avewright/proofwiki-math huggingface dataset:
Theorem: Moment Generating Function of Exponential Distribution

Let $X$ be a continuous random variable with an exponential distribution with parameter $\beta$ for some $\beta \in \R_{> 0}$.
Then the moment generating function $M_X$ of $X$ is given by: 
:$\map {M_X} t = \dfrac 1 {1 - \beta t}$
for $t < \dfrac 1 \beta$, and is undefined otherwise.

Proof:
From the definition of the exponential distribution, $X$ has probability density function:
:$\map {f_X} x = \dfrac 1 \beta e^{-\frac x \beta}$
From the definition of a moment generating function:
:$\ds \map {M_X} t = \expect {e^{t X} } = \int_0^\infty e^{t x} \map {f_X} x \rd x$
Then: 
{{begin-eqn}}
{{eqn | l = \map {M_X} t
      | r = \frac 1 \beta \int_0^\infty e^{x \paren {-\frac 1 \beta + t} } \rd x
      | c = Exponential of Sum
}}
{{eqn | r = \frac 1 {\beta \paren {-\frac 1 \beta + t} } \intlimits {e^{x \paren {-\frac 1 \beta + t} } } 0 \infty
      | c = Primitive of Exponential Function
}}
{{end-eqn}}
Note that if $t > \dfrac 1 \beta$, then $e^{x \paren {-\frac 1 \beta + t} } \to \infty$ as $x \to \infty$ by Exponential Tends to Zero and Infinity, so the integral diverges in this case. 
If $t = \dfrac 1 \beta$ then the integrand is identically $1$, so the integral similarly diverges in this case. 
If $t < \dfrac 1 \beta$, then $e^{x \paren {-\frac 1 \beta + t} } \to 0$ as $x \to \infty$ from Exponential Tends to Zero and Infinity, so the integral converges in this case. 
Therefore, the function is only well defined for $t < \dfrac 1 \beta$.
Proceeding:
{{begin-eqn}}
{{eqn | l = \frac 1 {\beta \paren {-\frac 1 \beta + t} } \intlimits {e^{x \paren {-\frac 1 \beta + t} } } 0 \infty
      | r = \frac 1 {\beta \paren {-\frac 1 \beta + t} } \paren {0 - 1}
      | c = Exponential Tends to Zero and Infinity, Exponential of Zero
}}
{{eqn | r = \frac 1 {\beta \paren {\frac 1 \beta - t} }
}}
{{eqn | r = \frac 1 {1 - \beta t}
}}
{{end-eqn}}
{{qed}}
Theorem: Moment Generating Function of Discrete Uniform Distribution

Let $X$ be a discrete random variable with a discrete uniform distribution with parameter $n$ for some $n \in \N$.
Then the moment generating function $M_X$ of $X$ is given by: 
:$\map {M_X} t = \dfrac {e^t \paren {1 - e^{n t} } } {n \paren {1 - e^t} }$

Proof:
From the definition of the discrete uniform distribution, $X$ has probability mass function:
:$\map \Pr {X = N} = \dfrac 1 n$
From the definition of a moment generating function:
:$\ds \map {M_X} t = \expect {e^{t X} } = \sum_{N \mathop = 1}^n \map \Pr {X = N} e^{N t}$
So: 
{{begin-eqn}}
{{eqn	| l = \map {M_X} t
	| r = \frac 1 n \sum_{N \mathop = 1}^n \paren {e^t}^N
}}
{{eqn	| r = \frac {e^t} n \sum_{N \mathop = 0}^{n - 1} \paren {e^t}^N
}}
{{eqn	| r = \frac {e^t \paren {1 - e^{n t} } } {n \paren {1 - e^t} }
	| c = Sum of Geometric Sequence with $r = e^t$
}}
{{end-eqn}}
{{qed}}
Category:Moment Generating Functions
Category:Discrete Uniform Distribution
2rk2fqdy8gvhthd9qhb5qjz7z9d6i26
Theorem: Maximal Ideal iff Quotient Ring is Field/Proof 1/Maximal Ideal implies Quotient Ring is Field

Let $\struct {R, +, \circ}$ be a commutative ring with unity whose zero is $0_R$ and whose unity is $1_R$.
Let $J$ be an ideal of $R$.
Let $J$ be a maximal ideal.
Then the quotient ring $R / J$ is a field.

