nyuuzyou/wikis huggingface dataset:
Principle of Stationary Action with Standard Lagrangian implies Newton's Laws of Motion

== Theorem ==
Let $\MM$ be an $n$-dimensional Euclidean manifold.
Let $P$ be a physical system composed of a countable number of classical particles with inertial masses $m_i$ with $i \in \N$.
Let $\mathbf x = \map {\mathbf x} t$ be twice-differentiable vector-valued function embedded in $\MM$.
Suppose ${\mathbf x}_i$ represents the position of the $i$th particle of $P$.
Suppose the action of $P$ is of the following form:
:$\ds S = \int_{t_0}^{t_1} L \rd t$
where $L$ is the standard Lagrangian.
Suppose, all (internal or external) forces ${\mathbf F}_i$ acting upon $P$ are of the form:
:${\mathbf F}_i = -\dfrac {\partial U} {\partial \mathbf x_i}$
where:
:$U = \map U {t, \set {\mathbf x_i} }$ is a differentiable real function
:$\set {\mathbf x_i}$ denotes dependence on the positions of all the particles.
Then the stationary point of $S$ implies Newton's Second Law of Motion.
== Proof ==
Standard Lagrangian is of the following form:
:$\ds L = \sum_{i \mathop = 1}^n \frac {m_i} 2 \dot {\mathbf x}_i^2 - U$
By the Principle of Stationary Action, equations of motion of $P$ are:
:$\forall i \in \N: m_i {\ddot {\mathbf x} }_i + \dfrac {\partial U} {\partial \mathbf x_i} = 0$
By definition of velocity:
:${\dot {\mathbf x} }_i = {\mathbf v}_i$
By assumption:
:${\mathbf F}_i = -\dfrac {\partial U} {\partial {\mathbf x}_i}$
Then it follows that the equations of motion of $P$ can be rewritten as:
:$\map {\dfrac \d {\d t} } {m_i {\mathbf v}_i } = {\mathbf F}_i$
By definition, these equations together represent Newton's Second Law of Motion.
== Sources ==
* : $\S 4.21$: The Principle of Least Action
Principle of Structural Induction

== Theorem ==
Let $\LL$ be a formal language.
Let the formal grammar of $\LL$ be a bottom-up grammar.
Let $\map P \phi$ be a statement (in the metalanguage of $\LL$) about well-formed formulas $\phi$ of $\LL$.
Then $P$ is true for all WFFs of $\LL$ both:
:$\map P a$ is true for all letters $a$ of $\LL$,
and, for each rule of formation of $\LL$, if $\phi$ is a WFF resulting from WFFs $\phi_1, \ldots, \phi_n$ by applying that rule, then:
:$\map P \phi$ is true only if $\map P {\phi_1}, \ldots, \map P {\phi_n}$ are all true.
== Proof ==
Let $\phi$ be a WFF of $\LL$.
Then $\phi$ is the result of applying finitely many rules of formation of $\LL$.
If we can show that the result of each rule of formation satisfies $P$, we will have finished.
Suppose now that for a rule of formation $\mathbf R$, all preceding rules have produced WFFs satisfying $P$.
By the bottom-up nature of the formal grammar of $\LL$, $\mathbf R$ either:
:Introduces a letter as a WFF; or
:Constructs a new WFF from previously constructed ones.
The two given hypotheses precisely ensure that the WFF resulting from $\mathbf R$ must also satisfy $P$.
For the first rule of formation applied, all preceding rules have vacuously produced WFFs satisfying $P$.
But now we see that any subsequent rule of formation will satisfy this premise.
In particular, it applies to the final rule of formation, and hence $\map P \phi$ is true.
== Remark ==
Although the proof is strongly reminiscent of the Principle of Mathematical Induction, it can be seen to be a finite, algorithmic procedure:
:Given any WFF $\phi$, we have a definite procedure for verifying $\map P \phi$, which involves only finitely many operations.
The assumptions of the theorem ensure that each of these operations will affirm the truth of $\map P \phi$.
It is therefore justifiable to accept this proof in the metalanguage.
== Sources ==
* : $\S 2.1.4$: Theorem $2.12$
Category:Formal Languages
Principle of Sufficient Reason/Historical Note

