nyuuzyou/wikis huggingface dataset:
Evolute of Circle is its Center

== Theorem ==
The evolute of a circle is a single point: its center.
== Proof ==
By definition, the evolute of $C$ is the locus of the centers of curvature of each point on $C$
, take the circle $C$ of radius $a$ whose center is positioned at the origin of a cartesian plane.
From Equation of Circle, $C$ has the equation:
:$x^2 + y^2 = a^2$
From the definition of curvature in cartesian form:
:$k = \dfrac {y' '} {\paren {1 + y'^2}^{3/2} }$
Here we have:
So:
Thus the curvature of $C$ is constant.
The radius of curvature of $C$ is likewise constant:
:$\rho = a$
From Radius at Right Angle to Tangent, the normal to $C$ at all points on $C$ passes through the center of $C$.
We have that $a$ is the distance from $C$ to the center of $C$.
Thus it follows that the center of curvature of $C$ is the center of $C$ at all points.
Hence the result by definition of evolute.
== Sources ==
* : Chapter $\text {B}.23$: Evolutes and Involutes. The Evolute of a Cycloid
* 
* 
Category:Examples of Evolutes
Category:Circles
Evolute of Cycloid is Cycloid

== Theorem ==
The evolute of a cycloid is another cycloid.
== Proof ==
Let $C$ be the cycloid defined by the equations:
:$\begin {cases} x = a \paren {\theta - \sin \theta} \\ y = a \paren {1 - \cos \theta} \end {cases}$
From Parametric Equations for Evolute: Formulation 2:
:$(1): \quad \begin {cases} X = x - \dfrac {y' \paren {x'^2 + y'^2} } {x' y' ' - y' x' '} \\ Y = y + \dfrac {x' \paren {x'^2 + y'^2} } {x' y' ' - y' x' '} \end {cases}$
where:
:$\tuple {x, y}$ denotes the Cartesian coordinates of a general point on $C$
:$\tuple {X, Y}$ denotes the Cartesian coordinates of a general point on the evolute of $C$
:$x'$ and $x' '$ denote the derivative and second derivative respectively of $x$ $\theta$
:$y'$ and $y' '$ denote the derivative and second derivative respectively of $y$ $\theta$.
Thus we have:
and:
Thus:
and so:
and:
The cycloid $C$ (blue) and its evolute (red) are illustrated below:
== Sources ==
* : Chapter $\text {B}.23$: Evolutes and Involutes. The Evolute of a Cycloid: Example $2$
Category:Examples of Evolutes
Category:Cycloids
Evolute of Ellipse

== Theorem ==
Let $E$ be an ellipse embedded in a Cartesian plane with the equation:
:$\dfrac {x^2} {a^2} + \dfrac {y^2} {b^2} = 1$
=== Cartesian Form ===
=== Parametric Form ===
:
== Proof ==
== Sources ==
* 
Category:Evolute of Ellipse
Category:Examples of Evolutes
Category:Ellipses
Evolute of Ellipse/Cartesian Form

== Theorem ==
Let $E$ be an ellipse embedded in a Cartesian plane with the equation:
:$\dfrac {x^2} {a^2} + \dfrac {y^2} {b^2} = 1$
The evolute of $E$ is given by the Cartesian equation:
:$\paren {a x}^{2 / 3} + \paren {b y}^{2 / 3} = \paren {a^2 - b^2}^{2 / 3}$
:
== Proof ==
== Sources ==
* : $\S 11$: Special Plane Curves: Evolute of an Ellipse: $11.29$
* : $\S 9$: Special Plane Curves: Evolute of an Ellipse: $9.29.$
* 
Category:Evolute of Ellipse
Evolute of Ellipse/Parametric Form

== Theorem ==
Let $E$ be an ellipse embedded in a Cartesian plane with the equation:
:$\dfrac {x^2} {a^2} + \dfrac {y^2} {b^2} = 1$
The evolute of $E$ can be expressed using the parametric equation:
:$\begin {cases} a x = \paren {a^2 - b^2} \cos^3 \theta \\ b y = \paren {a^2 - b^2} \sin^3 \theta \end {cases}$
:
== Proof ==
== Sources ==
* : $\S 11$: Special Plane Curves: Evolute of an Ellipse: $11.30$
* : $\S 9$: Special Plane Curves: Evolute of an Ellipse: $9.30.$
* 
Category:Evolute of Ellipse
Evolute of Parabola