Proof:
Let $J$ be a maximal ideal.
Because $J \subset R$, it follows from Quotient Ring of Commutative Ring is Commutative and Quotient Ring of Ring with Unity is Ring with Unity that $R / J$ is a commutative ring with unity.
We now need to prove that every non-zero element of $\struct {R / J, +,  \circ}$ has a product inverse in $R / J$.
Let $x \in R$ such that $x + J \ne J$, that is: $x \notin J$.
Thus $x + J \in R / J$ is not the zero element of $R / J$.
Take $K \subseteq R$ such that:
:$K = \set {j + r \circ x: j \in J, r \in R}$
that is, the subset of $R$ which can be expressed as a sum of an element of $J$ and a product in $R$ of $x$.
Now $0_R \in K$ as $0_R \in J$ and $0_R \in R$, giving $0_R + 0_R \circ x = 0_R$.
So:
:$(1): \quad K \ne \O$
Now let $g, h \in K$.
That is:
:$g = j_1 + r_1 \circ x, h = j_2 + r_2 \circ x$
Then:
:$-h = -j_2 + \paren {-r_2} \circ x$
But $j_1 - j_2 \in J$ from Test for Ideal.
Similarly $-r_2 \in R$.
So $-h \in K$ and we have:
:$(2) \quad g + \paren {-h} = \paren {j_1 - j_2} + \paren {r_1 - r_2} \circ x$
Now consider $g \in J, y \in R$.
Then:
:$g \circ y = \paren {j_1 + r_1 \circ x} \circ y = \paren {j_1 \circ y} + \paren {r_1 \circ y} \circ x$
which is valid by the fact that $R$ is a commutative ring.
But as $J$ is an ideal:
:$\paren {j_1 \circ y} \in J$
while $r_1 \circ y \in R$.
Thus:
:$(3) \quad g \circ y \in K$
and similarly:
:$(3) \quad y \circ g \in K$
So Test for Ideal can be applied to statements $(1)$ to $(3)$, and it is seen that $K$ is an ideal of $R$.
Now:
{{begin-eqn}}
{{eqn | l = j
      | o = \in
      | r = J
      | c = 
}}
{{eqn | ll= \leadsto
      | l = j + 0_R \circ x
      | o = \in
      | r = K
      | c = 
}}
{{eqn | ll= \leadsto
      | l = j
      | o = \in
      | r = K
      | c = 
}}
{{eqn | ll= \leadsto
      | l = J
      | o = \subseteq
      | r = K
      | c = 
}}
{{end-eqn}}
and because $x = 0_R + 1_R \circ x$ (remember $0_R \in J$), then $x \in K$ too.
So, because $x \notin J$, $K$ is an ideal such that $J \subset K \subseteq R$.
Because $J$ is a maximal ideal, then $K = R$.
Thus $1_R \in K$ and thus:
:$\exists j_0 \in J, s \in R: 1_R = j_0 + s \circ x$
So:
:$1_R + \paren {-s \circ x} = j_0 \in J$
Hence:
:$1_R + J = s \circ x + J = \paren {s + J} \circ \paren {x + J}$
So in the commutative ring $\struct {R / J, +,  \circ}$, the product inverse of $x + J$ is $s + J$.
The result follows.
Theorem: Maximal Ideal iff Quotient Ring is Field/Proof 1/Quotient Ring is Field implies Ideal is Maximal

Let $\struct {R, +, \circ}$ be a commutative ring with unity whose zero is $0_R$ and whose unity is $1_R$.
Let $J$ be an ideal of $R$.
Let the quotient ring $R / J$ be a field.
Then $J$ is a maximal ideal.

