avewright/proofwiki-math huggingface dataset:
Theorem: Center of Conic is Center of Symmetry

Let $\KK$ be a central conic.
Let $C$ be the center of $\KK$.
Then $C$ is a center of symmetry for $\KK$.

Proof:
First we note from Parabola has No Center that we do not need to consider the parabola.
Then we note from Circle is Ellipse with Equal Major and Minor Axes that a circle is a special case of the ellipse.
Hence there is no need to investigate the circle separately.
It remains to demonstrate the result for the ellipse and the hyperbola.
Let the central conic be expressed in reduced form.
By definition of reduced form, the center of $\KK$ lies at the origin.
From Equation of Ellipse in Reduced Form and Equation of Hyperbola in Reduced Form, $\KK$ can be expressed as:
:$\dfrac {x^2} {a^2} \pm \dfrac {y^2} {b^2} = 1$
Let $\tuple {p, q}$ be a point on $\KK$.
Then:
{{begin-eqn}}
{{eqn | l = \dfrac {p^2} {a^2} \pm \dfrac {q^2} {b^2}
      | r = 1
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \dfrac {\paren {-p}^2} {a^2} \pm \dfrac {\paren {-q}^2} {b^2}
      | r = 1
      | c = as $p^2 = \paren {-p}^2$ and $q^2 = \paren {-q}^2$
}}
{{end-eqn}}
and so $\tuple {-p, -q}$ is also a point on $\KK$.
Hence the result by definition of center of symmetry.
{{qed}}
Category:Centers of Conic Sections
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Theorem: Equation of Hyperbola in Reduced Form/Polar Frame

Let $K$ be aligned in a polar plane in reduced form.
$K$ can be expressed by the equation:
:$\dfrac {\cos^2 \theta} {a^2} - \dfrac {\sin^2 \theta} {b^2} = \dfrac 1 {r^2}$

Proof:
Let the polar plane be aligned with its corresponding Cartesian plane in the conventional manner.
We have that
:$\dfrac {x^2} {a^2} - \dfrac {y^2} {b^2} = 1$
From Conversion between Cartesian and Polar Coordinates in Plane:
{{begin-eqn}}
{{eqn | l = x
      | r = r \cos \theta
}}
{{eqn | l = y
      | r = r \sin \theta
}}
{{end-eqn}}
Hence:
{{begin-eqn}}
{{eqn | l = \dfrac {\paren {r \cos \theta}^2} {a^2} - \dfrac {\paren {r \sin \theta}^2} {b^2}
      | r = 1
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \dfrac {r^2 \cos^2 \theta} {a^2} - \dfrac {r^2 \sin^2 \theta} {b^2}
      | r = 1
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \dfrac {\cos^2 \theta} {a^2} - \dfrac {\sin^2 \theta} {b^2}
      | r = \dfrac 1 {r^2}
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Asymptotes to Hyperbola in Reduced Form

Let $\KK$ be a hyperbola embedded in a cartesian plane in reduced form with the equation:
:$\dfrac {x^2} {a^2} - \dfrac {y^2} {b^2} = 1$
$\KK$ has two asymptotes which can be described by the equation:
:$\dfrac {x^2} {a^2} - \dfrac {y^2} {b^2} = 0$
that is: 
:$y = \pm \dfrac b a x$

Proof:
From Equation of Hyperbola in Reduced Form: Polar Frame, $\KK$ can be described in polar coordinates as:
:$\dfrac {\cos^2 \theta} {a^2} - \dfrac {\sin^2 \theta} {b^2} = \dfrac 1 {r^2}$
When $\theta = 0$ we have that $r = a$.
As $\theta$ increases, $\cos^2 \theta$ decreases and $\sin^2 \theta$ increases.
Hence $\dfrac 1 {r^2}$ decreases and $r$ increases as a consequence.
This continues until:
:$\dfrac {\cos^2 \theta} {a^2} - \dfrac {\sin^2 \theta} {b^2} = 0$
that is:
{{begin-eqn}}
{{eqn | l = \tan^2 \theta
      | r = \dfrac {b^2} {a^2}
      | c = 
}}
{{eqn | l = \dfrac {y^2} {x^2}
      | r = \dfrac {b^2} {a^2}
      | c = 
}}
{{eqn | l = \dfrac {x^2} {a^2} - \dfrac {y^2} {b^2}
      | r = 0
      | c = 
}}
{{end-eqn}}
This can be expressed in the form:
:$b^2 x^2 - a^2 y^2 = 0$
This is a homogeneous quadratic equation in $2$ variables of the form:
:$a' x^2 + 2 h' x y + b' y^2 0 $
where:
:$h' = 0$
:$a' = b^2$
:$b' = a^2$
From Homogeneous Quadratic Equation represents Two Straight Lines through Origin, this is the equation for $2$ straight lines through the origin:
:$y = \pm \dfrac b a x$
{{qed}}
Theorem: Slope of Tangent to Curve at Point equals Value of Derivative

