The Joint Distribution Of The Sample Minimum And Maximum

Why Study the Minimum and Maximum Together?

In a sample, the smallest and largest observations often describe the range of possible outcomes. A question may constrain both at once: for example, it may require that the smallest recorded loss fall in one interval while the largest falls in another. The marginal distributions of the minimum and maximum alone do not generally answer such a question. Their joint distribution keeps track of how the two endpoints relate.

Let \(X_1,\ldots,X_n\) be independent and identically distributed (iid) continuous random variables with common cdf \(F\) and density \(f\), where \(n\geq 2\). Write \(U=X_{(1)}=\min(X_1,\ldots,X_n)\) and \(V=X_{(n)}=\max(X_1,\ldots,X_n)\). Continuity means ties occur with probability zero, so \(U<V\) with probability one.

Definition: The joint cdf of the sample minimum \(U\) and maximum \(V\) is \(G(u,v)=P(U\leq u,\ V\leq v)\). It describes the probability that the entire sample is at most \(v\), while at least one observation is at most \(u\).

The Joint CDF

The event \(V\leq v\) means all \(n\) observations are at most \(v\), so its probability is \(F(v)^n\). Within that event, the samples that fail \(U\leq u\) have every observation greater than \(u\). When \(u<v\), those observations must all lie in \((u,v]\), an interval with probability \(F(v)-F(u)\) for each draw. Independence makes the probability for all \(n\) draws \([F(v)-F(u)]^n\). Subtracting gives the joint cdf.

G(u,v)=P(U\leq u,V\leq v) =F(v)^n-\bigl(\max\{F(v)-F(u),0\}\bigr)^n.

The positive-part term handles the case \(u\geq v\): then \(F(v)-F(u)\leq 0\), and \(V\leq v\) already guarantees \(U\leq u\). In that case \(G(u,v)=F(v)^n\). For \(u<v\), the formula is \(F(v)^n-[F(v)-F(u)]^n\). The formula works whether or not the distribution has bounded support; the cdf itself accounts for the support.

Worked Example: Evaluating the Joint CDF

A hypothetical quality-control study records \(n=3\) iid measurements, each with the uniform distribution on \((0,1)\). Thus \(F(t)=t\) for \(0\leq t\leq 1\). Find the probability that the minimum is at most \(0.3\) and the maximum is at most \(0.8\).

Method. Use the joint cdf, subtracting the event that all measurements lie in \((0.3,0.8]\) from the event that all are at most \(0.8\).

First, \(P(V\leq 0.8)=F(0.8)^3=0.8^3=0.512\). If all three measurements are at most \(0.8\) but the minimum exceeds \(0.3\), all three lie in \((0.3,0.8]\). The probability for one measurement to be in that interval is \(0.8-0.3=0.5\), so the probability for all three is \(0.5^3=0.125\).

Therefore,

P(U\leq 0.3,V\leq 0.8)=0.512-0.125=0.387.

As a check, this is positive and no larger than \(P(V\leq0.8)=0.512\), as required. The calculation uses the cdf of the endpoints together; multiplying their marginal probabilities would not be justified.

The Joint Density and Its Support

For \(x<y\), the cdf can be written \(G(x,y)=F(y)^n-[F(y)-F(x)]^n\). Differentiating once with respect to \(x\) and then with respect to \(y\) gives the joint density. There is no two-dimensional density outside the region \(x<y\), because a sample minimum cannot exceed its maximum.

Theorem: For an iid continuous sample of size \(n\geq2\) with density \(f\) and cdf \(F\), the joint density of \(U=X_{(1)}\) and \(V=X_{(n)}\) is

f_{U,V}(x,y)= \begin{cases} n(n-1)\,[F(y)-F(x)]^{n-2}f(x)f(y),&x<y,\\ 0,&x\geq y. \end{cases}

The formula includes \(n=2\), with the interval-probability factor raised to the zero power, so the density on \(x<y\) becomes \(2f(x)f(y)\). The density is zero wherever either endpoint has zero parent density. For \(n=1\), the minimum and maximum are the same random variable, and their joint distribution is concentrated on the diagonal rather than having a two-dimensional density; that is why the theorem assumes \(n\geq2\).

