advanced30 min studyLesson 194 in Mathematics

Expectation, integrability, and legitimate limit exchanges

Define expectations through positive and negative parts, connect moment assumptions to tail behavior, and choose the right convergence theorem.

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What you’ll understand

  • Determine when a signed expectation is finite, extended, or undefined.
  • Use tail probabilities to diagnose moment assumptions.
  • Select monotone or dominated convergence from stated hypotheses.

An expectation is a Lebesgue integral, so a numerical average should not be manipulated until its existence has been checked. This is especially important for rare, large outcomes: most observations may be close to zero while a small tail carries substantial average mass. Integrability explains when linearity produces finite answers, when limiting operations are legitimate, and why convergence of typical values can coexist with failure of convergence of means. The issue is control of tails, not merely the number of possible outcomes.

Positive and negative contributions

For a nonnegative random variable, its expectation is defined by increasing simple approximations and may equal positive infinity. A real variable splits into its positive and negative parts. Its signed expectation is their difference only when the difference is not infinity minus infinity. If exactly one part has infinite integral, an extended expectation exists; if both do, the signed expectation is undefined. Declaring a distribution symmetric does not repair this undefined subtraction by cancellation.

Integrability means finite expectation of the absolute value. It guarantees that both parts have finite integrals and that the usual finite-valued linearity and triangle inequality apply. Square integrability is stronger on a probability space, by Cauchy–Schwarz, but integrability need not imply a finite variance. These distinctions are worth checking before using a variance-based estimate: a valid finite mean may coexist with a tail so heavy that the second moment diverges.

X=X+−X−,∣X∣=X++X−,X∈L1 ⟺ E[X+]+E[X−]<∞X=X^+-X^-,\quad |X|=X^++X^-,\qquad X\in L^1\ \Longleftrightarrow\ \mathbb E[X^+]+\mathbb E[X^-]<\infty
The signed expectation is finite when X is integrable. An infinity-minus-infinity expression is not a defined expectation.↔ Scroll wide equations horizontally.

Read moments from tails

For a nonnegative variable, integrating its upper-tail probability over all nonnegative thresholds gives its expectation. This follows by writing its value as the length of an interval beneath it and applying Tonelli to the corresponding nonnegative indicator. The formula turns an integrability question into an area question. A slowly decaying tail can have infinite area even though the probability of exceeding any individual large threshold is small and tends to zero.

For example, a variable with density proportional to x to the power minus three on [1,infinity) has a finite first moment but an infinite second moment after normalization. Weighting its tail by another power of x changes which integral converges. The same analysis guides engineering models of large losses or service times, but a mathematical tail model must be stated explicitly. A finite data set cannot by itself establish that an unobserved tail has any particular asymptotic exponent.

E[X]=∫0∞P(X>t) dt(X≥0),P(X≥a)≤E[X]a(a>0)\mathbb E[X]=\int_0^\infty\mathbb P(X>t)\,dt\quad(X\ge0),\qquad \mathbb P(X\ge a)\le\frac{\mathbb E[X]}{a}\quad(a>0)
The tail formula permits an infinite value. Markov’s inequality is useful when the nonnegative mean on its right is finite.↔ Scroll wide equations horizontally.

Limits require a theorem, not a picture

Monotone convergence says that increasing nonnegative variables have expectations increasing to the expectation of their pointwise limit, allowing an infinite answer. Dominated convergence instead allows signed variables when they converge almost surely and are bounded in absolute value by one common integrable variable. That single dominator must work for the entire sequence. A bound depending on the index, or an almost-sure limit without tail control, does not satisfy the hypothesis.

The counterexample X_n=n times the indicator of U≤1/n for a uniform U makes the danger visible. Outside the null event U=0, the sequence eventually vanishes, yet every mean equals one. Increasing heights compensate for shrinking support. These variables are not dominated by an integrable common envelope: their supremum has a divergent tail integral. The lesson is not that expectations ignore typical values, but that a vanishing fraction of the probability space can retain all the integral mass.

Xn→X a.s.,∣Xn∣≤Y,E[Y]<∞⟹E[∣Xn−X∣]→0X_n\to X\ \text{a.s.},\quad |X_n|\le Y,\quad \mathbb E[Y]<\infty\quad\Longrightarrow\quad\mathbb E[|X_n-X|]\to0
This dominated-convergence formulation gives L1 convergence and hence convergence of means; the common integrable Y is essential.↔ Scroll wide equations horizontally.

A finite mean with an infinite variance

Suppose X has density f(x)=2/x³ on x≥1 and zero elsewhere. Verify normalization, compute its mean, and determine whether its second moment exists finitely.

  1. Integrate the density: the integral of 2x⁻³ from one to infinity is one, so it defines a probability law with no negative values.

  2. Multiply the density by x. The mean is the integral of 2x⁻² over the same interval, which equals two.

  3. For the second moment, multiplying by x² leaves the integral of 2/x. Its logarithmic growth diverges, so the second moment and the variance are infinite.

E[X]=∫1∞2x2 dx=2,E[X2]=∫1∞2x dx=∞\mathbb E[X]=\int_1^\infty\frac{2}{x^2}\,dx=2,\qquad \mathbb E[X^2]=\int_1^\infty\frac{2}{x}\,dx=\infty
Normalization and both moment integrals are computed from the stated density on [1,infinity).↔ Scroll wide equations horizontally.
The result

X is integrable with mean two but is not square-integrable. A calculation requiring a finite variance cannot be justified from its finite mean alone.

Explore the model

Interactive lab

Interactive lab · probability

A rare event can keep the same mean

Set n=10 and p=0.1, then compare n=20 and p=0.05. Predict that the exact success-count mean stays one because np is unchanged, even though the individual-trial success probability halves. These are two binomial laws, not the concentrating-spike sequence in the text.

A rare event can keep the same meanThese are exact binomial probabilities for independent trials with one fixed success probability. Playback tours the distribution; it does not resample outcomes. Current readings: Mean · np: 5; Variance · np(1−p): 2.5; Total probability: 1. All parameters are adjustable with the labeled controls below.03581000.070.150.220.3Successes kP(X = k)k = 0 · P = 9.77e-4
Exact probabilityHighlighted outcomeMean np
Mean · np
5
Variance · np(1−p)
2.5
Total probability
1
t = 0 s

These are exact binomial probabilities for independent trials with one fixed success probability. Playback tours the distribution; it does not resample outcomes. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forRead the means and variances separately. Equal means do not identify a distribution, and finite binomial plots do not establish dominated convergence for an arbitrary sequence.

Check your understanding

Which condition is sufficient for dominated convergence of signed variables X_n to an almost-sure limit X?

Try it yourself

Let U be uniform on [0,1] and X_n=n² times the indicator of U≤1/n². Find the almost-sure limit and every expectation. Can an integrable random variable dominate all X_n?

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

First fix a strictly positive value of U, then evaluate the height times the support probability. Apply dominated convergence to test the proposed envelope.

Compare with the solution

For every U>0, eventually 1/n²<U, so X_n becomes zero; the limit is zero almost surely. Each expectation is n²·(1/n²)=1. If an integrable common dominator existed, dominated convergence would force the means to approach zero, contradicting their constant value one. Therefore no such dominator exists. Almost-sure convergence alone has not controlled the moving rare tail.

Take it with you

  • Finite signed expectations require control of both positive and negative parts.
  • A finite first moment need not imply a finite second moment.
  • Limit exchanges need the hypotheses of a specific convergence theorem.

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