== Historical Note on Principle of Sufficient Reason ==
The Principle of Sufficient Reason was the main philosophical starting point of .
He made this statement in article $32$ of his $1714$ work :
::[N]o fact can be real or existent, no statement true, unless there be a sufficient reason why it is so and not otherwise, although usually these reasons cannot be known to us.
== Sources ==
* : Chapter $\text {A}.19$: Leibniz ()
Category:Historical Notes
Principle of Superinduction

== Theorem ==
Let $A$ be a class.
Let $g: A \to A$ be a mapping on $A$.
Let $A$ be minimally superinductive under $g$.
Let $P: A \to \set {\T, \F}$ be a propositional function on $A$.
Suppose that:
Then:
:$\forall x \in A: \map P x = \T$
== Proof ==
We are given that $A$ is a minimally superinductive class under $g$.
That is, $A$ is a superinductive class under $g$ with the extra property that for $A$ has no proper class which is also superinductive under $g$.
Let $P$ be a propositional function on $A$ which has the properties specified:
Then by definition, the class $S$ of all elements of $A$ such that $\map P x = \T$ is a superinductive class under $g$.
But because $A$ is minimally superinductive under $g$, $S$ contains all elements of $A$.
That is:
:$\forall x \in A: \map P x = \T$
as we were to show.
== Terminology ==
== Also see ==
* Proof by General Induction
== Sources ==
* : Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text I$ -- Superinduction and Well Ordering: $\S 2$ Superinduction and double superinduction
Category:Mathematical Induction
Category:Proof Techniques
Category:Proof by Superinduction
Category:Closure under Chain Unions
Principle of Superposition/Examples/Electric Field

== Example of Use of Principle of Superposition ==
The Principle of Superposition applies to an electric field:
The total electric field caused by an assemblage of point charges is equal to the sum of the electric fields caused by each of the point charges individually.
== Proof ==
This is apparent from Electric Field Strength from Assemblage of Point Charges:
:$\ds \map {\mathbf E} {\mathbf r} = \dfrac 1 {4 \pi \epsilon_0} \sum_i \dfrac {q_i} {\size {\mathbf r - \mathbf r_i}^3} \paren {\mathbf r - \mathbf r_i}$
where:
:$q_1, q_2, \ldots, q_n$ are point charges
:$\mathbf r_1, \mathbf r_2, \ldots, \mathbf r_n$ are the position vectors of $q_1, q_2, \ldots, q_n$ respectively
:$\map {\mathbf E} {\mathbf r}$ is the electric field strength at a point $P$ whose position vector is $\mathbf r$.
It follows that the electric field strength caused by $q_i$ alone is: 
:$\ds \map {\mathbf E_i} {\mathbf r} = \dfrac 1 {4 \pi \epsilon_0} \dfrac {q_i \paren {\mathbf r - \mathbf r_i} } {\size {\mathbf r - \mathbf r_i}^3} $
as shown in Electric Field caused by Point Charge.
Thus we have:
:$\ds \map {\mathbf E} {\mathbf r} = \sum_i \ds \map {\mathbf E_i} {\mathbf r}$
Hence the result.
== Sources ==
* : Chapter $1$: Force and energy in electrostatics: $1.2$ The Electric Field
Category:Principle of Superposition
Category:Electric Fields
Pringsheim's Theorem

== Theorem ==
Let $f$ be a holomorphic function defined on a unit disc centered at the origin of the complex plane and is denoted by its Taylor series:
:$\map f z = \ds \sum_{n \mathop = 0}^{\infty} c_n z^n$
Let:
:$(1): \quad \forall n \ge 0: c_n \ge 0$
:$(2): \quad$ the radius of convergence of the Taylor series of function $f$ is $1$.
Then $z = 1$ is an isolated singularity of $f$.
== Proof ==
Prism on Triangular Base divided into Three Equal Tetrahedra

