nyuuzyou/wikis huggingface dataset:
Rules for Manipulating Summations

== Theorem ==
Let $R: \Z \to \set {\T, \F}$ and $S: \Z \to \set {\T, \F}$ and so on be propositional functions on the set of integers $\Z$.
Let $\ds \sum_{\map R i} x_i$ denote a summation over $R$.
Let $\pi$ be a permutation on the fiber of truth of $R$.
=== Change of Index Variable of Summation ===
=== Permutation of Indices of Summation ===
=== General Distributivity Theorem ===
=== Exchange of Order of Summation ===
=== Exchange of Order of Summation with Dependency on Both Indices ===
=== Sum of Summations over Overlapping Domains ===
Category:Summations
Rules of Quality

== Theorem ==
=== Conclusion of Valid Categorical Syllogism is Negative iff one Premise is Negative ===
=== No Valid Categorical Syllogism contains two Negative Premises ===
== Also see ==
* Rules of Quantity
== Historical Note ==
== Sources ==
* : Chapter $4$: The Predicate Calculus $2$: $4$ The Syllogism: Exercises: $\text{B}$: Rules of Quality
Category:Rules of Quality
Category:Categorical Syllogisms
Rules of Quantity

== Theorem ==
=== Distributed Term of Conclusion of Valid Categorical Syllogism is Distributed in Premise ===
=== Middle Term of Valid Categorical Syllogism is Distributed at least Once ===
== Also see ==
* Rules of Quality
== Historical Note ==
== Sources ==
* : Chapter $4$: The Predicate Calculus $2$: $4$ The Syllogism: $\text{A}$: Rules of Quantity
Category:Rules of Quantity
Category:Distributed Terms of Categorical Syllogisms
Run (Statistics)/Examples/Arbitrary Example 1

== Examples of Runs (Statistics) ==
Consider the following sequence of observations from the integer interval $\closedint 0 9$:
:$\tuple {2, 3, 4, 2, 7, 6, 3, 0, 3, 2, 1}$
:The numbers $2, 3, 4$ constitute an up run
:The numbers $7, 6, 3, 0$ constitute a down run
:The numbers $2, 3, 4, 2$ constitute a run below the median of $4 \cdotp 5$.
== Sources ==
* 
* 
Category:Examples of Runs (Statistics)
Run (Statistics)/Examples/Arbitrary Example 2

== Examples of Runs (Statistics) ==
Consider the following sequence of observations from coin-tossing:
:$\tuple {\text {H H H T T H H T H T} }$
There are $3$ runs of heads, defined as a sequence of one or more heads followed by either a tail or no further observation.
== Sources ==
* 
* 
Category:Examples of Runs (Statistics)
Russell's Paradox

== Theorem ==
The leads to a contradiction.
=== Corollary ===
== Proof 1 ==
== Proof 2 ==
== Also presented as ==
== Also known as ==
== Also see ==
* Barber Paradox
== Historical Note ==
== Sources ==
* : $\S 1$: Sets and Set Inclusion
* : $\S 2$: Equivalence of Sets. The Power of a Set: Problem $10$
* 
* : $2$. Graphs: Paradox
* : Introduction
* 
* 
* 
* : $1$. Notation, Conventions: $6$: Theorem $1.2$
* 
* 
* 
* : Chapter $1$: General Background: $\S 8$ Russell's paradox
* 
* : $\S 4.14$
* : Chapter $1$: Why Axiomatic Set Theory?
Category:Russell's Paradox
Category:Axiom of Abstraction
Category:Naive Set Theory
Category:Subsets
Category:Antinomies
Category:Logical Paradoxes
Russell's Paradox/Also presented as

== Russell's Paradox: Also presented as == 
Russell's paradox can also be presented as:
:There is no set $A$ that has every set as its elements.
Its proof follows the same lines: by assuming that such an $A$ exists, and considering the set $\set {x \in A: \map R x}$ where $\map R x$ is the property $x \notin x$.
The same conclusion is reached.
== Sources ==
* : $1$. Notation, Conventions: $6$: Theorem $1.2$
* 
Category:Russell's Paradox
Russell's Paradox/Corollary

