Why Calculate an Expectation From Its Tail?
The usual expectation calculation weights each possible value by its probability. Sometimes the probabilities of exceeding thresholds are easier to find than the individual point probabilities. A tail formula reorganizes the same expectation in terms of those exceedance probabilities: for an integer-valued variable, it adds the probabilities of reaching successive integer levels; for a continuous variable, it integrates the probability of exceeding each nonnegative threshold.
These formulas are useful both for calculation and for understanding what an expectation measures. Each tail probability contributes to the expected value: a large value contributes to many thresholds, while a small value contributes to fewer. The formulas also apply when the expectation is infinite, provided the variable is nonnegative.
Tail-Sum Formula for Integer-Valued Variables
The sum begins at \(k=1\), not \(k=0\): the event \(X\geq 0\) has probability one, but it corresponds to no positive unit of the value of \(X\). Each event \(X\geq k\) records whether \(X\) reaches at least the \(k\)th unit.
Proof. For every integer \(j\geq 0\), there are exactly \(j\) integers \(k\geq 1\) satisfying \(j\geq k\). Therefore, pointwise, $$ X=\sum_{k=1}^{\infty}\mathbf{1}_{\{X\geq k\}}. $$ For an outcome where \(X=j\), the sum on the right has \(j\) terms equal to one and all remaining terms equal to zero, so it equals \(j\). The indicators are nonnegative measurable random variables. By the monotone convergence theorem, expectation commutes with the limit of their increasing partial sums. Thus $$ E[X]=\sum_{k=1}^{\infty}E[\mathbf{1}_{\{X\geq k\}}] =\sum_{k=1}^{\infty}P(X\geq k). $$ The equality remains valid if the series diverges, in which case \(E[X]=+\infty\). \(\square\)
Worked Example: A Geometric Waiting Time
Let \(X\) be the number of failures before the first success in independent Bernoulli trials, each with success probability \(p=0.3\). Then \(X\) is nonnegative integer-valued. Write \(q=1-p=0.7\).
For \(k\geq 1\), the event \(X\geq k\) means that the first \(k\) trials are all failures. Independence gives $$ P(X\geq k)=q^k=0.7^k. $$
Using the tail-sum formula and the geometric series, $$ E[X]=\sum_{k=1}^{\infty}0.7^k =\frac{0.7}{1-0.7} =\frac{7}{3}. $$ The result is about \(2.33\) failures. The indexing is consistent with the model: \(P(X\geq 1)=0.7\), the probability the first trial fails.
Worked Example: A Binomial Count
Let \(X\sim\operatorname{Binomial}(3,0.4)\), the number of successes in three independent trials. Here \(X\) can equal 0, 1, 2, or 3, so only three tail probabilities are needed.
With failure probability \(0.6\), $$ P(X\geq 1)=1-P(X=0)=1-(0.6)^3=0.784. $$ For at least two successes, either exactly two or exactly three occur: $$ P(X\geq 2)=\binom{3}{2}(0.4)^2(0.6)+(0.4)^3 =3(0.16)(0.6)+0.064=0.352. $$ Finally, $$ P(X\geq 3)=P(X=3)=(0.4)^3=0.064. $$
Adding these tails gives $$ E[X]=0.784+0.352+0.064=1.2. $$ This agrees with the expected count of successes, \(3(0.4)=1.2\). The tail calculation reaches the same answer by counting the probability of attaining each of the first, second, and third success levels.
Tail-Integral Formula for Nonnegative Continuous Variables
For a continuous random variable with a density, the corresponding tail formula replaces the sum over integer levels by an integral over all nonnegative thresholds. The event in the integrand is \(X>t\): it asks whether the value exceeds the threshold \(t\).
Proof. By the density formula for expectation and the pointwise identity \(x=\int_0^\infty \mathbf{1}_{\{t<x\}}\,dt\) for \(x\geq 0\), $$ E[X]=\int_{0}^{\infty}x f(x)\,dx =\int_{0}^{\infty}\int_{0}^{\infty}\mathbf{1}_{\{t<x\}}f(x)\,dt\,dx. $$ The integrand is nonnegative and measurable on \([0,\infty)^2\), so Tonelli’s theorem permits reversing the order of integration, even if the integral is infinite. The region \(\{(t,x):0\leq t<x\}\), when described at a fixed \(t\), has \(x\) ranging from \(t\) to infinity. Consequently, $$ E[X]=\int_{0}^{\infty}\int_{t}^{\infty}f(x)\,dx\,dt =\int_{0}^{\infty}P(X>t)\,dt. $$ Since \(P(X>t)=1-F(t)\), the second expression follows. \(\square\)
The same derivation can be read as a layer-by-layer accounting identity: for each fixed value \(x\), the interval of thresholds \(0\leq t<x\) has length \(x\). Integrating the probability that a threshold is below \(X\) therefore recovers the average length.