Proof:
Let $R / J$ be a field.
Let $K$ be a left ideal of $R$ such that $J \subsetneq K \subseteq R$.
We have that $J$ is the zero of $R / J$.
Let $x \in K \setminus J$.
Because $x \notin J$ then $x + J \ne J$.
Because $R / J$ is a field then $x + J \in R / J$ has a product inverse, say $s + J$.
Hence:
:$1_R + J = \paren {s + J} \circ \paren {x + J} = \paren {s \circ x } + J$
By Left Cosets are Equal iff Product with Inverse in Subgroup:
:$1_R - s \circ x \in J \subsetneq K$
By the definition of an ideal:
:$x \in K, s \in R \implies s \circ x \in K$
:$1_R - s \circ x \in K, s \circ x \in K \implies \paren {1_R - s \circ x} + \paren {s \circ x} = 1_R \in K$
:$1_R \in K \implies \forall y \in R:  y \circ 1_R = y \in K$
Hence $K = R$.
The result follows.
Theorem: Ring of Polynomial Forms is not necessarily Isomorphic to Ring of Polynomial Functions

Let $D$ be an integral domain.
Let $D \sqbrk X$ be the ring of polynomial forms in $X$ over $D$.
Let $\map P D$ be the ring of polynomial functions over $D$.
Then it is not necessarily the case that $D \sqbrk X$ is isomorphic with $\map P D$.

Proof:
Proof by Counterexample:
Consider the integral domain $\struct {\Z_2, +, \times}$.
From Ring of Integers Modulo Prime is Integral Domain, it is seen that $\struct {\Z_2, +, \times}$ is indeed an integral domain.
Consider the ring of polynomial forms $\Z_2 \sqbrk X$.
This is an infinite ring, as it can be seen that $S \subseteq \Z_2 \sqbrk X$ where:
:$S = \set {1, X, X^2, X^3, \dotsc}$
But the ring of polynomial functions $\map P D$ is finite, as:
:$\map P D \subseteq \Z_2^{Z_2}$
where $\Z_2^{Z_2}$ is the set of all mappings from $\Z_2$ to $\Z_2$, and has $4$ elements.
{{qed}}
Theorem: Moment Generating Function of Bernoulli Distribution

Let $X$ be a discrete random variable with a Bernoulli distribution with parameter $p$ for some $0 \le p \le 1$. 
Then the moment generating function $M_X$ of $X$ is given by: 
:$\map {M_X} t = q + p e^t$
where $q = 1 - p$.

Proof:
From the definition of the Bernoulli distribution, $X$ has probability mass function:
:<nowiki>$\map \Pr {X = n} = \begin{cases}
q & : n = 0 \\
p & : n = 1 \\
0 & : n \notin \set {0, 1} \\
\end{cases}$</nowiki>
From the definition of a moment generating function:
:$\ds \map {M_X} t = \expect {e^{t X} } = \sum_{n \mathop = 0}^1 \map \Pr {X = n} e^{t n}$
So:
{{begin-eqn}}
{{eqn	| l = \map {M_X} t
	| r = \map \Pr {X = 0} e^0 + \map \Pr {X = 1} e^t
}}
{{eqn	| r = q + p e^t
	| c = Exponential of Zero
}}
{{end-eqn}}
{{qed}}
Category:Moment Generating Functions
Category:Bernoulli Distribution
en8cic7v1y1c1jy5klf5v7tzgute8py
Theorem: Field of Quotients of Ring of Polynomial Forms on Reals that yields Complex Numbers

Let $\struct {\R, +, \times}$ denote the field of real numbers.
Let $X$ be transcendental over $\R$.
Let $\R \sqbrk X$ be the ring of polynomials in $X$ over $F$.
Consider the field of quotients:
:$\R \sqbrk X / \ideal p$
where:
:$p = X^2 + 1$
:$\ideal p$ denotes the ideal generated by $p$.
Then $\R \sqbrk X / \ideal p$ is the field of complex numbers.