== Corollary to Russell's Paradox ==
:$\not \exists x: \forall y: \paren {\map \RR {x, y} \iff \neg \map \RR {y, y} }$
Given a relation $\RR$, there cannot exist an element $x$ that bears $\RR$ to all $y$ which do not bear $\RR$ to itself.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $1$: General Background: $\S 8$ Russell's paradox
Category:Russell's Paradox
Russell's Paradox/Corollary/Proof 1

== Theorem ==
== Proof ==
 there does exist such an $x$.
Let $\RR$ be such that $\map \RR {x, x}$.
Then $\neg \map \RR {x, x}$.
Hence it cannot be the case that $\map \RR {x, x}$.
Now suppose that $\neg \map \RR {x, x}$.
Then by definition of $x$ it follows that $\map \RR {x, x}$.
In both cases a contradiction results.
Hence there can be no such $x$.
Category:Russell's Paradox
Russell's Paradox/Corollary/Proof 2

== Theorem ==
== Proof ==
:
:$\exists x: \forall y: \paren {\map \RR {x, y} \iff \neg \map \RR {y, y} }$
By Existential Instantiation:
:$\forall y: \paren {\map \RR {x, y} \iff \neg \map \RR {y, y} }$
By Universal Instantiation:
:$\map \RR {x, x} \iff \neg \map \RR {x, x} $
But this contradicts Biconditional of Proposition and its Negation.
We thus conclude:
:$\not \exists x: \forall y: \paren {\map \RR {x, y} \iff \neg \map \RR {y, y} }$
Category:Russell's Paradox
Russell's Paradox/Historical Note

== Historical Note on Russell's Paradox ==
Russell's paradox was devised by in $1901$.
He communicated it in a letter to on $16$ June $1902$, and apparently also to some time earlier.
It was first published in .
This paradox is one of a series of antinomies arising from the inherent contradiction in allowing unrestricted use of the . In this case, it is being used to obtain a set which contains itself it does not contain itself.
A desire to avoid such antinomies was a motivation towards the development of various systems of axiomatic set theory.
The best-known system of axiomatic set theory, ZFC, includes only (relatively) restrictive methods of generating new sets by using properties
In particular:
:The allows us to use properties to define only subsets of sets which have already been demonstrated as being allowed to exist.
:The uses only those properties which define functions.
These restrictions both make the above argument invalid in ZFC, since the justification for the existence of the set $R$ is removed.
Some authors, for example , sidestep this issue:
:This logical impasse can be avoided by restricting the notion of set, so that 'very large' collections ... are not counted as sets. However this is done at some cost in simplicity, and in this book we shall do no more than keep to sets which appear to be harmless, and hope that paradoxes will not appear.
Again, from :
:It is well known that an informal point of view in the theory of sets leads to contradictions. These difficulties all arise in operations with very large sets. We shall never need to deal with any sets large enough to cause trouble in this way, and, consequently, we may put aside all such worries.
 puts it as:
:This is not a serious problem since fortunately most sets encountered in mathematics are of a harmless nature ... It is very fortunate that most of the properties dealt with in mathematics (and indeed all the properties we will deal with) are "set forming" in the sense that there is a set whose elements satisfy the property in question.
 remarks:
:The paradox is resolved by restricting the ways in which sets can be formed, so that it becomes inadmissible to consider 'the set of all those sets that are not members of themselves'. There are several ways of doing this, but they need not concern us here; they will not play a role in the rather simple set-theoretical arguments we shall meet.
== Sources ==
* : $\S 1.8$. Sets of sets
* : Chapter $1$: Set Theory
* : $\S 2$. Sets of sets
* : Chapter $1$: Sets and mappings: $\S 1.2$: Sets
* : Introduction
* 
* : $1$. Notation, Conventions: $6$: Theorem $1.2$
* 
Category:Historical Notes
Russell's Paradox/Proof 1

