nyuuzyou/wikis huggingface dataset:
Expectation of Real-Valued Discrete Random Variable/Lemma

== Lemma ==
Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space. 
Let $X$ be a discrete real-valued random variable such that: 
:$\map X \omega \ge 0$ for all $\omega \in \Omega$. 
Then: 
:$\ds \int X \rd \Pr = \sum_{x \in \Img X} x \map \Pr {X = x}$
== Proof ==
Since $X$ is a discrete random variable, there exists a sequence $\sequence {x_i}_{i \in \N}$ of distinct real numbers such that: 
:$\Img X = \set {x_i : i \in \N}$
For each $i$, let: 
:$E_i = \set {X = x_i}$
Then, we can write: 
:$\ds \map X \omega = \sum_{i \mathop = 1}^\infty x_i \map {\chi_{E_i} } \omega$
for each $\omega \in \Omega$.
Since $X$ is $\Sigma$-measurable, we have: 
:$E_i$ is $\Sigma$-measurable for each $i$
from Measurable Functions Determine Measurable Sets. 
For each $n \in \N$, define $X_n : \Omega \to \R$ by: 
:$\ds \map {X_n} \omega = \sum_{i \mathop = 1}^n x_i \map {\chi_{E_i} } \omega$
for each $\omega \in \Omega$.
Then $X_n$ is a positive simple function for each $n$. 
So, we have: 
From Integral of Positive Measurable Function Extends Integral of Positive Simple Function, we have: 
:$\ds \map {I_\Pr} {X_n} = \int X_n \rd \Pr$
That is: 
:$\ds \int X_n \rd \Pr = \sum_{i \mathop = 1}^n x_i \map \Pr {X = x}$ 
We aim to apply the Monotone Convergence Theorem for Positive Simple Functions. 
We have, from the definition of infinite series: 
:$\ds \map X \omega = \lim_{n \mathop \to \infty} \sum_{i \mathop = 1}^n x_i \map {\chi_{E_i} } \omega = \lim_{n \mathop \to \infty} \map {X_n} \omega$
for each $\omega \in \Omega$, so: 
:$X_n \to X$ pointwise.
Finally, for each $n \in \N$, we have: 
for each $\omega \in \Omega$.
So:
:$\map {X_{n + 1} } \omega \ge \map {X_n} \omega$ for each $n \in \N$ and $\omega \in \Omega$.
So, from Monotone Convergence Theorem for Positive Simple Functions, we have: 
Category:Expectation of Discrete Random Variable
Expectation of Real-Valued Measurable Function composed with Absolutely Continuous Random Variable

== Theorem ==
Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space. 
Let $X$ be an absolutely continuous random variable on $\struct {\Omega, \Sigma, \Pr}$.
Let $P_X$ be the probability distribution of $X$.
Let $\map \BB \R$ be the Borel $\sigma$-algebra of $\R$.
Let $h : \R \to \R$ be a $\map \BB \R$-measurable function.
Let $f_X$ be a probability density function for $X$.
Let $\lambda$ be the Lebesgue measure on $\struct {\R, \map \BB \R}$.
Then $\map h X$ is integrable :
:$\ds \int_\R \size {\map h x} \map {f_X} x \rd \map \lambda x < \infty$
and in this case: 
:$\ds \expect {\map h X} = \int_\R \map h x \map {f_X} x \rd \map \lambda x$
== Proof ==
From Composition of Measurable Mappings is Measurable:
:$\map h X$ is $\Sigma$-measurable.
So:
:$\map h X$ is a real-valued random variable.
From Characterization of Integrable Functions, we have that:
:$\map h X$ is integrable $\size {\map h X}$ is integrable.
We have:
Then: 
:$\ds \int \size {\map h X} \rd \Pr$ $\ds \int_\R \size {\map h x} \map {f_X} x \rd \map \lambda x < \infty$
so:
:$\size {\map h X}$ is integrable $\ds \int_\R \size {\map h x} \map {f_X} x \rd \map \lambda x < \infty$.
So:
:$\map h X$ is integrable $\ds \int_\R \size {\map h x} \map {f_X} x \rd \map \lambda x < \infty$.
If $\map h X$ is integrable, then we have:
== Sources ==
* : $6.12$: The 'elementary formula' for expectation
Category:Expectation
Category:Absolutely Continuous Random Variables
Expectation of Sample Distribution Function

