Distributions of Minima and Maxima

Why Minima and Maxima Need Care

The minimum and maximum of two measurements answer different questions. For example, in a hypothetical system that fails as soon as either of two independent components fails, the system lifetime is the minimum of the component lifetimes. If a task is complete only after both of two independent stages finish, its completion time is the maximum of the stage times. The distributions are related, but the events that define them are not the same.

Let $X$ and $Y$ be independent continuous random variables with densities $f_X$ and $f_Y$, cumulative distribution functions $F_X$ and $F_Y$, and survival functions $S_X(t)=1-F_X(t)$ and $S_Y(t)=1-F_Y(t)$. Define $U=\min(X,Y)$ and $V=\max(X,Y)$. The formulas below use independence; without it, the joint probabilities generally cannot be obtained by multiplying marginal probabilities.

Definition: For two independent continuous random variables $X$ and $Y$, the minimum is $U=\min(X,Y)$ and the maximum is $V=\max(X,Y)$. Their CDFs are $$ F_U(t)=1-S_X(t)S_Y(t)=1-\bigl(1-F_X(t)\bigr)\bigl(1-F_Y(t)\bigr) $$ and $$ F_V(t)=F_X(t)F_Y(t). $$ When the variables have densities, these CDFs can be differentiated to obtain the densities of $U$ and $V$.

Deriving the CDFs from Events

The event that the minimum exceeds $t$ occurs exactly when both variables exceed $t$. By independence, its probability is the product of their survival probabilities. Taking the complement gives the minimum CDF:

\begin{aligned} P(U>t)&=P(X>t,\;Y>t)\\ &=P(X>t)P(Y>t)=S_X(t)S_Y(t),\\ F_U(t)&=P(U\le t)=1-S_X(t)S_Y(t). \end{aligned}

For the maximum, both variables must be at most $t$ for the maximum to be at most $t$. Independence gives:

\begin{aligned} F_V(t)&=P(V\le t)=P(X\le t,\;Y\le t)\\ &=P(X\le t)P(Y\le t)=F_X(t)F_Y(t). \end{aligned}

If the marginal densities exist, differentiating these CDFs gives $f_U(t)=f_X(t)S_Y(t)+f_Y(t)S_X(t)$ and $f_V(t)=f_X(t)F_Y(t)+f_Y(t)F_X(t)$, wherever the derivatives exist. These expressions also follow by adding the two ways the variable attaining the minimum or maximum can be $X$ or $Y$.

Result: Under independence, the minimum CDF is formed from the product of the survival functions and then complemented. The maximum CDF is the product of the CDFs. If $X$ and $Y$ have densities, the corresponding densities are $$ f_U(t)=f_X(t)S_Y(t)+f_Y(t)S_X(t),\qquad f_V(t)=f_X(t)F_Y(t)+f_Y(t)F_X(t). $$ Use each density only where its terms are defined; outside the relevant supports it is zero.

The Joint Density of the Minimum and Maximum

The pair $(U,V)$ records the two observations in increasing order. For $u<v$, values near $(u,v)$ can arise in two ways: $X$ is near $u$ and $Y$ near $v$, or $Y$ is near $u$ and $X$ near $v$. The original variables have joint density $f_X(x)f_Y(y)$ by independence, so both contributions must be included.

More precisely, for small positive widths $du$ and $dv$, the probability that $U$ lies near $u$ and $V$ lies near $v$ is, to first order, $\{f_X(u)f_Y(v)+f_X(v)f_Y(u)\}\,du\,dv$. Dividing by the area $du\,dv$ and taking the limit gives the joint density. Since continuous variables have probability zero of being equal, the diagonal $u=v$ carries no probability.

f_{U,V}(u,v)= \begin{cases} f_X(u)f_Y(v)+f_X(v)f_Y(u),&u<v,\\ 0,&u\ge v. \end{cases}

The density is understood to be zero whenever either factor is outside its own support. Integrating this joint density over $v>u$ gives the minimum density; integrating over $u<v$ gives the maximum density. For example, $\int_u^\infty f_X(u)f_Y(v)\,dv=f_X(u)S_Y(u)$, which is one of the two terms in $f_U(u)$. The other term comes from the reversed assignment. This gives a joint-density justification for the marginal formulas as well as the event-probability derivation.

