avewright/proofwiki-math huggingface dataset:
Theorem: Differential Entropy of Normal Distribution

Let $X \sim \Gaussian \mu {\sigma^2}$ for some $\mu \in \R, \sigma \in \R_{> 0}$, where $N$ is the normal distribution.
Then the differential entropy $\map h X$ of $X$ is given by: 
:$\map h X = \map \ln {\sigma \sqrt {2 \pi} } + \dfrac 1 2$

Proof:
From the definition of the normal distribution, $X$ has probability density function:
:$\map {f_X} x = \dfrac 1 {\sigma \sqrt {2 \pi} } \, \map \exp {-\dfrac {\paren {x - \mu}^2} {2 \sigma^2} }$
From the definition of differential entropy: 
:$\ds \map h X = -\int_{-\infty}^\infty \map {f_X} x \ln \map {f_X} x \rd x$
So:
{{begin-eqn}}
{{eqn | l = \map h X
      | r = - \frac 1 {\sigma \sqrt{2 \pi} } \int_{-\infty}^\infty \map \exp {-\frac {\paren {x - \mu}^2} {2 \sigma^2} } \map \ln {\frac 1 {\sigma \sqrt {2 \pi} } \map \exp {-\frac {\paren {x - \mu}^2} {2 \sigma^2} } } \rd x
}}
{{eqn | r = \frac 1 {\sigma \sqrt {2 \pi} } \int_{-\infty}^\infty \map \exp {-\frac {\paren {x - \mu}^2} {2 \sigma^2} } \map \ln {\sigma \sqrt {2 \pi} \, \map \exp {\frac {\paren {x - \mu}^2} {2 \sigma^2} } } \rd x
      | c = Logarithm of Reciprocal
}}
{{eqn | r = \frac {\sqrt 2 \sigma} { \sigma \sqrt {2 \pi} } \int_{-\infty}^\infty \map \exp {-t^2} \map \ln {\sigma \sqrt {2 \pi} \, \map \exp {t^2} } \rd t
      | c = substituting $t = \dfrac {x - \mu} {\sqrt 2 \sigma}$
}}
{{eqn | r = \frac 1 {\sqrt \pi} \int_{-\infty}^\infty \paren {\map \ln {\sigma \sqrt {2 \pi} } + \map \ln {\map \exp {t^2} } } \, \map \exp {-t^2} \rd t
      | c = Sum of Logarithms
}}
{{eqn | r = \frac {\map \ln {\sigma \sqrt {2 \pi} } } {\sqrt \pi} \int_{-\infty}^\infty \map \exp {-t^2} \rd t + \frac 1 {\sqrt \pi} \int_{-\infty}^\infty t^2 \map \exp {-t^2} \rd t
}}
{{eqn | r = \frac {\sqrt \pi \, \map \ln {\sigma \sqrt {2 \pi} } } {\sqrt \pi} + \frac 1 {\sqrt \pi} \paren {\intlimits {-\frac t 2 \, \map \exp {-t^2} } {-\infty} \infty + \frac 1 2 \int_{-\infty}^\infty \map \exp {-t^2} \rd t}
      | c = Gaussian Integral, Integration by Parts, Fundamental Theorem of Calculus
}}
{{eqn | r = \map \ln {\sigma \sqrt {2 \pi} } + \frac 1 {2 \sqrt \pi} \int_{-\infty}^\infty \map \exp {-t^2} \rd t
      | c = Exponential Tends to Zero and Infinity
}}
{{eqn | r = \map \ln {\sigma \sqrt {2 \pi} } + \frac {\sqrt \pi} {2 \sqrt \pi}
      | c = Gaussian Integral
}}
{{eqn | r = \map \ln {\sigma \sqrt {2 \pi} } + \frac 1 2
}}
{{end-eqn}}
{{qed}}
Category:Normal Distribution
Category:Differential Entropy
ly24g2qu39qw8yi5gd8qvzaaaapwqke
Theorem: Magnitude of Scalar Triple Product equals Volume of Parallelepiped Contained by Vectors

Let $\mathbf a, \mathbf b, \mathbf c$ be vectors in a vector space of $3$ dimensions:
Let $\mathbf a \cdot \paren {\mathbf b \times \mathbf c}$ denote the scalar triple product of $\mathbf a, \mathbf b, \mathbf c$.
Then $\size {\mathbf a \cdot \paren {\mathbf b \times \mathbf c} }$ equals the volume of the parallelepiped contained by $\mathbf a, \mathbf b, \mathbf c$.