== Theorem ==
Let $S$ be a sample of ordinal data.
Let $\map {S_n} x$ denote the sample distribution function on $S$.
Then:
:$\expect {\map {S_n} x} = \map F x$
where:
:$\expect {\map {S_n} x}$ denotes the expectation of $\map {S_n} x$
:$\map F x$ denotes the cumulative distribution function of the population from which $S$ was taken.
== Proof ==
== Sources ==
* 
Category:Sample Distribution Functions
Category:Expectation
Expectation of Shifted Geometric Distribution

== Theorem ==
Let $X$ be a discrete random variable with the shifted geometric distribution with parameter $p$.
Then the expectation of $X$ is given by:
:$\expect X = \dfrac 1 p$
== Proof 1 ==
== Proof 2 ==
== Sources ==
* 
* : Appendix $13$: Probability distributions
* : Appendix $15$: Probability distributions
Category:Expectation of Shifted Geometric Distribution
Category:Expectation
Category:Geometric Distribution
Expectation of Shifted Geometric Distribution/Proof 1

== Theorem ==
== Proof ==
From the definition of expectation:
:$\expect X = \ds \sum_{x \mathop \in \Omega_X} x \map \Pr {X = x}$
By definition of shifted geometric distribution:
:$\expect X = \ds \sum_{k \mathop \in \Omega_X} k p \paren {1 - p}^{k - 1}$
Let $q = 1 - p$:
== Sources ==
* : $\S 2.4$: Expectation: Example $24$
Category:Expectation of Shifted Geometric Distribution
Expectation of Shifted Geometric Distribution/Proof 2

== Theorem ==
== Proof ==
From the Probability Generating Function of Shifted Geometric Distribution, we have:
:$\map {\Pi_X} s = \dfrac {p s} {1 - q s}$
where $q = 1 - p$.
From Expectation of Discrete Random Variable from PGF, we have:
:$\expect X = \map {\Pi'_X} 1$
We have:
Plugging in $s = 1$:
Hence the result.
Category:Expectation of Shifted Geometric Distribution
Expectation of Snedecor's F-Distribution

== Theorem ==
Let $n, m$ be strictly positive integers. 
Let $X \sim F_{n, m}$ where $F_{n, m}$ is Snedecor's $F$-distribution with $\tuple {n, m}$ degrees of freedom.
Then the expectation of $X$ is given by:
:$\expect X = \dfrac m {m - 2}$
for $m > 2$, and does not exist otherwise.
== Proof ==
Let $Y$ and $Z$ be independent random variables.
Let $Y \sim \chi^2_n$ where $\chi^2_n$ is the chi-squared distribution with $n$ degrees of freedom. 
Let $Z \sim \chi^2_m$ where $\chi^2_m$ is the chi-squared distribution with $m$ degrees of freedom.
Then, from Quotient of Independent Random Variables with $\chi^2$ Distribution Divided by Degrees of Freedom has Snedecor's $F$-Distribution:
:$\dfrac {Y / n} {Z / m} \sim F_{n, m}$
So we aim to compute:
:$\expect {\dfrac {Y / n} {Z / m} }$
Let $f_Y$ and $f_Z$ be the probability density functions of $Y$ and $Z$ respectively.
Let $f_{Y, Z}$ be the joint probability density function of $Y$ and $Z$. 
From Condition for Independence from Joint Probability Density Function, we have for each $y, z \in \R_{\ge 0}$: 
:$\map {f_{Y, Z} } {y, z} = \map {f_Y} y \map {f_Z} z$
We therefore have: 
Note that the integral converges: 
:$\ds \int_0^\infty z^{m / 2 - 2} e^{-z / 2} \rd z$
: 
:$\dfrac m 2 - 2 > -1$
That is: 
:$m > 2$
With that, we have: 
The integral: 
:$\ds \int_0^\infty y^{n / 2} e^{-y / 2} \rd y$
converges $n > -2$. 
This is ensured by the fact that $n \in \N$. 
With that, we have: 
So: 
Category:Snedecor's F-Distribution
Category:Expectation
Expectation of Square of Discrete Random Variable