Let $\CC$ be a curve embedded in the Cartesian plane described using the equation:
:$y = \map f x$
where $f$ is a real function.
Let there exist a unique tangent $\TT$ to $\CC$ at a point $P = \tuple {x_0, y_0}$ on $\CC$.
Then the slope of $\CC$ at $P$ is equal to the derivative of $f$ at $P$.

Proof:
We have been given that there exists a unique tangent $\TT$ to $\CC$ at $P$.
By definition of tangent, $\TT$ has a slope $M$ given by:
:$m = \ds \lim_{h \mathop \to 0} \frac {\map f {x_0 + h} - \map f {x_0} } h$
This is the definition of the derivative of $f$ at $P$.
{{qed}}
Theorem: Derivatives of Moment Generating Function of Gamma Distribution

The $n$th derivative of $M_X$ is given by:
:${M_X}^{\paren n} = \dfrac {\alpha^{\overline n} \beta^\alpha} {\paren {\beta - t}^{\alpha + n} }$
where $\alpha^{\overline n}$ denotes the $n$th rising factorial of $\alpha$.

Proof:
The proof proceeds by induction on $n$.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:${M_X}^{\paren n} = \dfrac {\alpha^{\overline n} \beta^\alpha} {\paren {\beta - t}^{\alpha + n} }$
Theorem: Expectation of Power of Gamma Distribution

Let $X \sim \map \Gamma {\alpha, \beta}$ for some $\alpha, \beta > 0$, where $\Gamma$ is the Gamma distribution.
Then:
:$\expect {X^n} = \dfrac {\alpha^{\overline n} } {\beta^n}$
where:
:$\expect {X^n}$ denotes the expectation of $X^n$
:$\alpha^{\overline n}$ denotes the $n$th rising factorial of $\alpha$.

Proof:
From Moment in terms of Moment Generating Function:
:$\expect {X^n} = \map { {M_X}^{\paren n} } 0$
where ${M_X}^{\paren n}$ denotes the $n$th derivative of $M_X$.
Then:
{{begin-eqn}}
{{eqn | l = \expect {X^n}
      | r = \map { {M_X}^{\paren n} } 0
      | c = 
}}
{{eqn | r = \valueat {\dfrac {\alpha^{\overline n} \beta^\alpha} {\paren {\beta - t}^{\alpha + n} } } {t \mathop = 0}
      | c = Derivatives of Moment Generating Function of Gamma Distribution where $t < \beta$
}}
{{eqn | r = \dfrac {\alpha^{\overline n} \beta^\alpha} {\beta^{\alpha + n} }
      | c = setting $t = 0$
}}
{{eqn | r = \dfrac {\alpha^{\overline n} } {\beta^n}
      | c = dividing top and bottom by $\beta^\alpha$
}}
{{end-eqn}}
{{Qed}}
Category:Gamma Distribution
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Theorem: Dirac Comb is Distribution

Let $\phi \in \map \DD \R$ be a test function.
Suppose $\map {\operatorname {III} } 0$ is a Dirac comb such that:
:$\ds \map {\map {\operatorname {III} } 0} \phi := \sum_{n \mathop \in \Z} \map \phi n$
Then $\map {\operatorname {III} } 0$ is a Schwartz distribution.