Worked Example: Ordering Two Uniform Observations

Let $X$ and $Y$ be independent and each uniformly distributed on $[0,1]$. Find the joint density and marginal densities of $U=\min(X,Y)$ and $V=\max(X,Y)$. Check that the joint density integrates to one.

Method. Each original density is 1 on $[0,1]$ and 0 elsewhere. For $0<u<v<1$, the joint-density formula has two contributions: $f_X(u)f_Y(v)=1\cdot1=1$ and $f_X(v)f_Y(u)=1\cdot1=1$. Thus $f_{U,V}(u,v)=2$ on that triangular region, and it is zero elsewhere.

For a fixed $u$ between 0 and 1, integrate over all possible $v$ from $u$ to 1:

f_U(u)=\int_u^1 2\,dv=2(1-u),\qquad 0<u<1.

For a fixed $v$ between 0 and 1, integrate over $u$ from 0 to $v$:

f_V(v)=\int_0^v 2\,du=2v,\qquad 0<v<1.

The total probability is $\int_0^1\int_u^1 2\,dv\,du=\int_0^1 2(1-u)\,du=[2u-u^2]_0^1=1$. The marginal densities also each integrate to one: $\int_0^1 2(1-u)\,du=1$ and $\int_0^1 2v\,dv=1$. As a further check, the CDF formulas give $F_U(t)=1-(1-t)^2$ and $F_V(t)=t^2$ for $0\le t\le1$; differentiating gives the same densities.

Common mistake avoided: The ordered pair can arise from either assignment of $X$ and $Y$. Including only one product would give joint density 1 on the triangle and total probability $1/2$, not 1.

Worked CDFs and Densities

Worked Example: Minimum and Maximum of Exponential Lifetimes

Suppose $X$ and $Y$ are independent exponential lifetimes with rates 1 and 2 per unit time, respectively. Thus $f_X(t)=e^{-t}$ and $F_X(t)=1-e^{-t}$ for $t\ge0$, while $f_Y(t)=2e^{-2t}$ and $F_Y(t)=1-e^{-2t}$ for $t\ge0$. Find both densities, then calculate $P(U>0.5)$ and $P(V\le1)$.

Minimum density. For $t\ge0$, the survival functions are $S_X(t)=e^{-t}$ and $S_Y(t)=e^{-2t}$. Therefore $F_U(t)=1-e^{-t}e^{-2t}=1-e^{-3t}$, and differentiation gives $f_U(t)=3e^{-3t}$. For $t<0$, $F_U(t)=0$ and $f_U(t)=0$.

Maximum density. Its CDF for $t\ge0$ is $F_V(t)=(1-e^{-t})(1-e^{-2t})$. Differentiating by the product rule gives

f_V(t)=e^{-t}(1-e^{-2t})+2e^{-2t}(1-e^{-t}) =e^{-t}+2e^{-2t}-3e^{-3t},\qquad t\ge0.

Both densities are zero for $t<0$. The requested probabilities are:

\begin{aligned} P(U>0.5)&=S_X(0.5)S_Y(0.5) =e^{-0.5}e^{-1}=e^{-1.5}\approx0.2231,\\ P(V\le1)&=F_X(1)F_Y(1) =(1-e^{-1})(1-e^{-2})\\ &\approx(0.6321)(0.8647)=0.5466. \end{aligned}

Probabilities are rounded to four decimal places. As a check, $P(V\le1)$ is the probability that both lifetimes are at most 1, while $P(U>0.5)$ is the probability that both exceed 0.5. These descriptions match the events used in the calculations.

Common mistake avoided: The minimum exceeds a threshold only if both variables exceed it; the maximum is at most a threshold only if both are at most it. Reversing either event changes the answer.

Worked Example: Different Uniform Supports

Let $X\sim \operatorname{Uniform}(0,2)$ and $Y\sim \operatorname{Uniform}(1,3)$, independently. Find the piecewise CDFs and densities of $U=\min(X,Y)$ and $V=\max(X,Y)$. Then evaluate both CDFs and densities at $t=1.5$.

The marginal CDFs are $F_X(t)=t/2$ on $0\le t<2$, with values 0 below 0 and 1 at or above 2; $F_Y(t)=(t-1)/2$ on $1\le t<3$, with values 0 below 1 and 1 at or above 3.