Proof:
It is taken as read that $X^2 + 1$ is irreducible in $\R \sqbrk X$.
Hence by Polynomial Forms over Field form Principal Ideal Domain: Corollary 1, $\R \sqbrk X / \ideal p$ is indeed a field.
Let $\nu$ be the quotient epimorphism from $\R \sqbrk X$ onto $\R \sqbrk X / \ideal p$.
From Quotient Ring Epimorphism is Epimorphism:
:$\map \ker {\map \nu \R} = \R \cap \ideal p = \set 0$
So $\map \nu \R$ is a monomorphism from $\R \sqbrk X$ to $\R \sqbrk X / \ideal p$.
Thus $\map \nu \R$ is an isomorphic copy of $\R$ inside $\R \sqbrk X / \ideal p$.
We identify this isomorphic copy of $\R$ with $\R$ itself, ignoring the difference between $\map \nu x$ and $x$ when $x = \R$.
Hence:
:$(1): \quad \R \subseteq \R \sqbrk X / \ideal p$
Let $f \in \R \sqbrk X$ be arbitrary.
By Division Theorem for Polynomial Forms over Field:
:$\exists q, r \in \R \sqbrk X: f = q p + r$
where $r =a + b X$ for some $a, b \in \R$.
Hence:
{{begin-eqn}}
{{eqn | l = \map \nu f
      | r = \map \nu q \, \map \nu p + \map \nu r
      | c = as $\nu$ is a ring honomorphism
}}
{{eqn | r = \map \nu r
      | c = as $p \in \map \ker \nu$
}}
{{eqn | r = \map \nu a + \map \nu b \, \map \nu X
      | c = 
}}
{{eqn | r = a + b \, \map \nu X
      | c = as we have identified $\map \nu \R$ with $\R$
}}
{{eqn | r = a + b i
      | c = where $i := \map \nu X$
}}
{{end-eqn}}
As $\nu$ is an epimorphism, it is a fortiori surjection.
Hence:
:$(2): \quad$ Every element $w$ of $\R \sqbrk X / \ideal p$ can be expressed in the form:
::$w = a + b i$
:for some $a, b \in \R$.
Because $X^2 + 1 \in \ker \nu$, we have:
{{begin-eqn}}
{{eqn | l = 0
      | r = \map \nu {X^2 + 1}
      | c = 
}}
{{eqn | r = \paren {\map \nu X}^2 + \map \nu 1
      | c = 
}}
{{eqn | r = i^2 + 1
      | c = 
}}
{{end-eqn}}
Hence we have that:
:$(3): \quad$ In $\R \sqbrk X / \ideal p$, $i^2 = -1$
Thus $(2)$ can be improved to:
:$(4): \quad$ Every element $w$ of $\R \sqbrk X / \ideal p$ can be expressed '''uniquely''' in the form:
::$w = a + b i$
:for some $a, b \in \R$.
From $(1)$, $(3)$ and $(4)$, the field $\R \sqbrk X / \ideal p$ is recognised as the field of complex numbers.
{{Qed}}
Theorem: Double of Antiperiodic Element is Periodic

Let $f: \R \to \R$ be a real function.
Let $L \in \R_{>0}$ be an anti-periodic element of $f$.
Then $2 L$ is a periodic element of $f$.
In other words, every anti-periodic function is also periodic.

Proof:
By Non-Zero Real Numbers Closed under Multiplication we have that $2 L \in \R_{>0}$.
Then:
{{begin-eqn}}
{{eqn | l = \map f {x + 2 L}
      | r = \map f {x + \paren {L + L} }
}}
{{eqn | r = \map f {\paren {x + L} + L}
      | c = Real Addition is Associative
}}
{{eqn | r = -\map f {x + L}
}}
{{eqn | r = \map f x
      | c = Negative of Negative Real Number
}}
{{end-eqn}}
{{qed}}
Category:Antiperiodic Functions
tb9pnuqm9ub1x4vfyezfo6j08kq35wv
Theorem: Idempotent Ring has Characteristic Two/Corollary