== Theorem ==
== Proof ==
Sets have elements.
Some of those elements may themselves be sets.
So, given two sets $S$ and $T$, we can ask the question:
:Is $S$ an element of $T$?
The answer will either be yes or no.
In particular, given any set $S$, we can ask the question:
:Is $S$ an element of $S$?
Again, the answer will either be yes or no.
Recall the definitions for a set to be:
:ordinary it is not an element of itself
:extraordinary it is an element of itself.
Thus, $\map P S = S \in S$ is a property on which we can use the to build the set $T$ of all extraordinary:
:$T = \set {S: S \in S}$
which is the set of all sets which contain themselves.
Or we can apply the to build the set $T$ of all ordinary sets:
:$R = \set {S: S \notin S}$
($R$ for , of course.)
We ask the question:
:Is $R$ itself an element of $R$?
There are two possible answers: yes or no.
If $R \in R$, then $R$ must satisfy the property that $R \notin R$.
So from that contradiction we know that $R \in R$ does not hold.
So the only other answer, $R \notin R$, must hold instead.
But now we see that $R$ satisfies the conditions of the property that $R \in R$.
So we can see that $R \notin R$ does not hold either.
Thus we have generated a contradiction from the .
== Sources ==
* : $\S 2$: The Axiom of Specification
* : $\S 1$: Sets and Set Inclusion
* . Sets of sets: $\S 1.8$
* : $\S 2$. Sets of sets
* : Chapter $1$: Sets and mappings: $\S 1.2$: Sets
* : $2$. Graphs: Paradox
* : Introduction
* 
* : $1$. Notation, Conventions: $6$: Theorem $1.2$
* : Chapter $1$: Elementary, my dear Watson: $\S 1.2$: Elements, my dear Watson: Remark $1.2.1$
* 
* : Chapter $1$: General Background: $\S 8$ Russell's paradox
* 
Category:Russell's Paradox
Russell's Paradox/Proof 2

== Theorem ==
== Proof ==
 the , that for all predicates $P$ where $S$ is not free:
:$\exists S : \forall x : \paren {x \in S \iff \map P x}$
Since $x \notin x$ is a predicate where $S$ is not free, it follows that:
:$\exists S : \forall x : \paren {x \in S \iff x \notin x}$
is an instance of the .
By Existential Instantiation:
:$\forall x: \paren {x \in S \iff x \notin x}$
By Universal Instantiation:
:$S \in S \iff S \notin S$
But this contradicts Biconditional of Proposition and its Negation.
Thus, the yields a contradiction.
Category:Russell's Paradox
Rutherford's Experiment that led to Nuclear Model of Atom

== Physics Experiment ==
When swiftly-moving $\alpha$ particles are directed at gold atoms, sometimes they are deflected as much as $180 \degrees$.
It is deduced that the atoms consist of a heavy central nucleus surrounded by a lot of what appears to be empty space.
=== Strength of Force ===
== Historical Note ==
== Sources ==
* : Chapter $1$: Force and energy in electrostatics: $1.1$ Electric Charge
Category:Rutherford's Experiment that led to Nuclear Model of Atom
Category:Nuclear Physics
Rutherford's Experiment that led to Nuclear Model of Atom/Strength of Force

== Rutherford's Experiment that led to Nuclear Model of Atom: Strength of Force ==
Consider 's experiment that led him to propose the nuclear model of the atom: that most of its mass is concentrated in one place.
The $\alpha$ particles that were deflected through $180 \degrees$ were coming within $10^{-14} \ \mathrm m$ of the nucleus.
The electric charge on the $\alpha$ particle is $2 \E$, where $\E$ is the elementary charge.
The electric charge on the nucleus of gold is $79 \E$.
Hence by Coulomb's Law of Electrostatics, the force $F$ on the $\alpha$ particle at this range is given by:
given the values:
:$\E = 1.60217 \, 6634 \times 10^{−19} \ \mathrm C$
:$\varepsilon_0 = 8 \cdotp 85418 \, 78128 (13) \times 10^{-12} \, \mathrm F \, \mathrm m^{-1}$
Hence we have a force acting within a single atom of the weight of nearly $10 \mathrm {kg}$.
== Sources ==
* : Chapter $1$: Force and energy in electrostatics: $1.1$ Electric Charge
Category:Rutherford's Experiment that led to Nuclear Model of Atom
Sam Loyd's Missing Square