Proof:
By definition of test function, $\phi$ is supported on a compact subset of $\R$.
Hence:
:$\exists N \in \N : \forall x \in \R \setminus \closedint {-N} N : \map \phi x = 0$
Therefore:
:$\ds \sum_{n \mathop \in \Z} \map \phi n = \sum_{n \mathop = - N}^N \map \phi n$
This is a finite sequence of real numbers.
Thus, the sequence converges, and $\map {\operatorname {III} } 0$ is a mapping such that $\map {\operatorname {III} } 0 : \map \DD \R \to \R$.
Theorem: Jadhav Theorem

Let $a, b, c$ be real numbers in arithmetic sequence.
Let the common difference of this arithmetic sequence be $d$.
Then:
:$b^2 - a c = d^2$

Proof:
{{begin-eqn}}
{{eqn | l = b^2 - a c
      | r = b^2 - \paren {b + d} \paren {b - d}
      | c = {{Defof|Arithmetic Sequence}}: $a + d = b$, $b + d = c$
}}
{{eqn | r = b^2 - \paren {b^2 - d^2}
      | c = Difference of Two Squares
}}
{{eqn | r = d^2
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Equivalence of Definitions of P-adic Norms

Let $p \in \N$ be a prime.
Let $\Q$ denote the rational numbers.
{{TFAE|def = P-adic Norm|view = $p$-adic norm on $\Q$}} 
=== Definition 1 ===
{{:Definition:P-adic Norm/Rational Numbers/Definition 1}}
=== Definition 2 ===
{{:Definition:P-adic Norm/Rational Numbers/Definition 2}}

Proof:
From Negative Powers of Group Elements, Definition 2 can be rewritten as:
:$\forall r \in \Q: \norm r_p = \begin {cases} 0 & : r = 0 \\ p^{-k} & : r = p^k \dfrac m n: k, m, n \in \Z, p \nmid m, n \end {cases}$
Hence if follows that Definition 1 and Definition 2 are equivalent if it is shown:
:$\forall r \in \Q_{\ne 0}: \map {\nu_p} r = k  : r = p^k \dfrac m n: k, m, n \in \Z, p \nmid m, n$
Let $r \in \Q_{\ne 0}$.
Let $r = \dfrac a b : a, b \in \Z_{\ne 0}$
We have:
{{begin-eqn}}
{{eqn| l = \map {\nu_p} r 
     | r = \map {\nu_p} {\dfrac a b}
     | c =
}}
{{eqn| r = \map {\nu_p} a - \map {\nu_p} b
     | c = {{Defof|P-adic Valuation}}
}}
{{end-eqn}}
Let:
:$k_a := \map {\nu_p} a$
:$k_b := \map {\nu_p} b$
Theorem: Equivalence of Definitions of P-adic Norms/Lemma 1

:$\forall x \in Z_{\ne 0}: \map {\nu_p} x = k : x = p^k y  : p \nmid y$

Proof:
Let $x \in \Z_{\ne 0}$.
By definition of the $p$-adic valuation:
:$\map {\nu_p} x = \sup \set {v \in \N: p^v \divides x}$
Let $\map {\nu_p} x = k$.
Then:
:$p^k \nmid x$
By definition of a divisor:
:$\exists y \in Z : x = p^k y$
{{AimForCont}}:
:$p \divides y$
By definition of a divisor:
:$\exists y' \in Z : y = p y'$
Hence:
{{begin-eqn}}
{{eqn | l = x
      | r = p^k \paren{p y'}
}}
{{eqn | r = p^{k + 1} y'
}}
{{end-eqn}}
This contradicts:
:$k = \sup \set {v \in \N: p^v \divides x}$
Hence:
:$p \nmid y$
Since the choice of $x \in \Z_{\ne 0}$ was arbitrary:
:$\forall x \in Z_{\ne 0}: \map {\nu_p} x = k : x = p^k y : p \nmid y$
{{qed}}
Category:Equivalence of Definitions of P-adic Norms
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Theorem: Distributional Derivative of Absolute Value Function

Let $H: \R \to \closedint 0 1$ be the Heaviside step function.
Let $\size x$ be the absolute value of $x$.
Let $T_{\size x}$ be the Schwartz distribution associated with $\size x$.
Then the distributional derivative of $T_{\size x}$ is $T_{2 H - 1}$