Worked Example: An Exponential Lifetime
Let \(X\) have an exponential distribution with rate \(\lambda=2\), so its density is \(f(x)=2e^{-2x}\) for \(x\geq 0\). Its survival probability is $$ P(X>t)=\int_t^\infty 2e^{-2x}\,dx=e^{-2t},\qquad t\geq 0. $$
The tail-integral formula gives $$ E[X]=\int_0^\infty e^{-2t}\,dt =\left[-\frac{1}{2}e^{-2t}\right]_{0}^{\infty} =0-\left(-\frac{1}{2}\right) =\frac{1}{2}. $$ Thus the expected lifetime is \(0.5\) in the time units used by the model. The rate \(2\) appears in the survival curve’s exponential decay, and the mean is its reciprocal.
Worked Example: A Uniform Lifetime
Suppose \(X\sim\operatorname{Uniform}(0,3)\), with density \(f(x)=1/3\) on \([0,3]\). For thresholds between zero and three, $$ P(X>t)=\int_t^3\frac{1}{3}\,dx=\frac{3-t}{3},\qquad 0\leq t\leq 3. $$ For \(t>3\), the survival probability is zero. Therefore, $$ E[X]=\int_0^3\frac{3-t}{3}\,dt =\frac{1}{3}\left[3t-\frac{t^2}{2}\right]_0^3 =\frac{1}{3}\left(9-\frac{9}{2}\right) =\frac{3}{2}. $$ This matches the midpoint of the interval \([0,3]\), providing a check on the calculation.
What the Tail Formulas Tell Us
The tail formulas are not restricted to cases where the expectation is finite. Each formula equates two nonnegative quantities, and the proof uses a theorem that allows the value \(+\infty\). For example, if the sum of the integer tails diverges, the nonnegative integer-valued variable has infinite expectation. If the area under a continuous survival curve is infinite, the variable’s expectation is infinite. No cancellation between positive and negative contributions is involved.
In applications, survival probabilities can be convenient when a model directly describes exceedances, reliability, or waiting beyond a threshold. The formulas also provide a useful way to check a calculation: a nonnegative variable’s tail probabilities cannot increase as the threshold increases, and the resulting sum or integral must be nonnegative.
Common Mistake: Replacing Tails With Point Probabilities
A tempting but incorrect calculation for an integer-valued \(X\) is \(E[X]=\sum_{k=1}^{\infty}P(X=k)\). The terms on the right add to \(P(X\geq 1)\), not to the expected value. They count whether \(X\) is positive once; they do not count how many units large \(X\) is.
For example, in the geometric model above, \(\sum_{k=1}^{\infty}P(X=k)=P(X\geq 1)=0.7\), whereas \(E[X]=7/3\). The correct tail sum uses \(P(X\geq k)\): an outcome \(X=3\) contributes to the terms for \(k=1,2,3\), so it is counted three times, as required by its value.
For continuous variables, a related error is integrating the density itself over thresholds and treating that as an expectation. The density integrates to one; the tail-integral formula instead integrates the survival probability \(P(X>t)\) over threshold \(t\). Also, the integer formula uses the discrete levels \(k=1,2,\ldots\), while the continuous formula integrates over every real threshold \(t\geq 0\). Keep the variable’s support and the tail event in view.
Check Your Understanding
Use the support of the variable to choose the appropriate tail formula.
- Why does the tail-sum formula start at \(k=1\) rather than \(k=0\)?
- If \(P(X\geq k)=0.4^k\) for every positive integer \(k\), calculate \(E[X]\).
- For a nonnegative continuous variable with \(P(X>t)=e^{-3t}\), calculate \(E[X]\).
- Explain why summing \(P(X=k)\) over positive integers does not generally give \(E[X]\).
- What does a divergent tail integral imply about the expectation of a nonnegative continuous variable?