== Theorem ==
Let $X$ be a discrete random variable whose probability generating function is $\map {\Pi_X} s$.
Then the square of the expectation of $X$ is given by the expression:
:$\expect {X^2} = \map { {\Pi_X}} 1 + \map { {\Pi_X}'} 1$
where $\map { {\Pi_X}} 1$ and $\map { {\Pi_X}'} 1$ denote the second and first derivative respectively of the PGF $\map {\Pi_X} s$ evaluated at $1$.
== Proof ==
From Derivatives of Probability Generating Function at One:
:$\map { {\Pi_X}} 1 = \expect {X \paren {X - 1} }$
and from Expectation of Discrete Random Variable from PGF:
:$\map { {\Pi_X}'} 1 = \expect X$
So:
== Sources ==
* : $\S 4.3$: Moments: $(19)$
Category:Expectation
Category:Probability Generating Functions
Expectation of Student's t-Distribution

== Theorem ==
Let $k$ be a strictly positive integer. 
Let $X \sim t_k$ where $t_k$ is the $t$-distribution with $k$ degrees of freedom.
Then the expectation of $X$ is equal to $0$ for $k > 1$, and does not exist otherwise.
== Proof == 
From the definition of the Student's t-Distribution, $X$ has probability density function:
:$\map {f_X} x = \dfrac {\map \Gamma {\frac {k + 1} 2} } {\sqrt {\pi k} \map \Gamma {\frac k 2} } \paren {1 + \dfrac {x^2} k}^{-\frac {k + 1} 2}$
with $k$ degrees of freedom for some $k \in \R_{>0}$.
From the definition of the expected value of a continuous random variable: 
:$\ds \expect X = \int_{-\infty}^\infty x \map {f_X} x \rd x$
So for $k > 1$:
When $k = 1$, we have the natural log evaluated at infinity where it is undefined.
Hence: 
Therefore, the expectation of $X$ does not exist:
== Sources ==
* : Appendix $15$: Probability distributions
Category:Expectation
Category:Student's t-Distribution
Expectation of Truncated Distribution

== Theorem ==
Let $X$ be a probability distribution.
Let $x_0$ be a given value of $X$.
Let $Y$ be a truncated distribution formed from $X$ by ignoring the part either to the left or right of $x_0$.
Then the expectation of $Y$ is dependent upon $X$ and $x_0$.
== Proof ==
== Sources ==
* 
* 
Category:Truncated Distributions
Category:Expectation
Expectation of Weibull Distribution

== Theorem == 
Let $X$ be a continuous random variable with the Weibull distribution with $\alpha, \beta \in \R_{> 0}$.
The expectation of $X$ is given by:
:$\expect X = \beta \, \map \Gamma {1 + \dfrac 1 \alpha}$
where $\Gamma$ is the Gamma function.
== Proof == 
From Raw Moment of Weibull Distribution, we have:
The $n$th raw moment $\expect {X^n}$ of $X$ is given by: 
:$\expect {X^n} = \beta^n \map \Gamma {1 + \dfrac n \alpha}$
Therefore, for $n = 1$ we have:
:$\expect X = \beta^1 \map \Gamma {1 + \dfrac 1 \alpha}$
Hence the result.
Category:Weibull Distribution
Category:Expectation
Expected-Utility Maximization Theorem

== Theorem ==
Let $P$ be a rational decision-maker.
Let $A$ be a set of possible moves available to $P$.
Then there exists a way of assigning values to a payoff function $u: A \to \R$
so that $P$ will always choose the move such that $u$ is a maximum.
== Proof ==
== Historical Note ==
== Sources ==
* : $1.1$ Game Theory, Rationality, and Intelligence
Category:Game Theory
Category:Named Theorems
Expected-Utility Maximization Theorem/Historical Note

== Historical Note on Expected-Utility Maximization Theorem ==
The idea behind the Expected-Utility Maximization Theorem goes back at least as far as , who raised the question in his of $1738$ (translated ).
It was proved by and in their of $1947$.
== Sources ==
* : $1.1$ Game Theory, Rationality, and Intelligence
Category:Historical Notes
Expected Value of Martingale is Constant in Time/Continuous Time

