avewright/proofwiki-math huggingface dataset:
Theorem: Intersection of Subrings is Largest Subring Contained in all Subrings

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

Proof:
Let $\ds L = \bigcap \mathbb L$.
From Intersection of Subrings is Subring, $L$ is indeed a subring of $R$.
By Intersection of Subgroups is Subgroup, $\struct {L, +}$ is the largest subgroup of $\struct {R, +}$ contained in each member of $\mathbb L$.
By Intersection of Subsemigroups, $\struct {L, \circ}$ is the largest subsemigroup of $\struct {R, \circ}$ contained in each member of $\mathbb L$.
Let $S$ be a subring of $R$ such that:
:$\forall K \in \mathbb L: S \subseteq K$
By Intersection is Largest Subset, $S \subseteq L$.
Thus $L$ is the largest subring of $R$ contained in each member of $\mathbb L$.
{{qed}}
Theorem: Median of Continuous Uniform Distribution

Let $X$ be a continuous random variable which is uniformly distributed on a closed real interval $\closedint a b$. 
Then the median $M$ of $X$ is given by: 
:$M = \dfrac {a + b} 2$

Proof:
From the definition of the continuous uniform distribution, $X$ has probability density function:
:$\map {f_X} x = \begin{cases} \dfrac 1 {b - a} & : a \leq x \leq b \\ 0 & : \text{otherwise} \end{cases}$
Note that $f_X$ is non-zero, so the median is unique. 
We have by the definition of a median: 
{{begin-eqn}}
{{eqn | l = \map \Pr {X < M}
      | r = \frac 1 2
}}
{{eqn | ll= \leadsto
      | l = \int_{-\infty}^M \frac 1 {b - a} \rd x
      | r = \frac 1 2
}}
{{eqn | ll= \leadsto
      | l = \int_a^M \frac 1 {b - a} \rd x
      | r = \frac 1 2
      | c = $\map {f_X} x = 0$ when $x < a$
}}
{{eqn | ll= \leadsto
      | l = \dfrac {M - a} {b - a}
      | r = \frac 1 2
      | c = Primitive of Constant
}}
{{end-eqn}}
So:
{{begin-eqn}}
{{eqn | l = M
      | r = a + \frac {b - a} 2
}}
{{eqn | r = \frac {b - a + 2 a} 2
}}
{{eqn | r = \frac {b + a} 2
}}
{{end-eqn}}
{{qed}}
Category:Continuous Uniform Distribution
Category:Medians
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Theorem: Intersection of All Ring Ideals Containing Subset is Smallest

Let $\struct {R, +, \circ}$ be a ring
Let $S \subseteq R$ be a subset of $R$.
Let $L$ be the intersection of the set of all ideals of $R$ containing $S$.
Then $L$ is the smallest ideal of $R$ containing $S$.

Proof:
From Intersection of All Subrings Containing Subset is Smallest, $L$ is the smallest subring of $R$ containing $S$.
From Intersection of Ring Ideals is Ideal, $L$ is an ideal of $R$.
As $L$ is the smallest subring of $R$ containing $S$, and it is an ideal of $R$, there can be no smaller ideal of $R$ containing $S$ as it would then not be a subring.
So the intersection of the set of all ideals of $R$ containing $S$ is the smallest ideals of $R$ containing $S$. 
{{qed}}
Theorem: Intersection of All Subrings Containing Subset is Smallest

Let $\struct {R, +, \circ}$ be a ring.
Let $S \subseteq R$ be a subset of $R$.
Let $L$ be the intersection of the set of all subrings of $R$ containing $S$.
Then $L$ is the smallest subring of $R$ containing $S$.

Proof:
From Intersection of Subrings is Subring, $L$ is indeed a subring of $R$.
Let $T$ be a subring of $R$ containing $S$.
Let $x, y \in L$.
By the Subring Test, we have that:
{{begin-eqn}}
{{eqn | l = x - y
      | o = \in
      | r = L
}}
{{eqn | l = x \circ y
      | o = \in
      | r = L
}}
{{end-eqn}}
By Intersection is Largest Subset, it follows that $x, y \in T$.
But $T$ is also a subring of $R$.
So, by the Subring Test again, we have that:
{{begin-eqn}}
{{eqn | l = x - y
      | o = \in
      | r = T
}}
{{eqn | l = x \circ y
      | o = \in
      | r = T
}}
{{end-eqn}}
So by definition of subset, $L \subseteq T$.
{{qed}}
Theorem: Intersection of Division Subrings is Division Subring