Proof:
{{begin-eqn}}
{{eqn | l = \dfrac {\d \size x} {\d x}
      | r =  <nowiki>\begin{cases}
                    1 & : x > 0 \\
                   -1 & : x < 0
             \end{cases} </nowiki>
}}
{{eqn | r = -1 + <nowiki>\begin{cases}
                    2 & : x > 0 \\
                    0 & : x < 0
             \end{cases} </nowiki>
}}
{{eqn | r = -1 + 2 <nowiki> \begin{cases}
                    1 & : x > 0 \\
                    0 & : x < 0
             \end{cases} </nowiki>
}}
{{eqn | r = 2 \map H x - 1
      | c = {{defof|Heaviside Step Function}}
}}
{{end-eqn}}
Furthermore:
:$\ds \lim_{x \mathop \to 0^+} \size x = \lim_{x \mathop \to 0^-} \size x = 0$
By the Jump Rule:
:$T_{\size x}' = T_{2 H - 1}$
{{qed}}
Theorem: Moment Generating Function of Geometric Distribution/Formulation 1

:$\map X \Omega = \set {0, 1, 2, \ldots} = \N$
:$\map \Pr {X = k} = \paren {1 - p} p^k$
Then the moment generating function $M_X$ of $X$ is given by: 
:$\map {M_X} t = \dfrac {1 - p} {1 - p e^t}$
for $t < -\map \ln p$, and is undefined otherwise.

Proof:
From the definition of the geometric distribution, $X$ has probability mass function:
:$\map \Pr {X = k} = \paren {1 - p} p^k$
From the definition of a moment generating function:
:$\ds \map {M_X} t = \expect {e^{t X} } = \sum_{k \mathop = 0}^\infty \map \Pr {X = k} e^{k t}$
So:
{{begin-eqn}}
{{eqn	| l = \sum_{k \mathop = 0}^\infty \map \Pr {X = k} e^{k t}
	| r = \sum_{k \mathop = 0}^\infty \paren {1 - p} p^k e^{k t}
}}
{{eqn	| r = \paren {1 - p} \sum_{k \mathop = 0}^\infty \paren {p e^t}^k
}}
{{end-eqn}}
By Sum of Infinite Geometric Sequence, for this sum to be convergent we must have: 
:$\size {p e^t} < 1$
In the case $p = 0$, this demand is satisfied immediately regardless of $t$.
Otherwise, as both $e^t$ and $p$ are positive: 
:$e^t < \dfrac 1 p$
So, by Logarithm of Power: 
:$t < -\map \ln p$
is the range of $t$ for which the expectation is well-defined.
Now applying Sum of Infinite Geometric Sequence, we have for this range of $t$: 
:$\ds \map {M_X} t = \paren {1 - p} \sum_{k \mathop = 0}^\infty \paren {p e^t}^k = \frac {1 - p} {1 - p e^t}$
{{qed}}
Theorem: Moment Generating Function of Geometric Distribution/Formulation 2

:$\map X \Omega = \set {0, 1, 2, \ldots} = \N$
:$\map \Pr {X = k} = p \paren {1 - p}^k$
Then the moment generating function $M_X$ of $X$ is given by: 
:$\map {M_X} t = \dfrac p {1 - \paren {1 - p} e^t}$
for $t < -\map \ln {1 - p}$, and is undefined otherwise.

Proof:
From the definition of the geometric distribution, $X$ has probability mass function:
:$\map \Pr {X = k} = p \paren {1 - p}^k$
From the definition of a moment generating function:
:$\ds \map {M_X} t = \expect {e^{t X} } = \sum_{k \mathop = 0}^\infty \map \Pr {X = k} e^{k t}$
So:
{{begin-eqn}}
{{eqn	| l = \sum_{k \mathop = 0}^\infty \map \Pr {X = k} e^{k t}
	| r = \sum_{k \mathop = 0}^\infty p \paren {1 - p}^k e^{k t}
}}
{{eqn	| r = p \sum_{k \mathop = 0}^\infty \paren {\paren {1 - p } e^t}^k
}}
{{end-eqn}}
By Sum of Infinite Geometric Sequence, for this sum to be convergent we must have: 
:$\size {\paren {1 - p} e^t} < 1$
In the case $p = 1$, this demand is satisfied immediately regardless of $t$.
Otherwise, as both $e^t$ and $p$ are positive: 
:$e^t < \dfrac 1 {1 - p}$
So, by Logarithm of Power: 
:$t < -\map \ln {1 - p}$
is the range of $t$ for which the expectation is well-defined.
Now applying Sum of Infinite Geometric Sequence, we have for this range of $t$: 
:$\ds \map {M_X} t = p \sum_{k \mathop = 0}^\infty \paren {\paren {1 - p} e^t}^k = \frac p {1 - \paren {1 - p } e^t}$
{{qed}}
Theorem: Expectation of Geometric Distribution/Formulation 1

