Jun 8, 2026
Here are some things to know when reading about the Simple Unified Model.
An algebra of sets on a set S is a collection of subsets of S that is closed under union and complement. We assume the empty set belongs to any algebra.
Exercise. Show algebras of sets are closed under intersection.
Hint. Use De Morgan’s law.
If the algebra \mathcal{A} is finite then the atoms of the algebra form a partition of S. A set A\in\mathcal{A} is an atom of \mathcal{A} if B\subseteq A and B\in\mathcal{A} imply B = \emptyset or B = A.
Exercise. Show A_s = \cap\{A\in\mathcal{A}\mid s\in A\} is an atom of \mathcal{A}.
For s\in S the atom containing s is the intersection of all sets in the algebra containing s, A_s = \cap\{A\in\mathcal{A}\mid s\in A\}.
Exercise. Show if B\subseteq A_s and B\in\mathcal{A} then either B=\emptyset or B = A_s.
Hint: If s\in B then A_s\subseteq B. If s\not\in B then s\in A_s\setminus B so A_s\subseteq A_s\setminus B\subseteq A_s. It follows A_s = A_s\setminus B and B = \emptyset.
Clearly S = \cup_{s\in S} A_s.
Exercise. Show A_s\cap A_t\not=\emptyset implies A_s = A_t.
Hint: Use A_s\subseteq A_t and A_t\subseteq A_s then apply the previous exercise.
A partition models partial information about a sample space S. No information corresponds to the singleton \{S\}. Full information corresponds to the partition of singletons \{\{s\}\mid s\in S\}. Partial information means knowing only which atom contains the element s.
For example, the partition \{[0,1/2),[1/2,1)\} of [0, 1) represents knowledge of the first digit in the base 2 expansion of s\in[0,1). If s\in[0,1/2) its first binary digit is 0 and if s\in[1/2,1) its first binary digit is 1.
A function f\colon S\to\boldsymbol{{R}} is measurable with respect to an algebra if and only if it is constant on atoms. In particular, it is a function on the atoms. This is useful for software implementation of measurable functions.
The set of bounded functions on a set S is {B(S) = \{f\colon S\to\boldsymbol{{R}}\mid \|f\| < \infty \}} where {\|f\| = \sup_{s\in S}|f(s)|} is the norm of f.
The vector space dual of bounded functions, B(S)^*, is isometrically isomorphic to ba(S), the space of finitely additive measures on S. Given a linear functional L\colon B(S)\to\boldsymbol{{R}} define {\lambda(A) = L1_A}. The algebraic equality {1_{A\cup B} = 1_A + 1_B - 1_{A\cap B}} shows \lambda is a finitely additive measure.
Exercise. Show \lambda(\emptyset) = 0.
Hint: 1_\emptyset(s) = 0 for all s\in S so 1_\emptyset is the zero function in B(S).
Exercise. If T\colon V\to W is a linear operator then T0_V = 0_W.
Hint First show v + v = v for any v\in V implies v = 0_v. Consider T(0_V) = T(0_V + 0_v).
Given a finitely additive measure \lambda\in ba(S) we can define a linear functional on the vector space of finite linear combinations {f = \sum_j a_j 1_{A_j}} by {Lf = \sum_j a_j \lambda(A_j)}. The A_j can be chosen to be disjoint.
Exercise. Write A\cup B as a disjoint union.
Hint: Use A\setminus B = \{a\in A\mid a\not\in B\}.
This makes the definition well-defined. The norm of \lambda is the supremum over all disjoint \{A_j\} of {\sum_j |\lambda(A_j)|}. A short argument shows {\|\lambda\| = \|L\|}. Elementary functions are uniformly dense in B(S) so we can extend L to an isometry from B(S) to ba(S).
For g\in B(S) define M_g\colon B(S)\to B(S) by {M_gf = fg}. Its adjoint {M_g^*\colon ba(S)\to ba(S)} defines multiplication of a measure by a bounded function {g\lambda = M_g^*\lambda}.
If P is a positive measure having mass 1 (a “probability” measure) and \mathcal{A} is an algebra of sets on the sample space then the conditional expectation {Y = E[X\mid\mathcal{A}]} is equivalent to {Y(P|_\mathcal{A}) = (XP)|_\mathcal{A}} where |_\mathcal{A} indicates restriction of measure. It is trivial to implement restriction of a function in software.
If F\colon X\to Y is a function between normed linear spaces the Fréchet derivative {DF\colon X\to\mathcal{B}(X,Y)} is defined by F(x + h) = F(x) + DF(x)h + o(h) for x,h\in X where \mathcal{B}(X,Y) is the set of bounded linear operators from X to Y. The “little o” notation means \|F(x + h) - F(x) - DF(x)h\|/\|h\| \to 0 as \|h\| \to 0. The function F can be approximated near x by the linear operator DF(x).
Exercise. If X is a Banach algebra and F(x) = x^2 show DF(x)h = xh + hx.
Hint: Don’t assume multiplication is commutative and use h^2 = o(h).
If we define left multiplication by x\in X, L_x\colon X\to X, as L_xy = xy and right multiplication by x as R_xy = yx then the above exercise shows {DF(x^2) = L_x + R_x}.
Exercise Show D(x^n) = \sum_{k = 0}^n L_x^{n-k} R_x^k.
For x\in\boldsymbol{{R}}^n define the linear functional x^*\colon\boldsymbol{{R}}^n\to\boldsymbol{{R}} by x^*(y) = y\cdot x.
Exercise. Show x\mapsto x^* is an isometric isomorphism from \boldsymbol{{R}}^n to its dual (\boldsymbol{{R}}^n)^* = \mathcal{B}(\boldsymbol{{R}}^n,\boldsymbol{{R}}).
Exercise Let \|x\|^p = \sum_i |x_i|^p for x = (x_1,\ldots,x_n)\in\boldsymbol{{R}}^n. Show D\|x\|^p = p\|x\|^{p-2}x^*.
Hint: Start with p = 2 and use \|x\|^p = \exp((p/2)\log\|x\|^2). Note x \mapsto \|x\|^p is a function \boldsymbol{{R}}^n\to\boldsymbol{{R}} so its Fréchet derivative goes from \boldsymbol{{R}}^n\to\mathcal{B}(\boldsymbol{{R}}^n,\boldsymbol{{R}}) = (\boldsymbol{{R}}^n)^*.