== Theorem ==
The evolute of the parabola $y = x^2$ is the curve:
:$27 X^2 = 16 \paren {Y - \dfrac 1 2}^3$
== Proof ==
From Parametric Equations for Evolute: Formulation 1:
:$\begin {cases}
X = x - \dfrac {y' \paren {1 + y'^2} } {y' '} \\
Y = y + \dfrac {1 + y'^2} {y}
\end{cases}$
where:
:$\tuple {x, y}$ denotes the Cartesian coordinates of a general point on $C$
:$\tuple {X, Y}$ denotes the Cartesian coordinates of a general point on the evolute of $C$
:$y'$ and $y' '$ denote the derivative and second derivative respectively of $y$ $x$.
Thus we have:
and so:
and:
Then:
The parabola (blue) and its evolute (red) are illustrated below:
== Sources ==
* : Chapter $\text {B}.23$: Evolutes and Involutes. The Evolute of a Cycloid: Example $1$
Category:Examples of Evolutes
Category:Parabolas
Category:Semicubical Parabola
Exact Differential Equation/Examples

== Examples of Exact Differential Equation ==
This page gathers examples of exact differential equations of the first order:
:$\map M {x, y} \rd x + \map N {x, y} \rd y = 0$
where there exists a function $\map f {x, y}$ such that:
:$\dfrac {\partial f} {\partial x} = M, \dfrac {\partial f} {\partial y} = N$
such that the second partial derivatives of $f$ exist and are continuous.
=== $e^y \rd x + \paren {x e^y + 2 y} \rd y = 0$ ===
=== $\paren {x + \dfrac 2 y} \rd y + y \rd x = 0$ ===
=== $\paren {y - x^3} \rd x + \paren {x + y^3} \rd y = 0$ ===
=== $\paren {y + y \cos x y} \rd x + \paren {x + x \cos x y} \rd y = 0$ ===
=== $\paren {\sin x \sin y - x e^y} \rd y = \paren {e^y + \cos x \cos y} \rd x$ ===
=== $-\dfrac 1 y \sin \dfrac x y \rd x + \dfrac x {y^2} \sin \dfrac x y \rd y = 0$ ===
=== $\paren {2 x y^3 + y \cos x} \rd x + \paren {3 x^2 y^2 + \sin x} \rd y = 0$ ===
=== $\d x = \dfrac y {1 - x^2 y^2} \rd x + \dfrac x {1 - x^2 y^2} \rd y$ ===
Category:Exact Differential Equations
Exact Form of Prime-Counting Function

== Theorem ==
Let: 
:$\ds \map \Pi x = \map \Li x - \sum_\rho \map \Li {x^\rho} - \map \ln 2 + \int_x^\infty \frac {\d t} {t \paren {t^2 - 1} \map \ln t}$ 
where:
:$\map \Li x$ is the offset logarithmic integral
:the sum $\ds \sum_\rho$ is taken over all $0 < \rho \in \R$ such that the zeta function $\map \zeta {\alpha + i \rho} = 0$ for some $\alpha \in \R$.
Then the prime-counting function is precisely:
:$\ds \map \pi x = \sum_{n \mathop = 1}^\infty \paren {\frac {\map \mu n} n \map \Pi {x^{1 / n} } }$
where $\map \mu n$ denotes the Möbius function.
== Proof ==
Category:Number Theory
Example:Antiassociative Structure

== Theorem ==
Let $\left({\R_{>0}, \circ}\right)$ be an algebraic structure where $\R_{>0}$ denotes the strictly positive real numbers.
Define:
:$\forall x, y \in \R_{>0}: x \circ y = x y + y$
Then $\circ$ is antiassociative on $\R_{>0}$:
:$\forall x, y, z \in \R_{>0}: \left({x \circ y}\right) \circ z \ne x \circ \left({y \circ z}\right)$
== Proof ==
Let $a, b, c \in \R_{>0}$:
Then:
and:
As $ac \ne 0$:
:$\forall x, y, z \in \R_{>0}: \left({x \circ y}\right) \circ z \ne x \circ \left({y \circ z}\right)$
Hence the result.
Category:Examples of Abstract Algebra
Example of Set Intersection with Union/Mistake