:$\map X \Omega = \set {0, 1, 2, \ldots} = \N$
:$\map \Pr {X = k} = \paren {1 - p} p^k$
Then the expectation of $X$ is given by:
:$\expect X = \dfrac p {1 - p}$

Proof:
From the Probability Generating Function of Geometric Distribution:
:$\map {\Pi_X} s = \dfrac q {1 - p s}$
where $q = 1 - p$.
From Expectation of Discrete Random Variable from PGF:
:$\expect X = \map {\Pi'_X} 1$
We have:
{{begin-eqn}}
{{eqn | l = \map {\Pi'_X} s
      | r = \map {\frac \d {\d s} } {\frac q {1 - p s} }
}}
{{eqn | r = \frac {q p} {\paren {1 - p s}^2}
      | c = Derivatives of PGF of Geometric Distribution
}}
{{end-eqn}}
Plugging in $s = 1$:
{{begin-eqn}}
{{eqn | l = \map {\Pi'_X} 1
      | r = \frac {q p} {\paren {1 - p}^2}
}}
{{eqn | r = \frac p {1 - p}
      | c = as $q = 1 - p$
}}
{{end-eqn}}
Hence the result.
{{qed}}
Theorem: Expectation of Geometric Distribution/Formulation 1

:$\map X \Omega = \set {0, 1, 2, \ldots} = \N$
:$\map \Pr {X = k} = \paren {1 - p} p^k$
Then the expectation of $X$ is given by:
:$\expect X = \dfrac p {1 - p}$

Proof:
From the definition of expectation:
:$\ds \expect X = \sum_{x \mathop \in \Omega_X} x \map \Pr {X = x}$
Then
{{begin-eqn}}
{{eqn | l = \expect X
      | r = \sum_{k \mathop \in \N} k p^k \paren {1 - p}
      | c = {{Defof|Geometric Distribution}}
}}
{{eqn | r = \sum_{k \mathop \ge 1} k p^k \paren {1 - p}
      | c = as the $k = 0$ term is zero
}}
{{eqn | r = \sum_{k \mathop \ge 1} k p^k - k p^{k+1}
      | c = Real Multiplication Distributes over Addition
}}
{{end-eqn}}
By the Ratio Test, both $\sum_{k \mathop \ge 1} k p^k$ and $\sum_{k \mathop \ge 1} k p^{k+1}$ converge absolutely.
From Absolutely Convergent Real Series is Convergent, both series converge.
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop \ge 1} k p^k - k p^{k+1}
      | r = \sum_{k \mathop \ge 1} k p^k - \sum_{k \mathop \ge 1} k p^{k+1}
      | c = Convergent Series can be Added Term by Term
}}
{{eqn | r = \sum_{k \mathop \ge 1} k p^k - \sum_{k \mathop \ge 2} \paren {k - 1} p^k
      | c = Translation of Index Variable of Summation
}}
{{eqn | r = p + \sum_{k \mathop \ge 2} k p^k - \sum_{k \mathop \ge 2} \paren {k - 1} p^k
      | c = Moving out the first term
}}
{{eqn | r = p + \sum_{k \mathop \ge 2} k p^k - \paren {k - 1} p^k
      | c = Recombining the two convergent series
}}
{{eqn | r = p + \sum_{k \mathop \ge 2} p^k
}}
{{eqn | r = p + \frac {p^2}{1-p}
      | c = Sum of Infinite Geometric Sequence
}}
{{eqn | r = \frac {p \paren {1-p} }{1-p} + \frac {p^2} {1-p}
}}
{{eqn | r = \frac p {1-p}
}}
{{end-eqn}}
{{qed}}
Theorem: Variance of Geometric Distribution/Formulation 1