== Paradox ==
Consider a square of side length $13$.
Let it be divided into:
:a $13 \times 5$ rectangle, divided into two right triangles by its diagonal
:a $13 \times 8$ rectangle, divided into trapezia with side lengths $8$ and $5$:
:
Let the pieces be rearranged to form two long right triangles arranged to form a $21 \times 8$ rectangle.
The area of the square is $13 \times 13 = 169$.
The area of the rectangle is $21 \times 8 = 168$.
Where did the missing $1 \times 1$ square go?
=== Variant ===
== Resolution ==
This is a falsidical paradox.
When you place the $13 \times 5$ right triangle against the trapezium, supposedly to make a $21 \times 8$ right triangle, the hypotenuse of that figure is not actually straight.
When the $21 \times 8$ rectangle is drawn accurately, you will see an overlap:
:
It is noted that $5$, $8$ and $13$ are consecutive Fibonacci numbers.
From Cassini's Identity:
:$F_n^2 = F_{n - 1} F_{n + 1} \pm 1$
Hence the resolution.
== Historical Note ==
== Sources ==
* : $5$
* : The Vanishing Square Paradox: $143$
* : $5$
Category:Sam Loyd's Missing Square
Category:Falsidical Paradoxes
Category:Puzzles
Category:Classic Problems
Sam Loyd's Missing Square/Mistake

== Source Work ==
:
:The Dictionary
:: $5$
:
:The Dictionary
:: $5$
== Mistake ==
: also discovered the identity $F_{n - 1} F_{n + 1} - {F_n}^2 = \paren {-1}^n$, which is the basis of a puzzling trick first presented by .
In fact the puzzle was first first published in $1774$ in by 
== Sources ==
* : $5$
* : $5$
Category:Mistakes/Curious and Interesting Numbers
Sam Loyd's Missing Square/Variant

== Puzzle ==
You have a square which is made from $4$ large triangles, $4$ small triangles, $4$ irregular octagons and $4$ small squares.
You jumble them up and reassemble the pieces once again into that same large square, but this time there is a hole in the middle.
:
Where did that hole come from?
== Solution ==
The original shape that was made from the smaller shapes is not actually a square.
Each of the larger triangles has legs which are $3$ and $8$ units.
Hence it has area $12$ square units.
Each of the smaller triangles has legs which are $2$ and $5$ units.
Hence it has area $5$ square units.
Each of the small squares has sides which are $2$ units.
Hence it has area $4$ square units.
Each of the octagons has area $27$ units, which is determined by counting and measuring.
Hence the total area $\AA$ is given by:
The side of the large square, if it were a square, would be the hypotenuse of a right triangle whose legs are $13$ and $5$ units.
Hence by Pythagoras's Theorem, the side of that large square is $\sqrt {194}$ units.
That is, its area is $194$ square units.
So the shapes do not completely fill the large square.
That is because the hypotenuses of the large triangle and small triangle, when assembled into this shape, do not make a straight line.
== Sources ==
* 
* : The Vanishing Square Paradox: $144$
Category:Sam Loyd's Missing Square
Same-Matrix Product of Matrix Exponentials