Justification: On \(x<y\), differentiate \(G(x,y)=F(y)^n-[F(y)-F(x)]^n\). Differentiating first with respect to \(x\) gives \(n[F(y)-F(x)]^{n-1}f(x)\). Differentiating that expression with respect to \(y\) gives \(n(n-1)[F(y)-F(x)]^{n-2}f(x)f(y)\), as stated.

The factors also have a direct interpretation. Choose which of the \(n\) observations supplies the minimum, then which of the remaining \(n-1\) supplies the maximum. Their densities contribute \(f(x)f(y)\), and each of the other \(n-2\) observations must fall between \(x\) and \(y\), with probability \(F(y)-F(x)\). This explains both the coefficient \(n(n-1)\) and the interval-probability factor.

Probabilities for Two Endpoint Intervals

The joint cdf gives probabilities for rectangles by inclusion-exclusion. For \(a<b\) and \(c<d\), the probability that \(a<U\leq b\) and \(c<V\leq d\) is obtained from the four corners of the rectangle. This method works whether the intervals overlap or are separated, provided the cdf is evaluated correctly.

P(a<U\leq b,\ c<V\leq d) =G(b,d)-G(a,d)-G(b,c)+G(a,c).

When the intervals are separated, with \(b\leq c\), the minimum must come from the lower interval and the maximum from the upper one. For \(n=2\), the remaining sample has no observations to place between them: either of the two labeled observations can supply the minimum. For \(n>2\), all remaining observations must lie between the selected endpoints, which is why simply multiplying the two interval probabilities and the number of endpoint assignments is generally incomplete.

Worked Example: Separated Endpoint Intervals for Two Observations

Two iid hypothetical measurements have cdf \(F(t)=t^2\) on \(0\leq t\leq1\), with density \(f(t)=2t\) there. Find \(P(0.2<U\leq0.5,\ 0.6<V\leq0.9)\), where \(n=2\).

Method. Since the intervals are disjoint and the lower one lies entirely below the upper one, one observation must fall in each interval. Calculate both interval probabilities from the cdf, then count the two possible assignments to the labeled observations.

The lower-interval probability is

P(0.2<X_i\leq0.5)=F(0.5)-F(0.2)=0.5^2-0.2^2=0.25-0.04=0.21.

The upper-interval probability is

P(0.6<X_i\leq0.9)=F(0.9)-F(0.6)=0.9^2-0.6^2=0.81-0.36=0.45.

There are two assignments: the first observation can be the minimum and the second the maximum, or vice versa. Independence gives

P(0.2<U\leq0.5,\ 0.6<V\leq0.9) =2(0.21)(0.45)=0.1890.

The result is between zero and one, and it does not exceed the probability that both observations fall in the union of the two intervals. The factor of two is essential because the observations are labeled even though the endpoints are ordered.

Worked Example: Integrating the Joint Density for Three Observations

A hypothetical system produces \(n=3\) iid readings from a uniform distribution on \((0,1)\), so \(F(t)=t\) and \(f(t)=1\) on that interval. Find the probability that \(0.2<U\leq0.5\) and \(0.7<V\leq0.9\).

Method. Integrate the joint density over the endpoint rectangle. Since \(x\) is in \((0.2,0.5]\) and \(y\) is in \((0.7,0.9]\), every point in the rectangle satisfies \(x<y\). For \(n=3\), the density is \(6(y-x)\).

Thus the probability is

\int_{0.7}^{0.9}\int_{0.2}^{0.5}6(y-x)\,dx\,dy.

For fixed \(y\), the inner integral is

\int_{0.2}^{0.5}6(y-x)\,dx =6\left[xy-\frac{x^2}{2}\right]_{0.2}^{0.5} =6\left((0.5y-0.125)-(0.2y-0.02)\right) =1.8y-0.63.

Now integrate over \(y\):

\int_{0.7}^{0.9}(1.8y-0.63)\,dy =\left[0.9y^2-0.63y\right]_{0.7}^{0.9} =(0.729-0.567)-(0.441-0.441) =0.162.