== Theorem ==
=== Porism ===
== Proof ==
:
Let $ABCDEF$ be a prism whose base is $\triangle ABC$ and whose opposite is $\triangle DEF$.
Then it is to be demonstrated that $ABCDEF$ can be divided into three equal tetrahedra.
Let $BD, EC, CD$ be joined.
We have that $ABED$ is a parallelogram.
Therefore $BD$ is the diameter of $ABED$.
So from :
:$\triangle ABD = \triangle EBD$
Therefore from :
:the tetrahedron $ABDC$ whose base is $\triangle ABD$ and whose apex is $C$ equals the tetrahedron $DEBC$ whose base is $\triangle DEB$ and whose apex is $C$.
But the tetrahedron $DEBC$ is the same as the tetrahedron $EBCD$ whose base is $\triangle EBC$ and whose apex is $D$.
Therefore the tetrahedron $ABDC$ equals the tetrahedron $EBCD$.
We have that $FCBE$ is a parallelogram.
Therefore $CE$ is the diameter of $FCBE$.
So from :
:$\triangle CEF = \triangle CBE$
Therefore from :
:the tetrahedron $BCED$ whose base is $\triangle BCE$ and whose apex is $D$ equals the tetrahedron $ECFD$ whose base is $\triangle ECF$ and whose apex is $D$.
But the tetrahedron $BCED$ was proved equal to the tetrahedron $ABDC$ whose base is $\triangle ABD$ and whose apex is $C$.
Therefore the tetrahedron $CEFD$ whose base is $\triangle CEF$ and whose apex is $D$ equals the tetrahedron $ABDC$.
Therefore the prism $ABCDEF$ has been divided into three equal tetrahedra.
We have that the tetrahedron $ABDC$ is the same as the tetrahedron $CABD$ whose base is $\triangle CAB$ and whose apex is $D$.
We have that the tetrahedron $ABDC$ is a third of the prism $ABCDEF$ whose base is $\triangle ABC$ and whose opposite is $\triangle DEF$.
Therefore tetrahedron $ABCD$ whose base is $\triangle ABC$ and whose apex is $D$ is a third of the prism $ABCDEF$ whose base is $\triangle ABC$ and whose opposite is $\triangle DEF$.
== Historical Note ==
== Sources ==
* : Book $\text{XII}$. Propositions
* : Chapter $\text {A}.3$: Democritus (ca. B.C.)
Category:Euclid Book XII
Category:Tetrahedra
Category:Prisms
Prism on Triangular Base divided into Three Equal Tetrahedra/Historical Note

== Historical Note ==
This theorem is Proposition $7$ of Book $\text{XII}$ of .
The treatise of , discovered in $1906$, tells us that this result was discovered by .
== Sources ==
* : Chapter $\text {A}.3$: Democritus (ca. B.C.)
Category:Euclid Book XII
Category:Historical Notes
Prismoidal Formula/Formulation 1

== Theorem ==
The Prismoidal Formula is a formula for calculating the volume $V$ of a prismatoid $\PP$.
:$V = \dfrac {h \paren {B + 3 A} } 4$
where:
:$h$ is the altitude of $\PP$
:$B$ is the area of one of the bases of $\PP$
:$A$ is the area of a cross-section of $\PP$ parallel to the bases of $\PP$ at $\dfrac 2 3$ of the distance to the other base of $\PP$.
== Proof ==
== Sources ==
* 
* 
Category:Prismoidal Formula
Prismoidal Formula/Formulation 2

== Theorem ==
The Prismoidal Formula is a formula for calculating the volume $V$ of a prismatoid $\PP$.
:$V = \dfrac {h \paren {B_1 + B_2 + 4 A_m} } 6$
where:
:$h$ is the altitude of $\PP$
:$B_1$ and $B_2$ are the areas of the bases of $\PP$
:$A_m$ is the area of a cross-section of $\PP$ parallel to the bases of $\PP$ midway between the bases of $\PP$.
== Proof ==
== Sources ==
* 
* 
Category:Prismoidal Formula
Prisms of equal Height with Parallelogram and Triangle as Base