== Theorem ==
Let $\mathbf A$ be a square matrix.
Let $s, t \in \R$ be real numbers.
Let $e^{\mathbf A t}$ denote the matrix exponential of $\mathbf A$.
Then:
:$e^{\mathbf A t} e^{\mathbf A s} = e^{\mathbf A \paren {t + s} }$
== Proof ==
Let
:$\map \Phi t = e^{\mathbf A t} e^{\mathbf A s} - e^{\mathbf A \paren {t + s} }$
for some fixed $s \in \R$.
From Derivative of Matrix Exponential:
Since $\map \Phi 0 = e^{\mathbf A s} - e^{\mathbf A s} = 0$, it follows that:
:$\map \Phi t = e^{\mathbf A t} \map \Phi 0 = 0$
independent of $s$.
Hence the result.
Category:Matrix Exponential
Same Degrees of Vertices does not imply Graph Isomorphism

== Theorem ==
Let $G = \struct {\map V G, \map E G}$ and $H = \struct {\map V H, \map E H}$ be graphs such that:
:$\card {\map V G} = \card {\map V H}$
where $\card {\map V G}$ denotes the order of $G$.
Let $\phi: G \to H$ be a mapping which preserves the degrees of the vertices:
:$\forall v \in \map V G: \map {\deg_H} {\map \phi v} = \map {\deg_G} v$
Then it is not necessarily the case that $\phi$ is an isomorphism.
== Proof ==
Proof by Counterexample:
:
Consider a bijection $\phi: \map V {G_1} \to \map V {G_2}$, where $G_1$ is the graph on the left and $G_2$ is the graph on the right.
The vertices $v_1$, $v_2$ and $v_5$ of $G_2$ are each adjacent to both of the others.
Because $\phi$ is a bijection, it must map $3$ vertices of $G_1$ to $v_1$, $v_2$ and $v_5$.
For $\phi$ to be an isomorphism, two of the vertices of $G_1$ are adjacent the two image vertices in $G_2$ udner $\phi$ are also adjacent.
So the $3$ vertices of $G_1$ which map to $v_1$, $v_2$ and $v_5$ of $G_2$ must also each be adjacent to both of the others.
But $G_1$ does not contain $3$ such vertices.
It follows that there is no such isomorphism from $\map V {G_1}$ to $\map V {G_2}$.
That is, $G_1$ and $G_2$ are not isomorphic.
== Also see ==
* Vertex Condition for Isomorphic Graphs, where it is shown that if $\phi$ is an isomorphism then the degrees of the vertices are preserved.
== Sources ==
* : Chapter $2$: Elementary Concepts of Graph Theory: $\S 2.2$: Isomorphic Graphs
Category:Degrees of Vertices
Category:Graph Isomorphisms
Same Dimensional Vector Spaces are Isomorphic