:$\map X \Omega = \set {0, 1, 2, \ldots} = \N$
:$\map \Pr {X = k} = \paren {1 - p} p^k$
Then the variance of $X$ is given by:
:$\var X = \dfrac p {\paren {1-p}^2}$

Proof:
From Variance of Discrete Random Variable from PGF, we have:
:$\var X = \map {\Pi' '_X} 1 + \mu - \mu^2$
where $\mu = \map E x$ is the expectation of $X$.
From the Probability Generating Function of Geometric Distribution, we have:
:$\map {\Pi_X} s = \dfrac q {1 - ps}$
where $q = 1 - p$.
From Expectation of Geometric Distribution, we have:
:$\mu = \dfrac p q$
From Derivatives of PGF of Geometric Distribution, we have:
:$\map {\Pi' '_X} s = \dfrac {2 q p^2} {\paren {1 - ps}^3}$
Putting $s = 1$ using the formula $\map {\Pi' '_X} 1 + \mu - \mu^2$:
{{begin-eqn}}
{{eqn 	| l = \var X
	| r = \dfrac {2 q p^2} {\paren {1 - p}^3} + \dfrac p q - \paren {\dfrac p q}^2
        | c =  
}}
{{eqn 	| r = \dfrac {2 \paren {1 - p} p^2} {\paren {1 - p}^3} + \dfrac p {1 - p} \paren {\dfrac {\paren {1 - p}^2 } {\paren {1 - p}^2} } - \dfrac {p^2} {\paren {1 - p}^2 } \paren {\dfrac {1 - p} {1 - p} }
        | c = multiplying by $1$
}}
{{eqn 	| r = \dfrac {2 p^2 - 2 p^3 + p - 2 p^2 + p^3 - p^2 + p^3} {\paren {1 - p}^3}
        | c =
}}
{{eqn 	| r = \dfrac {p - p^2} {\paren {1 - p}^3}
        | c =
}}
{{eqn 	| r = \dfrac p {\paren {1 - p}^2}
        | c =
}}
{{end-eqn}}
{{qed}}
Theorem: Distributional Derivative of Floor Function

Let $\floor x$ be the floor function.
Let $\map {\operatorname {III} } x$ be the Dirac comb.
Then the distributional derivative of $\floor x$ is $\map {\operatorname {III} } 0$.

Proof:
By definition:
:$\floor x := \sup \set {m \in \Z: m \le x}$
Hence, $\forall m \in \Z : \forall x \in \openint m {m + 1}$ the floor function is constant.
Therefore:
:$\forall m \in \Z : \forall x \in \openint m {m + 1} : \dfrac {\d \floor x} {\d x} = 0$
Every $x \in \Z$ is a discontinuity of $\floor x$.
Hence, the jump rule has to be applied to each such $x$.
Suppose $k \in \Z$.
Then:
:$\ds \lim_{x \mathop \to k^+} \floor x = k$
:$\ds \lim_{x \mathop \to k^-} \floor x = k - 1$
By the Jump Rule:
{{begin-eqn}}
{{eqn | l = T'_{\floor x} 
      | r = \sum_{k \mathop \in \Z} \delta_k
}}
{{eqn | r = \map {\operatorname {III} } 0
      | c = {{defof|Dirac Comb}}
}}
{{end-eqn}}
{{qed}}
Theorem: Variance of Geometric Distribution/Formulation 2

:$\map X \Omega = \set {0, 1, 2, \ldots} = \N$
:$\map \Pr {X = k} = p \paren {1 - p}^k$
Then the variance of $X$ is given by:
:$\var X = \dfrac {1 - p} {p^2}$

