nyuuzyou/wikis huggingface dataset:
Domain of Injection to Countable Set is Countable

== Theorem ==
Let $X$ be a set, and let $Y$ be a countable set.
Let $f: X \to Y$ be an injection.
Then $X$ is also countable.
== Proof ==
Since $Y$ is countable, there exists an injection $g: Y \to \N$.
From Composite of Injections is Injection, $g \circ f: X \to \N$ is also an injection.
That is, $X$ is countable.
Category:Injections
Category:Countable Sets
Domain of Real Natural Logarithm

== Theorem ==
Let $\ln$ be the natural logarithm function on the real numbers.
Then the domain of $\ln$ is the set of strictly positive real numbers:
:$\Dom \ln = \R_{>0}$
== Proof ==
By definition of natural logarithm:
:$\ln^{-1} = \exp$
From Exponential Tends to Zero and Infinity:
:$\Dom \exp = \openint {-\infty} {+\infty}$
:$\Img \exp = \openint 0 {+\infty}$
From Exponential is Strictly Increasing:
:$\exp$ is strictly increasing.
From Strictly Monotone Real Function is Bijective, $\exp: \R \to \R_{>0}$ is a bijection.
Thus:
:$\Dom \ln = \Img \exp$
and so $\Dom \ln = \R_{>0}$.
Category:Natural Logarithms
Domain of Real Square Function

== Theorem ==
The domain of the real square function is the entire set of real numbers $\R$.
== Proof ==
The operation of real multiplication is defined on all real numbers.
Thus:
:$\forall x \in \R: \exists y \in \R: x^2 = y$
Hence the result by definition of domain.
== Sources ==
* : $\S 2.2$: Functions
Category:Square Function
Domain of Real Square Root Function

== Theorem ==
The domain of the real square root function is the set of positive real numbers $\R_{\ge 0}$:
:$\set{x \in \R: x \ge 0}$
== Proof ==
From Square of Real Number is Non-Negative:
:$\forall x \in \R: x^2 \ge 0$
Hence the result by definition of domain.
== Sources ==
* : $\S 2.2$: Functions
Category:Real Functions
Domain of Relation is Image of Inverse Relation

== Theorem ==
Let $\RR \subseteq S \times T$ be a relation.
Let $\RR^{-1} \subseteq T \times S$ be the inverse of $\RR$.
Then:
:$\Dom \RR = \Img {\RR^{-1} }$
That is, the domain of a relation is the image of its inverse.
== Proof ==
By definition:
== Also see ==
* Image of Relation is Domain of Inverse Relation
== Sources ==
* : Chapter $0$: Relations
* : $\S 10$: Inverses and Composites
* : $\S 1.11$: Relations
Category:Relation Theory
Category:Inverse Relations
Domain of Relation is Subclass of Union of Union of Relation

== Theorem ==
Let $V$ be a basic universe.
Let $\RR \subseteq V \times V$ be a relation.
Let $\Dom \RR$ denote the domain of $\RR$.
Then:
:$\Dom \RR \subseteq \map \bigcup {\bigcup \RR}$
where $\bigcup \RR$ denotes the union of $\RR$.
== Proof ==
== Sources ==
* : Chapter $2$: Some Basics of Class-Set Theory: $\S 8$ Relations: Lemma $8.1$: $(1)$
Category:Class Union
Category:Relations
Category:Domain of Relation is Subclass of Union of Union of Relation
Domain of Small Relation is Small

== Theorem ==
Let $a$ be a small class.
Let $a$ also be a relation.
Then the domain of $a$ is small.
== Proof ==
Let $A$ equal:
:$\set {\tuple {\tuple {x, y}, x}: \tuple {x, y} \in a}$
Then, $A$ maps $a$ to its domain.
Thus, the domain of $a$ is the image of $a$ under $A$.
By Image of Small Class under Mapping is Small, the domain of $a$ is small.
== Sources ==
* : $\S 6.8 \ (2)$
Category:Zermelo-Fraenkel Class Theory
Domain of Union of Nest of Mappings is Union of Class of Domains

== Theorem ==
Let $N$ be a nest of mappings.
Let $\bigcup N$ denote the union of $N$.
Then:
:$\Dom {\bigcup N} = \ds \bigcup_{f \mathop \in N} \Dom f$
where $\Dom f$ denotes the domain of $f$.
== Proof ==
== Sources ==
* : Chapter $6$: Order Isomorphism and Transfinite Recursion: $\S 1$ A few preliminaries: Proposition $1.4$
Category:Domain of Union of Nest of Mappings is Union of Class of Domains
Category:Nests of Mappings
Domain of Union of Nest of Mappings is Union of Class of Domains/Proof