== Source Work ==
: Chapter $1$: Complex Numbers
:: Supplementary Problems: $123 \ \text{(c)}$
This mistake can be seen in the 1981 printing of the second edition (1974) as published by Schaum: ISBN 0-070-84382-1
== Mistake ==
:If $A$, $B$ and $C$ are the point sets defined by $\cmod {z + i} < 3$, $\cmod z < 5$, $\cmod {z + 1} < 4$, represent graphically ... :
::$\textit {(c)} \quad A \cap B \cup C$
== Correction ==
Without any definition of binding priority between set intersection and set union, $A \cap B \cup C$ is ambiguous.
It can mean either:
:$(1): \quad A \cap \paren {B \cup C}$
which can be represented graphically as follows:
:
or as:
:$(2): \quad \paren {A \cap B} \cup C$
which can be represented graphically as follows:
:
As can be seen, the two are not the same.
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Point Sets: $123 \ \text{(c)}$
Category:Mistakes/Theory and Problems of Complex Variables
Examples of Algebra Problems/Hindu Problem

== Example of Problem in Algebra ==
:The first man has $16$ azure-blue gems,
:the second man has $10$ emeralds,
:and the third has $8$ diamonds.
:Each among them gives to each of the others $2$ gems of the kind owned by himself;
:and then all $3$ men come to be possessed of equal wealth.
:What are the prices of those azure-blue gems, emeralds and diamonds?
== Solution ==
Let $b$, $e$ and $d$ denote the value of the azure-blue, emerald and diamond respectively.
Then:
:$b : e : d = 2 : 5 : 10$
Without knowing the price of any of the gems in any given monetary units, the best that can be done is to give their ratios.
== Proof ==
We are given:
That is:
and the result follows.
== Historical Note ==
== Sources ==
* : Indian Puzzles: $58$
Category:Examples of Algebra Problems
Examples of Set Intersection and Set Union/Mistake

== Source Work ==
: Chapter $1$: Complex Numbers
:: Supplementary Problems: $123 \ \text{(c)}$
This mistake can be seen in the 1981 printing of the second edition (1974) as published by Schaum: ISBN 0-070-84382-1
== Mistake ==
:If $A$, $B$ and $C$ are the point sets defined by $\cmod {z + i} < 3$, $\cmod z < 5$, $\cmod {z + 1} < 4$, represent graphically each of the following:
::$\textit {(a)} \quad A \cap B \cap C$, $\quad \textit {(b)} \quad A \cup B \cup C$, $\quad \textit {(c)} \quad A \cap B \cup C$, $\quad \textit {(d)} \quad C \paren {A + B}$, $\quad \textit {(d)} \quad \paren {A \cup B} \cap \paren {B \cup C}$, $\quad \textit {(e)} \quad AB + BC + CA$, $\quad \textit {(f)} \quad A \tilde B + B \tilde C + C \tilde A$
== Correction ==
It's a minor point, but shows lack of attention to detail:
The exercises are lettered: $\text {(a)}$, $\text {(b)}$, $\text {(c)}$, $\text {(d)}$, $\text {(d)}$, $\text {(e)}$, $\text {(f)}$
To wit: two instances of $\text {(d)}$.
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Point Sets: $123$
Category:Mistakes/Theory and Problems of Complex Variables
Excenters and Incenter of Orthic Triangle

== Theorem ==
=== Acute Triangle ===
=== Obtuse Triangle ===
== Sources ==
* : $\text V$. Trigonometry: The pedal triangle
Category:Excenters and Incenter of Orthic Triangle
Category:Incircles of Triangles
Category:Excircles of Triangles
Category:Orthic Triangles
Excenters and Incenter of Orthic Triangle/Acute Triangle