== Theorem ==
Let $\struct {\Omega, \Sigma, \sequence {\FF_t}_{t \ge 0}, \Pr}$ be a continuous-time filtered probability space.
Let $\sequence {X_t}_{t \ge 0}$ be a $\sequence {\FF_t}_{t \ge 0}$-martingale.
Then: 
:$\expect {X_t} = \expect {X_0}$
for each $t \in \hointr 0 \infty$. 
== Proof ==
From the definition of a continuous-time martingale, we have: 
:$\expect {X_t \mid \FF_0} = X_0$ almost surely
for each $t \in \hointr 0 \infty$. 
So:
:$\expect {\expect {X_t \mid \FF_0} } = \expect {X_0}$
From Expectation of Conditional Expectation, we have:
:$\expect {\expect {X_t \mid \FF_0} } = \expect {X_t}$
So we have: 
:$\expect {X_t} = \expect {X_0}$
for each $t \in \hointr 0 \infty$.
== Sources ==
* : $3.3$: Continuous Time Martingales and Supermartingales
Category:Expected Value of Martingale is Constant in Time
Expected Value of Martingale is Constant in Time/Discrete Time

== Theorem ==
Let $\struct {\Omega, \Sigma, \sequence {\FF_n}_{n \mathop \ge 0}, \Pr}$ be a discrete-time filtered probability space.
Let $\sequence {X_n}_{n \mathop \ge 0}$ be a martingale.
Then: 
:$\expect {X_n} = \expect {X_0}$
for each $n \in \Z_{\ge 0}$. 
== Proof ==
From Definition 2 of a discrete time martingale, we have:
:$\expect {X_n \mid \FF_0} = X_0$ almost surely.
So:
:$\expect {\expect {X_n \mid \FF_0} } = \expect {X_0}$
From Expectation of Conditional Expectation, we have:
:$\expect {\expect {X_n \mid \FF_0} } = \expect {X_n}$
So:
:$\expect {X_n} = \expect {X_0}$
for each $n \in \Z_{\ge 0}$.
Category:Expected Value of Martingale is Constant in Time
Expected Value of Submartingale Greater Than or Equal To Initial Expected Value

== 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 a submartingale.
Then: 
:$\expect {X_n} \ge \expect {X_0}$
for each $n \in \Z_{\ge 0}$. 
== Proof ==
From Definition 2 of a discrete time submartingale, we have:
:$\expect {X_n \mid \FF_0} \ge X_0$ almost surely.
So from Expectation is Monotone:
:$\expect {\expect {X_n \mid \FF_0} } \ge \expect {X_0}$
From Expectation of Conditional Expectation, we have:
:$\expect {\expect {X_n \mid \FF_0} } \ge \expect {X_n}$
So:
:$\expect {X_n} \ge \expect {X_0}$
Category:Submartingales
Expected Value of Submartingale is Increasing in Time/Continuous Time

== Theorem ==
Let $\struct {\Omega, \Sigma, \sequence {\FF_t}_{t \ge 0}, \Pr}$ be a continuous-time filtered probability space.
Let $\sequence {X_t}_{t \ge 0}$ be a $\sequence {\FF_t}_{t \ge 0}$-submartingale.
Let $t, s \in \hointr 0 \infty$ with $0 \le s < t$.
Then, we have: 
:$\expect {X_s} \le \expect {X_t}$
== Proof ==
From the definition of a submartingale, we have: 
:$\expect {X_t \mid \FF_s} \ge X_s$ almost surely.
From Expectation is Monotone, we have: 
:$\expect {\expect {X_t \mid \FF_s} } \ge \expect {X_s}$
From Expectation of Conditional Expectation, we have: 
:$\expect {X_s} \le \expect {X_t}$
== Sources ==
* : $3.3$: Continuous Time Martingales and Supermartingales
Category:Expected Value of Submartingale is Increasing in Time
Expected Value of Supermartingale Less Than or Equal To Initial Expected Value

== 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 a supermartingale.
Then: 
:$\expect {X_n} \le \expect {X_0}$
for each $n \in \Z_{\ge 0}$. 
== Proof ==
From Definition 2 of a discrete time supermartingale, we have:
:$\expect {X_n \mid \FF_0} \le X_0$ almost surely.
So from Expectation is Monotone:
:$\expect {\expect {X_n \mid \FF_0} } \le \expect {X_0}$
From Expectation of Conditional Expectation, we have:
:$\expect {\expect {X_n \mid \FF_0} } \le \expect {X_n}$
So:
:$\expect {X_n} \le \expect {X_0}$
Category:Supermartingales
Expected Value of Supermartingale is Decreasing in Time/Continuous Time