== Theorem ==
== Proof ==
:
Let $ABCDEF$ and $GHKLMN$ be triangular prisms.
Let $ABCDEF$ be positioned so that the parallelogram $AF$ can be identified as its base, so that its height is defined as the height of one of its triangular faces.
Let $GHK$ be the triangular (conventionally defined) lower base of $GHKLMN$.
Let the height of $ABCDEF$ equal the height of $GHKLMN$.
Let the area of parallelogram $AF$ be twice the area of $GHK$.
It is to be demonstrated that $ABCDEF$ and $GHKLMN$ are equal in volume.
Let the parallelepipeds $AO$ and $GP$ be completed.
We have that $AF$ is twice $\triangle GHK$.
From :
:the parallelogram $HK$ is twice $\triangle GHK$.
Therefore $AF = HK$.
From :
:the parallelepipeds $AO$ is equal to the parallelepipeds $GP$.
But:
:$ABCDEF$ is half of $AO$
and from :
:$GHKLMN$ is half of $GP$.
Hence $ABCDEF$ is equal to $GHKLMN$.
== Sources ==
* : Book $\text{XI}$. Propositions
Category:Prisms
Probability Density Function of Chi-Squared Distribution

== Definition ==
Let $X$ be a continuous random variable on a probability space $\struct {\Omega, \Sigma, \Pr}$.
Let $\Img X = \hointr 0 \infty$. 
Let $r$ be a strictly positive integer.
Let $X$ have a $\chi$-squared distribution with $r$ degrees of freedom.
Then the probability density function of $X$ is given by:
:$\map {f_X} x = \dfrac 1 {2^{r / 2} \map \Gamma {r / 2} } x^{\paren {r / 2} - 1} e^{-x / 2}$
where $\Gamma$ denotes the gamma function.
== Proof == 
== Sources ==
* : $\S 39$: Probability Distributions: Chi Square Distribution: $39.6$
* : Appendix $13$: Probability distributions
* : Appendix $15$: Probability distributions
* 
Category:Chi-Squared Distribution
Probability Density Function of Continuous Triangular Distribution

== Theorem == 
Let $X$ be a continuous random variable on a probability space $\struct {\Omega, \Sigma, \Pr}$.
Let $X$ have a triangular distribution over the interval $\closedint a b$.
The probability density function of $X$ is:
:$\map {f_X} x = \begin {cases} 0 & : x \le 2 a \\ \\ \dfrac {x - 2 a} {\paren {b - a}^2} & : 2 a < x < a + b \\ \\ \dfrac {2 b - x} {\paren {b - a}^2} & : a + b < x < 2 b \\ \\ 0 & : x > 2 b \end {cases}$
== Proof == 
== Sources == 
* 
Category:Continuous Triangular Distributions
Category:Examples of Probability Density Functions
Probability Density Function of Convolution of Probability Distributions

== Theorem == 
Let $X$ and $Y$ be continuous independent random variables. 
Let $Z = X + Y$. 
Let $f_X$ be a probability density function for $X$.
Let $f_Y$ be a probability density function for $Y$.
Then a probability density function for $Z$, $f_Z$, is given by: 
:$\ds \map {f_Z} z = \int_{-\infty}^\infty \map {f_X} x \map {f_Y} {z - x} \rd x$ 
== Proof == 
Category:Probability Density Functions
Probability Density Function of Exponential Distribution

== Theorem ==
Let $X$ be a continuous random variable with the exponential distribution with parameter $\beta$.
Then the probability density function of $X$ is given by:
:$\map {f_X} x = \begin{cases} \dfrac 1 \beta e^{-\frac x \beta} & : x \ge 0 \\ 0 & : \text{otherwise} \end{cases}$
== Proof ==
By definition of exponential distribution:
:$\map {F_X} \Omega = \R_{\ge 0}$
:$\map \Pr {X < x} = 1 - e^{-\frac x \beta}$
where $0 < \beta$.
By definition of probability density function:
:$\forall x \in \R: \map {f_X} x = \begin {cases} \map {F_X'} x & : x \in \Sigma \\ 0 & : x \notin \Sigma \end {cases}$
where $\map {F_X'} x$ denotes the derivative of $F_X$ at $x$.
Then:
Hence the result.
== Sources ==
* 
* 
* 
* 
* : Appendix $13$: Probability distributions
* : Appendix $15$: Probability distributions
Category:Exponential Distribution
Category:Examples of Probability Density Functions
Probability Density Function of Snedecor's F-Distribution

