Tail Integrals for Expected Sample Extremes

Why Tail Probabilities Give Expected Extremes

An expected value can sometimes be easier to calculate from the probability that a variable exceeds each threshold than from its density or mass function. This is especially useful for sample extremes: independence gives simple formulas for the probability that a minimum or maximum exceeds a threshold, and a single integral then yields its expectation. The integral also gives a precise answer to a different question: when is that expectation finite?

Throughout, let $X_1,\ldots,X_n$ be independent and identically distributed nonnegative random variables, with distribution function $F$ and survival function $S(t)=P(X_1>t)=1-F(t)$ for $t\geq 0$. Write $X_{(1)}=\min_{1\leq i\leq n}X_i$ and $X_{(n)}=\max_{1\leq i\leq n}X_i$. Expectations below are allowed to be $+\infty$.

The Tail-Integral Identity

Theorem: If $Y$ is a nonnegative random variable, then $$ E[Y]=\int_0^\infty P(Y>t)\,dt, $$ where the integral and expectation may both be infinite. Thus $E[Y]<\infty$ if and only if the area under the survival function is finite.

The identity follows by writing the value of $Y$ as the length of the interval of thresholds it exceeds. For each outcome,

Y=\int_0^\infty \mathbf{1}_{\{Y>t\}}\,dt.

This holds also when $Y=0$, and in the extended sense if $Y=+\infty$. The integrand is nonnegative and jointly measurable in the outcome and $t$: the set of pairs $(\omega,t)$ for which $Y(\omega)>t$ is measurable. Tonelli's theorem therefore permits interchanging expectation and integration without assuming that either is finite:

E[Y] =E\!\left[\int_0^\infty \mathbf{1}_{\{Y>t\}}\,dt\right] =\int_0^\infty E[\mathbf{1}_{\{Y>t\}}]\,dt =\int_0^\infty P(Y>t)\,dt.

The strict inequality in $P(Y>t)$ works for both continuous and discrete distributions. Atoms can change the survival probability at individual thresholds, but values at a single point do not change the integral.

Tail Formulas for the Sample Minimum and Maximum

The minimum exceeds $t$ exactly when every observation exceeds $t$. The maximum exceeds $t$ unless every observation is at most $t$. Independence thus gives their survival probabilities directly.

Theorem: For an IID nonnegative sample of size $n$, $$ P(X_{(1)}>t)=S(t)^n,\qquad P(X_{(n)}>t)=1-F(t)^n,\qquad t\geq 0. $$ Consequently, $$ E[X_{(1)}]=\int_0^\infty S(t)^n\,dt,\qquad E[X_{(n)}]=\int_0^\infty \bigl(1-F(t)^n\bigr)\,dt. $$ Each formula holds as an equality of extended nonnegative quantities.

For the minimum, the event is $\bigcap_{i=1}^n\{X_i>t\}$, whose probability is $S(t)^n$. For the maximum, the complement of $\{X_{(n)}>t\}$ is $\bigcap_{i=1}^n\{X_i\leq t\}$, which has probability $F(t)^n$. Applying the tail-integral identity to each extreme proves the expectation formulas. No density is required, and the argument allows atoms in $F$. $\square$

These formulas give exact finiteness criteria: the expected minimum is finite precisely when $\int_0^\infty S(t)^n\,dt<\infty$, and the expected maximum is finite precisely when $\int_0^\infty [1-F(t)^n]\,dt<\infty$. For the maximum, this criterion is equivalent to finiteness of the parent mean. Indeed, for $0\leq s\leq1$,

s\leq 1-(1-s)^n\leq ns.

The first inequality follows from $(1-s)^n\leq 1-s$; the second follows by the union bound, or by expanding the probability that at least one of $n$ events occurs. Taking $s=S(t)$ shows that the maximum's survival probability is bounded below by $S(t)$ and above by $nS(t)$. The two tail integrals are therefore finite together. Since $E[X_1]=\int_0^\infty S(t)\,dt$, it follows that

E[X_{(n)}]<\infty \quad\Longleftrightarrow\quad E[X_1]<\infty.

The minimum has no such general equivalence: raising $S(t)$ to the $n$th power can make its tail integrable even when $S(t)$ itself is not.

Worked Calculations

Worked Example: Exponential Sample Extremes

Suppose $X_1,\ldots,X_3$ are IID exponential random variables with rate $\lambda>0$. Their survival function is $S(t)=e^{-\lambda t}$ for $t\geq0$. The minimum has survival function $e^{-3\lambda t}$, so

E[X_{(1)}] =\int_0^\infty e^{-3\lambda t}\,dt =\left[-\frac{1}{3\lambda}e^{-3\lambda t}\right]_0^\infty =\frac{1}{3\lambda}.