== Theorem ==
Let $\struct {\Omega, \Sigma, \sequence {\FF_t}_{t \ge 0}, \Pr}$ be a continuous-time filtered probability space.
Let $\sequence {X_t}_{t \ge 0}$ be a $\sequence {\FF_t}_{t \ge 0}$-supermartingale.
Let $t, s \in \hointr 0 \infty$ with $0 \le s < t$.
Then, we have: 
:$\expect {X_t} \le \expect {X_s}$
== Proof ==
From the definition of a supermartingale, we have: 
:$\expect {X_t \mid \FF_s} \le X_s$ almost surely.
From Expectation is Monotone, we have: 
:$\expect {\expect {X_t \mid \FF_s} } \le \expect {X_s}$
From Expectation of Conditional Expectation, we have: 
:$\expect {X_t} \le \expect {X_s}$ 
== Sources ==
* : $3.3$: Continuous Time Martingales and Supermartingales
Category:Expected Value of Supermartingale is Decreasing in Time
Experiment/Examples/Throwing a 6-Sided Die

== Example of Experiment ==
Let $\EE$ be the experiment of throwing a standard $6$-sided die.
:The sample space of $\EE$ is $\Omega = \set {1, 2, 3, 4, 5, 6}$.
:Various events can be identified:
:$(1): \quad$ The result is $3$:
::::The event space of $\EE$ is: $\Sigma = \set 3$.
:$(2): \quad$ The result is at least $4$:
::::The event space of $\EE$ is: $\Sigma = \set {\forall \omega \in \Omega: \omega \ge 4}$.
:$(3): \quad$ The result is a prime number:
::::The event space of $\EE$ is: $\Sigma = \set {2, 3, 5}$.
:The probability measure is defined as:
::$\forall \omega \in \Omega: \map \Pr \omega = \dfrac 1 6$
== Sources ==
* : $1$: Events and probabilities: $1.1$: Experiments with chance
Category:Examples of Experiments
Experiment/Examples/Tossing 2 Coins

== Example of Experiment ==
Let $\EE$ be the experiment of tossing $2$ coins.
The sample space of $\EE$ is:
:$\Omega = \set {\tuple {\mathrm H, \mathrm H}, \tuple {\mathrm H, \T}, \tuple {\T, \mathrm H}, \tuple {\T, \T} }$
where $\mathrm H$ denotes heads and $\T$ denotes tails.
Suppose we are interested only in whether the coins fall alike ($\mathrm A$) or different ($\mathrm D$).
Then the sample space of $\EE$ is:
:$\Omega = \set {\mathrm A, \mathrm D}$
== Sources ==
* : Chapter $2$: Probability and Discrete Sample Spaces: $2.2$ Sample spaces and events
Category:Examples of Experiments
Explicit Form for Generated C*-Algebra

== Theorem ==
Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a $\text C^\ast$-algebra.
Let $S \subseteq A$. 
Define:
:$S^\ast = \set {x^\ast : x \in A}$
Let $\map {\text C^\ast} S$ be the $\text C^\ast$-algebra generated by $S$. 
Then:
:$\ds \map {\text C^\ast} S = \span \set {x_1^{k_1} x_2^{k_2} \ldots x_n^{k_n} : x_1, \ldots, x_n \in S \cup S^\ast, \, k_1, \ldots, k_n \ge 0}^-$
where $\paren {\cdot}^-$ denotes closure in $\struct {A, \norm {\, \cdot \,} }$. 
== Proof ==
Let $C$ be the subalgebra generated by $S \cup S^\ast$.
From Explicit Form for Generated Subalgebra, we have:
:$\span \set {x_1^{k_1} x_2^{k_2} \ldots x_n^{k_n} : x_1, \ldots, x_n \in S, \, k_1, \ldots, k_n \ge 1}$
We show that $\map {\text C^\ast} S = C^-$.
Note that $S \cup S^\ast$ is self-adjoint.
From Subalgebra Generated by Self-Adjoint Set is Self-Adjoint, $C$ is self-adjoint.
From Closure of Self-Adjoint Subset of Banach *-Algebra is Self-Adjoint, $C^-$ is self-adjoint hence a $\ast$-subalgebra of $A$.
Hence from Closed *-Subalgebra of C*-Algebra is C*-Algebra, $C^-$ is a $\text C^\ast$-subalgebra of $A$ containing $S \cup S^\ast$. 
So $\map {\text C^\ast} S \subseteq C^-$.
To conclude we just need to show that $C^- \subseteq \map {\text C^\ast} S$.
By definition, $C$ is the $\subseteq$-minimal subalgebra with $S \subseteq C$. 
Hence $C \subseteq \map {\text C^\ast} S$.
So $C^- \subseteq \map {\text C^\ast} S^-$ from Set Closure Preserves Set Inclusion.
From Subspace of Complete Metric Space is Closed iff Complete, we have that $\map {\text C^\ast} S$ is closed in $A$. 
So from Set is Closed iff Equals Topological Closure we have $C^- \subseteq \map {\text C^\ast} S$.
Hence:
:$C^- = \span \set {x_1^{k_1} x_2^{k_2} \ldots x_n^{k_n} : x_1, \ldots, x_n \in S, \, k_1, \ldots, k_n \ge 1}^- = \map {\text C^\ast} S$
Category:C*-Algebras
Explicit Form for Generated Subalgebra