== Theorem ==
== Proof ==
From Union of Nest of Mappings is Mapping we have that $\bigcup N$ is a mapping.
Let $x \in \Dom {\bigcup N}$.
Then by definition of mapping:
:$\exists \tuple {x, y} \in \bigcup N$
Then by definition of union of class:
:$\exists f \subseteq \bigcup N: \tuple {x, y} \in f$
Hence:
:$\exists f \subseteq \bigcup N: x \in \Dom f$
That is:
:$x \in \ds \bigcup_{f \mathop \in N} \Dom f$
That is:
:$\Dom {\bigcup N} \subseteq \ds \bigcup_{f \mathop \in N} \Dom f$
Let $x \in \ds \bigcup_{f \mathop \in N} \Dom f$.
Then:
:$\exists f \subseteq \bigcup N: x \in \Dom f$
Then by definition of mapping:
:$\exists f \subseteq \bigcup N: \tuple {x, y} \in f$
Thus by definition of union of class:
:$\exists \tuple {x, y} \in \bigcup N$
It follows that:
:$x \in \Dom {\bigcup N}$
That is:
:$\ds \bigcup_{f \mathop \in N} \Dom f \subseteq \Dom {\bigcup N}$
Hence by definition of set equality:
:$\Dom {\bigcup N} = \ds \bigcup_{f \mathop \in N} \Dom f$
== Sources ==
* : Chapter $6$: Order Isomorphism and Transfinite Recursion: $\S 1$ A few preliminaries: Exercise $1.1$
Category:Domain of Union of Nest of Mappings is Union of Class of Domains
Domain of Variable/Examples/Arbitrary Example 1

== Example of Domain of Variable ==
Let $f$ be the function:
:$\map f x = \dfrac 1 {\sqrt {4 - x^2} }$
The domain of $f$ is the open interval $\openint {-2} 2$, that is, where $-2 < x < 2$.
== Sources ==
* : Chapter $1$: Real Numbers and Functions of a Real Variable: $1.3$ Functions of a Real Variable: $\text {(b)}$ Interval of Definition $(1)$
Category:Examples of Domains of Variables
Domain of Variable/Examples/Books on Library Shelf

== Example of Domain of Variable ==
Let $B$ be the number of books on a library shelf.
The domain of $B$ is the closed interval $\closedint 0 C$, where $C$ is the largest number of books that can be held on a shelf.
$B$ is a discrete variable.
== Sources ==
* : Chapter $1$: Variables and Graphs: Solved Problems: Variables: $1.2 \ \text{(b)}$
Category:Examples of Domains of Variables
Domain of Variable/Examples/Countries in Europe

== Example of Domain of Variable ==
Let $C$ be a country in Europe.
The domain of $C$ is the set $\set {\text {France}, \text {Germany}, \text {Spain}, \text {Italy}, \ldots}$
These can be represented numerically if desired, by assigning an integer to each of the countries in Europe, for example:
:$1: \text {France}$
:$2: \text {Germany}$
:$3: \text {Spain}$
:$4: \text {Italy}$
:$\vdots$
$C$ is a discrete variable.
== Sources ==
* : Chapter $1$: Variables and Graphs: Solved Problems: Variables: $1.2 \ \text{(e)}$
Category:Examples of Domains of Variables
Domain of Variable/Examples/Diameter of Sphere

== Example of Domain of Variable ==
Let $d$ be the diameter of a sphere.
The domain of $d$ is the open interval $\openint 0 \to$.
$d$ is a continuous variable.
== Sources ==
* : Chapter $1$: Variables and Graphs: Solved Problems: Variables: $1.2 \ \text{(d)}$
Category:Examples of Domains of Variables
Domain of Variable/Examples/Litres of Water in Washing Machine

== Example of Domain of Variable ==
Let $V$ be the number of litres of water in a washing machine.
The domain of $V$ is the closed interval $\closedint 0 C$, where $C$ is the capacity of the washing machine.
$V$ is a continuous variable.
== Sources ==
* : Chapter $1$: Variables and Graphs: Solved Problems: Variables: $1.2 \ \text{(a)}$
Category:Examples of Domains of Variables
Domain of Variable/Examples/Points on Pair of Dice