Proof:
Let us construct the parallelepiped $P$ contained by $\mathbf a, \mathbf b, \mathbf c$.
:500px
We have by Magnitude of Vector Cross Product equals Area of Parallelogram Contained by Vectors that:
:$\mathbf b \times \mathbf c$ is a vector area equal to and normal to the area of the bottom face $S$ of $P$.
The dot product $\mathbf a \cdot \paren {\mathbf b \times \mathbf c}$ is equal to the product of this vector area and the projection of $\mathbf a$ along $\mathbf b \times \mathbf c$.
Depending on the relative orientations of $\mathbf a$, $\mathbf b$ and $\mathbf c$, $\mathbf a \cdot \paren {\mathbf b \times \mathbf c}$ may or may not be negative.
So, taking its absolute value, $\size {\mathbf a \cdot \paren {\mathbf b \times \mathbf c} }$ is the volume of the parallelepiped which has $\mathbf a$, $\mathbf b$ and $\mathbf c$ as edges.
{{qed}}
Theorem: Lagrange's Formula/Corollary

:$\paren {\mathbf a \times \mathbf b} \times \mathbf c = \paren {\mathbf a \cdot \mathbf c} \mathbf b - \paren {\mathbf b \cdot \mathbf c} \mathbf a$

Proof:
{{begin-eqn}}
{{eqn | l = \mathbf c \times \paren {\mathbf a \times \mathbf b}
      | r = \paren {\mathbf c \cdot \mathbf b} \mathbf a - \paren {\mathbf c \cdot \mathbf a} \mathbf b
      | c = Lagrange's Formula
}}
{{eqn | ll= \leadsto
      | l = \paren {\mathbf a \times \mathbf b} \times \mathbf c
      | r = -\paren {\paren {\mathbf c \cdot \mathbf b} \mathbf a - \paren {\mathbf c \cdot \mathbf a} \mathbf b}
      | c = Vector Cross Product is Anticommutative
}}
{{eqn | r = -\paren {\mathbf b \cdot \mathbf c} \mathbf a + \paren {\mathbf a \cdot \mathbf c} \mathbf b
      | c = Dot Product Operator is Commutative
}}
{{eqn | r = \paren {\mathbf a \cdot \mathbf c} \mathbf b - \paren {\mathbf b \cdot \mathbf c} \mathbf a
      | c = Real Addition is Commutative
}}
{{end-eqn}}
{{qed}}
{{Namedfor|Joseph Louis Lagrange}}
Theorem: Differential Entropy of Continuous Uniform Distribution

Let $X \sim \ContinuousUniform a b$ for some $a, b \in \R$, $a \ne b$, where $\operatorname U$ is the continuous uniform distribution. 
Then the differential entropy of $X$, $\map h X$, is given by: 
:$\map h X = \map \ln {b - a}$

Proof:
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 differential entropy: 
:$\ds \map h X = - \int_{-\infty}^\infty \map {f_X} x \map \ln {\map {f_X} x} \rd x$
So: 
{{begin-eqn}}
{{eqn | l = \map h X
      | r = -\int_a^b \frac 1 {b - a} \map \ln {\frac 1 {b - a} } \rd x
}}
{{eqn | r = \frac {\map \ln {b - a} } {b - a} \int_a^b \rd x
      | c = Logarithm of Reciprocal
}}
{{eqn | r = \frac {\map \ln {b - a} } {b - a} \bigintlimits x a b 
      | c = Primitive of Constant, Fundamental Theorem of Calculus
}}
{{eqn | r = \frac {\paren {b - a} \map \ln {b - a} } {b - a}
}}
{{eqn | r = \map \ln {b - a}
}}
{{end-eqn}}
{{qed}}
Category:Continuous Uniform Distribution
Category:Differential Entropy
lv9517a0o7kuy2ko2wqysgpckrydkqd
Theorem: Dot Product of Vector Cross Products