== Theorem ==
Let $\triangle ABC$ be an acute triangle.
Let $\triangle DEF$ be the orthic triangle of $\triangle ABC$ such that:
:$D$ is on $BC$
:$E$ is on $AC$
:$F$ is on $AB$
Then:
:the excenter of $\triangle DEF$ $EF$ is $A$
:the excenter of $\triangle DEF$ $DF$ is $B$
:the excenter of $\triangle DEF$ $DE$ is $C$
and:
:the incenter of $\triangle DEF$ is the orthocenter of $\triangle ABC$.
== Proof ==
:
From Altitudes of Triangle Bisect Angles of Orthic Triangle, $AD$ is the angle bisector of $\angle FDE$.
From Bisectors of Adjacent Angles between Straight Lines Meeting at Point are Perpendicular, the angle bisector of $\angle PDE$ is perpendicular to $AD$.
The line perpendicular to $AD$ is $BC$.
Similarly, from Altitudes of Triangle Bisect Angles of Orthic Triangle, $BE$ is the angle bisector of $\angle FED$.
From Bisectors of Adjacent Angles between Straight Lines Meeting at Point are Perpendicular, the angle bisector of $\angle DEQ$ is perpendicular to $BE$.
The line perpendicular to $BE$ is $AC$.
From Construction of Excircle to Triangle, the intersection of $AC$ and $BC$ is the excenter of $\triangle DEF$ $DE$.
The same argument can be used to demonstrate the locations of the excenter of $\triangle DEF$ $DF$ and $EF$.
:
From Altitudes of Triangle Bisect Angles of Orthic Triangle:
:$AD$ is the angle bisector of $\angle FDE$
:$BE$ is the angle bisector of $\angle DEF$
:$FC$ is the angle bisector of $\angle EFD$
From Line from Vertex of Triangle to Incenter is Angle Bisector it follows that $H$ is the incenter of $\triangle DEF$.
Category:Excenters and Incenter of Orthic Triangle
Excenters and Incenter of Orthic Triangle/Obtuse Triangle

== Theorem ==
Let $\triangle ABC$ be an obtuse triangle such that $A$ is the obtuse angle.
Let $\triangle DEF$ be the orthic triangle of $\triangle ABC$ such that:
:$D$ is on $BC$
:$E$ is on $AC$ produced
:$F$ is on $AB$ produced.
Let $H$ be the orthocenter of $\triangle ABC$.
:
Then:
:the excenter of $\triangle DEF$ $EF$ is $H$
:the excenter of $\triangle DEF$ $DF$ is $B$
:the excenter of $\triangle DEF$ $DE$ is $C$
and:
:the incenter of $\triangle DEF$ is $A$.
== Proof ==
From Orthic Triangle of Obtuse Triangle:
:$\triangle DEF$ is also the orthic triangle of $\triangle HBC$, which is an acute triangle.
It follows immediately from Excenters and Incenter of Orthic Triangle of Acute Triangle that:
:the excenter of $\triangle DEF$ $EF$ is $H$
:the excenter of $\triangle DEF$ $DF$ is $B$
:the excenter of $\triangle DEF$ $DE$ is $C$
and:
:$A$ is the incenter of $\triangle DEF$
Category:Excenters and Incenter of Orthic Triangle
Excess Kurtosis of Bernoulli Distribution

== Theorem ==
Let $X$ be a discrete random variable with a Bernoulli distribution with parameter $p$.
Then the excess kurtosis $\gamma_2$ of $X$ is given by: 
:$\gamma_2 = \dfrac {1 - 6 p q} {p q}$
where $q = 1 - p$.
 
== Proof == 
From the definition of excess kurtosis, we have: 
:$\gamma_2 = \expect {\paren {\dfrac {X - \mu} \sigma}^4} - 3$
where: 
:$\mu$ is the expectation of $X$. 
:$\sigma$ is the standard deviation of $X$.
By Expectation of Bernoulli Distribution, we have: 
:$\mu = p$
By Variance of Bernoulli Distribution, we have: 
:$\sigma = \sqrt {p q}$ 
So: 
Category:Kurtosis
Category:Bernoulli Distribution
Excess Kurtosis of Beta Distribution