== Example of Domain of Variable ==
Let $S$ be the total number of points that are obtained when tossing a pair of dice.
The domain of $S$ is the set $\set {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}$.
$S$ is a discrete variable.
== Proof ==
The set of points that can be obtained on a single die is $\set {1, 2, 3, 4, 5, 6}$.
Hence the set of points that can be obtained on a pair of die is found by adding each of the elements of $\set {1, 2, 3, 4, 5, 6}$ with each of the elements of the same set.
== Sources ==
* : Chapter $1$: Variables and Graphs: Solved Problems: Variables: $1.2 \ \text{(c)}$
Category:Examples of Domains of Variables
Dominance Relation is Ordering

== Theorem ==
Let $S$ and $T$ be cardinals.
Let $S \preccurlyeq T$ denote that $S$ is dominated by $T$.
Let $\mathbb S$ be any set of cardinals.
Then the relational structure $\struct {\mathbb S, \preccurlyeq}$ is an ordered set.
That is, $\preccurlyeq$ is an ordering (at least partial) on $\mathbb S$.
== Proof ==
From the definition, a cardinal is a set, so standard set theoretic results apply.
So, checking in turn each of the criteria for an ordering:
=== Reflexivity ===
For any cardinal $S$, the identity mapping $I_S: S \to S$ is an injection.
Thus:
:$\forall S \in \mathbb S: S \preccurlyeq S$
So $\preccurlyeq$ is reflexive.
=== Transitivity ===
Let $S_1, S_2, S_3 \in \mathbb S$ such that $f: S_1 \to S_2$ and $g: S_2 \to S_3$ be injections.
From Composite of Injections is Injection, $g \circ f$ is an injection and so $S_1 \preccurlyeq S_3$.
So $\preccurlyeq$ is transitive.
=== Antisymmetry ===
Suppose $S \preccurlyeq T$ and $T \preccurlyeq S$.
Then from the Cantor-Bernstein-Schröder Theorem, $S \sim T$ and, as $S$ and $T$ are cardinals, $S = T$ by definition.
So $\preccurlyeq$ is antisymmetric on a set of cardinals.
Hence the result.
== Sources ==
* : Introduction to Set Theory: $2$. Set Theoretical Equivalence and Denumerability
* : $\S 8$: Theorem $8.2$ (Corollary)
Category:Set Theory
Category:Order Theory
Dominated Strategy may be Optimal

== Theorem ==
A dominated strategy of a game may be the optimal strategy for a player of that game.
== Proof ==
Consider the game defined by the following payoff table:
This has two solutions:
:$(1): \quad A: \tuple {1, 0}, B: \tuple {1, 0}$
:$(2): \quad A: \tuple {0, 1}, B: \tuple {1, 0}$
Thus both pure strategies for $A$ are optimal, but $A_1$ is dominated by $A_1$.
== Sources ==
* : Chapter $\text{I}$: An Outline of the Theory of Games: $3$
Category:Dominating Strategies
Category:Optimal Strategies
Doob's Maximal Inequality/Discrete Time

== Theorem ==
Let $\struct {\Omega, \Sigma, \sequence {\FF_n}_{n \mathop \ge 0}, \Pr}$ be a filtered probability space.
Let $\sequence {X_n}_{n \mathop \ge 0}$ be a non-negative $\sequence {\FF_n}_{n \mathop \ge 0}$-submartingale.
Let:
:$\ds X_n^\ast = \max_{0 \mathop \le k \mathop \le n} X_k$
where $\max$ is the pointwise maximum. 
Let $\lambda > 0$.
Then:
:$\lambda \map \Pr {X_n^\ast \ge \lambda} \le \expect {X_n}$
== Proof 1 ==
== Proof 2 ==
Category:Doob's Maximal Inequality
Doob's Maximal Inequality/Discrete Time/Proof 1

