Minimum and Maximum of a Uniform Sample

Why the Sample Extremes Have Their Own Distributions

A random sample from a uniform distribution has observations that are equally likely across the population interval. Its minimum and maximum do not have uniform distributions when the sample contains more than one observation: the maximum tends to lie toward the upper endpoint, while the minimum tends to lie toward the lower endpoint. Their distributions can be derived by asking whether all sample observations fall on one side of a proposed value.

This tutorial considers a fixed sample size $n$ of independent, identically distributed observations from a continuous uniform distribution on a bounded interval. There is no interest or mortality assumption in these probability models; the only distributional assumption is the stated uniform distribution and independence.

Definition: Let $X_1,\ldots,X_n$ be IID observations with $X_i\sim\operatorname{Uniform}(a,b)$, where $a<b$ and $n$ is a positive integer. The sample minimum and maximum are $$ X_{(1)}=\min(X_1,\ldots,X_n), \qquad X_{(n)}=\max(X_1,\ldots,X_n). $$ Write $L=b-a$ for the interval length. Each observation has density $1/L$ on $[a,b]$, and zero outside that interval.

The support of both extremes is $[a,b]$. Because the parent distribution is continuous, the probability that an observation equals any particular endpoint or threshold is zero. Thus inequalities that differ only by including or excluding a threshold have the same probability here.

Distribution Functions and Densities

For $a\leq x\leq b$, the minimum exceeds $x$ precisely when every observation exceeds $x$. Independence turns that event into a product of individual probabilities. The maximum is at most $x$ precisely when every observation is at most $x$. These event descriptions give the two CDFs.

Theorem: For $a\leq x\leq b$, $$ F_{X_{(1)}}(x)=1-\left(\frac{b-x}{L}\right)^n, \qquad F_{X_{(n)}}(x)=\left(\frac{x-a}{L}\right)^n. $$ For $x<a$, both CDFs are zero; for $x>b$, both are one. Their densities, for $a<x<b$, are $$ f_{X_{(1)}}(x)=\frac{n(b-x)^{n-1}}{L^n}, \qquad f_{X_{(n)}}(x)=\frac{n(x-a)^{n-1}}{L^n}, $$ and are zero outside $[a,b]$.

To justify the minimum formula, observe that $P(X_i>x)=(b-x)/L$. The observations are independent, so $$ P(X_{(1)}>x)=P(X_1>x,\ldots,X_n>x) =\left(\frac{b-x}{L}\right)^n. $$ Since $F_{X_{(1)}}(x)=P(X_{(1)}\leq x)=1-P(X_{(1)}>x)$, this gives its CDF. For the maximum, the event $X_{(n)}\leq x$ means that all $n$ observations are at most $x$, each with probability $(x-a)/L$. Independence therefore gives the maximum CDF. Differentiating each CDF on the interior of the support gives the displayed densities.

When $n>1$, the maximum density increases across the interval and the minimum density decreases across it. When $n=1$, both densities are constant at $1/L$: the minimum and maximum are simply the single observation. That case is an important check on the formulas, since $X_{(1)}=X_{(n)}=X_1$ when there is only one observation.

Worked Example: A Maximum on a Shifted Interval

A hypothetical laboratory draws $n=3$ independent measurements from $\operatorname{Uniform}(2,14)$. Find the probability that the sample maximum is greater than 6 and at most 10, its density at 8, and its mean and variance. The model has no interest or mortality component.

Method. Use the maximum CDF to find the interval probability, differentiate it for the density, and use the general moment formulas established below. Here $a=2$, $b=14$, $L=14-2=12$, and $n=3$.

The maximum CDF at 10 is $$ F_{X_{(3)}}(10)=\left(\frac{10-2}{12}\right)^3 =\left(\frac{8}{12}\right)^3 =\left(\frac{2}{3}\right)^3 =\frac{8}{27}. $$ At 6 it is $$ F_{X_{(3)}}(6)=\left(\frac{6-2}{12}\right)^3 =\left(\frac{4}{12}\right)^3 =\frac{1}{27}. $$ Therefore, $$ P(6<X_{(3)}\leq 10)=F_{X_{(3)}}(10)-F_{X_{(3)}}(6) =\frac{8}{27}-\frac{1}{27} =\frac{7}{27} \approx 0.2593. $$ The density at 8 is $$ f_{X_{(3)}}(8)=\frac{3(8-2)^2}{12^3} =\frac{3(6^2)}{1728} =\frac{108}{1728} =\frac{1}{16} =0.0625. $$ The mean and variance are $$ E[X_{(3)}]=b-\frac{L}{n+1} =14-\frac{12}{4}=11, $$ $$ \operatorname{Var}(X_{(3)}) =\frac{nL^2}{(n+1)^2(n+2)} =\frac{3(12^2)}{4^2(5)} =\frac{432}{80}=5.4. $$ As a check, the maximum’s support is $[2,14]$, and its mean of 11 is between the sample interval’s midpoint, 8, and its upper endpoint, as expected for a maximum from three observations. The common mistake to avoid is using the minimum CDF in the interval-probability calculation; the maximum CDF is $((x-a)/L)^n$, not $1-((b-x)/L)^n$.