== Theorem ==
Let $K$ be a field. 
Let $A$ be an algebra over $K$.
Let $S \subseteq A$ be a non-empty set. 
Let $K \sqbrk S$ be the subalgebra generated by $A$.
Then:
:$\ds K \sqbrk S = \span \set {x_1^{k_1} x_2^{k_2} \ldots x_n^{k_n} : x_1, \ldots, x_n \in S, \, k_1, \ldots, k_n \ge 1}$
== Proof ==
Let:
:$B = \span \set {x_1^{k_1} x_2^{k_2} \ldots x_n^{k_n} : x_1, \ldots, x_n \in S, \, k_1, \ldots, k_n \ge 1}$
First, for each $x \in S$ we have:
:$x = {\mathbf 1}_K x^1 \in \span \set {x_1^{k_1} x_2^{k_2} \ldots x_n^{k_n} : x_1, \ldots, x_n \in S, \, k_1, \ldots, k_n \ge 1}$
Hence $S \subseteq B$.
We show that $B$ is a subalgebra.
From Linear Span is Linear Subspace, $B$ is a linear subspace of $A$.
Now let $x, y \in B$.
Since $B$ is a linear subspace, it suffices to consider $x, y$ of the form:
:$x = x_1^{k_1} x_2^{k_2} \ldots x_n^{k_n}$
and:
:$y = y_1^{s_1} y_2^{s_2} \ldots y_m^{s_m}$
We have:
:$x y = x_1^{k_1} x_2^{k_2} \ldots x_n^{k_n} y_1^{s_1} y_2^{s_2} \ldots y_m^{s_m}$
By relabelling $x_{n + t} = y_t$ and $k_{n + t} = s_t$ for $1 \le t \le m$ we obtain that $x y \in B$.
Hence $B$ is subalgebra of $A$.
Since $S \subseteq B$, we obtain $K \sqbrk S \subseteq B$ from the definition of the subalgebra generated by $A$.
Conversely, note that for each $x_1, \ldots, x_n \in S$ we have $x_1^{k_1} x_2^{k_2} \ldots x_n^{k_n} \in S$ from the definition of a subalgebra.
Since $K \sqbrk S$ is a linear subspace of $A$, we have $B = \span \set {x_1^{k_1} x_2^{k_2} \ldots x_n^{k_n} : x_1, \ldots, x_n \in S, \, k_1, \ldots, k_n \ge 1} \subseteq K \sqbrk S$.
Hence $B = K \sqbrk S$. 
Category:Algebras
Explicit Set Definition/Examples

== Examples of Explicit Set Definition ==
Some examples of explicit set definition are:
=== Example 1 ===
=== Example 2 ===
=== Example 3 ===
=== Example 4 ===
=== Example 5 ===
=== Example 6 ===
=== Example 7 ===
=== Example 8 ===
=== Example 9 ===
=== Example 10 ===
=== Example 11 ===
Category:Examples of Sets
Explicit Set Definition/Examples/Example 1

== Example of Explicit Set Definition ==
An example of explicit set definition is:
:$A := \set {\dfrac 1 2, 1, \sqrt 2, e, \pi}$
This denotes the set whose elements are the numbers $\dfrac 1 2$, $1$, $\sqrt 2$, $e$ and $\pi$.
== Sources ==
* : $\S 1$: Real Numbers: $\S 1.1$: Set Notation
Category:Examples of Sets
Explicit Set Definition/Examples/Example 2