Proof:
By Moment Generating Function of Geometric Distribution, the moment generating function of $X$ is given by: 
:$\map {M_X} t = \dfrac p {1 - \paren {1 - p} e^t}$
for $t < -\map \ln {1 - p}$, and is undefined otherwise.
From Variance as Expectation of Square minus Square of Expectation: 
:$\ds \var X = \expect {X^2} - \paren {\expect X}^2$
From Expectation of Geometric Distribution/Formulation 2, we have:
:$\expect X = \dfrac {1 - p} p$
From Moment Generating Function of Geometric Distribution: Second Moment: 
:$\map { {M_X}' '} t = \dfrac {p \paren {1 - p} e^t + p \paren {1 - p}^2 e^{2 t} } {\paren {1 - \paren {1 - p} e^t}^3}$
From Moment in terms of Moment Generating Function, we also have: 
:$\expect {X^2} = \map { {M_X}' '} 0$
Setting $t = 0$, we obtain the second moment: 
{{begin-eqn}}
{{eqn | l = \map {M' '_X} 0 
      | r = \expect {X^2}
}}
{{eqn | r = \dfrac {p \paren {1 - p} + p \paren {1 - p}^2} {\paren {1 - \paren {1 - p} }^3}
}}
{{eqn | r = \dfrac {p \paren {1 - p} + p \paren {1 - p}^2} {p^3}
}}
{{end-eqn}}
So: 
{{begin-eqn}}
{{eqn | l = \var X
      | r = \dfrac {p \paren {1 - p} + p \paren {1 - p}^2} {p^3} - \dfrac {\paren {1 - p}^2} {p^2}
}}
{{eqn | r = \dfrac {\paren {\paren {1 - p} + p \paren {1 - p}^2} - \paren {1 - p}^2} {p^2}
      | c = simplifying
}}
{{eqn | r = \dfrac {1 - p} {p^2}
      | c = simplifying
}}
{{end-eqn}}
{{qed}}
Theorem: Union of Union of Cartesian Product

Let $A$ and $B$ be sets such that $A \ne \O$ and $B \ne \O$.
Let the ordered pair $\tuple {a, b}$ be defined using the Kuratowski formalization:
:$\tuple {a, b} := \set {\set a, \set {a, b} }$
Then:
:$\ds \bigcup \bigcup \paren {A \times B} = A \cup B$
where:
:$\cup$ denotes union
:$\times$ denotes Cartesian product.

Proof:
{{begin-eqn}}
{{eqn | l = \bigcup \bigcup \paren {A \times B}
      | r = \bigcup \bigcup \set {\tuple {a, b}: a \in A, b \in B}
      | c = {{Defof|Cartesian Product}}
}}
{{eqn | r = \bigcup \paren {\bigcup \set {\set {\set a, \set {a, b} }: a \in A, b \in B} }
      | c = {{Defof|Kuratowski Formalization of Ordered Pair}}
}}
{{eqn | r = \bigcup \set {\set {a, b}: a \in A, b \in B}
      | c = {{Defof|Set Union}}: $\ds \bigcup \set {\set a, \set {a, b} } = \set {a, b}$
}}
{{eqn | r = \set {x: x \in A \text { or } x \in B}
      | c = {{Defof|Set Union}}
}}
{{eqn | r = A \cup B
      | c = {{Defof|Set Union}}
}}
{{end-eqn}}
{{qed}}
Theorem: Union of Union of Cartesian Product with Empty Factor

Let $A$ and $B$ be sets such that either $A = \O$ or $B = \O$.
Let the ordered pair $\tuple {a, b}$ be defined using the Kuratowski formalization:
:$\tuple {a, b} := \set {\set a, \set {a, b} }$
Then:
:$\ds \bigcup \bigcup \paren {A \times B} = A \cup B \iff A = B = \O$
where:
:$\cup$ denotes union
:$\times$ denotes Cartesian product.
That is, if either $A$ or $B$ is empty:
:$\ds \bigcup \bigcup \paren {A \times B} = A \cup B$
holds {{iff}} they are ''both'' empty

Proof:
Let $A = \O$ or $B = \O$.
From Cartesian Product is Empty iff Factor is Empty:
:$A \times B = \O$
Hence from Union of Empty Set:
:$\ds \bigcup \bigcup \paren {A \times B} = \O$
However, from Union is Empty iff Sets are Empty:
:$A \cup B = \O \iff A = \O \text { and } B = \O$
The result follows.
{{qed}}
Theorem: Skewness of Geometric Distribution/Formulation 2