For the maximum, $F(t)=1-e^{-\lambda t}$. Hence

\begin{aligned} E[X_{(3)}] &=\int_0^\infty \left[1-(1-e^{-\lambda t})^3\right]dt\\ &=\int_0^\infty \left(3e^{-\lambda t}-3e^{-2\lambda t}+e^{-3\lambda t}\right)dt\\ &=\frac{3}{\lambda}-\frac{3}{2\lambda}+\frac{1}{3\lambda} =\frac{11}{6\lambda}. \end{aligned}

Both expectations are finite, as the exponential tail is integrable. The maximum has the larger expectation, as it must because $X_{(3)}\geq X_{(1)}$ for every sample.

Worked Example: A Finite-Mean Minimum with an Infinite-Mean Parent

Let $X_1,X_2$ be IID Pareto random variables with lower endpoint $2$ and shape parameter $3/4$. In survival-function form,

S(t)= \begin{cases} 1, & 0\leq t<2,\\ (2/t)^{3/4}, & t\geq 2. \end{cases}

The parent mean is infinite because its tail integral includes $\int_2^\infty (2/t)^{3/4}\,dt$, whose integrand has power $t^{-3/4}$ and whose integral diverges. In contrast, the minimum has survival function $S(t)^2$, so its expectation is

\begin{aligned} E[X_{(1)}] &=\int_0^2 1\,dt+\int_2^\infty (2/t)^{3/2}\,dt\\ &=2+2^{3/2}\left[-2t^{-1/2}\right]_2^\infty\\ &=2+2^{3/2}\left(0-(-2\cdot 2^{-1/2})\right)\\ &=2+2^{3/2}\sqrt{2} =2+4 =6. \end{aligned}

Thus the sample minimum has finite expectation even though each observation has infinite expectation. The maximum, by contrast, has infinite expectation: it is at least $X_1$, and $E[X_1]=\infty$. The direct maximum-tail criterion gives the same conclusion.

Worked Example: Discrete Sample Extremes

Suppose each observation equals $0$ with probability $1/2$ and $4$ with probability $1/2$, and take an IID sample of size $3$. For $0\leq t<4$, $S(t)=1/2$, while for $t\geq4$, $S(t)=0$. Therefore

E[X_{(1)}] =\int_0^4 (1/2)^3\,dt =4\cdot\frac18 =\frac12.

For the maximum, $F(t)=1/2$ for $0\leq t<4$, and $F(t)=1$ for $t\geq4$. Thus

E[X_{(3)}] =\int_0^4 \left(1-(1/2)^3\right)dt =4\cdot\frac78 =\frac72.

These results also agree with the possible outcomes. The minimum is $4$ only if all three observations equal $4$, an event of probability $1/8$, giving mean $4/8=1/2$. The maximum is $0$ only if all three equal $0$, also with probability $1/8$, giving mean $4(7/8)=7/2$. This check illustrates why the tail formula applies without requiring a continuous distribution.

Using the Formulas and Avoiding a Common Error

For independent but not identically distributed nonnegative observations, with survival functions $S_i(t)=P(X_i>t)$ and distribution functions $F_i(t)$, the same event argument gives

P(X_{(1)}>t)=\prod_{i=1}^n S_i(t),\qquad P(X_{(n)}>t)=1-\prod_{i=1}^n F_i(t).

Integrating these expressions gives the corresponding expectations, again with infinite values allowed. Identical distributions are needed for the powers $S(t)^n$ and $F(t)^n$, not for the tail-integral identity itself.

A common mistake is to use the parent survival function directly for a sample minimum, forgetting that all observations must exceed the threshold. For an IID sample the correct probability is $S(t)^n$, not $S(t)$. Another is to infer that a finite expected minimum requires a finite parent mean. The Pareto calculation shows why that inference fails: the minimum's tail is the parent tail raised to a power. Conversely, for a finite sample maximum, the event that at least one observation exceeds a threshold has probability at least $S(t)$, so the maximum cannot have finite mean when the parent mean is infinite.

Key takeaway: For a nonnegative random variable, its expectation is the area under its survival curve. Independence turns the minimum's survival into a product and the maximum's survival into one minus a product; integrating those tails gives both the expected extremes and their exact finiteness criteria.

Check Your Understanding

Use the tail formulas and justify each conclusion.

For an IID nonnegative sample, explain why $P(X_{(1)}>t)=S(t)^n$.
State the integral criterion for $E[X_{(n)}]<\infty$, and explain why it is equivalent to $E[X_1]<\infty$.
If $S(t)=(1+t)^{-2}$ for $t\geq0$, write the integral for the expected minimum of a sample of size $3$. Is it finite?
For a nonnegative variable taking values $0$ and $5$ with equal probabilities, find the expected maximum of two IID observations using a tail integral.
Can a sample minimum have finite expectation when the parent mean is infinite? Explain using the tail-integral criterion.