== Theorem == 
Let $X$ be a continuous random variable on a probability space $\struct {\Omega, \Sigma, \Pr}$.
Let $\Img X = \hointr 0 \infty$. 
Let $n, m$ be strictly positive integers.
Let $X$ have Snedecor's $F$-distribution with $\tuple {n, m}$ degrees of freedom.
=== Formulation 1 ===
=== Formulation 2 ===
== Proof == 
== Sources == 
* 
Category:Probability Density Function of Snedecor's F-Distribution
Category:Snedecor's F-Distribution
Category:Probability Density Functions
Probability Density Function of Snedecor's F-Distribution/Formulation 1

== Theorem == 
Let $X$ be a continuous random variable on a probability space $\struct {\Omega, \Sigma, \Pr}$.
Let $\Img X = \hointr 0 \infty$. 
Let $n, m$ be strictly positive integers.
Let $X$ have Snedecor's $F$-distribution with $\tuple {n, m}$ degrees of freedom.
The probability density function of $X$ is:
:$\map {f_X} x = \dfrac {m^{m / 2} n^{n / 2} x^{\paren {n / 2} - 1} } {\paren {m + n x}^{\paren {n + m} / 2} \map \Beta {n / 2, m / 2} }$
where $\Beta$ denotes the beta function. 
== Proof == 
== Sources == 
* 
Category:Probability Density Function of Snedecor's F-Distribution
Probability Density Function of Snedecor's F-Distribution/Formulation 2

== Theorem == 
Let $X$ be a continuous random variable on a probability space $\struct {\Omega, \Sigma, \Pr}$.
Let $\Img X = \hointr 0 \infty$. 
Let $n, m$ be strictly positive integers.
Let $X$ have Snedecor's $F$-distribution with $\tuple {n, m}$ degrees of freedom.
The probability density function of $X$ is:
:$\map {f_X} x = \dfrac {\map \Gamma {\dfrac {n + m} 2} n^{n / 2} m^{m / 2} } {\map \Gamma {n / 2} \map \Gamma {m / 2} } \ds \int_0^x t^{\paren {n / 2} - 1} \paren {m + n t}^{-\paren {n + m} / 2} \rd t$
where $\Gamma$ denotes the gamma function. 
== Proof == 
== Sources == 
* : $\S 39$: Probability Distributions: $F$ Distribution: $39.7$
* 
Category:Probability Density Function of Snedecor's F-Distribution
Probability Density Function of Student's t-Distribution

== Definition ==
Let $X$ be a continuous random variable on a probability space $\struct {\Omega, \Sigma, \Pr}$.
Let $X$ have a Student's $t$-distribution with $k$ degrees of freedom.
Then the probability density function of $X$ is given by:
:$\map {f_X} x = \dfrac {\map \Gamma {\frac {k + 1} 2} } {\sqrt {\pi k} \map \Gamma {\frac k 2} } \paren {1 + \dfrac {x^2} k}^{-\frac {k + 1} 2}$
where $\Gamma$ denotes the gamma function.
== Proof == 
== Sources ==
* : $\S 39$: Probability Distributions: Student's $t$ Distribution: $39.5$
* : Appendix $15$: Probability distributions
* 
Category:Student's t-Distribution
Probability Distribution is Probability Measure

== Theorem ==
Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space. 
Let $\struct {S, \Sigma'}$ be a measurable space.
Let $X$ be a random variable on $\struct {\Omega, \Sigma, \Pr}$ taking values in $\struct {S, \Sigma'}$. 
Let $P_X$ be the probability distribution of $X$.
Then: 
:$P_X$ is a probability measure on $\struct {S, \Sigma'}$. 
== Proof ==
From the definition of probability distribution, we have: 
:$P_X = X_* \Pr$
where $X_* \Pr$ is the pushforward $X_* \Pr$ of $\Pr$, under $X$, on $\Sigma'$.
From Pushforward Measure is Measure, we have: 
:$P_X$ is a measure.
We then have: 
so:
:$P_X$ is a probability measure. 
Category:Probability Distributions
Probability Generating Function as Expectation