Let $\mathbf a, \mathbf b, \mathbf c, \mathbf d$ be vectors in a vector space $\mathbf V$ of $3$ dimensions.
Let $\mathbf a \times \mathbf b$ denote the vector cross product of $\mathbf a$ with $\mathbf b$.
Let $\mathbf a \cdot \mathbf b$ denote the dot product of $\mathbf a$ with $\mathbf b$.
Then:
:$\paren {\mathbf a \times \mathbf b} \cdot \paren {\mathbf c \times \mathbf d} = \paren {\mathbf a \cdot \mathbf c} \paren {\mathbf b \cdot \mathbf d} - \paren {\mathbf a \cdot \mathbf d} \paren {\mathbf b \cdot \mathbf c}$

Proof:
{{begin-eqn}}
{{eqn | l = \paren {\mathbf a \times \mathbf b} \cdot \paren {\mathbf c \times \mathbf d}
      | r = \sqbrk {\mathbf a, \mathbf b, \mathbf c \times \mathbf d}
      | c = {{Defof|Scalar Triple Product}}
}}
{{eqn | r = \sqbrk {\mathbf b, \mathbf c \times \mathbf d, \mathbf a}
      | c = Equivalent Expressions for Scalar Triple Product
}}
{{eqn | r = \paren {\mathbf b \times \paren {\mathbf c \times \mathbf d} } \cdot \mathbf a
      | c = {{Defof|Scalar Triple Product}}
}}
{{eqn | r = \paren {\paren {\mathbf b \cdot \mathbf d} \mathbf c - \paren {\mathbf b \cdot \mathbf c} \mathbf d} \cdot \mathbf a
      | c = Lagrange's Formula on $\mathbf b \times \paren {\mathbf c \times \mathbf d}$
}}
{{eqn | r = \paren {\mathbf b \cdot \mathbf d} \paren {\mathbf c \cdot \mathbf a} - \paren {\mathbf b \cdot \mathbf c} \paren {\mathbf d \cdot \mathbf a}
      | c = Dot Product Distributes over Addition
}}
{{eqn | r = \paren {\mathbf a \cdot \mathbf c} \paren {\mathbf b \cdot \mathbf d} - \paren {\mathbf a \cdot \mathbf d} \paren {\mathbf b \cdot \mathbf c}
      | c = Dot Product Operator is Commutative
}}
{{end-eqn}}
Hence the result.
{{qed}}
Theorem: Vector Cross Product of Vector Cross Products

Let $\mathbf a, \mathbf b, \mathbf c, \mathbf d$ be vectors in a vector space $\mathbf V$ of $3$ dimensions:
Let $\mathbf a \times \mathbf b$ denote the vector cross product of $\mathbf a$ with $\mathbf b$.
Let $\sqbrk {\mathbf a, \mathbf b, \mathbf c}$ denote the scalar triple product of $\mathbf a$, $\mathbf b$ and $\mathbf c$.
Then:
{{begin-eqn}}
{{eqn | l = \paren {\mathbf a \times \mathbf b} \times \paren {\mathbf c \times \mathbf d}
      | r = \sqbrk {\mathbf a, \mathbf b, \mathbf d} \mathbf c - \sqbrk {\mathbf a, \mathbf b, \mathbf c} \mathbf d
      | c = 
}}
{{eqn | r = \sqbrk {\mathbf a, \mathbf c, \mathbf d} \mathbf b - \sqbrk {\mathbf b, \mathbf c, \mathbf d} \mathbf a
      | c = 
}}
{{end-eqn}}