== Theorem == 
Let $X \sim \BetaDist \alpha \beta$ for some $\alpha, \beta > 0$, where $\operatorname{Beta}$ is the Beta distribution. 
Then the excess kurtosis $\gamma_2$ of $X$ is given by:
:$\gamma_2 = \dfrac {6 \paren {\paren {\alpha - \beta}^2 \paren {\alpha + \beta + 1} - \alpha \beta \paren {\alpha + \beta + 2} } } {\alpha \beta \paren {\alpha + \beta + 2} \paren {\alpha + \beta + 3} }$
=== Lemma 1 ===
=== Lemma 2 ===
=== Lemma 3 ===
=== Lemma 4 ===
== Proof == 
From Kurtosis in terms of Non-Central Moments, we have: 
:$\gamma_2 = \dfrac {\expect {X^4} - 4 \mu \expect {X^3} + 6 \mu^2 \expect {X^2} - 3 \mu^4} {\sigma^4} - 3$
where: 
:$\mu$ is the expectation of $X$. 
:$\sigma$ is the standard deviation of $X$.
We have, by Expectation of Beta Distribution:
:$\expect X = \dfrac {\alpha} {\alpha + \beta}$
By Variance of Beta Distribution:
:$\var X = \sigma^2 = \dfrac {\alpha \beta} {\paren {\alpha + \beta}^2 \paren {\alpha + \beta + 1} }$
so:
:$\sigma = \dfrac {\sqrt {\alpha \beta} } {\paren {\alpha + \beta} \paren {\sqrt {\alpha + \beta + 1 } } }$
From Raw Moment of Beta Distribution, we have:
Hence:
Relying upon Lemma 1, Lemma 2, Lemma 3 and Lemma 4 to simplify the $80$ terms in the numerator, we obtain:
Putting our simplified numerator back in, we obtain:
Category:Kurtosis
Category:Beta Distribution
Category:Excess Kurtosis of Beta Distribution
Excess Kurtosis of Binomial Distribution

== Theorem ==
Let $X$ be a discrete random variable with a binomial distribution with parameters $n$ and $p$ for some $n \in \N$ and $0 \le p \le 1$:
Then the excess kurtosis $\gamma_2$ of $X$ is given by: 
:$\gamma_2 = \dfrac {1 - 6 p q} {n p q}$
where $q = 1 - p$.
 
== Proof == 
From the definition of excess kurtosis, we have: 
:$\gamma_2 = \expect {\paren {\dfrac {X - \mu} \sigma}^4} - 3$
where: 
:$\mu$ is the expectation of $X$. 
:$\sigma$ is the standard deviation of $X$.
By Expectation of Binomial Distribution, we have: 
:$\mu = n p$
By Variance of Binomial Distribution, we have: 
:$\sigma = \sqrt {n p \paren {1 - p} }$ 
So: 
To calculate $\gamma_2$, we must calculate $\expect {X^4}$.
We find this using the moment generating function of $X$, $M_X$. 
By Moment Generating Function of Binomial Distribution, this is given by: 
:$\map {M_X} t = \paren {1 - p + p e^t}^n$
From Moment in terms of Moment Generating Function:
:$\expect {X^4} = \map {M_X'} 0$
So:
Setting $t = 0$:
Plugging this result back into our equation above:
Category:Kurtosis
Category:Binomial Distribution
Excess Kurtosis of Chi-Squared Distribution

== Theorem == 
Let $n$ be a strictly positive integer. 
Let $X \sim \chi^2_n$ where $\chi^2_n$ is the chi-squared distribution with $n$ degrees of freedom. 
Then the excess kurtosis $\gamma_2$ of $X$ is given by: 
:$\gamma_2 = \dfrac {12} n$
== Proof ==
From the definition of excess kurtosis, we have: 
:$\gamma_2 = \expect {\paren {\dfrac {X - \mu} \sigma}^4} - 3$
where: 
:$\mu$ is the expectation of $X$. 
:$\sigma$ is the standard deviation of $X$.
By Expectation of Chi-Squared Distribution:
:$\mu = n$
By Variance of Chi-Squared Distribution:
:$\sigma = \sqrt {2 n}$
As is shown in Skewness of Chi-Squared Distribution, we have: 
:$\expect {X^3} = n^3 + 6 n^2 + 8 n$
We also have: 
So:
Category:Kurtosis
Category:Chi-Squared Distribution
Excess Kurtosis of Continuous Uniform Distribution

== Theorem ==
Let $X$ be a continuous random variable which is uniformly distributed on a closed real interval $\closedint a b$. 
Then the excess kurtosis $\gamma_2$ of $X$ is equal to $-\dfrac 6 5$.
== Proof ==
From the definition of excess kurtosis, we have: 
:$\gamma_2 = \expect {\paren {\dfrac {X - \mu} \sigma}^4} - 3$
From the definition of the continuous uniform distribution, $X$ has probability density function: 
:$\map {f_X} x = \begin {cases} \dfrac 1 {b - a} & : a \le x \le b \\ 0 & : \text {otherwise} \end {cases}$
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:
Category:Kurtosis
Category:Continuous Uniform Distribution
Excess Kurtosis of Erlang Distribution