Means and Variances

The moment formulas can be understood by rescaling the interval to $[0,1]$. Define $U_i=(X_i-a)/L$. Each $U_i$ is uniform on $[0,1]$, and the rescaled minimum and maximum are $(X_{(1)}-a)/L$ and $(X_{(n)}-a)/L$. Their densities are the corresponding order-statistic densities on $[0,1]$. In particular, the rescaled minimum has density $n(1-u)^{n-1}$, and the rescaled maximum has density $nu^{n-1}$, for $0<u<1$.

For example, integrating the rescaled maximum density gives $$ E[U_{(n)}]=\int_0^1 u\cdot nu^{n-1}\,du =n\int_0^1u^n\,du=\frac{n}{n+1}. $$ Likewise, its second moment is $$ E[U_{(n)}^2]=n\int_0^1u^{n+1}\,du=\frac{n}{n+2}. $$ Subtracting the square of its mean gives $$ \operatorname{Var}(U_{(n)}) =\frac{n}{n+2}-\left(\frac{n}{n+1}\right)^2 =\frac{n}{(n+1)^2(n+2)}. $$ The rescaled minimum has the same variance by reflection: if $U_i$ is uniform on $[0,1]$, then $1-U_i$ is also uniform, and $1-U_{(1)}$ has the same distribution as the maximum of the reflected sample. Its rescaled mean is $1/(n+1)$, either by integration of the minimum density or by the same reflection.

Theorem: For an IID sample from $\operatorname{Uniform}(a,b)$, with $L=b-a$, $$ E[X_{(1)}]=a+\frac{L}{n+1}, \qquad E[X_{(n)}]=b-\frac{L}{n+1}, $$ and $$ \operatorname{Var}(X_{(1)}) =\operatorname{Var}(X_{(n)}) =\frac{nL^2}{(n+1)^2(n+2)}. $$ The means follow by applying $X=a+LU$ to the rescaled extremes. Variance scales by $L^2$ under this transformation.

The two extreme values have equal variances because the uniform interval is symmetric under reflection about its midpoint: reflecting every observation exchanges the roles of the minimum and maximum. Their distributions are not the same unless $n=1$, but their spreads, measured by variance, are equal. The formulas also pass the $n=1$ check: each mean is the midpoint $(a+b)/2$, and the variance is $L^2/12$, the variance of a single uniform observation.

Worked Example: Minimum Exceeding a Threshold

A hypothetical sensor system takes $n=4$ independent readings from $\operatorname{Uniform}(0,20)$. Find the probability that the minimum exceeds 5, the CDF and density of the minimum at 5, and its mean and variance. No interest or mortality assumption is involved.

Method. The minimum’s survival probability is the probability that all four readings exceed the threshold. Here $a=0$, $b=20$, $L=20$, and $n=4$.

For one reading, $P(X_i>5)=(20-5)/20=15/20=3/4$. By independence, $$ P(X_{(1)}>5)=\left(\frac{3}{4}\right)^4 =\frac{81}{256} \approx 0.3164. $$ Consequently, $$ F_{X_{(1)}}(5)=1-\frac{81}{256} =\frac{175}{256} \approx 0.6836. $$ The density at 5 is $$ f_{X_{(1)}}(5) =\frac{4(20-5)^3}{20^4} =\frac{4(15^3)}{160000} =\frac{13500}{160000} =0.084375. $$ The mean and variance are $$ E[X_{(1)}]=0+\frac{20}{5}=4, \qquad \operatorname{Var}(X_{(1)}) =\frac{4(20^2)}{5^2(6)} =\frac{1600}{150} =\frac{32}{3} \approx 10.6667. $$ The result checks out: the probability that the minimum exceeds 5 is the complement of its CDF at 5, and the mean 4 lies near the lower endpoint, as expected for the minimum of four readings. The trap is to calculate the probability using only one observation; the event that the minimum exceeds 5 requires all four readings to exceed 5.