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

Proof:
Let $\ds L = \bigcap \mathbb K$.
Let $0$ be the zero of $\struct {D, +, \circ}$.
By Intersection of Subgroups is Subgroup: General Result, $\struct {L, +}$ is a subgroup of $\struct {D, +}$.
By the One-Step Subgroup Test:
:$\forall x, y \in L: x + \paren {-y} \in L$
By Non-Zero Elements of Division Ring form Group:
:$\struct {D \setminus \set 0, \circ}$ is a group
:$\struct {K \setminus \set 0, \circ}$ is a group for each $K \in \mathbb K$
By Set Difference over Subset, $\struct {K \setminus \set 0, \circ}$ is a subgroup of $\struct {D \setminus \set 0, \circ}$ for each $K \in \mathbb K$.
By Set Difference is Right Distributive over Set Intersection:
:$\ds L \setminus \set 0 = \bigcap_{K \mathop \in \mathbb K} \paren {K \setminus \set 0}$
By Intersection of Subgroups is Subgroup: General Result, $\struct {L \setminus \set 0, \circ}$ is a subgroup of $\struct {D \setminus \set 0, \circ}$.
By the Two-Step Subgroup Test and Ring Product with Zero:
{{begin-eqn}}
{{eqn | q = \forall x, y \in L
      | l = x \circ y
      | o = \in
      | r = L
}}
{{eqn | q = \forall x \in L \setminus \set 0
      | l = x^{-1}
      | o = \in
      | r = L \setminus \set 0
}}
{{end-eqn}}
By the Division Subring Test it follows that $\struct {L, +, \circ}$ is a division subring of $\struct {D, +, \circ}$.
{{qed}}
Category:Division Subrings
Category:Set Intersection
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Theorem: Intersection of Subfields is Subfield

Let $\struct {F, +, \circ}$ be a field.
Let $\mathbb K$ be a non-empty set of subfields of $F$.
Then the intersection $\ds \bigcap \mathbb K$ of the members of $\mathbb K$ is itself a subfield of $F$.

Proof:
Let $\ds L = \bigcap \mathbb K$.
A field is by definition also a division ring.
From Intersection of Division Subrings is Division Subring, $L$ is itself a division subring of $F$.
As $\struct {F, +, \circ}$ is a field, $\circ$ is commutative on $F$.
By Restriction of Commutative Operation is Commutative, it follows that $\circ$ is also commutative on $L$.
Thus $\struct {L, +, \circ}$ is a division ring where $\circ$ is commutative.
Thus, by definition, of $\struct {L, +, \circ}$ is a field.
Hence the result.
{{qed}}
Theorem: Intersection of Division Subrings is Largest Division Subring Contained in all Division Subrings

Let $\struct {D, +, \circ}$ be a division ring.
Let $\mathbb K$ be a non-empty set of division subrings of $D$.
Let $\ds \bigcap \mathbb K$ be the intersection of the elements of $\mathbb K$.
Then $\ds \bigcap \mathbb K$ is the largest division subring of $D$ contained in each element of $\mathbb K$.

Proof:
Let $\ds L = \bigcap \mathbb K$.
Let $0$ be the zero of $\struct {D, +, \circ}$.
From Intersection of Division Subrings is Division Subring, $\struct {L, +, \circ}$ is a division subring of $\struct {D, +, \circ}$.
By Intersection of Subgroups is Subgroup, $\struct {L, +}$ is the largest subgroup of $\struct {D, +}$ contained in each element of $\mathbb K$.
By Intersection of Subgroups is Subgroup, $\struct {L \setminus \set 0, \circ}$ is the largest subgroup of $\struct {D \setminus \set 0, \circ}$ contained in each element of $\mathbb K$.
Let $S$ be a subring of $D$ such that:
:$\forall K \in \mathbb K: S \subseteq K$
By Intersection is Largest Subset, $S \subseteq L$.
Thus $L$ is the largest division subring of $D$ contained in each element of $\mathbb L$.
{{qed}}
Theorem: Intersection of Division Subrings Containing Subset is Smallest

Let $\struct {D, +, \circ}$ be a division ring.
Let $S \subseteq D$ be a subset of $D$.
Let $L$ be the intersection of the set of all division subrings of $D$ containing $S$.
Then $L$ is the smallest division subring of $D$ containing $S$.