:$\forall x \in R: -x = x$

Proof:
Let $0_R$ denote the zero of $R$.
Let $x \in R$.
Then:
{{begin-eqn}}
{{eqn | l = x + x
      | r = 0_R
      | c = Idempotent Ring has Characteristic Two
}}
{{eqn | ll= \leadsto
      | l = -x + x + x
      | r = -x + 0_R
      | c = 
}}
{{eqn | ll= \leadsto
      | l = x
      | r = -x
      | c = 
}}
{{end-eqn}}
Hence the result.
{{qed}}
Theorem: Skewness of Normal Distribution

Let $X$ be a continuous random variable with a normal distribution with parameters $\mu$ and $\sigma^2$ for some $\mu \in \R$ and $\sigma \in \R_{> 0}$. 
Then the skewness $\gamma_1$ of $X$ is equal to $0$.

Proof:
From the definition of skewness: 
:$\gamma_1 = \expect {\paren {\dfrac {X - \mu} \sigma}^3}$
From the definition of the normal distribution, $X$ has probability density function:
:$\map {f_X} x = \dfrac 1 {\sigma \sqrt{2 \pi} } \, \map \exp {-\dfrac { \paren {x - \mu}^2} {2 \sigma^2} }$
So, from Expectation of Function of Continuous Random Variable:
:$\ds \gamma_1 = \expect {\paren {\dfrac {X - \mu} \sigma}^3} = \dfrac 1 {\sigma \sqrt{2 \pi} } \int_{-\infty}^\infty \paren {\dfrac {x - \mu} \sigma}^3 \map \exp {-\dfrac { \paren {x - \mu}^2} {2 \sigma^2} } \rd x$
Making a substitution of $u = x - \mu$:
:$\ds \gamma_1 = \dfrac 1 {\sigma \sqrt{2 \pi} } \int_{-\infty}^\infty \paren {\dfrac u \sigma}^3 \map \exp {-\dfrac {u^2} {2 \sigma^2} } \rd u$
We have that: 
:$\paren {-u}^3 \map \exp {-\dfrac {\paren {-u}^2} {2 \sigma^2} } = -u^3 \map \exp {-\dfrac {u^2} {2 \sigma^2} }$
So we can see that the integrand is odd.
So, by Definite Integral of Odd Function:
:$\ds \dfrac 1 {\sigma \sqrt{2 \pi} } \int_{-\infty}^\infty \paren {\dfrac u \sigma}^3 \map \exp {-\dfrac {u^2} {2 \sigma^2} } \rd u = 0$
giving: 
:$\gamma_1 = 0$ 
{{qed}}
Theorem: Skewness of Normal Distribution

Let $X$ be a continuous random variable with a normal distribution with parameters $\mu$ and $\sigma^2$ for some $\mu \in \R$ and $\sigma \in \R_{> 0}$. 
Then the skewness $\gamma_1$ of $X$ is equal to $0$.

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 Normal Distribution, we have: 
:$\mu = \mu$
By Variance of Normal Distribution, we have: 
:$\sigma = \sigma$ 
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 \mu \paren {\sigma^2 + \mu^2} + 3 \mu^2 \paren \mu - \mu^3} {\sigma^3}
      | c = Variance of Normal Distribution
}}
{{end-eqn}}
To calculate $\gamma_1$, we must calculate $\expect {X^3}$.
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 Normal Distribution: Third Moment: 
:$\map { {M_X}' ' '} t = \paren {3 \sigma^2 \paren {\mu + \sigma^2 t} + \paren {\mu + \sigma^2 t}^3} \map \exp {\mu t + \dfrac 1 2 \sigma^2 t^2}$
Setting $t = 0$:
{{begin-eqn}}
{{eqn | l = \expect {X^3}
      | r = \paren {3 \sigma^2 \paren {\mu + \sigma^2 0} + \paren {\mu + \sigma^2 0}^3} \map \exp {\mu 0 + \dfrac 1 2 \sigma^2 0^2}
}}
{{eqn | r = 3 \mu \sigma^2 + \mu^3
      | c = Exponential of Zero
}}
{{end-eqn}}
So:
{{begin-eqn}}
{{eqn | l = \gamma_1
      | r = \frac {\expect {X^3} - 3 \mu \paren {\sigma^2 + \mu^2} + 3 \mu^2 \paren {\mu} - \mu^3} {\sigma^3}
}}
{{eqn | r = \frac {\paren {3 \mu \sigma^2 + \mu^3 } - 3 \mu \paren {\sigma^2 + \mu^2} + 3 \mu^2 \paren {\mu} - \mu^3} {\sigma^3}
}}
{{eqn | r = 0
}}
{{end-eqn}}
{{qed}}
Theorem: Inverse of Unit in Centralizer of Ring is in Centralizer