== Theorem ==
== Proof ==
Let:
:$\map T \omega = \inf \set {k \ge 0 : \map {X_k} \omega \ge \lambda} \wedge n$
for each $\omega \in \Omega$. 
From:
:Least Time at which Discrete-Time Adapted Stochastic Process equals or exceeds Real Number is Stopping Time
:Constant Function is Stopping Time
:Pointwise Minimum of Stopping Times is Stopping Time
we have: 
:$T$ is a stopping time with respect to $\sequence {\FF_n}_{n \mathop \ge 0}$.
Further:
:$T \le n$ 
Note that $X_T \ge \lambda$ :
:$X_k \ge \lambda$ for some $0 \le k \le n$.
This is equivalent to: 
:$\ds \sup_{0 \mathop \le k \mathop \le n} X_k = X_n^\ast \ge \lambda$ 
Let $\FF_T$ be the stopped $\sigma$-algebra associated with $T$. 
By Doob's Optional Stopping Theorem for Stopped Sigma-Algebra of Bounded Stopping Time: Discrete Time: Submartingale, we have:
:$\expect {X_n \mid \FF_T} \ge X_T$ almost surely.
From Adapted Stochastic Process at Stopping Time is Measurable with respect to Stopped Sigma-Algebra:
:$X_T$ is $\FF_T$-measurable. 
From Conditional Expectation of Measurable Random Variable, we have:
:$X_T = \expect {X_T \mid \FF_T}$ almost surely.
So, by Conditional Expectation is Linear we have:
:$\expect {X_n - X_T \mid \FF_T} \ge 0$ almost surely.
So from Condition for Conditional Expectation to be Almost Surely Non-Negative, we have:
:$\expect {X_n \cdot \chi_A} \ge \expect {X_T \cdot \chi_A}$
for all $A \in \FF_T$.
We can now calculate: 
Multiplying through $\lambda > 0$ allows us to conclude.
Category:Doob's Maximal Inequality
Doob's Maximal Inequality/Discrete Time/Proof 2

== Theorem ==
== Proof ==
Let $\lambda > 0 $ and $n \ge 0$.
Let:
:$E := \set {X^\ast _n \ge \lambda}$.
That is, $E$ is a disjoint union:
:$(1):\quad \ds E = \bigsqcup _{0 \mathop \le k \mathop \le n} E_k$
where:
:$\ds E_k := \set {X_k \ge \lambda} \cap \bigcap _{0 \mathop \le j \mathop \le k-1} \set {X_j < \lambda}$
By construction, we have:
:$\forall k \in \set {0, \ldots, n} : E_k \in \FF_k$
In particular:
where $\chi_A$ denotes the characteristic function of $A \subseteq \Omega$.
Therefore:
Category:Doob's Maximal Inequality
Doob's Optional Stopping Theorem

== Theorem ==
=== Discrete Time ===
== Also see ==
* Integrable Adapted Stochastic Process is Supermartingale iff Optional Stopping Theorem Holds establishes the converse to this result.
* Doob's Optional Stopping Theorem for Stopped Sigma-Algebra of Bounded Stopping Time generalises case $(1)$. 
Category:Doob's Optional Stopping Theorem
Category:Stopping Times
Category:Adapted Stochastic Processes
Doob's Optional Stopping Theorem/Discrete Time/Martingale

== Theorem ==
Let $\struct {\Omega, \Sigma, \sequence {\FF_n}_{n \ge 0}, \Pr}$ be a filtered probability space.
Let $\sequence {X_n}_{n \ge 0}$ be an $\sequence {\FF_n}_{n \ge 0}$-martingale.
Let $T$ be a stopping time with respect to $\sequence {\FF_n}_{n \ge 0}$.
Let:
:$\map {X_T} \omega = \map {X_{\map T \omega} } \omega \map {\chi_{\set {\omega \in \Omega : \map T \omega < \infty} } } \omega$
for each $\omega \in \Omega$.
Suppose one of the following conditions holds: 
:$(1) \quad$ $T$ is bounded
:$(2) \quad$ $T$ is finite almost surely, and there exists an integrable random variable $Y$ with $\size {X_n} \le Y$ for $n \in \Z_{\ge 0}$
:$(3) \quad$ $T$ is integrable, and there exists a real number $M > 0$ such that for each $n \in \Z_{\ge 0}$ we have $\size {X_{n + 1} - X_n} \le M$ almost surely.
Then $X_T$ is integrable and in particular:
:$\expect {X_T} = \expect {X_0}$
== Proof ==
From Adapted Stochastic Process is Martingale iff Supermartingale and Submartingale:
:$\sequence {X_n}_{n \ge 0}$ is a $\sequence {\FF_n}_{n \ge 0}$-supermartingale and $\sequence {\FF_n}_{n \ge 0}$-submartingale.
From Doob's Optional Stopping Theorem: Discrete Time: Supermartingale and Doob's Optional Stopping Theorem: Discrete Time: Submartingale:
:$\expect {X_T} \le \expect {X_0}$
and:
:$\expect {X_T} \ge \expect {X_0}$ 
respectively, with $X_T$ integrable.
So we have:
:$\expect {X_T} = \expect {X_0}$
in any of the three cases.
== Sources ==
* : $10.10$: Doob's Optional Stopping Theorem
Category:Doob's Optional Stopping Theorem
Category:Martingales
Doob's Optional Stopping Theorem/Discrete Time/Submartingale