Proof:
From Intersection of Division Subrings is Division Subring, $L$ is indeed a division subring of $D$.
Let $T$ be a division subring of $D$ containing $S$.
Let $x, y \in L$.
By the Division Subring Test, we have that:
{{begin-eqn}}
{{eqn | l = x - y
      | o = \in
      | r = L
}}
{{eqn | l = x \circ y
      | o = \in
      | r = L
}}
{{eqn | l = x^{-1} \circ y
      | o = \in
      | r = L
}}
{{end-eqn}}
By Intersection is Largest Subset, it follows that $x, y \in T$.
But $T$ is also a division subring of $D$.
So, by the Division Subring Test again, we have that:
{{begin-eqn}}
{{eqn | l = x - y
      | o = \in
      | r = T
}}
{{eqn | l = x \circ y
      | o = \in
      | r = T
}}
{{eqn | l = x^{-1} \circ y
      | o = \in
      | r = T
}}
{{end-eqn}}
So by definition of subset, $L \subseteq T$.
{{qed}}
Theorem: Intersection of Subfields Containing Subset is Smallest

Let $\struct {F, +, \circ}$ be a field.
Let $S \subseteq F$ be a subset of $F$.
Let $L$ be the intersection of the set of all subfields of $F$ containing $S$.
Then $L$ is the smallest subfield of $F$ containing $S$.

Proof:
A field is by definition also a division subring.
Thus $L$ is the intersection of the set of all division subrings of $F$ containing $S$.
From Intersection of Division Subrings Containing Subset is Smallest, $L$ is the smallest division subring of $F$ containing $S$.
From Intersection of Subfields is Subfield, $L$ is a subfield of $F$.
As $L$ is the smallest division subring of $F$ containing $S$, and it is a subfield of $F$, there can be no smaller subfield of $F$ containing $S$ as it would then not be a division subring.
So the intersection of the set of all subfields of $F$ containing $S$ is the smallest subfield of $F$ containing $S$. 
{{qed}}
Theorem: Intersection of Subfields is Largest Subfield Contained in all Subfields

Let $\struct {F, +, \circ}$ be a field.
Let $\mathbb K$ be a non-empty set of subfields of $F$.
Let $\bigcap \mathbb K$ be the intersection of the elements of $\mathbb K$.
Then $\bigcap \mathbb K$ is the largest subfield of $F$ contained in each element of $\mathbb K$.

Proof:
Let $L = \bigcap \mathbb K$.
From Intersection of Subfields is Subfield, $\struct {L, +, \circ}$ is itself a subfield of $\struct {F, +, \circ}$.
A field is by definition also a division subring.
Thus $\struct {L, +, \circ}$ is the largest division subring of $F$ contained in each element of $\mathbb K$.
But as $\struct {F, +, \circ}$ is a field, $\circ$ is commutative.
So by Restriction of Commutative Operation is Commutative, $\struct {L, +, \circ}$ is also a field.
Hence the result.
{{qed}}
Theorem: Equivalent Norms on Rational Numbers/Necessary Condition

Let $\norm {\, \cdot \,}_1$ and $\norm {\, \cdot \,}_2$ be norms on the rational numbers $\Q$.
Let $\norm {\, \cdot \,}_1$ and $\norm {\, \cdot \,}_2$ be equivalent norms.
Then:
:$\exists \alpha \in \R_{\gt 0}: \forall n \in \N: \norm n_1 = \norm n_2^\alpha$

Proof:
Let $\norm {\, \cdot \,}_1$ and $\norm {\, \cdot \,}_2$ be equivalent.
By Norm is Power of Other Norm then:
:$\exists \alpha \in \R_{\gt 0}: \forall q \in \Q: \norm q_1 = \norm q_2^\alpha$
In particular:
:$\exists \alpha \in \R_{\gt 0}: \forall n \in \N: \norm n_1 = \norm n_2^\alpha$
Theorem: Equivalent Norms on Rational Numbers/Sufficient Condition

Let $\norm {\, \cdot \,}_1$ and $\norm {\, \cdot \,}_2$ be norms on the rational numbers $\Q$.
Let $\norm {\, \cdot \,}_1$ and $\norm {\, \cdot \,}_2$ satisfy:
:$\exists \alpha \in \R_{\gt 0}: \forall n \in \N: \norm n_1 = \norm n_2^\alpha$
Then $\norm {\, \cdot \,}_1$ and $\norm {\, \cdot \,}_2$ are equivalent

