probability is $1-1/4=3/4$, independent of $u$.

Dependence and a Common Pitfall

The minimum and maximum are generally dependent: their possible values are restricted by $X_{(1)}<X_{(n)}$, and their joint density contains the factor $[F(v)-F(u)]^{n-2}$, which does not usually separate into a function of $u$ times a function of $v$. In the Uniform$(0,1)$ example, the joint density is zero for $u\geq v$, even though both marginal densities are positive at many such pairs. They therefore cannot be independent.

Pitfall: treating the extremes as independent. Multiplying their marginal probabilities ignores the restriction that the minimum cannot exceed the maximum. Use the joint density or joint CDF when a question involves both extremes.
Pitfall: using $n$ rather than $n(n-1)$. The minimum and maximum are attained by distinct observations. There are $n$ choices for the minimum's label and $n-1$ choices for the maximum's label.
Pitfall: allowing pairs outside the support. The joint density is zero unless $u<v$. This support restriction must be included in integrals and probability calculations.
Pitfall: assuming every continuous distribution has a density. The density formula requires absolute continuity. A continuous but singular distribution does not satisfy that assumption.

The joint law is useful whenever an analysis depends on both endpoints of the observed sample, rather than on either extreme alone. Its structure separates three ingredients: the choices of which observations attain the extremes, the densities at those endpoint values, and the probability that the remaining observations fall between them. The conditional formulas then translate the same structure into questions about one extreme when the other is known.

Check Your Understanding

Use the joint density and conditional distribution formulas to answer these questions.

For an iid absolutely continuous sample of size $n$, explain why the joint density of the minimum and maximum is zero when $u>v$.
Identify the three factors in $n(n-1)[F(v)-F(u)]^{n-2}f(u)f(v)$, and explain what each represents.
For an iid Uniform$(0,1)$ sample of size $n$, write the joint density of the minimum and maximum on its support.
Why must the intersection term be included when using complements to calculate $P(X_{(1)}\leq a,\ X_{(n)}\geq b)$ for $a<b$?
State the conditional CDF of the maximum given that the minimum equals $u$, and specify its range of validity.

The Joint Law of Sample Minimum and Maximum

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