== Example of Explicit Set Definition ==
An example of explicit set definition is:
:$B := \set {\textrm {Romeo}, \textrm {Juliet} }$
This denotes the set whose elements are (presumably) the characters and .
== Sources ==
* : $\S 1$: Real Numbers: $\S 1.1$: Set Notation
Category:Examples of Sets
Explicit Set Definition/Examples/Example 5

== Example of Explicit Set Definition ==
An example of explicit set definition is:
:$V = \set {\textrm {red}, \textrm {orange}, \textrm {yellow}, \textrm {green}, \textrm {blue}, \textrm {indigo}, \textrm {violet} }$
This denotes the set whose elements are the colours of the rainbow.
Category:Examples of Sets
Explicit Set Definition/Examples/Example 6

== Example of Explicit Set Definition ==
An example of explicit set definition is:
:$W := \set {a, b, c}$
which can also be written as:
:$W := \set {b, a, c}$
or:
:$W := \set {c, b, a}$
or even as:
:$W := \set {a, a, b, c, a, c, c, b, a}$
== Sources ==
* : $\S 1.1$. Sets: Example $4$
* : $\S 1$. Sets; inclusion; intersection; union; complementation; number systems
* 
* 
Category:Examples of Sets
Exponent Base of One

== Theorem ==
Let $x$ be an ordinal.
Then:
:$1^x = 1$
== Proof ==
The proof shall proceed by Transfinite Induction on $x$.
=== Basis for the Induction ===
This proves the basis for the induction.
=== Induction Step ===
The inductive hypothesis supposes that $1^x = 1$ for some $x$.
This proves the induction step.
=== Limit Case ===
The inductive hypothesis supposes that $\forall y \in x: 1^y = 1$.
It follows that:
This proves the limit case.
== Sources ==
* : $\S 8.31$
Category:Ordinal Arithmetic
Exponent Combination Laws

== Theorem ==
Let $a, b \in \R_{>0}$ be strictly positive real numbers.
Let $x, y \in \R$ be real numbers.
Let $a^x$ be defined as $a$ to the power of $x$.
Then:
=== Product of Powers ===
=== Power of Product ===
=== Negative Power ===
=== Power of Power ===
=== Quotient of Powers ===
=== Power of Quotient ===
=== Negative Power of Quotient ===
== Also known as ==
== Also see ==
* Powers of Group Elements
== Sources ==
* 
* 
Category:Exponent Combination Laws
Category:Exponents
Category:Powers
Category:Analysis
Exponent Combination Laws/Negative Power

== Theorem ==
Let $a \in \R_{>0}$ be a strictly positive real number.
Let $x \in \R$ be a real number.
Let $a^x$ be defined as $a$ to the power of $x$.
Then:
: $a^{-x} = \dfrac 1 {a^x}$
== Proof ==
== Sources ==
* : $\S 7$: Laws of Exponents: $7.5$
* : $\S 14.7 \ (1) \ \text{(iii)}$
* : $\S 1.2.2$: Numbers, Powers, and Logarithms
* 
* 
* : $\S 13$: Laws of Exponents: $13.5.$
Category:Exponent Combination Laws
Exponent Combination Laws/Negative Power of Quotient

== Theorem ==
Let $a, b \in \R_{>0}$ be (strictly) positive real numbers.
Let $x \in \R$ be a real number.
Let $a^x$ be defined as $a$ to the power of $x$.
Then:
:$\paren {\dfrac a b}^{-x} = \paren {\dfrac b a}^x$
== Proof ==
Category:Exponent Combination Laws
Exponent Combination Laws/Power of Power

== Theorem ==
Let $a \in \R_{>0}$ be a (strictly) positive real number.
Let $x, y \in \R$ be real numbers.
Let $a^x$ be defined as $a$ to the power of $x$.
Then:
:$\paren {a^x}^y = a^{x y}$
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $\S 2$. Geometrical Representations
* : $\S 7$: Laws of Exponents: $7.3$
* : $\S 1.2.2$: Numbers, Powers, and Logarithms
* 
* 
* : $\S 13$: Laws of Exponents: $13.3.$
* 
Category:Exponent Combination Laws