Proof:
Let $\norm {\, \cdot \,}_1$ and $\norm {\, \cdot \,}_2$ satisfy:
:$\exists \alpha \in \R_{\gt 0}: \forall n \in \N: \norm n_1 = \norm n_2^\alpha$
By Norm of Negative then:
:$\forall n \in \N: \norm {-n}_1 = \norm n_1 = \norm n_2^\alpha = \norm {-n}_2^\alpha$
Hence:
:$\forall k \in \Z: \norm k_1 = \norm k_2^\alpha$
By Norm of Quotient then:
:$\forall \dfrac a b \in \Q: \norm {\dfrac a b}_1 = \dfrac {\norm a_1} {\norm b_1} = \dfrac {\norm a_2^\alpha} {\norm b_2^\alpha} = \norm {\dfrac a b}_2^\alpha$
By Norm is Power of Other Norm then  $\norm {\, \cdot \,}_1$ and $\norm {\, \cdot \,}_2$ are equivalent.
Theorem: Commutative and Unitary Ring with 2 Ideals is Field

Let $\struct {R, +, \circ}$ be a commutative ring with unity whose zero is $0_R$.
Let $\struct {R, +, \circ}$ be such that the only ideals of $\struct {R, +, \circ}$ are:
:$\set {0_R}$
and:
$\struct {R, +, \circ}$ itself.
That is, such that $\struct {R, +, \circ}$ has no non-null proper ideals.
Then $\struct {R, +, \circ}$ is a field.

Proof:
From Null Ring is Ideal and Ring is Ideal of Itself, it is always the case that $\set {0_R}$ and $\struct {R, +, \circ}$ are ideals of $\struct {R, +, \circ}$.
Let $a \in R^*$, where $R^* := R \setminus \set {0_R}$.
Let $\ideal a$ be the principal ideal of $R$ generated by $a$.
We have that $\ideal a$ is a non-null ideal and hence $\ideal a = R$.
Thus $1_R \in \ideal a$.
Thus $\exists x \in R: x \circ a = 1_R$ by the definition of principal ideal.
Therefore $a$ is invertible.
As $a$ is arbitrary, it follows that all such $a$ are invertible.
Thus by definition $\struct {R, +, \circ}$ is a division ring such that $\circ$ is commutative.
The result follows by definition of field.
{{Qed}}
Theorem: Field has 2 Ideals

Let $\struct {F, +, \circ}$ be a field whose zero is $0_F$ and whose unity is $1_F$.
Then the only ideals of $\struct {F, +, \circ}$ are $\struct {F, +}$ and $\set {0_F}$.
That is, $\struct {F, +, \circ}$ has no non-null proper ideals.

Proof:
By definition, a field is a division ring.
From Null Ring is Ideal and Ring is Ideal of Itself, it is always the case that $\set {0_F}$ and $\struct {F, +}$ are ideals of $\struct {F, +, \circ}$.
From Ideals of Division Ring, it follows that the only ideals of $\struct {F, +, \circ}$ are $\struct {F, +}$ and $\set {0_F}$.
{{qed}}
Theorem: Non-Commutative Ring with Unity and 2 Ideals not necessarily Division Ring

Let $\struct {R, +, \circ}$ be a ring with unity whose zero is $0_F$ and whose unity is $1_F$.
Let $\struct {R, +, \circ}$ specifically not be commutative.
Let $\struct {R, +, \circ}$ be such that the only ideals of $\struct {R, +, \circ}$ are $\set {0_R}$ and $R$ itself.
Then it is not necessarily the case that $\struct {R, +, \circ}$ is a division ring.