The probability is \(0.1620\) to four decimal places. As a check on the integral, the rectangle has area \(0.3(0.2)=0.06\), and the average of \(y-x\) over it is \(0.8-0.35=0.45\). Multiplying \(6(0.06)(0.45)\) also gives \(0.162\).

Using Complements for Threshold Events

Not every endpoint event is most conveniently computed by integrating a density. Events such as “the minimum is at most a lower threshold and the maximum exceeds an upper threshold” can be found by subtracting the failure events and adding back their overlap. Independence makes the overlap particularly simple: if every observation must lie in an interval, raise that interval's probability to the \(n\)th power.

Worked Example: Requiring Both a Low Minimum and a High Maximum

A hypothetical sample consists of \(n=4\) iid uniform measurements on \((0,1)\). Find the probability that \(U\leq0.2\) and \(V>0.8\).

Method. Let \(A\) be the event that at least one measurement is at most \(0.2\), and \(B\) the event that at least one measurement exceeds \(0.8\). The desired event is \(A\cap B\). Apply inclusion-exclusion to the complements.

The complement of \(A\) requires all four measurements to exceed \(0.2\), with probability \(0.8^4=0.4096\). The complement of \(B\) requires all four to be at most \(0.8\), also with probability \(0.8^4=0.4096\). Both complements occur when every measurement lies in \((0.2,0.8]\), an interval of probability \(0.6\), giving probability \(0.6^4=0.1296\).

P(A\cap B) =1-P(A^c)-P(B^c)+P(A^c\cap B^c) =1-0.4096-0.4096+0.1296 =0.3104.

The probability is \(0.3104\). This is no larger than either endpoint event alone: \(P(A)=1-0.8^4=0.5904\), and \(P(B)=1-0.8^4=0.5904\). This provides a basic consistency check.

Common Exam Trap

A frequent error is to treat the minimum and maximum as independent because the original observations are independent. They are not generally independent: knowing the maximum is large changes the probability that the minimum is small. A related counting error occurs with separated intervals: for two observations there are two endpoint assignments, but for a larger sample the other observations must also be accounted for between the endpoints.

Wrong: For the two-observation example above, use \(0.21(0.45)=0.0945\), as if there were only one assignment.
Correction: Either labeled observation can be the minimum, so the probability is \(2(0.21)(0.45)=0.1890\).
Wrong: For \(n>2\), multiply by \(n(n-1)\) and stop after multiplying the two endpoint interval probabilities.
Correction: The other \(n-2\) observations must lie between the selected minimum and maximum. Use the joint density or the joint cdf rectangle formula so this restriction is included.

Key takeaway: For an iid continuous sample, the joint cdf is \(F(v)^n-[F(v)-F(u)]^n\) when \(u<v\), and the joint density on \(x<y\) is \(n(n-1)[F(y)-F(x)]^{n-2}f(x)f(y)\). Choose between cdf inclusion-exclusion, density integration, and complements according to the event's shape.

Check Your Understanding

For each problem, assume the observations are iid and continuous. Check both the endpoint restrictions and the sample size before choosing a method.

Three observations are uniform on \((0,1)\). Find \(P(U\leq0.4,\ V\leq0.7)\). (Answer: 0.3160)
Two observations have cdf \(F(t)=t^2\) on \([0,1]\). Find \(P(0.1<U\leq0.4,\ 0.7<V\leq0.9)\). (Answer: 0.0960)
Four observations are uniform on \((0,1)\). Find the probability that all observations lie in \((0.2,0.8]\), equivalently \(U>0.2\) and \(V\leq0.8\). (Answer: 0.1296)
Three observations are uniform on \((0,1)\). Find \(P(0.1<U\leq0.3,\ 0.6<V\leq0.9)\) using the joint density. (Answer: 0.1980)
Five observations are uniform on \((0,1)\). Find the probability that \(U\leq0.2\) and \(V\geq0.8\). (Answer: 0.4224)

What You'll Learn