Quant Academy Math · Probability · Quant Finance
Phase 7 - Lesson 7.2

Expectation, Moments, and Key Inequalities

Expectation as the Lebesgue integral of a random variable, its moments, and the Markov, Chebyshev, Jensen, and Cauchy-Schwarz inequalities that power all of probability.

⏱ 50 min● Advanced🔗 Prereqs: 7.1
↖ Phase 7 hub
Builds on: 7.1 gave laws, densities, and CDFs.
Leads to: Chebyshev drives the weak law (7.5); Jensen underlies risk-averse pricing and the CLT scaling.

Learning Objectives

Click a status chip to cycle: Not started → In progress → Studied → Practiced → Needs review → Mastered.

Key Vocabulary

Expectation
\(\E[X]=\int_\Omega X\,d\Prob=\int_\R x\,d\mu_X\); the Lebesgue integral / probability-weighted average.
Variance
\(\Var(X)=\E[(X-\E X)^2]=\E[X^2]-(\E X)^2\); mean-squared spread.
Moment
\(\E[X^k]\) (raw) or \(\E[(X-\mu)^k]\) (central); the \(k\)-th shape descriptor.
Markov inequality
For \(X\ge0\), \(\Prob(X\ge a)\le \E[X]/a\).
Chebyshev inequality
\(\Prob(|X-\mu|\ge k\sigma)\le 1/k^2\); a distribution-free spread bound.
Jensen inequality
For convex \(\varphi\), \(\varphi(\E X)\le \E[\varphi(X)]\); reversed for concave \(\varphi\).

Intuition & Motivation

Intuition
Expectation is just the integral \(\int X\,d\Prob\): a weighted average where the weights are probabilities. The inequalities are the reason probability is useful without knowing the full distribution. Markov says a non-negative quantity rarely lands far above its mean; Chebyshev refines this using the variance; Jensen tells you how averaging interacts with curvature; Cauchy-Schwarz bounds correlation. These four crude-looking bounds are the load-bearing walls of the laws of large numbers, concentration, and risk analysis.

Expectation and moments

Definition - Expectation
For a random variable \(X\) on \((\Omega,\F,\Prob)\), \(\E[X]=\int_\Omega X\,d\Prob\), provided \(\E|X|\lt \infty\). By the change-of-variables (law) formula,
\[\E[g(X)]=\int_\R g(x)\,d\mu_X(x)=\int_\R g(x)f_X(x)\,dx \ \text{ or }\ \sum_x g(x)p_X(x).\] (7.3)

Expectation is linear: \(\E[aX+bY]=a\E X+b\E Y\), always (independence not required). The variance is the second central moment,

\[\Var(X)=\E[(X-\E X)^2]=\E[X^2]-(\E X)^2\ \ge 0.\] (7.4)

Higher central moments describe shape: the third gives skewness (asymmetry), the fourth kurtosis (tail heaviness) - both central to modeling fat-tailed asset returns.

The four inequalities

Theorem - Markov
If \(X\ge0\) and \(a\gt 0\), then \(\Prob(X\ge a)\le \E[X]/a\).
Proof
Since \(X\ge a\,\ind\{X\ge a\}\), take expectations: \(\E[X]\ge a\,\Prob(X\ge a)\). Divide by \(a\).
Theorem - Chebyshev
For any \(X\) with mean \(\mu\), variance \(\sigma^2\lt \infty\), and \(k\gt 0\): \(\Prob(|X-\mu|\ge k\sigma)\le 1/k^2\).
Proof
Apply Markov to the non-negative variable \((X-\mu)^2\) at level \(a=k^2\sigma^2\): \(\Prob((X-\mu)^2\ge k^2\sigma^2)\le \E[(X-\mu)^2]/(k^2\sigma^2)=1/k^2\).
Theorem - Jensen
If \(\varphi\) is convex and \(\E|X|,\E|\varphi(X)|\lt \infty\), then \(\varphi(\E X)\le \E[\varphi(X)]\). Concave \(\varphi\) reverses the inequality.
Theorem - Cauchy-Schwarz
\(\big(\E[XY]\big)^2\le \E[X^2]\,\E[Y^2]\), with equality iff \(X,Y\) are linearly dependent a.s. This bounds covariance: \(|\Cov(X,Y)|\le \sqrt{\Var X\,\Var Y}\), i.e. \(|\rho|\le1\).
Worked Example - Chebyshev bounds a tail with no distributional assumption
1
Let \(X\) have mean \(\mu=100\) and standard deviation \(\sigma=10\); the law is otherwise unknown.
2
Bound \(\Prob(|X-100|\ge 30)\). Here \(30=3\sigma\), so \(k=3\).
3
Chebyshev: \(\Prob(|X-100|\ge 30)\le 1/3^2=1/9\approx 0.111\).
4
For a normal law the true value is \(\approx 0.0027\): Chebyshev is loose but requires nothing beyond the variance - that generality is the point.
5
Its looseness is exactly why it still suffices to prove the weak law of large numbers in 7.5.

Interactive: convexity and Jensen

Plot a convex function and compare \(\varphi(\E X)\) with \(\E[\varphi(X)]\). The gap is the Jensen premium - the mathematical origin of why volatility has value.

