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Real Analysis for Quants
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Lessons

  • The Real Number System & Completeness8m
  • Sequences & Cauchy Convergence8m
  • Series & Convergence Tests8m
  • Power Series8m
  • Open, Closed & Compact Sets8m
  • Continuity & Uniform Continuity8m
  • Differentiability & the Mean Value Theorem8m
  • Taylor's Theorem with Remainder8m
  • The Riemann Integral8m
  • Lebesgue Measure & Measurable Sets8m
  • The Lebesgue Integral & Convergence Theorems8m
  • L^p Spaces8m
  • Metric Spaces & Completeness8m
  • Banach & Hilbert Spaces8m
  • The Contraction Mapping Theorem8m
  • Linear Operators & the Riesz Representation8m

MatheLinux — quantitative finance, taught rigorously.

Course content credited to Ibrahim Lanre Adedimeji.

← Real Analysis for Quants

The rational numbers feel solid until you try to measure the diagonal of a unit square. You ask for a number whose square is 222, and the rationals have nothing to offer: there is a gap exactly where you reach. Real analysis begins by repairing that gap, and the repair is a single structural assumption called completeness. Everything else in the course, from convergent sequences to power series used in option pricing, rests on it.

Ordered fields

The real numbers form an ordered field. The field axioms say we can add, subtract, multiply, and divide (except by zero) with the usual rules. The order axioms add a relation that is total, transitive, compatible with addition (if aaa is less than bbb then a+ca+ca+c is less than b+cb+cb+), and compatible with multiplication by positives. The rationals satisfy all of these. So do the reals. The order axioms alone cannot distinguish them, which is exactly why we need one more axiom.

Knowledge check

Q1. Why does the set of rationals with square less than 2 fail to have a supremum within the rational numbers?

Q2. The Archimedean property is most directly used to prove which statement?

Q3. If u is the supremum of a nonempty set S, which statement must hold?

Problems

P1

Let S be the set of values 3 + 2/n for natural numbers n = 1, 2, 3, ... . Compute the infimum of S. Give a decimal.

P2

Prove that between any two distinct real numbers there exists an irrational number.

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c
Q\mathbb{Q}Q

Bounds, supremum, infimum

Let SSS be a nonempty set of real numbers. A number uuu is an upper bound for SSS if every element sss of SSS satisfies sss at most uuu. The set is bounded above if such a uuu exists. The supremum, written sup⁡S\sup SsupS, is the least upper bound: an upper bound that is at most every other upper bound. Symmetrically the infimum inf⁡S\inf SinfS is the greatest lower bound.

The defining property of the supremum has two halves. First, sup⁡S\sup SsupS is an upper bound. Second, nothing smaller works: for any ε\varepsilonε greater than 000, there is an element sss of SSS with sss greater than sup⁡S−ε\sup S - \varepsilonsupS−ε. That second half is the workhorse in proofs.

The completeness axiom

Completeness: every nonempty subset of R that is bounded above has a supremum in R.\textbf{Completeness: every nonempty subset of } \mathbb{R} \textbf{ that is bounded above has a supremum in } \mathbb{R}.Completeness: every nonempty subset of R that is bounded above has a supremum in R.

This is the axiom Q\mathbb{Q}Q fails. Consider the set of rationals whose square is less than 222. It is bounded above (by 222, say) but its least upper bound would have to be 2\sqrt{2}2​, which is not rational. Inside R\mathbb{R}R the supremum exists; inside Q\mathbb{Q}Q it does not. The reals are, up to isomorphism, the unique complete ordered field.

The Archimedean property

Claim. For any real xxx there is a natural number nnn with nnn greater than xxx.

Proof sketch. Suppose not. Then the natural numbers N\mathbb{N}N are bounded above by xxx. By completeness N\mathbb{N}N has a supremum α\alphaα. Since α\alphaα is the least upper bound, α−1\alpha - 1α−1 is not an upper bound, so some natural number mmm satisfies mmm greater than α−1\alpha - 1α−1. But then m+1m + 1m+1 is a natural number with m+1m + 1m+1 greater than α\alphaα, contradicting that α\alphaα bounds N\mathbb{N}N. Hence no such xxx exists, proving the claim. ■\blacksquare■

An immediate consequence: for every ε\varepsilonε greater than 000 there is an nnn with 1/n1/n1/n less than ε\varepsilonε. Apply the property to x=1/εx = 1/\varepsilonx=1/ε.

Density of the rationals

Claim. Between any two reals aaa less than bbb there is a rational.

Derivation. Since b−ab - ab−a is positive, the Archimedean property gives an nnn with 1/n1/n1/n less than b−ab - ab−a, that is, n(b−a)n(b-a)n(b−a) greater than 111. Now choose the integer mmm to be the smallest integer greater than nanana. Then m−1m - 1m−1 is at most nanana, so mmm is at most na+1na + 1na+1, which is less than na+n(b−a)=nbna + n(b-a) = nbna+n(b−a)=nb. Combining, nanana less than mmm less than nbnbnb, hence aaa less than m/nm/nm/n less than bbb. The rational m/nm/nm/n lands strictly between aaa and bbb. ■\blacksquare■

Worked example

Let SSS be the set of all numbers 1−1/n1 - 1/n1−1/n for natural nnn, namely 0,1/2,2/3,3/4,…0, 1/2, 2/3, 3/4, \ldots0,1/2,2/3,3/4,…. We claim sup⁡S=1\sup S = 1supS=1. First, 111 is an upper bound: each 1−1/n1 - 1/n1−1/n is less than 111. Second, take any ε\varepsilonε greater than 000. By the Archimedean corollary pick nnn with 1/n1/n1/n less than ε\varepsilonε. Then the element 1−1/n1 - 1/n1−1/n exceeds 1−ε1 - \varepsilon1−ε. So no number below 111 can be an upper bound, and 111 is the least one. Notice 111 itself is not an element of SSS: a supremum need not be attained.

Why quants care

Every convergence result a quant relies on is downstream of completeness. The fixed point that defines an implied volatility, the limiting price of a binomial tree as steps go to infinity, the value of a perpetual annuity as an infinite sum, and the existence of an optimal portfolio on a closed bounded constraint set all depend on suprema and limits actually existing. Completeness is the guarantee that when a Cauchy iteration or a monotone bisection search ought to converge, there is a genuine real number waiting at the end rather than a hole. Without it, numerical methods would chase limits that may not exist.

Adapted from MIT OpenCourseWare (18.100 Real Analysis), CC BY-NC-SA.