== Theorem ==
Let $X$ be a discrete random variable whose codomain, $\Omega_X$, is a subset of the natural numbers $\N$.
Let $p_X$ be the probability mass function for $X$.
Let $\map {\Pi_X} s$ be the probability generating function for $X$.
Then:
:$\map {\Pi_X} s = \expect {s^X}$
where $\expect {s^X}$ denotes the expectation of $s^X$.
== Proof ==
By definition of probability generating function:
:$\ds \map {\Pi_X} s := \sum_{n \mathop = 0}^\infty \map {p_X} n s^n = \map {p_X} 0 + \map {p_X} 1 s + \map {p_X} 2 s^2 + \cdots$
where $p_X$ is the probability mass function for $X$.
For any real function $g: \R \to \R$, by Expectation of Function of Discrete Random Variable:
:$\ds \expect {\map g X} = \sum_{x \mathop \in \Omega_X} \map g X \map \Pr {X = x}$
whenever the sum is absolutely convergent.
In this case:
:$\map g X = s^X$
Thus:
:$\ds \expect {s^X} = \sum_{x \mathop \in \Omega_X} s^x \map \Pr {X = x}$
or, using $\map {p_X} x := \map \Pr {X = x}$:
:$\ds \expect {s^X} = \sum_{x \mathop \in \Omega_X} \map {p_X} x s^x$
By definition of $X$, this can then be expressed as:
:$\ds \expect {s^X} = \sum_{n \mathop = 0}^\infty \map {p_X} n s^n$
Thus whenever $\ds \sum_{n \mathop = 0}^\infty \map {p_X} n s^n$ is absolutely convergent:
:$\map {\Pi_X} s = \expect {s^X}$
by definition of probability mass function.
As $p_X$ is a probability mass function for $X$:
:$\ds \sum_{k \mathop = 0}^\infty \map {p_X} k = 1$
Thus the condition on absolute convergence is satisfied by all $s$ such that $\size s < 1$ by the observation:
:$\ds \sum_{k \mathop = 0}^\infty \size {\map {p_X} k s^k} \le \sum_{k \mathop = 0}^\infty \map {p_X} k= 1$
== Sources ==
* : $\S 4.2$: Integer-valued random variables: $(7)$
Category:Probability Generating Functions
Category:Expectation
Probability Generating Function defines Probability Distribution

== Theorem ==
Let $X$ and $Y$ be discrete random variables whose codomain, $\Omega_X$, is a subset of the natural numbers $\N$.
Let the probability generating functions of $X$ and $Y$ be $\map {\Pi_X} s$ and $\map {\Pi_Y} s$ respectively.
Then:
:$\forall s \in \closedint {-1} 1: \map {\Pi_X} s = \map {\Pi_Y} s$
:
:$\forall k \in \N: \Pr \left({X = k}\right) = \map \Pr {Y = k}$
That is, discrete random variables which are integer-valued have the same PGFs they have the same PMF.
== Proof ==
By the definition of PGF, it follows that if:
:$\forall k \in \N: \map \Pr {X = k} = \map \Pr {Y = k}$
then:
:$\forall s \in \closedint {-1} 1: \map {\Pi_X} s = \map {\Pi_Y} s$
Suppose that $\map {\Pi_X} s = \map {\Pi_Y} s$ for all $s \in \closedint {-1} 1$.
From Probability Generating Function as Expectation the radius of convergence of both $\map {\Pi_X} s$ and $\Pi_Y \left({s}\right)$ is at least $1$.
Therefore they have unique power series expansions about $s = 0$:
:$\ds \map {\Pi_X} s = \sum_{n \mathop = 0}^\infty s^n \map \Pr {X = k}$
:$\ds \map {\Pi_Y} s = \sum_{n \mathop = 0}^\infty s^n \map \Pr {Y = k}$
As $\map {\Pi_X} s = \map {\Pi_Y} s$, these two power series have identical coefficients.
Hence the result.
== Sources ==
* : $\S 4.2$: Integer-valued random variables: Theorem $4 \ \text{A}$
Category:Probability Generating Functions
Probability Generating Function of Bernoulli Distribution