Proof:
Let $S$ be the set of square matrices of order $2$ over the real numbers $\R$.
$S$ is not a division ring, as for example:
:$\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$
and so both $\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$ and $\begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}$ are proper zero divisors of $S$.
Let $J$ be an ideal of $S$ containing a non-zero matrix $A$ whose $\tuple {r, s}$th entry is $\lambda \ne 0$.
Let $E_{i j}$ be the matrix of order $2$ defined as:
:$e_{a b} = \begin{cases} 1 & : a = i, b = j \\ 0 & : \text {otherwise} \end{cases}$
Thus for example:
:$E_{1 2} = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$
Then for each $i, j \in \set {1, 2}$, consider the matrix $\mathbf M = \dfrac 1 \lambda E_{i r} A E_{s j}$.
In Matrix Multiplication is Associative it is shown that, if $\mathbf A, \mathbf B, \mathbf C \in S$ and $\mathbf S = \mathbf A \mathbf B \mathbf C$:
:$\ds s_{m n} = \sum_{k \mathop = 1}^2 \sum_{l \mathop = 1}^2 a_{m l} b_{l k} c_{k n}$
Notice that each $E_{i j}$ only has one non-zero entry $e_{i j} = 1$.
From the equation above we see that only $m_{i j}$ is can be non-zero.
Moreover the summand is non-zero only when $l = r$ and $k = s$.
Hence:
:$m_{i j} = \dfrac 1 \lambda \times 1 \times a_{r s} \times 1 = 1$
and thus $\mathbf M = E_{i j}$.
It follows that:
:$\forall i, j \in \set {1, 2}: E_{i j} \in J$
It remains to be shown that $J$ is the whole of $S$. 
Since $J$ is an ideal, we have:
:$\forall r \in \R, i, j \in \set {1, 2}: r E_{i j} \in J$
:$\struct {J, +}$ is a subgroup of $\struct {S, +}$, in particular, $\struct {J, +}$ is closed under $+$.
Since every element of $S$ can be written as:
:$\begin{pmatrix} a & b \\ c & d \end{pmatrix} = a E_{1 1} + b E_{1 2} + c E_{2 1} + d E_{2 2}$
for some $a, b, c, d \in \R$, by the above every element of $S$ is an element of $J$.
Therefore we have $S = J$, and thus the only ideals of $S$ are $\set {\mathbf 0}$ and $S$ itself.
{{qed}}
Theorem: General Antiperiodicity Property

Let $f: X \to X$ be an antiperiodic function, where $X$ is either $\R$ or $\C$.
Let $L$ be an antiperiodic element of $f$.
Let $n \in \Z$ be an integer.
:If $n$ is even, then $n L$ is a periodic element of $f$.
:If $n$ is odd, then $n L$ is an antiperiodic element of $f$.

Proof:
Suppose that $X = \C$.
Theorem: Antiperiodic Element is Multiple of Antiperiod

Let $f: \R \to \R$ be a real anti-periodic function with anti-period $A$.
Let $L$ be an anti-periodic element of $f$.
Then $A \divides L$.

Proof:
{{AimForCont}} that $A \nmid L$.
By the Division Theorem we have:
:$\exists! q \in \Z, r \in \R: L = q A + r, 0 < r < A$
By Even and Odd Integers form Partition of Integers, it follows that $q$ must be either even or odd.
Theorem: Differential Entropy 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 differential entropy of $X$, $\map h X$, is given by: 
:$\map h X = 1 + \map \ln \beta$

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 differential entropy: 
:$\ds \map h X = -\int_0^\infty \map {f_X} x \map \ln {\map {f_X} x} \rd x$
So:
{{begin-eqn}}
{{eqn | l = \map h X
      | r = -\frac 1 \beta \int_0^\infty e^{-\frac x \beta} \map \ln {\frac 1 \beta e^{-\frac x \beta} } \rd x
}}
{{eqn | r = \frac 1 \beta \int_0^\infty e^{-\frac x \beta} \map \ln {\beta e^{\frac x \beta} } \rd x
      | c = Reciprocal of Logarithm
}}
{{eqn | r = \frac {\map \ln \beta} \beta \int_0^\infty e^{-\frac x \beta} \rd x + \frac 1 {\beta^2} \int_0^\infty x e^{-\frac x \beta} \rd x
      | c = Logarithm of Product, {{Defof|Natural Logarithm}}
}}
{{eqn | r = \frac {\map \ln \beta} \beta \intlimits {-\beta e^{-\frac x \beta} } 0 \infty + \frac 1 {\beta^2} \paren {\intlimits {-\beta x e^{-\frac x \beta} } 0 \infty + \beta \int_0^\infty e^{-\frac x \beta} \rd x}
      | c = Primitive of Exponential Function, Integration by Parts
}}
{{eqn | r = \frac {\beta \map \ln \beta} \beta + \frac 1 \beta \int_0^\infty e^{-\frac x \beta} \rd x
      | c = Exponential Tends to Zero and Infinity, Limit at Infinity of Polynomial over Real Exponential
}}
{{eqn | r = \map \ln \beta + \frac 1 \beta \intlimits {-\beta e^{-\frac x \beta} } 0 \infty
      | c = Primitive of Exponential Function
}}
{{eqn | r = \map \ln \beta + \frac \beta \beta
      | c = Exponential Tends to Zero and Infinity, Exponential of Zero
}}
{{eqn | r = 1 + \map \ln \beta
}}
{{end-eqn}}
{{qed}}
Category:Differential Entropy
Category:Exponential Distribution
djopwq27y7f4xg1ihm45r8sb6op53th
Theorem: Skewness of Poisson Distribution