Let $\struct {R, +, \circ}$ be a ring.
Let $S$ be a subset of $R$.
Let $\map {C_R} S$ denote the centralizer of $S$ in $R$
Let $u \in R$ be a unit of $R$.
Then:
:$u \in \map {C_R} S \implies u^{-1} \in \map {C_R} S$

Proof:
Let $u \in R$ be a unit of $R$.
Let $u \in \map {C_R} S$.
Then from Commutation with Inverse in Monoid:
:$u^{-1} \in \map {C_R} S$
{{qed}}
Theorem: Inverse of Central Unit of Ring is in Center

Let $R$ be a ring.
Let $\map Z R$ denote the center of $R$.
Let $u \in R$ be a unit of $R$.
Then:
:$u \in \map Z R \implies u^{-1} \in \map Z R$

Proof:
Follows directly from the definition of center and Inverse of Unit in Centralizer of Ring is in Centralizer.
{{qed}}
Theorem: Skewness in terms of Non-Central Moments

Let $X$ be a random variable with mean $\mu$ and standard deviation $\sigma$.
Then the skewness $\gamma_1$ of $X$ is given by: 
:$\gamma_1 = \dfrac {\expect {X^3} - 3 \mu \sigma^2 - \mu^3} {\sigma^3}$

Proof:
{{begin-eqn}}
{{eqn | l = \gamma_1
      | r = \expect {\paren {\dfrac {X - \mu} \sigma}^3}
      | c = {{Defof|Skewness}}
}}
{{eqn | r = \frac {\expect {X^3 - 3 \mu X^2 + 3 \mu^2 X - \mu^3} } {\sigma^3} 
      | c = Expectation is Linear, 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 \mu \paren {\expect {X^2} - \mu \expect X} - \mu^3} {\sigma^3}
}}
{{eqn | r = \frac {\expect {X^3} - 3 \mu \paren {\expect {X^2} - \paren {\expect X}^2} - \mu^3} {\sigma^3}
      | c = $\mu = \expect X$
}}
{{eqn | r = \frac {\expect {X^3} - 3 \mu \sigma^2 - \mu^3} {\sigma^3}
      | c = Variance as Expectation of Square minus Square of Expectation
}}
{{end-eqn}}
{{qed}}
Category:Skewness
e0smnr31tb3xkgmkuewwu26ytqynuro
Theorem: Intersection of Ring Ideals is Ideal

Let $\struct {R, +, \circ}$ be a ring
Let $\mathbb L$ be a non-empty set of ideals of $R$.
Then the intersection $\bigcap \mathbb L$ of the members of $\mathbb L$ is itself an ideal of $R$.

Proof:
Let $L = \bigcap \mathbb L$.
From Intersection of Subrings is Subring, we have that $L$ is a subring of $R$.
Let $x \in L$ and $y \in R$.
Then:
:$\forall T \in \bigcap \mathbb L: x \circ y \in T, y \circ x \in T$
as every element of $\bigcap \mathbb L$, including $T$, is an ideal of $R$.
If $y \in R$, then $x \circ y$ and $y \circ x$ are in every element of $\mathbb L$, and hence in $L$.
So $L$ is an ideal of $R$.
{{qed}}
Theorem: Set of Ring Elements forming Zero Product with given Element is Ideal

Let $\struct {R, +, \circ}$ be a commutative ring whose zero is $0_R$.
Let $a \in R$ be an arbitrary element of $R$.
Let $A$ be the subset of $R$ defined as:
:$A = \set {x \in R: x \circ a = 0_R}$
Then $A$ is an ideal of $A$.