== Theorem ==
Let $X$ be a discrete random variable with the Bernoulli distribution with parameter $p$.
Then the p.g.f. of $X$ is:
:$\map {\Pi_X} s = q + p s$
where $q = 1 - p$.
== Proof ==
From the definition of p.g.f:
:$\ds \map {\Pi_X} s = \sum_{x \mathop \ge 0} \map {p_X} x s^x$
From the definition of the Bernoulli distribution:
:$\map {p_X} x = \begin{cases}
p & : x = a \\
1 - p & : x = b \\
0 & : x \notin \set {a, b} \\
\end{cases}$
So:
Hence the result.
== Sources ==
* : $\S 4.2$: Integer-valued random variables: $(10)$
Category:Probability Generating Functions
Category:Bernoulli Distribution
Probability Generating Function of Binomial Distribution

== Theorem ==
Let $X$ be a discrete random variable with the binomial distribution with parameters $n$ and $p$.
Then the p.g.f. of $X$ is:
:$\map {\Pi_X} s = \paren {q + p s}^n$
where $q = 1 - p$.
== Proof ==
From the definition of p.g.f:
:$\ds \map {\Pi_X} s = \sum_{k \mathop \ge 0} \map {p_X} k s^k$
From the definition of the binomial distribution:
:$\map {p_X} k = \dbinom n k p^k \paren {1 - p}^{n - k}$
So:
Hence the result.
== Sources ==
* : $\S 4.2$: Integer-valued random variables: $(11)$
* 
* 
Category:Binomial Distribution
Category:Examples of Probability Generating Functions
Probability Generating Function of Degenerate Distribution

== Theorem ==
Let $X$ be the degenerate distribution:
:$\forall x \in \N: \map {p_X} x = \begin{cases}
1 & : x = k \\
0 & : x \ne k
\end{cases}$
where $k \in \N$.
Then the p.g.f. of $X$ is:
:$\map {\Pi_X} x = s^k$
== Proof ==
Follows directly from the definition:
:$\ds \map {\Pi_X} s = \sum_{x \mathop \ge 0} \map {p_X} x s^x$
As $\map {p_X} x \ne 0$ for only one value of $x$, all the terms vanish except that one.
Hence the result.
Category:Probability Generating Functions
Category:Degenerate Distribution
Probability Generating Function of Discrete Uniform Distribution

== Theorem ==
Let $X$ be a discrete random variable with the discrete uniform distribution with parameter $n$.
Then the p.g.f. of $X$ is:
:$\map {\Pi_X} s = \dfrac {s \paren {1 - s^n} } {n \paren {1 - s} }$
== Proof ==
From the definition of p.g.f:
:$\ds \map {\Pi_X} s = \sum_{x \mathop \ge 0} \map {p_X} x s^x$
From the definition of the discrete uniform distribution:
:$\forall k \in \N, 1 \le k \le n: \map {p_X} k = \dfrac 1 n$
So:
Hence the result.
Category:Probability Generating Functions
Category:Discrete Uniform Distribution
Probability Generating Function of Geometric Distribution

== Theorem ==
Let $X$ be a discrete random variable with the geometric distribution with parameter $p$.
Then the p.g.f. of $X$ is:
:$\map {\Pi_X} s = \dfrac q {1 - p s}$
where $q = 1 - p$.
== Proof ==
From the definition of p.g.f:
:$\ds \map {\Pi_X} s = \sum_{x \mathop \ge 0} \map {p_X} x s^x$
From the definition of the geometric distribution:
:$\forall k \in \N, k \ge 0: \map {p_X} k = q p^k$
So:
Hence the result.
Category:Probability Generating Functions
Category:Geometric Distribution