Proof:
{{begin-eqn}}
{{eqn | l = \paren {\mathbf a \times \mathbf b} \times \paren {\mathbf c \times \mathbf d}
      | r = \paren {\paren {\mathbf a \times \mathbf b} \cdot \mathbf d} \mathbf c - \paren {\paren {\mathbf a \times \mathbf b} \cdot \mathbf c} \mathbf d
      | c = Lagrange's Formula
}}
{{eqn | r = \sqbrk {\mathbf a, \mathbf b, \mathbf d} \mathbf c - \sqbrk {\mathbf a, \mathbf b, \mathbf c} \mathbf d
      | c = {{Defof|Scalar Triple Product}}
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \paren {\mathbf a \times \mathbf b} \times \paren {\mathbf c \times \mathbf d}
      | r = \paren {\mathbf a \cdot \paren {\mathbf c \times \mathbf d} } \mathbf b - \paren {\mathbf b \cdot \paren {\mathbf c \times \mathbf d} } \mathbf a
      | c = Lagrange's Formula: Corollary
}}
{{eqn | r = \sqbrk {\mathbf a, \mathbf c, \mathbf d} \mathbf b - \sqbrk {\mathbf b, \mathbf c, \mathbf d} \mathbf a
      | c = {{Defof|Scalar Triple Product}}
}}
{{end-eqn}}
{{qed}}
Theorem: Derivative of Scalar Triple Product of Vector-Valued Functions

Let $\mathbf a$, $\mathbf b$ and $\mathbf c$ be differentiable vector-valued functions in Cartesian $3$-space.
The derivative of their scalar triple product is given by:
:$\map {\dfrac \d {\d x} } {\mathbf a \cdot \paren {\mathbf b \times \mathbf c} } = \dfrac {\d \mathbf a} {\d x} \cdot \paren {\mathbf b \times \mathbf c} + \mathbf a \cdot \paren {\dfrac {\d \mathbf b} {\d x} \times \mathbf c} + \mathbf a \cdot \paren {\mathbf b \times \dfrac {\d \mathbf c} {\d x} }$

Proof:
{{begin-eqn}}
{{eqn | l = \map {\dfrac \d {\d x} } {\mathbf a \cdot \paren {\mathbf b \times \mathbf c} }
      | r = \dfrac {\d \mathbf a} {\d x} \cdot \paren {\mathbf b \times \mathbf c} + \mathbf a \cdot \map {\dfrac \d {\d x} } {\mathbf b \times \mathbf c}
      | c = Derivative of Dot Product of Vector-Valued Functions
}}
{{eqn | r = \dfrac {\d \mathbf a} {\d x} \cdot \paren {\mathbf b \times \mathbf c} + \mathbf a \cdot \paren {\dfrac {\d \mathbf b} {\d x} \times \mathbf c + \mathbf b \times \dfrac {\d \mathbf c} {\d x} }
      | c = Derivative of Vector Cross Product of Vector-Valued Functions
}}
{{eqn | r = \dfrac {\d \mathbf a} {\d x} \cdot \paren {\mathbf b \times \mathbf c} + \mathbf a \cdot \paren {\dfrac {\d \mathbf b} {\d x} \times \mathbf c} + \mathbf a \cdot \paren {\mathbf b \times \dfrac {\d \mathbf c} {\d x} }
      | c = Dot Product Distributes over Addition
}}
{{end-eqn}}
{{qed}}
Theorem: Dot Product of Vector-Valued Function with its Derivative

Let:
:$\map {\mathbf f} x = \ds \sum_{k \mathop = 1}^n \map {f_k} x \mathbf e_k$
be a differentiable vector-valued function.
The dot product of $\mathbf f$ with its derivative is given by:
:$\map {\mathbf f} x \cdot \dfrac {\map {\d \mathbf f} x} {\d x} = \size {\map {\mathbf f} x} \dfrac {\d \size {\map {\mathbf f} x} } {\d x}$
where $\size {\map {\mathbf f} x} \ne 0$.