Proof:
By definition of ring zero:
:$\forall x \in R: x \circ 0_R = 0_R$
Hence $0_R \in A$ and so $A \ne \O$.
Let $a, b \in A$.
{{begin-eqn}}
{{eqn | q = \forall x \in R
      | l = x \circ b
      | r = 0_R
      | c = 
}}
{{eqn | ll= \leadsto
      | l = -\paren {x \circ b}
      | r = 0_R
      | c = 
}}
{{eqn | ll= \leadsto
      | l = x \circ \paren {-b}
      | r = 0_R
      | c = 
}}
{{end-eqn}}
Thus:
{{begin-eqn}}
{{eqn | q = \forall x \in R
      | l = x \circ a
      | r = 0_R
      | c = 
}}
{{eqn | lo= \land
      | l = x \circ \paren {-b}
      | r = 0_R
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \paren {x \circ a} + \paren {x \circ \paren {-b} }
      | r = 0_R
      | c = as $0_R$ is identity of $\struct {R, +}$
}}
{{eqn | ll= \leadsto
      | l = x \circ \paren {a + \paren {-b} }
      | r = 0_R
      | c = {{Ring-axiom|D}}
}}
{{eqn | ll= \leadsto
      | l = a + \paren {-b}
      | o = \in
      | r = A
      | c = Definition of $A$
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | q = \forall x \in R
      | l = x \circ a
      | r = 0_R
      | c = 
}}
{{eqn | lo= \land
      | l = x \circ b
      | r = 0_R
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \paren {x \circ a} \circ b
      | r = 0_R
      | c = {{Defof|Ring Zero}}
}}
{{eqn | ll= \leadsto
      | l = x \circ \paren {a \circ b}
      | r = 0_R
      | c = {{Ring-axiom|M1}}
}}
{{eqn | ll= \leadsto
      | l = a \circ b
      | o = \in
      | r = A
      | c = Definition of $A$
}}
{{end-eqn}}
Hence the result, from Test for Ideal:
{{qed}}
Theorem: Skewness of Bernoulli Distribution

Let $X$ be a discrete random variable with a Bernoulli distribution with parameter $p$.
Then the skewness $\gamma_1$ of $X$ is given by: 
:$\gamma_1 = \dfrac {1 - 2 p} {\sqrt {p q} }$
where $q = 1 - p$.

Proof:
From Skewness in terms of Non-Central Moments: 
:$\gamma_1 = \dfrac {\expect {X^3} - 3 \mu \sigma^2 - \mu^3} {\sigma^3}$ 
where $\mu$ is the mean of $X$, and $\sigma$ the standard deviation.
We have, by Expectation of Bernoulli Distribution:
:$\mu = p$
By Variance of Bernoulli Distribution, we also have: 
:$\var X = \sigma^2 = p \paren {1 - p}$
so: 
:$\sigma = \sqrt {p \paren {1 - p} }$
By Raw Moment of Bernoulli Distribution, we have: 
:$\expect {X^3} = p$
So:
{{begin-eqn}}
{{eqn	| l = \gamma_1
	| r = \frac {p - 3 p^2 \paren {1 - p} - p^3} {p^{3/2} \paren {1 - p}^{3/2} }
}}
{{eqn	| r = \frac {p \paren {1 - p^2} - 3 p^2 \paren {1 - p} } {p^{3/2} \paren {1 - p}^{3/2} }
}}
{{eqn	| r = \frac {p \paren {1 - p} \paren {1 + p} - 3 p^2 \paren {1 - p} } {p^{3/2} \paren {1 - p}^{3/2} }
	| c = Difference of Two Squares
}}
{{eqn	| r = \frac {p \paren {1 - p} \paren {1 + p - 3p} } {p^{3/2} \paren {1 - p}^{3/2} }
}}
{{eqn	| r = \frac {1 - 2 p} {\sqrt {p \paren {1 - p} } }
}}
{{eqn	| r = \frac {1 - 2 p} {\sqrt {p q} }
	| c = $q = 1 - p$
}}
{{end-eqn}}
{{qed}}
Category:Bernoulli Distribution
Category:Skewness
sdipu0lqoiyvtqullidfadba1bjeyul
Theorem: Intersection of Subrings is Subring