Using Both Extremes in One Event

An event involving both extremes can often be translated into a statement about where every observation must fall. For example, $X_{(1)}\geq c$ and $X_{(n)}\leq d$ together mean that all observations lie in $[c,d]$. If $a\leq c\leq d\leq b$, each observation has probability $(d-c)/L$ of lying in that interval. Independence then gives the probability $((d-c)/L)^n$. This event translation is usually clearer than trying to combine separate marginal probabilities for the minimum and maximum.

Worked Example: Minimum and Maximum on Opposite Sides

A hypothetical calibration procedure takes $n=5$ IID readings from $\operatorname{Uniform}(10,30)$. Find the probability that the minimum is below 14 and the maximum is above 26. Also calculate the mean and variance of each extreme. There is no interest or mortality assumption.

Method. For the probability, use the complement of the event that either extreme fails its required condition. Here $a=10$, $b=30$, $L=20$, and $n=5$.

Let $A$ be the event $X_{(1)}\geq14$, and $B$ the event $X_{(n)}\leq26$. The desired event is the complement of $A\cup B$. Each event requires all five observations to fall in a specified interval: $$ P(A)=\left(\frac{30-14}{20}\right)^5 =\left(\frac{4}{5}\right)^5 =\frac{1024}{3125}, $$ $$ P(B)=\left(\frac{26-10}{20}\right)^5 =\left(\frac{4}{5}\right)^5 =\frac{1024}{3125}. $$ Their intersection means every observation lies between 14 and 26, an interval of length 12: $$ P(A\cap B)=\left(\frac{26-14}{20}\right)^5 =\left(\frac{3}{5}\right)^5 =\frac{243}{3125}. $$ By inclusion-exclusion, $$ P(X_{(1)}<14,\ X_{(n)}>26) =1-P(A)-P(B)+P(A\cap B) =\frac{1320}{3125} =0.4224. $$ For either extreme, the variance is $$ \frac{5(20^2)}{6^2(7)} =\frac{2000}{252} =\frac{500}{63} \approx 7.9365. $$ Their means are $$ E[X_{(1)}]=10+\frac{20}{6} =\frac{40}{3}\approx13.3333, \qquad E[X_{(n)}]=30-\frac{20}{6} =\frac{80}{3}\approx26.6667. $$ As a check, the means lie on the expected sides of the interval midpoint, 20, and the desired event has positive probability because five observations can include values near both endpoints. The common mistake is to treat $A$ and $B$ as disjoint; they overlap when all observations lie between 14 and 26, so that intersection must be added back.

Common Exam Traps and Interpretation

The sample size affects both the location and the shape of the extreme distributions. As $n$ increases, the expected minimum moves toward $a$, the expected maximum moves toward $b$, and the variance of each extreme decreases. These claims follow directly from the formulas: $L/(n+1)$ shrinks with $n$, and $n/((n+1)^2(n+2))$ shrinks as $n$ grows.

The CDFs are easy to swap if the defining events are not written down first. For the minimum, use the complement of “all observations exceed $x$.” For the maximum, use “all observations are at most $x$.” Applying the maximum formula to a minimum event, or vice versa, can produce a plausible-looking number that corresponds to the wrong event.

Key takeaway: For a uniform sample, derive an extreme’s probability from the event involving every observation. The minimum CDF is $1-((b-x)/L)^n$; the maximum CDF is $((x-a)/L)^n$. Their means move inward by $L/(n+1)$, and their variances are equal.

Check Your Understanding

For each problem, assume the stated observations are IID and the distribution is continuous uniform on the stated interval. There is no interest or mortality component.

For $n=2$ observations from $\operatorname{Uniform}(0,8)$, find $P(X_{(2)}\leq6)$. (Answer: $9/16=0.5625$)
For $n=3$ observations from $\operatorname{Uniform}(1,13)$, find $P(X_{(1)}\leq4)$. (Answer: $37/64\approx0.5781$)
For $n=5$ observations from $\operatorname{Uniform}(-2,18)$, find the mean and variance of the maximum. (Answer: $E[X_{(5)}]=44/3\approx14.6667$, $\operatorname{Var}(X_{(5)})=500/63\approx7.9365$)
For $n=4$ observations from $\operatorname{Uniform}(5,25)$, find $P(X_{(1)}>10,\ X_{(4)}\leq20)$. (Answer: $(10/20)^4=1/16=0.0625$)
For $n=3$ observations from $\operatorname{Uniform}(0,10)$, find $P(X_{(1)}<2,\ X_{(3)}>8)$. (Answer: $1-(0.8)^3-(0.8)^3+(0.6)^3=0.1920$)

What You'll Learn