== Theorem ==
Let $\struct {\Omega, \Sigma, \sequence {\FF_n}_{n \ge 0}, \Pr}$ be a filtered probability space.
Let $\sequence {X_n}_{n \ge 0}$ be an $\sequence {\FF_n}_{n \ge 0}$-submartingale.
Let $T$ be a stopping time with respect to $\sequence {\FF_n}_{n \ge 0}$.
Let:
:$\map {X_T} \omega = \map {X_{\map T \omega} } \omega \map {\chi_{\set {\omega \in \Omega : \map T \omega < \infty} } } \omega$
for each $\omega \in \Omega$.
Suppose one of the following conditions holds: 
:$(1) \quad$ $T$ is bounded
:$(2) \quad$ $T$ is finite almost surely, and there exists an integrable random variable $Y$ with $\size {X_n} \le Y$ for $n \in \Z_{\ge 0}$
:$(3) \quad$ $T$ is integrable, and there exists a real number $M > 0$ such that for each $n \in \Z_{\ge 0}$ we have $\size {X_{n + 1} - X_n} \le M$ almost surely.
Then $X_T$ is integrable and:
:$\expect {X_T} \ge \expect {X_0}$
== Proof ==
From Adapted Stochastic Process is Supermartingale iff Negative is Submartingale:
:$\sequence {-X_n}_{n \ge 0}$ is a $\sequence {\FF_n}_{n \ge 0}$-supermartingale. 
=== Case $(1)$ ===
Suppose that $T$ is bounded.
Then we apply Doob's Optional Stopping Theorem: Discrete Time: Supermartingale to obtain: 
:$\expect {-X_T} \le \expect {-X_0}$
and:
:$-X_T$ is integrable.
=== Case $(2)$ ===
Suppose that:
:$T$ is finite almost surely, and there exists an integrable random variable $Y$ with $\size {X_n} \le Y$ for $n \in \Z_{\ge 0}$
Then we have:
:$T$ is finite almost surely, and there exists an integrable random variable $Y$ with $\size {-X_n} \le Y$ for $n \in \Z_{\ge 0}$
So applying Doob's Optional Stopping Theorem: Discrete Time: Supermartingale to $\sequence {-X_n}_{n \ge 0}$ we obtain:
:$\expect {-X_T} \le \expect {-X_0}$
and:
:$-X_T$ is integrable.
=== Case $(3)$ ===
Suppose that: 
:$T$ is integrable, and there exists a real number $M > 0$ such that for each $n \in \Z_{\ge 0}$ we have $\size {X_{n + 1} - X_n} \le M$ almost surely.
Then we have:
:$T$ is integrable, and there exists a real number $M > 0$ such that for each $n \in \Z_{\ge 0}$ we have $\size {-X_{n + 1} - \paren {-X_n} } \le M$ almost surely.
So applying Doob's Optional Stopping Theorem: Discrete Time: Supermartingale to $\sequence {-X_n}_{n \ge 0}$ we obtain:
:$\expect {-X_T} \le \expect {-X_0}$
and:
:$-X_T$ is integrable.
In all three cases we have:
:$\expect {-X_T} \le \expect {-X_0}$
and so, by Expectation is Linear:
:$\expect {X_T} \ge \expect {X_0}$
We also have $-X_T$ is integrable in all three cases.
From Scalar Multiple of Integrable Function is Integrable Function, we then have that $X_T$ is integrable.
Category:Doob's Optional Stopping Theorem
Category:Submartingales
Doob's Optional Stopping Theorem for Stopped Sigma-Algebra of Bounded Stopping Time

== Theorem ==
=== Discrete Time ===
== Also see ==
* Doob's Optional Stopping Theorem
Category:Doob's Optional Stopping Theorem for Stopped Sigma-Algebra of Bounded Stopping Time
Category:Stopped Sigma-Algebras
Category:Stopping Times
Category:Adapted Stochastic Processes
Doob's Optional Stopping Theorem for Stopped Sigma-Algebra of Bounded Stopping Time/Discrete Time/Martingale