Proof:
{{begin-eqn}}
{{eqn | l = \map {\mathbf f} x \cdot \dfrac {\map {\d \mathbf f} x} {\d x}
      | r = \map {\mathbf f} x \cdot \sum_{k \mathop = 0}^n \dfrac {\map {\d f_k} x} {\d x} \mathbf e_k
      | c = {{Defof|Derivative of Vector-Valued Function}}
}}
{{eqn | r = \sum_{k \mathop = 0}^n \map {f_k} x \dfrac {\map {\d f_k} x} {\d x}
      | c = {{Defof|Dot Product}}
}}
{{eqn | n = 1
      | r = \sum_{k \mathop = 0}^n \map {f_k} x \dfrac {\map {\d f_k} x} {\d x}
      | c = {{Defof|Dot Product}}
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \dfrac {\d \size {\map {\mathbf f} x} } {\d x}
      | r = \dfrac \d {\d x} \sqrt {\sum_{k \mathop = 0}^n \paren {\map {f_k} x}^2}
      | c = {{Defof|Vector Length}}
}}
{{eqn | r = \frac 1 {2 \sqrt {\ds \sum_{k \mathop = 0}^n \paren {\map {f_k} x}^2} } \dfrac \d {\d x} \sum_{k \mathop = 0}^n \paren {\map {f_k} x}^2
      | c = Chain Rule for Derivatives, Derivative of Power
}}
{{eqn | r = \frac 1 {\size {\map {\mathbf f} x} } \frac 1 2 \sum_{k \mathop = 0}^n \dfrac \d {\d x} \paren {\paren {\map {f_k} x}^2}
      | c = {{Defof|Vector Length}}, Sum Rule for Derivatives: General Result
}}
{{eqn | r = \frac 1 {\size {\map {\mathbf f} x} } \frac 1 2 \sum_{k \mathop = 0}^n 2 \map {f_k} x \dfrac {\map {\d f_k} x} {\d x}
      | c = Chain Rule for Derivatives, Derivative of Power
}}
{{eqn | r = \frac 1 {\size {\map {\mathbf f} x} } \map {\mathbf f} x \cdot \dfrac {\map {\d \mathbf f} x} {\d x}
      | c = simplification, and from $(1)$
}}
{{end-eqn}}
Hence the result.
{{qed}}
Theorem: Expectation of Beta Distribution

Let $X \sim \BetaDist \alpha \beta$ for some $\alpha, \beta > 0$, where $\operatorname {Beta}$ denotes the beta distribution. 
Then:
:$\expect X = \dfrac \alpha {\alpha + \beta}$

Proof:
From the definition of the beta distribution, $X$ has probability density function:
:$\map {f_X} x = \dfrac {x^{\alpha - 1} \paren {1 - x}^{\beta - 1} } {\map \Beta {\alpha, \beta} }$
From the definition of the expected value of a continuous random variable: 
:$\ds \expect X = \int_0^1 x \map {f_X} x \rd x$
So:
{{begin-eqn}}
{{eqn	| l = \expect X
	| r = \frac 1 {\map \Beta {\alpha, \beta} } \int_0^1 x^\alpha \paren {1 - x}^{\beta - 1} \rd x
}}
{{eqn	| r = \frac {\map \Beta {\alpha + 1, \beta} } {\map \Beta {\alpha, \beta} } 
	| c = {{Defof|Beta Function|index = 1}}
}}
{{eqn	| r = \frac {\map \Gamma {\alpha + 1} \map \Gamma \beta} {\map \Gamma {\alpha + \beta + 1} } \cdot \frac {\map \Gamma {\alpha + \beta} } {\map \Gamma \alpha \map \Gamma \beta}
	| c = {{Defof|Beta Function|index = 3}}
}}
{{eqn	| r = \frac \alpha {\alpha + \beta} \cdot \frac {\map \Gamma \alpha \map \Gamma \beta \map \Gamma {\alpha + \beta} } {\map \Gamma \alpha \map \Gamma \beta \map \Gamma {\alpha + \beta} }
	| c = Gamma Difference Equation
}}
{{eqn	| r = \frac \alpha {\alpha + \beta}
}}
{{end-eqn}}
{{qed}}
Theorem: Expectation of Beta Distribution

Let $X \sim \BetaDist \alpha \beta$ for some $\alpha, \beta > 0$, where $\operatorname {Beta}$ denotes the beta distribution. 
Then:
:$\expect X = \dfrac \alpha {\alpha + \beta}$

Proof:
{{begin-eqn}}
{{eqn	| l = \expect X
	| r = \prod_{r \mathop = 0}^0 \frac {\alpha + r} {\alpha + \beta + r}
	| c = Raw Moment of Beta Distribution
}}
{{eqn	| r = \frac {\alpha + 0} {\alpha + \beta + 0}
}}
{{eqn	| r = \frac \alpha {\alpha + \beta}
}}
{{end-eqn}}
{{qed}}
Theorem: Variance of Beta Distribution

Let $X \sim \BetaDist \alpha \beta$ for some $\alpha, \beta > 0$, where $\operatorname {Beta}$ denotes the beta distribution. 
Then:
:$\var X = \dfrac {\alpha \beta} {\paren {\alpha + \beta}^2 \paren {\alpha + \beta + 1} }$