== Theorem ==
Let $k$ be a strictly positive integer. 
Let $\lambda$ be a strictly positive real number. 
Let $X$ be a continuous random variable with an Erlang distribution with parameters $k$ and $\lambda$.
Then the excess kurtosis $\gamma_2$ of $X$ is given by:
:$\gamma_2 = \dfrac 6 k$
== Proof ==
From the definition of excess kurtosis, we have: 
:$\gamma_2 = \expect {\paren {\dfrac {X - \mu} \sigma}^4} - 3$
where: 
:$\mu$ is the expectation of $X$. 
:$\sigma$ is the standard deviation of $X$.
By Expectation of Erlang Distribution we have: 
:$\mu = \dfrac k \lambda$
By Variance of Erlang Distribution we have: 
:$\sigma = \dfrac {\sqrt k} \lambda$
So: 
Category:Kurtosis
Category:Erlang Distribution
Excess Kurtosis of Exponential Distribution

== Theorem ==
Let $X$ be a continuous random variable of the exponential distribution with parameter $\beta$ for some $\beta \in \R_{> 0}$.
Then the excess kurtosis $\gamma_2$ of $X$ is equal to $6$.
 
== Proof ==
From the definition of excess kurtosis, we have: 
:$\gamma_2 = \expect {\paren {\dfrac {X - \mu} \sigma}^4} - 3$
where: 
:$\mu$ is the expectation of $X$. 
:$\sigma$ is the standard deviation of $X$.
By Expectation of Exponential Distribution we have: 
:$\mu = \beta$ 
By Variance of Exponential Distribution we have: 
:$\sigma = \beta$
So: 
Category:Kurtosis
Category:Exponential Distribution
Excess Kurtosis of Gamma Distribution

== Theorem == 
Let $X \sim \map \Gamma {\alpha, \beta}$ for some $\alpha, \beta > 0$, where $\Gamma$ is the Gamma distribution.
Then the excess kurtosis $\gamma_2$ of $X$ is given by: 
:$\gamma_2 = \dfrac 6 \alpha$
== Proof 1 == 
== Proof 2 == 
Category:Kurtosis
Category:Gamma Distribution
Category:Excess Kurtosis of Gamma Distribution
Excess Kurtosis of Gamma Distribution/Proof 1

== Theorem == 
== Proof == 
From the definition of excess kurtosis, we have: 
:$\gamma_2 = \expect {\paren {\dfrac {X - \mu} \sigma}^4} - 3$
where: 
:$\mu$ is the expectation of $X$. 
:$\sigma$ is the standard deviation of $X$.
By Expectation of Gamma Distribution, we have: 
:$\mu = \dfrac \alpha \beta$
By Variance of Gamma Distribution, we have: 
:$\sigma = \dfrac {\sqrt \alpha} \beta$ 
So: 
To calculate $\gamma_2$, we must calculate $\expect {X^4}$.
We find this using the moment generating function of $X$, $M_X$. 
By Moment Generating Function of Gamma Distribution, this is given by: 
:$\map {M_X} t = \paren {1 - \dfrac t \beta}^{-\alpha} = \beta^\alpha \paren {\beta - t}^{-\alpha}$
From Moment in terms of Moment Generating Function:
:$\expect {X^4} = \map {M_X'} 0$
So:
Setting $t = 0$:
Plugging this result back into our equation above:
Category:Excess Kurtosis of Gamma Distribution
Excess Kurtosis of Gamma Distribution/Proof 2

== Theorem == 
== Proof == 
From the definition of excess kurtosis, we have: 
:$\gamma_2 = \expect {\paren {\dfrac {X - \mu} \sigma}^4} - 3$
where: 
:$\mu = \expect X$ is the expectation of $X$
:$\sigma = \sqrt {\var X}$ is the standard deviation of $X$.
By Expectation of Gamma Distribution, we have: 
:$\mu = \dfrac \alpha \beta$
By Variance of Gamma Distribution, we have: 
:$\sigma^2 = \dfrac \alpha {\beta^2}$ 
From Expectation of Power of Gamma Distribution‎, we have:
:$\expect {X^n} = \dfrac {\alpha^{\overline n} } {\beta^n}$
where $\alpha^{\overline n}$ denotes the $n$th rising factorial of $\alpha$.
Hence:
Category:Excess Kurtosis of Gamma Distribution
Excess Kurtosis of Geometric Distribution/Formulation 1