For the minimum, when $0\le t<1$, $F_Y(t)=0$, so $F_U(t)=F_X(t)=t/2$. When $1\le t<2$, both CDFs vary, giving $F_U(t)=1-(1-t/2)(1-(t-1)/2)$. Since $1-t/2=(2-t)/2$ and $1-(t-1)/2=(3-t)/2$, this is $1-(2-t)(3-t)/4$. At and above 2, $F_X(t)=1$, so $F_U(t)=1$. Differentiating the two nonconstant pieces gives

f_U(t)= \begin{cases} \frac12,&0<t<1,\\ \frac{5-2t}{4},&1<t<2,\\ 0,&\text{otherwise}. \end{cases}

For the maximum, $F_V(t)=0$ below 1. On $1\le t<2$, $F_V(t)=(t/2)((t-1)/2)=t(t-1)/4$. On $2\le t<3$, $F_X(t)=1$, so $F_V(t)=(t-1)/2$; at and above 3 it equals 1. Differentiation gives

f_V(t)= \begin{cases} \frac{2t-1}{4},&1<t<2,\\ \frac12,&2<t<3,\\ 0,&\text{otherwise}. \end{cases}

At $t=1.5$, the minimum CDF and density are $F_U(1.5)=1-(0.5)(0.75)=0.625$ and $f_U(1.5)=(5-3)/4=0.5$. The maximum CDF and density are $F_V(1.5)=(1.5)(0.5)/4=0.1875$ and $f_V(1.5)=(3-1)/4=0.5$. These CDF values are consistent: the minimum is at most 1.5 with probability 0.625, while the maximum is at most 1.5 only when both observations are at most 1.5, an event with probability 0.1875.

Common mistake avoided: Applying one formula across the entire real line without respecting where each marginal CDF changes would give incorrect densities. The support endpoints determine the piecewise cases.

Common Errors and Checks

Swapping the minimum and maximum formulas. For independent $X,Y$, $F_U(t)=1-S_X(t)S_Y(t)$, while $F_V(t)=F_X(t)F_Y(t)$. For example, with two independent Uniform$(0,1)$ variables at $t=0.5$, the incorrect expression $1-(1-F_X)(1-F_Y)$ for the maximum gives 0.75. The correct maximum CDF is $0.5\cdot0.5=0.25$; 0.75 is the minimum CDF.
Forgetting independence. In general, $P(U>t)=P(X>t,Y>t)$ and $P(V\le t)=P(X\le t,Y\le t)$. These joint probabilities equal products of marginal probabilities only when $X$ and $Y$ are independent.
Omitting one assignment in the joint density. For $u<v$, either $X=u,Y=v$ or $X=v,Y=u$ can produce the ordered pair. Both terms are needed, even when the variables have different distributions.
Ignoring support boundaries. A density formula is zero where its variables cannot occur. For nonidentical supports, the resulting CDFs and densities may need several intervals.

A useful final check is that each CDF is nondecreasing, begins at 0, and approaches 1; each density is nonnegative and integrates to 1. For independent continuous variables, $U\le V$ always, and the joint density is supported only where $u<v$, apart from the diagonal of probability zero. The event formulas also provide a direct interpretation in reliability or completion-time models without requiring the joint-density calculation.

Check Your Understanding

For each problem, identify the relevant event before applying a formula.

$X$ and $Y$ are independent Uniform$(0,4)$ variables. Find $P(\min(X,Y)>1)$ and $F_{\max(X,Y)}(2)$. (Answer: $9/16=0.5625$ and $1/4=0.25$.)
Independent exponential variables have rates 0.5 and 1.5 per unit time. Give the density of their minimum for $t\ge0$, and find the probability their maximum is at most 1. (Answer: $f_{\min}(t)=2e^{-2t}$; $(1-e^{-0.5})(1-e^{-1.5})\approx0.3057$.)
$X\sim\operatorname{Uniform}(0,2)$ and $Y\sim\operatorname{Uniform}(1,3)$ independently. Find $F_{\min}(1.25)$ and $f_{\min}(1.25)$. (Answer: $0.671875$ and $0.625$.)
For independent identically distributed continuous variables with CDF $F$, express $P(\min(X,Y)\le t<\max(X,Y))$ in terms of $F(t)$. (Answer: $2F(t)(1-F(t))$.)
$X\sim\operatorname{Uniform}(0,1)$ and $Y\sim\operatorname{Uniform}(0,2)$ independently. Find the joint density of $(U,V)=(\min(X,Y),\max(X,Y))$ at $(0.5,1.5)$. (Answer: $0.5$.)

What You'll Learn