Common Mistakes to Avoid
  • Assuming \(\E[g(X)]=g(\E X)\). This holds only for affine \(g\); Jensen shows the gap for convex/concave \(g\).
  • Using \(\Var(X+Y)=\Var X+\Var Y\) without independence - the cross term \(2\Cov(X,Y)\) is generally present.
  • Applying Markov to a variable that can be negative; it requires \(X\ge0\).
  • Reading Chebyshev as tight; it is a worst-case bound and is usually very conservative.
Quant Practitioner Tips
  • A concave utility plus Jensen is the whole content of risk aversion: \(\E[u(W)]\le u(\E W)\).
  • For option payoffs (convex), Jensen implies more dispersion in the underlying raises expected payoff - vega is positive.
  • When you know only a mean and variance, Chebyshev gives an honest, if loose, guarantee.

Practice this in the Euler Lab

Computational problems that exercise exactly this technique. Each opens in the Euler Lab with a Python workbench, a progressive hint ladder, and answer checking. Tier A/B run at full scale in the browser.

Warm-up:
#59 XOR Decryption (4%, tier A) #79 Passcode Derivation (7%, tier A) #102 Triangle Containment (8%, tier A) #54 Poker Hands (9%, tier A)

Applied:
#938 Exhausting a Colour (16%, tier B) #816 Shortest Distance Among Points (17%, tier B) #323 Bitwise-OR Operations on Random In (18%, tier B)

Challenge:
#375 Minimum of Subsequences (41%, tier C) #666 Polymorphic Bacteria (41%, tier C)

124 Project Euler problems in total are mapped to this lesson. Open the Euler Lab to filter them all.

Knowledge Check

Q1 Easy
Markov’s inequality \(\Prob(X\ge a)\le\E[X]/a\) requires which assumption?
\(X\) is normal
\(X\ge 0\)
\(X\) has finite variance
\(X\) is discrete
Q2 Medium
For a convex function \(\varphi\), Jensen’s inequality states:
\(\varphi(\E X)\ge\E[\varphi(X)]\)
\(\varphi(\E X)\le\E[\varphi(X)]\)
\(\varphi(\E X)=\E[\varphi(X)]\)
No general relation holds
Q3 Medium
A variable has mean 50 and standard deviation 5. The best distribution-free bound on \(\Prob(|X-50|\ge20)\) is:
\(\le 1/4\)
\(\le 1/16\)
\(\le 1/2\)
exactly 0.05

Practical Exercise

Let \(X\) be exponential with rate \(\lambda\), so \(f_X(x)=\lambda e^{-\lambda x}\) on \(x\ge0\). (a) Compute \(\E[X]\) and \(\Var(X)\). (b) Use Markov to bound \(\Prob(X\ge 3/\lambda)\) and compare to the exact value.

▶ Show full solution

(a) \(\E[X]=\int_0^\infty x\lambda e^{-\lambda x}dx=1/\lambda\). \(\E[X^2]=\int_0^\infty x^2\lambda e^{-\lambda x}dx=2/\lambda^2\), so \(\Var(X)=2/\lambda^2-1/\lambda^2=1/\lambda^2\).

(b) Markov with \(a=3/\lambda\): \(\Prob(X\ge3/\lambda)\le \E[X]/a=(1/\lambda)/(3/\lambda)=1/3\).

Exact: \(\Prob(X\ge3/\lambda)=e^{-\lambda\cdot3/\lambda}=e^{-3}\approx0.0498\). Markov’s \(1/3\) is valid but very loose - typical of a mean-only bound.

After the reveal, answer for yourself: How much does the bound tighten if you switch to Chebyshev using the variance you computed?

Lesson Summary

Expectation is the Lebesgue integral of a random variable and is always linear; variance and higher moments describe spread and shape. Markov and Chebyshev turn moment information into tail bounds without knowing the full law, while Jensen and Cauchy-Schwarz govern how expectation interacts with convexity and correlation. These four inequalities are the analytic engine behind the limit theorems.

Formula Sheet Additions

Chebyshev
\[\Prob(|X-\mu|\ge k\sigma)\le 1/k^2\]
Distribution-free spread bound from the variance.
Jensen
\[\varphi\ \text{convex}\Rightarrow \varphi(\E X)\le \E[\varphi(X)]\]
Curvature makes averaging inputs an underestimate.
Error Log Checklist
  • Did I confirm \(X\ge0\) before invoking Markov?
  • Did I add the covariance term when computing \(\Var(X+Y)\)?
  • Did I get the direction of Jensen right for convex vs concave?

Retrieval Practice

Close the lesson and answer from memory before checking. This is deliberate, effortful recall - the single highest-yield study action.

▶ Show retrieval prompts & answers
Q: Prove Chebyshev from Markov in one line.
A: Apply Markov to \((X-\mu)^2\ge0\) at level \(k^2\sigma^2\): \(\Prob((X-\mu)^2\ge k^2\sigma^2)\le \E[(X-\mu)^2]/(k^2\sigma^2)=1/k^2\).
Q: Why is \(\E[u(W)]\le u(\E W)\) for a concave utility?
A: Jensen for concave \(u\): the certainty of the mean is preferred to the gamble, which is exactly risk aversion.
Q: State Cauchy-Schwarz and its correlation consequence.
A: \((\E[XY])^2\le\E[X^2]\E[Y^2]\); hence \(|\Cov(X,Y)|\le\sqrt{\Var X\,\Var Y}\), i.e. \(|\rho|\le1\).

Flashcards

Click to flip. These feed the site-wide spaced-repetition queue.

Markov inequality
\(X\ge0\Rightarrow \Prob(X\ge a)\le \E[X]/a\).
Variance identity
\(\Var(X)=\E[X^2]-(\E X)^2\).

Completion Checklist

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Source References

This lesson synthesizes and paraphrases concepts from the sources below. No copyrighted text, problem sets, or solutions are reproduced. Return to the originals for full depth.

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