== Theorem ==
Let $\struct {\Omega, \Sigma, \sequence {\FF_n}_{n \ge 0}, \Pr}$ be a filtered probability space.
Let $\sequence {X_n}_{n \ge 0}$ be an $\sequence {\FF_n}_{n \ge 0}$-martingale.
Let $S$ and $T$ be bounded stopping times with respect to $\sequence {\FF_n}_{n \ge 0}$ and $S \le T$. 
Let $\FF_S$ be the stopped $\sigma$-algebra associated with $S$. 
Let $X_T$ and $X_S$ be $X$ at the stopping times $T$ and $S$.
Then: 
:$\expect {X_T \mid \FF_S} = X_S$ almost surely.
== Proof ==
From Adapted Stochastic Process is Martingale iff Supermartingale and Submartingale:
:$\sequence {X_n}_{n \ge 0}$ is a $\sequence {\FF_n}_{n \ge 0}$-supermartingale and $\sequence {\FF_n}_{n \ge 0}$-submartingale.
From Doob's Optional Stopping Theorem for Stopped Sigma-Algebra of Bounded Stopping Time: Discrete Time: Supermartingale and Doob's Optional Stopping Theorem for Stopped Sigma-Algebra of Bounded Stopping Time: Discrete Time: Submartingale, we have:
:$\expect {X_T \mid \FF_S} \le X_S$ almost surely and $\expect {X_t \mid \FF_S} \ge X_S$ almost surely
respectively. 
So:
:$\expect {X_T \mid \FF_S} = X_S$ almost surely.
Category:Doob's Optional Stopping Theorem for Stopped Sigma-Algebra of Bounded Stopping Time
Category:Martingales
Doob's Optional Stopping Theorem for Stopped Sigma-Algebra of Bounded Stopping Time/Discrete Time/Submartingale

== Theorem ==
Let $\struct {\Omega, \Sigma, \sequence {\FF_n}_{n \ge 0}, \Pr}$ be a filtered probability space.
Let $\sequence {X_n}_{n \ge 0}$ be an $\sequence {\FF_n}_{n \ge 0}$-submartingale.
Let $S$ and $T$ be bounded stopping times with respect to $\sequence {\FF_n}_{n \ge 0}$ and $S \le T$. 
Let $\FF_S$ be the stopped $\sigma$-algebra associated with $S$. 
Let $X_T$ and $X_S$ be $X$ at the stopping times $T$ and $S$.
Then: 
:$\expect {X_T \mid \FF_S} \ge X_S$ almost surely.
== Proof ==
From Adapted Stochastic Process is Supermartingale iff Negative is Submartingale:
:$\sequence {-X_n}_{n \ge 0}$ is a $\sequence {\FF_n}_{n \ge 0}$-supermartingale. 
From Doob's Optional Stopping Theorem for Stopped Sigma-Algebra of Bounded Stopping Time: Discrete Time: Supermartingale:
:$\expect {-X_T \mid \FF_S} \le -X_S$ almost surely.
From Expectation is Linear:
:$\expect {X_T \mid \FF_S} \ge X_S$ almost surely.
Category:Doob's Optional Stopping Theorem for Stopped Sigma-Algebra of Bounded Stopping Time
Category:Submartingales
Doob's Optional Stopping Theorem for Stopped Sigma-Algebra of Bounded Stopping Time/Discrete Time/Supermartingale

