CONTENTS

Solution: The Sigma-Algebra Generated by an Indicator

Exercise: The Sigma-Algebra Generated by an Indicator

Part 1: σ(1A)\sigma(\mathbf{1}_A) explicitly

By definition, σ(X)={X−1(B):B∈B(R)}\sigma(X) = \{X^{-1}(B) : B \in \mathcal{B}(\mathbb{R})\}. Since XX only takes the values 00 and 11, every Borel set B⊆RB \subseteq \mathbb{R} falls into one of four cases depending on which of {0,1}\{0, 1\} it contains:

B∩{0,1}B \cap \{0,1\}X−1(B)X^{-1}(B)
∅\emptyset∅\emptyset
{1}\{1\}AA
{0}\{0\}AcA^c
{0,1}\{0, 1\}Ω\Omega

So the preimage of any Borel set under XX is one of exactly four subsets of Ω\Omega:

σ(1A)={∅, A, Ac, Ω}\sigma(\mathbf{1}_A) = \{\emptyset,\, A,\, A^c,\, \Omega\}

Part 2: size and degenerate cases

Generically ∣σ(1A)∣=4|\sigma(\mathbf{1}_A)| = 4.

The degenerate cases collapse:

  • If A=∅A = \emptyset: then X≡0X \equiv 0, Ac=ΩA^c = \Omega, and σ(X)={∅,Ω}\sigma(X) = \{\emptyset, \Omega\} — the trivial σ\sigma-algebra with 22 elements.
  • If A=ΩA = \Omega: then X≡1X \equiv 1, Ac=∅A^c = \emptyset, and again σ(X)={∅,Ω}\sigma(X) = \{\emptyset, \Omega\}.

In both degenerate cases XX is constant, carries no information, and its generated σ\sigma-algebra reduces to the trivial one. This is the sharpest possible illustration of "constant random variables generate the trivial σ\sigma-algebra" — they distinguish no outcomes.

Part 3: digital call

Take A={ST>K}A = \{S_T > K\}, so D=1AD = \mathbf{1}_A. By Part 1:

σ(D)={∅, {ST>K}, {ST≤K}, Ω}\sigma(D) = \{\emptyset,\, \{S_T > K\},\, \{S_T \leq K\},\, \Omega\}

Four events. An observer who knows the value of DD (but nothing else) knows whether the call finished in the money or not — and nothing finer. They cannot distinguish ST=K+1S_T = K + 1 from ST=2KS_T = 2K, nor ST=0S_T = 0 from ST=KS_T = K.

Contrast with σ(ST)\sigma(S_T). The σ\sigma-algebra generated by the full stock price is vastly richer:
σ(ST)={ST−1(B):B∈B(R)}={{ST∈B}:B∈B(R)}\sigma(S_T) = \{S_T^{-1}(B) : B \in \mathcal{B}(\mathbb{R})\} = \bigl\{ \{S_T \in B\} : B \in \mathcal{B}(\mathbb{R})\bigr\}
It contains every event of the form {ST∈B}\{S_T \in B\} for any Borel set B⊆RB \subseteq \mathbb{R} — including {a<ST≤b}\{a < S_T \leq b\} for any a<ba < b, so the observer can resolve the exact value of STS_T up to Borel-measurable resolution.

Since D=1{ST>K}D = \mathbf{1}_{\{S_T > K\}} is a (Borel-measurable) function of STS_T, we have σ(D)⊆σ(ST)\sigma(D) \subseteq \sigma(S_T) — strict inclusion, typically. The coarsening is substantial: σ(ST)\sigma(S_T) distinguishes infinitely many price levels; σ(D)\sigma(D) keeps just two bins.

Finance interpretation. A trader holding only a digital call observes DD alone at expiry. They know the outcome of the bet — in or out — but not the realised price. A delta-one investor holding the stock observes STS_T itself and has access to the full σ(ST)\sigma(S_T). This is the formal reason why exotic payoffs generally reveal less information than the underlying, and why pricing/hedging exotics requires careful bookkeeping of the available information filtration.

Takeaways

  • An indicator generates a 4-element σ\sigma-algebra (or 2 elements in the constant case). This is the smallest non-trivial σ\sigma-algebra you ever see generated by a random variable.
  • Functions of random variables generate coarser σ\sigma-algebras. If Y=g(X)Y = g(X) for some Borel gg, then σ(Y)⊆σ(X)\sigma(Y) \subseteq \sigma(X) — measuring YY gives at most the information of measuring XX, often strictly less.
  • Coarser σ\sigma-algebra ⇔\Leftrightarrow less information. This exercise makes precise what it means for a payoff like a digital call to "reveal only binary information" about the underlying.
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Solution — The Sigma-Algebra Generated by an Indicator | q4quant.studio