:$\map X \Omega = \set {0, 1, 2, \ldots} = \N$
:$\map \Pr {X = k} = p \paren {1 - p}^k$
Then the skewness of $X$ is given by:
:$\gamma_1 = \dfrac {2 - p} {\sqrt {1 - p} }$

Proof:
From the definition of skewness, we have: 
:$\gamma_1 = \expect {\paren {\dfrac {X - \mu} \sigma}^3}$
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: 
{{begin-eqn}}
{{eqn | l = \gamma_1
      | r = \frac {\expect {X^3 - 3 X^2 \mu + 3 X \mu^2 - \mu^3} } {\sigma^3}
      | c = Cube of Difference
}}
{{eqn | r = \frac {\expect {X^3} - 3 \mu \expect {X^2} + 3 \mu^2 \expect {X} - \mu^3} {\sigma^3}
      | c = Expectation is Linear
}}
{{eqn | r = \frac {\expect {X^3} - 3 \paren {\dfrac {1 - p} p } \expect {X^2} + 2 \paren {\dfrac {\paren {1 - p} \paren {1 - p}^2 } {p^3} } } {\paren {\dfrac {\paren {1 - p} \sqrt {1 - p} } {p^3} } } 
      | c = 
}}
{{end-eqn}}
To calculate $\gamma_1$, we must calculate both $\expect {X^2}$ and $\expect {X^3}$.
From Moment in terms of Moment Generating Function:
:$\expect {X^n} = \map { {M_X}^{\paren n} } 0$
where $M_X$ is the moment generating function of $X$.
From Moment Generating Function of Geometric Distribution: Second Moment: 
:$\map { {M_X}' '} t = p \paren {1 - p} e^t \paren {\dfrac {1  + \paren {1 - p} e^t } {\paren {1 - \paren {1 - p} e^t}^3} }$
From Moment Generating Function of Geometric Distribution: Third Moment: 
:$\map { {M_X}' ' '} t = p \paren {1 - p} e^t \paren {\dfrac {1 + 4 \paren {1 - p} e^t + \paren {1 - p}^2 e^{2t} } {\paren {1 - \paren {1 - p} e^t}^4} }$
Setting $t = 0$ and from Exponential of Zero, we have:
:$\map { {M_X}' '} 0 = \dfrac {\paren {1 - p} \paren {2 - p} } {p^2}$
:$\map { {M_X}' ' '} 0 = \dfrac {\paren {1 - p} \paren {6 - 6 p + p^2} } {p^3}$
So:
{{begin-eqn}}
{{eqn | l = \gamma_1
      | r = \frac {\expect {X^3} - 3 \paren {\dfrac {1 - p} p} \expect {X^2} + 2 \paren {\dfrac {\paren {1 - p} \paren {1 - p}^2 } {p^3} } } {\paren {\dfrac {\paren {1 - p} \sqrt {1 - p} } {p^3} } } 
}}
{{eqn | r = \frac {\dfrac {\paren {1 - p} \paren {6 - 6 p + p^2} } {p^3} - 3 \paren {\dfrac {1 - p} p } \dfrac {\paren {1 - p} \paren {2 - p} } {p^2 } + 2 \paren {\dfrac {\paren {1 - p} \paren {1 - p}^2} {p^3} } } {\paren {\dfrac {\paren {1 - p} \sqrt {1 - p} } {p^3} } } 
}}
{{eqn | r = \frac {\paren {6 - 6 p + p^2} - 3 \paren {1 - p} \paren {2 - p} + 2 \paren {1 - p}^2} {\sqrt {1 - p} } 
      | c = $\dfrac {1 - p} {p^3}$ cancels
}}
{{eqn | r = \frac {6 - 6 p + p^2 - 6 + 9 p - 3 p^2 + 2 - 4 p + 2 p^2} {\sqrt {1 - p} } 
      | c = 
}}
{{eqn | r = \dfrac {2 - p} {\sqrt {1 - p} }
}}
{{end-eqn}}
{{qed}}
Category:Skewness of Geometric Distribution
Category:Skewness
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