Let $\struct {R, +, \circ}$ be a ring.
Let $\mathbb L$ be a non-empty set of subrings of $R$.
Then the intersection $\bigcap \mathbb L$ of the members of $\mathbb L$ is itself a subring of $R$.

Proof:
Let $L = \bigcap \mathbb L$.
By Intersection of Subgroups is Subgroup, $\struct {L, +}$ is a subgroup of $\struct {R, +}$.
By the One-Step Subgroup Test:
:$\forall x, y \in \struct {L, +}: x + \paren {-y} \in L$
By Intersection of Subsemigroups, $\struct {L, \circ}$ a subsemigroup of $\struct {R, \circ}$.
So by definition of subsemigroup:
:$\forall x, y \in \struct {L, \circ}: x \circ y \in L$
By the Subring Test it follows that $\struct {L, +, \circ}$ is a subring of $\struct {R, +, \circ}$.
{{qed}}
Theorem: Median of Exponential Distribution

Let $X$ be a continuous random variable of the exponential distribution with parameter $\beta$ for some $\beta \in \R_{> 0}$. 
Then the median of $X$ is equal to $\beta \ln 2$.

Proof:
Let $M$ be the median of $X$.
From the definition of the exponential distribution, $X$ has probability density function: 
:$\map {f_X} x = \dfrac 1 \beta e^{-\frac x \beta}$
Note that $f_X$ is non-zero, so the median is unique.
{{Explain|Why the above follows}}
We have by the definition of a median: 
:$\ds \map \Pr {X < M} = \frac 1 \beta \int_0^M e^{-\frac x \beta} \rd x = \frac 1 2$
Evaluating this integral:
{{begin-eqn}}
{{eqn | l = \frac 1 \beta \int_0^M e^{-\frac x \beta} \rd x
      | r = \frac 1 \beta \intlimits {-\beta e^{-\frac x \beta} } 0 M
      | c = Primitive of Exponential Function
}}
{{eqn | r = 1 - e^{-\frac M \beta}
}}
{{end-eqn}}
So: 
:$1 - e^{-\frac M \beta} = \dfrac 1 2$
Then: 
:$e^{-\frac M \beta} = \dfrac 1 2$
giving: 
:$-\dfrac M \beta = \map \ln {\dfrac 1 2}$
So, by Logarithm of Reciprocal: 
:$M = \beta \ln 2$
{{qed}}
Category:Exponential Distribution
Category:Medians
c3e8njyhqzs5quqm5li9pwj6w8slk5e
Theorem: Intersection of Ring Ideals is Largest Ideal Contained in all Ideals

Let $\struct {R, +, \circ}$ be a ring
Let $\mathbb L$ be a non-empty set of ideals of $R$.
Then the intersection $\bigcap \mathbb L$ of the members of $\mathbb L$ is the largest ideal of $R$ contained in each member of $\mathbb L$.

Proof:
Let $L = \bigcap \mathbb L$.
From Intersection of Ring Ideals is Ideal  $L$ is indeed an ideal of $R$.
Let $L = \bigcap \mathbb L$.
From Intersection of Subrings is Largest Subring Contained in all Subrings, we have that $L$ is the largest subring of $R$ contained in each member of $\mathbb L$.
As $L$ is the largest subring of $R$ contained in each member of $\mathbb L$, and it is an ideal of $R$, there can be no larger ideal as it would then not be a subring.
So $L$ is the largest ideal of $R$ contained in each member of $\mathbb L$.
{{qed}}