Proof:
From the definition of the Beta distribution, $X$ has probability density function:
:$\map {f_X} x = \dfrac {x^{\alpha - 1} \paren {1 - x}^{\beta - 1} } {\map \Beta {\alpha, \beta} }$
From Variance as Expectation of Square minus Square of Expectation: 
:$\ds \var X = \int_0^1 x^2 \map {f_X} X \rd x - \paren {\expect X}^2$
So: 
{{begin-eqn}}
{{eqn	| l = \var X
	| r = \frac 1 {\map \Beta {\alpha, \beta} } \int_0^1 x^{\alpha + 1} \paren {1 - x}^{\beta - 1} \rd x - \frac {\alpha^2} {\paren {\alpha + \beta}^2}
	| c = Expectation of Beta Distribution
}}
{{eqn	| r = \frac {\map \Beta {\alpha + 2, \beta} } {\map \Beta {\alpha, \beta} } - \frac {\alpha^2} {\paren {\alpha + \beta}^2}
	| c = {{Defof|Beta Function|index = 1}}
}}
{{eqn	| r = \frac {\map \Gamma {\alpha + 2} \, \map \Gamma \beta} {\map \Gamma {\alpha + \beta + 2} } \cdot \frac {\map \Gamma {\alpha + \beta} } {\map \Gamma \alpha \, \map \Gamma \beta} - \frac {\alpha^2} {\paren {\alpha + \beta}^2}
	| c = {{Defof|Beta Function|index = 3}}
}}
{{eqn	| r = \frac {\alpha \paren {\alpha + 1} } {\paren {\alpha + \beta} \paren {\alpha + \beta + 1} } \cdot \frac {\map \Gamma \alpha \, \map \Gamma \beta \, \map \Gamma {\alpha + \beta} } {\map \Gamma \alpha \, \map \Gamma \beta \, \map \Gamma {\alpha + \beta} } - \frac {\alpha^2} {\paren {\alpha + \beta}^2}
	| c = Gamma Difference Equation
}}
{{eqn	| r = \frac {\paren {\alpha^2 + \alpha} \paren {\alpha + \beta} } {\paren {\alpha + \beta}^2 \paren {\alpha + \beta + 1} } - \frac {\alpha^2 \paren {\alpha + \beta + 1} } {\paren {\alpha + \beta}^2 \paren {\alpha + \beta + 1} }
}}
{{eqn	| r = \frac {\alpha ^3 + \alpha^2 \beta + \alpha^2 + \alpha \beta - \alpha^3 - \alpha^2 \beta - \alpha^2} {\paren {\alpha + \beta}^2 \paren {\alpha + \beta + 1} }
}}
{{eqn	| r = \frac {\alpha \beta} {\paren {\alpha + \beta}^2 \paren {\alpha + \beta + 1} }
}}
{{end-eqn}}
{{qed}}
Theorem: Variance of Beta Distribution

Let $X \sim \BetaDist \alpha \beta$ for some $\alpha, \beta > 0$, where $\operatorname {Beta}$ denotes the beta distribution. 
Then:
:$\var X = \dfrac {\alpha \beta} {\paren {\alpha + \beta}^2 \paren {\alpha + \beta + 1} }$