== Theorem ==
Let $K$ be a division ring.
Let $V$, $W$ be finite dimensional $K$-vector spaces.
Suppose that $\dim_K V = \dim_K W$.
Then:
:$V \cong W$
That is, $V$ and $W$ are isomorphic.
== Proof ==
Let $\mathbb V$, $\mathbb W$ be bases for $V$, $W$ respectively.
We have that:
:$\dim_K V = \dim_K W$
Thus by the definition of dimension:
:$\mathbb V \sim \mathbb W$
Therefore we can choose a bijection $\phi: \mathbb V \leftrightarrow \mathbb W$.
Define the mapping $\lambda: V \to W$ by:
:$\ds \map \lambda {\sum \limits_{\mathbf v \mathop \in \mathbb V} a_{\mathbf v} \mathbf v} = \sum \limits_{\mathbf v \mathop \in \mathbb V} a_\mathbf v \map \phi {\mathbf v}$
For $\mathbf v \in \mathbb V$ let $l_\mathbf v \in V^\star$ be the unique linear transformation defined on the basis $\mathbb V$ by:
:$\forall \mathbf v' \in \mathbb V: \map {l_\mathbf v} {\mathbf v'} = \delta_{\mathbf v, \mathbf v'}$
where $\delta : V \times V \to K$ is the Kronecker delta and $V^*$ is the dual of $V$.
Now:
::$\ds \map {l_{\mathbf v} } {\sum \limits_{\mathbf u \mathop \in \mathbb V} a_\mathbf u \mathbf u} = \map {l_\mathbf v} {\sum_{\mathbf u \mathop \in \mathbb V \mathop \setminus \set {\mathbf v} } a_\mathbf u \mathbf u + a_{\mathbf v} \mathbf v} = \sum_{\mathbf u \mathop \in \mathbb V \mathop \setminus \set {\mathbf v} } a_\mathbf u \map {l_\mathbf v} {\mathbf u} + a_{\mathbf v} \map {l_\mathbf v} {\mathbf v}$
By the definition of $l_{\mathbf v}$ and by Vector Scaled by Zero is Zero Vector, all the terms but the last vanish, and so:
:$\ds \forall \mathbf v \in \mathbb V : \map {l_\mathbf v} {\sum \limits_{\mathbf u \mathop \in \mathbb V} a_\mathbf u \mathbf u} = a_\mathbf v$
For all $\mathbf v, \mathbf v' \in V, c \in K$:
Thus $\lambda$ is linear.
Let $\mathbf x \in \ker \lambda$ where $\ker \lambda$ denotes the kernel of $\lambda$.
Then:
:$\ds \mathbf 0 = \map \lambda {\mathbf x} = \sum \limits_{\mathbf v \mathop \in \mathbb V} \map {l_\mathbf v} {\mathbf x} \mathbf v$
Therefore:
:$\forall \mathbf v \in \mathbb V: \map {l_\mathbf v} {\mathbf x} = 0$
because $\mathbb V$ is linearly independent.
By Vector Scaled by Zero is Zero Vector, $\mathbf x = \mathbf 0$.
That is:
:$\ker \lambda = \set {\mathbf 0}$
By Linear Transformation is Injective iff Kernel Contains Only Zero, it follows that $\lambda$ is injective.
Recall that $\phi$ is a bijection.
From Inverse of Bijection is Bijection, $\phi$ is invertible.
Suppose $\mathbf y \in W$.
Then:
where this last vector belongs to $\map \lambda V$.
Thus $\lambda$ is surjective.
$\lambda$ has been shown to be injective and surjective, and so is a bijection.
$\lambda$ has also been shown to be linear transformation.
Thus, by definition, $\lambda$ is an isomorphism.
== Also see ==
* Linear Transformation from Finite-Dimensional Vector Space is Injective iff Surjective‎
Category:Linear Algebra
Category:Dimension of Vector Space
Sample Distribution Function is Consistent Estimator of Cumulative Distribution Function

== Theorem ==
Let $S$ be a sample of ordinal data.
Let $x \in S$ be arbitrary.
Let $\map {S_n} x$ denote the sample distribution function on $S$.
Let $\map F x$ denote the cumulative distribution function of the population from which $S$ was taken.
Then $\map {S_n} x$ is a consistent estimator of $\map F x$.
== Proof ==
== Sources ==
* 
Category:Sample Distribution Functions
Category:Consistent Estimators
Sample Distribution Function tends to Cumulative Distribution Function

== Theorem ==
Let $S$ be a sample of ordinal data.
Let $\map {S_n} x$ denote the sample distribution function on $S$.
Let $\map F x$ denote the cumulative distribution function of the population from which $S$ was taken.
Then:
:$\ds \lim_{n \mathop \to \infty} \map {S_n} x = \map F x$
== Proof ==
== Sources ==
* 
Category:Sample Distribution Functions
Sample Matrix Independence Test