== Theorem ==
Let $\struct {\Omega, \Sigma, \sequence {\FF_n}_{n \ge 0}, \Pr}$ be a filtered probability space.
Let $\sequence {X_n}_{n \ge 0}$ be an $\sequence {\FF_n}_{n \ge 0}$-supermartingale.
Let $S$ and $T$ be bounded stopping times with respect to $\sequence {\FF_n}_{n \ge 0}$ and $S \le T$. 
Let $\FF_S$ be the stopped $\sigma$-algebra associated with $S$. 
Let $X_T$ and $X_S$ be $X$ at the stopping times $T$ and $S$.
Then: 
:$\expect {X_T \mid \FF_S} \le X_S$ almost surely.
== Proof ==
From Adapted Stochastic Process at Stopping Time is Measurable with respect to Stopped Sigma-Algebra:
:$X_S$ is $\FF_S$-measurable.
Then from Conditional Expectation of Measurable Random Variable:
:$\expect {X_S \mid \FF_S} = X_S$ almost surely.
Also from Doob's Optional Stopping Theorem: Discrete Time: Supermartingale:
:$X_T$ and $X_S$ are integrable.
Then from Conditional Expectation is Linear, we have:
:$\expect {X_T \mid \FF_S} \le X_S$ almost surely.
:
:$\expect {X_T - X_S \mid \FF_S} \le 0$ almost surely.
From Condition for Conditional Expectation to be Almost Surely Non-Negative, we aim to show that:
:$\expect {\paren {X_T - X_S} \cdot \chi_A} \le 0$ for each $A \in \FF_S$.
Suppose that $t \in \Z_{\ge 0}$ is such that: 
:$\map T \omega \le t$ for all $\omega \in \Omega$.
Then we have: 
Since $T \le t$, we have: 
:$\ds X_S + \sum_{k \mathop = 0}^\infty \paren {X_{k + 1} - X_k} \cdot \chi_{S \le k < T} = X_S + \sum_{k \mathop = 0}^{t - 1} \paren {X_{k + 1} - X_k} \cdot \chi_{S \le k < T}$
Then for $A \in \FF_S$ we have by Expectation is Linear: 
:$\ds \expect {X_T \cdot \chi_A} = \expect {X_S \cdot \chi_A} + \sum_{k \mathop = 0}^{t - 1} \expect {\paren {X_{k + 1} - X_k} \cdot \chi_{S \le k < T} \cdot \chi_A}$
By Characteristic Function of Intersection, we have: 
:$\chi_{S \le k < T} \cdot \chi_A = \chi_{\set {S \le k < T} \cap A}$
Since $A \in \FF_S$, we have: 
:$\set {S \le k} \cap A \in \FF_k$
We also have:
:$\set {T > k} = \set {T \le k - 1}^c \in \FF_{k - 1}$
Since $\sequence {\FF_n}_{n \ge 0}$ is a filtration:
:$\FF_{k - 1} \subseteq \FF_k$ 
So:
:$\set {S \le k < T} \cap A = \paren {\set {S \le k} \cap A} \cap \set {T > k} \in \FF_k$
Since $\sequence {X_n}_{n \ge 0}$ is a $\sequence {\FF_n}_{n \ge 0}$-supermartingale, we have: 
:$\expect {X_{k + 1} \mid \FF_k} \le X_k$
and so, from Condition for Conditional Expectation to be Almost Surely Non-Negative:
:$\expect {\paren {X_{k + 1} - X_k} \cdot \chi_A} \le 0$ 
for each $A \in \FF_k$.
So, we have:
:$\expect {\paren {X_{k + 1} - X_k} \cdot \chi_{S \le k < T} \cdot \chi_A} \le 0$
So:
:$\expect {X_T \cdot \chi_A} \le \expect {X_S \cdot \chi_A}$
that is:
:$\expect {\paren {X_T - X_S} \cdot \chi_A} \le 0$
for each $A \in \FF_S$, as required.
Category:Doob's Optional Stopping Theorem for Stopped Sigma-Algebra of Bounded Stopping Time
Category:Supermartingales
Dot Product/Examples/Arbitrary Example 1

== Example of Dot Product ==
Let $\cdot$ denote the dot product operator.
Let:
Then:
:$\mathbf A \cdot \mathbf B = -9$
Hence the angle between $\mathbf A$ and $\mathbf B$ is approximately $104.2 \degrees$.
== Proof ==
We have:
Hence:
== Sources ==
* : Chapter $1$ Vector Analysis $1.3$ Scalar or Dot Product: Example $1.3.1$
Category:Examples of Dot Product
Dot Product/Examples/Work Done

== Examples of Dot Product ==
Let $\mathbf F$ represent a force acting on a body $B$.
Let $\mathbf d$ denote the displacement effected on $B$ by $\mathbf F$.
Then the work done by $\mathbf F$ on $B$ is given by:
:$W = \mathbf F \cdot \mathbf d = \norm {\mathbf F} \norm {\mathbf d} \cos \theta$
where:
:$\cdot$ denotes dot product
:$\theta$ is the angle between the directions of $\mathbf F$ and $\mathbf d$.
== Sources ==
* : Chapter $\text {II}$: The Products of Vectors: $1$. General
* 
* 
Category:Examples of Dot Product