Proof:
From the definition of Variance as Expectation of Square minus Square of Expectation:
:$\var X = \expect {X^2} - \paren {\expect X}^2$
From Expectation of Beta Distribution:
:$\expect X = \dfrac \alpha {\alpha + \beta}$
From Raw Moment of Beta Distribution:
:$\ds \expect {X^n} = \prod_{r \mathop = 0}^{n - 1} \frac {\alpha + r} {\alpha + \beta + r}$
So:
{{begin-eqn}}
{{eqn	| l = \expect {X^2}
	| r = \prod_{r \mathop = 0}^1 \frac {\alpha + r} {\alpha + \beta + r}
	| c = Raw Moment of Beta Distribution
}}
{{eqn	| r = \paren {\frac {\alpha} {\alpha + \beta} } \paren {\frac {\alpha + 1} {\alpha + \beta + 1} }
}}
{{end-eqn}}
Therefore:
{{begin-eqn}}
{{eqn	| l = \expect {X^2} - \paren {\expect X}^2
	| r = \paren {\frac {\alpha} {\alpha + \beta} } \paren {\frac {\alpha + 1} {\alpha + \beta + 1} } - \paren {\dfrac \alpha {\alpha + \beta} }^2
	| c = 
}}
{{eqn	| r = \paren {\frac {\alpha} {\alpha + \beta} } \paren {\frac {\alpha + 1} {\alpha + \beta + 1} } \paren {\frac {\alpha + \beta} {\alpha + \beta} } - \paren {\dfrac \alpha {\alpha + \beta} }^2 \paren {\frac {\alpha + \beta + 1} {\alpha + \beta + 1} }
        | c = multiplying by $1$
}}
{{eqn	| r = \frac {\alpha^3 + \alpha^2 \beta + \alpha^2 + \alpha \beta - \alpha^3 - \alpha^2 \beta - \alpha^2} {\paren {\alpha + \beta}^2 \paren {\alpha + \beta + 1} }
        | c =
}}
{{eqn	| r = \frac {\alpha \beta} {\paren {\alpha + \beta}^2 \paren {\alpha + \beta + 1} }
        | c =
}}
{{end-eqn}}
{{qed}}
Theorem: Moment Generating Function of Gamma Distribution

Let $X \sim \map \Gamma {\alpha, \beta}$ for some $\alpha, \beta > 0$, where $\Gamma$ is the Gamma distribution.
Then the moment generating function of $X$ is given by: 
:$\map {M_X} t = \begin {cases} \paren {1 - \dfrac t \beta}^{-\alpha} & t < \beta \\ \text {does not exist} & t \ge \beta \end {cases}$

Proof:
From the definition of the Gamma distribution, $X$ has probability density function:
:$\map {f_X} x = \dfrac {\beta^\alpha x^{\alpha - 1} e^{-\beta x} } {\map \Gamma \alpha}$
From the definition of a moment generating function:
:$\ds \map {M_X} t = \expect {e^{t X} } = \int_0^\infty e^{t x} \map {f_X} x \rd x$
First take $t < \beta$. 
Then:
{{begin-eqn}}
{{eqn | l = \map {M_X} t
      | r = \frac {\beta^\alpha} {\map \Gamma \alpha} \int_0^\infty x^{\alpha - 1} e^{-\paren {\beta - t} x} \rd x
}}
{{eqn | r = \frac {\beta^\alpha} {\map \Gamma \alpha} \int_0^\infty \paren {\frac u {\beta - t} }^{\alpha - 1} e^{-u} \frac {\rd u} {\beta - t}
      | c = substituting $\paren {\beta - t} x = u$
}}
{{eqn | r = \frac {\beta^\alpha} {\map \Gamma \alpha \paren {\beta - t}^\alpha} \int_0^\infty u^{\alpha - 1} e^{-u} \rd u
}}
{{eqn | r = \frac {\beta^\alpha \map \Gamma \alpha} {\map \Gamma \alpha \paren {\beta - t}^\alpha}
      | c = {{Defof|Gamma Function}}
}}
{{eqn | r = \paren {\frac \beta {\beta - t} }^\alpha
}}
{{eqn | r = \paren {\frac {\beta - t} \beta}^{-\alpha}
}}
{{eqn | r = \paren {1 - \frac t \beta}^{-\alpha}
}}
{{end-eqn}}
{{qed|lemma}}
Now take $t = \beta$. 
Our integral becomes: 
{{begin-eqn}}
{{eqn | l = \frac {\beta^\alpha} {\map \Gamma \alpha} \int_0^\infty x^{\alpha - 1} \rd x
      | r = \frac {\beta^\alpha} {\alpha \map \Gamma \alpha} \bigintlimits {x^\alpha} 0 \infty
      | c = Primitive of Power, Fundamental Theorem of Calculus
}}
{{eqn | r = 0 + \frac {\beta^\alpha} {\alpha \map \Gamma \alpha} \lim_{x \mathop \to \infty} x^\alpha
}}
{{eqn | o = \to
      | r = \infty
      | c = for $\alpha > 0$, $x^\alpha$ increases without bound as $x \mathop \to \infty$
}}
{{end-eqn}}
So $\expect {e^{\beta X} }$ does not exist.
{{qed|lemma}}
Finally take $t > \beta$.
We have that $-\paren {\beta - t}$ is positive.
As a consequence of Exponential Dominates Polynomial, we have:
:$x^{\alpha - 1} < e^{-\paren {\beta - t} x}$
for sufficiently large $x$. 
Therefore, in this case, the integrand increases without bound. 
We conclude that the integral is divergent.
Hence $\expect {e^{t X} }$ does not exist for $t > \beta$. 
{{qed}}
Theorem: Gradient Operator Distributes over Addition