== Theorem ==
Let $X$ be a discrete random variable with the geometric distribution with parameter $p$ for some $0 < p < 1$.
:$\map X \Omega = \set {0, 1, 2, \ldots} = \N$
:$\map \Pr {X = k} = \paren {1 - p} p^k$
Then the excess kurtosis $\gamma_2$ of $X$ is given by: 
:$\gamma_2 = 6 + \dfrac {\paren {1 - p}^2} p$
== Proof == 
From the definition of excess kurtosis, we have: 
:$\gamma_2 = \expect {\paren {\dfrac {X - \mu} \sigma}^4} - 3$
where: 
:$\mu$ is the expectation of $X$. 
:$\sigma$ is the standard deviation of $X$.
By Expectation of Geometric Distribution: Formulation 1, we have: 
:$\mu = \dfrac p {1 - p}$
By Variance of Geometric Distribution: Formulation 1, we have: 
:$\sigma = \dfrac {\sqrt p } {1 - p}$ 
So: 
To calculate $\gamma_2$, we must calculate $\expect {X^4}$.
We find this using the moment generating function of $X$, $M_X$. 
From Moment in terms of Moment Generating Function:
:$\expect {X^4} = \map { {M_X}^{\paren 4} } 0$
From Moment Generating Function of Geometric Distribution: Fourth Moment: 
:$\map { {M_X}^{\paren 4} } t = p \paren {1 - p } e^t \paren {\dfrac {1 + 11p e^t + 11 p^2 e^{2t} + p^3 e^{3t} } {\paren {1 - p e^t}^5 } }$
Setting $t = 0$ and from Exponential of Zero, we have:
:$\expect {X^4} = p \paren {\dfrac {1 + 11p + 11 p^2 + p^3 } {\paren {1 - p}^4 } }$
So:
Category:Excess Kurtosis of Geometric Distribution
Excess Kurtosis of Geometric Distribution/Formulation 2

== Theorem ==
Let $X$ be a discrete random variable with the geometric distribution with parameter $p$ for some $0 < p < 1$.
:$\map X \Omega = \set {0, 1, 2, \ldots} = \N$
:$\map \Pr {X = k} = p \paren {1 - p}^k$
Then the excess kurtosis $\gamma_2$ of $X$ is given by: 
:$\gamma_2 = 6 + \dfrac {p^2} {1 - p}$
== Proof == 
From the definition of excess kurtosis, we have: 
:$\gamma_2 = \expect {\paren {\dfrac {X - \mu} \sigma}^4} - 3$
where: 
:$\mu$ is the expectation of $X$. 
:$\sigma$ is the standard deviation of $X$.
By Expectation of Geometric Distribution: Formulation 2, we have: 
:$\mu = \dfrac {1 - p} p$
By Variance of Geometric Distribution: Formulation 2, we have: 
:$\sigma = \dfrac {\sqrt {1 - p} } p$ 
So: 
To calculate $\gamma_2$, we must calculate $\expect {X^4}$.
We find this using the moment generating function of $X$, $M_X$. 
From Moment in terms of Moment Generating Function:
:$\expect {X^4} = \map { {M_X}^{\paren 4} } 0$
From Moment Generating Function of Geometric Distribution: Fourth Moment: 
:$\map { {M_X}^{\paren 4} } t = p \paren {1 - p } e^t \paren {\dfrac {1 + 11 \paren {1 - p} e^t + 11 \paren {1 - p}^2 e^{2t} + \paren {1 - p}^3 e^{3t} } {\paren {1 - \paren {1 - p} e^t}^5 } }$
Setting $t = 0$ and from Exponential of Zero, we have:
:$\expect {X^4} = \paren {1 - p } \paren {\dfrac {1 + 11 \paren {1 - p} + 11 \paren {1 - p}^2 + \paren {1 - p}^3 } {p^4} }$
So:
Category:Excess Kurtosis of Geometric Distribution