== Theorem ==
Let $V$ be a vector space of real or complex-valued functions on a set $J$.
Let $f_1, \ldots, f_n$ be functions in $V$.
Let $x_1, \ldots, x_n$ from $J$ be given.
Define the sample matrix:
:$S = \begin{bmatrix}
\map {f_1} {x_1} & \cdots & \map {f_n} {x_1} \\
\vdots & \ddots & \vdots \\
\map {f_1} {x_n} & \cdots & \map {f_n} {x_n} \\
\end{bmatrix}$
Let $S$ be nonsingular.
Then $f_1, \ldots, f_n$ are linearly independent in $V$.
== Proof ==
The definition of linear independence is applied.
Assume a linear combination of the functions $f_1, \ldots, f_n$ is the zero function:
Let $\vec c$ have components $c_1, \ldots, c_n$.
For $i = 1, \ldots, n$ replace $x = x_i$ in $(1)$.
There are $n$ linear homogeneous algebraic equations, written as:
:$S \vec c = \vec 0$
Because $S$ is nonsingular:
:$\vec c = \vec 0$
The functions are linearly independent.
== Examples ==
== Also see ==
* Zero Wronskian of Solutions of Homogeneous Linear Second Order ODE iff Linearly Dependent
* Definition:Linearly Independent Set of Real Vectors
* Two Linearly Independent Solutions of Homogeneous Linear Second Order ODE generate General Solution
* Linearly Independent Solutions of y - y = 0
Category:Sample Matrix Independence Test
Category:Sample Matrices
Category:Linear Independence
Sample Matrix Independence Test/Examples/Linear Independence of Powers

== Example of Sample Matrix Independence Test: Linear Independence of Powers ==
Let $V$ be the vector space of all continuous functions on $\R$.
Let $n$ be a strictly positive integer and define:
:$S = \set {1, x, \ldots, x^{n - 1} }$
Then $S$ is a linearly independent subset of $V$.
== Proof ==
Choose samples $x_j = j$, $j = 1, \ldots, n$ from $\R$. 
Define:
:$\map {f_j} x = x^{j - 1}$ for $1 \le j \le n$.
Then the sample matrix is:
:$S = \begin{bmatrix}
1 & 1 & \cdots & 1\\
1 & 2 & \cdots & 2^{n - 1} \\
\vdots & \vdots & \ddots & \vdots \\
1 & n & \cdots & n^{n - 1} \\
\end{bmatrix}$
Matrix $S$ is an invertible Vandermonde matrix.
Then functions $f_1, \ldots, f_n$ are linearly independent.
Category:Sample Matrix Independence Test
Sample Matrix Independence Test/Examples/Linearly Independent Solutions of y'' - y = 0

== Example of Sample Matrix Independence Test: Linearly Independent Solutions of $y' ' - y = 0$ ==
Prove independence of the solutions $e^x$, $e^{-x}$ to:
:$y' ' - y = 0$ 
== Proof ==
Choose $x_1 = 0$, $x_2 = 1$ from set $J = \R$. 
Define:
:$\map {f_1} x = e^x$
and:
:$\map {f_2} x = e^{-x}$
Then the sample matrix is:
:$S = \begin{bmatrix}
1 & 1 \\
e & 1/e \\
\end{bmatrix}$
Since $\det S \ne 0$, the matrix $S$ is invertible.
Hence, by Linearly Independent Solutions of y - y = 0, the solutions are linearly independent.
== Also see ==
* Linearly Independent Solutions of y - y = 0 by the Wronskian test
* Zero Wronskian of Solutions of Homogeneous Linear Second Order ODE iff Linearly Dependent
Category:Sample Matrix Independence Test
Sample Mean is Unbiased Estimator of Population Mean

== Theorem ==
Let $X_1, X_2, \ldots, X_n$ form a random sample from a population with mean $\mu$ and variance $\sigma^2$.
Then: 
:$\ds \bar X = \frac 1 n \sum_{i \mathop = 1}^n X_i$
is an unbiased estimator of $\mu$. 
== Proof ==
If $\bar X$ be an unbiased estimator of $\mu$, then: 
:$\expect {\bar X} = \mu$
We have: 
So $\bar X$ is an unbiased estimator of $\mu$. 
== Sources ==
* 
* 
Category:Inductive Statistics
Sample Space/Examples/Throwing 2 6-Sided Dice

== Example of Sample Space ==
Let $\EE$ be the experiment of throwing two standard $6$-sided dice and adding the numbers.
:The sample space of $\EE$ is $\Omega = \set {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}$.
== Sources ==
* : $1$: Events and probabilities: $1.2$: Outcomes and events
* 
* 
Category:Examples of Sample Spaces