Let $\mathbf V$ be a vector space of $n$ dimensions.
Let $\tuple {\mathbf e_1, \mathbf e_2, \ldots, \mathbf e_n}$ be the standard ordered basis of $\mathbf V$.
Let $\map f {x_1, x_2, \ldots, x_n}, \map g {x_1, x_2, \ldots, x_n}: \mathbf V \to \R$ be differentiable real-valued functions on $\mathbf V$.
Let $\nabla f$ denote the gradient of $f$.
Then:
:$\nabla \paren {f + g} = \nabla f + \nabla g$

Proof:
{{begin-eqn}}
{{eqn | l = \nabla \paren {f + g}
      | r = \sum_{k \mathop = 1}^n \frac {\partial \paren {f + g} } {\partial x_k} \mathbf e_k
      | c = {{Defof|Gradient Operator}}
}}
{{eqn | r = \sum_{k \mathop = 1}^n \paren {\frac {\partial f} {\partial x_k} \mathbf e_k + \frac {\partial g} {\partial x_k} \mathbf e_k}
      | c = Linear Combination of Partial Derivatives
}}
{{eqn | r = \sum_{k \mathop = 1}^n \frac {\partial f} {\partial x_k} \mathbf e_k + \sum_{k \mathop = 1}^n \frac {\partial g} {\partial x_k} \mathbf e_k
      | c = Sum of Summations equals Summation of Sum
}}
{{eqn | r = \nabla f + \nabla g
      | c = {{Defof|Gradient Operator}}
}}
{{end-eqn}}
{{qed}}
Theorem: Divergence Operator Distributes over Addition

Let $\map {\mathbf V} {x_1, x_2, \ldots, x_n}$ be a vector space of $n$ dimensions.
Let $\tuple {\mathbf e_1, \mathbf e_2, \ldots, \mathbf e_n}$ be the standard ordered basis of $\mathbf V$.
Let $\mathbf f$ and $\mathbf g: \mathbf V \to \mathbf V$ be vector-valued functions on $\mathbf V$:
:$\mathbf f := \tuple {\map {f_1} {\mathbf x}, \map {f_2} {\mathbf x}, \ldots, \map {f_n} {\mathbf x} }$
:$\mathbf g := \tuple {\map {g_1} {\mathbf x}, \map {g_2} {\mathbf x}, \ldots, \map {g_n} {\mathbf x} }$
Let $\nabla \cdot \mathbf f$ denote the divergence of $f$.
Then:
:$\nabla \cdot \paren {\mathbf f + \mathbf g} = \nabla \cdot \mathbf f + \nabla \cdot \mathbf g$

Proof:
{{begin-eqn}}
{{eqn | l = \nabla \cdot \paren {\mathbf f + \mathbf g}
      | r = \sum_{k \mathop = 1}^n \frac {\map \partial {f_k + g_k} } {\partial x_k}
      | c = {{Defof|Divergence Operator}}
}}
{{eqn | r = \sum_{k \mathop = 1}^n \paren {\frac {\partial f_k} {\partial x_k} + \frac {\partial g_k} {\partial x_k} }
      | c = Linear Combination of Partial Derivatives
}}
{{eqn | r = \sum_{k \mathop = 1}^n \frac {\partial f_k} {\partial x_k} + \sum_{k \mathop = 1}^n \frac {\partial g_k} {\partial x_k}
      | c = 
}}
{{eqn | r = \nabla \cdot \mathbf f + \nabla \cdot \mathbf g
      | c = 
}}
{{end-eqn}}
{{qed}}