Let $X$ be a discrete random variable with a Poisson distribution with parameter $\lambda$.
Then the skewness $\gamma_1$ of $X$ is given by: 
:$\gamma_1 = \dfrac 1 {\sqrt \lambda}$

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 Poisson Distribution:
:$\expect X = \lambda$
By Variance of Poisson Distribution:
:$\var X = \sigma^2 = \lambda$
so:
:$\sigma = \sqrt \lambda$
To now calculate $\gamma_1$, we must calculate $\expect {X^3}$.
We find this using the moment generating function of $X$, $M_X$. 
By Moment Generating Function of Poisson Distribution, this is given by: 
:$\ds \map {M_X} t = e^{\lambda \paren {e^t - 1} }$
From Moment in terms of Moment Generating Function:
:$\expect {X^3} = \map {M_X' ' '} 0$
In Variance of Poisson Distribution: Proof 3, it is shown that:
:$\ds \map {M_X' '} t = \lambda \paren {\lambda e^t + 1} e^{\lambda \paren {e^t - 1} + t}$
So:
{{begin-eqn}}
{{eqn | l = \map {M_X' ' '} t
      | r = \lambda^2 e^{\lambda \paren {e^t - 1} + 2t} + \lambda \paren {\lambda e^t + 1} \frac \d {\d t} \paren {\lambda \paren {e^t - 1} + t} \frac \d {\d \paren {\lambda \paren {e^t - 1} + t} } \paren {e^{\lambda \paren {e^t - 1} + t} }
      | c = Product Rule for Derivatives, Exponential of Sum, Chain Rule for Derivatives
}}
{{eqn | r = \lambda^2 e^{\lambda \paren {e^t - 1} + 2t} + \lambda \paren {\lambda e^t + 1}^2 e^{\lambda \paren {e^t - 1} + t}
      | c = Derivative of Exponential Function, Derivative of Power
}}
{{end-eqn}}
Setting $t = 0$:
{{begin-eqn}}
{{eqn | l = \expect {X^3}
      | r = \lambda^2 e^{\lambda \paren {e^0 - 1} + 0} + \lambda \paren {\lambda e^0 + 1}^2 e^{\lambda \paren {e^0 - 1} + 0}
}}
{{eqn | r = \lambda^2 + \lambda \paren {\lambda + 1}^2
      | c = Exponential of Zero
}}
{{eqn | r = \lambda^3 + 3 \lambda^2 + \lambda
      | c = Square of Sum
}}
{{end-eqn}}
So:
{{begin-eqn}}
{{eqn | l = \gamma_1
      | r = \frac {\lambda^3 + 3 \lambda^2 + \lambda - 3 \lambda^2 - \lambda^3} {\lambda^{3/2} }
}}
{{eqn | r = \frac \lambda {\lambda^{3/2} }
}}
{{eqn | r = \frac 1 {\sqrt \lambda}
}}
{{end-eqn}}
{{qed}}
Category:Skewness
Category:Poisson Distribution
oh9rlphop6q2g0rqyyp4uh022fhr9ca
Theorem: Periodic Element is Multiple of Antiperiod

Let $f: \R \to \R$ be a real anti-periodic function with anti-period $A$.
Let $L$ be a periodic element of $f$.
Then $A \divides L$.

Proof:
Consider $A + L$:
{{begin-eqn}}
{{eqn | l = \map f {x + \paren {A + L} }
      | r = \map f {\paren {x + A} + L}
}}
{{eqn | r = \map f {x + A}
}}
{{eqn | r = -\map f x
}}
{{end-eqn}}
Hence $A + L$ is an anti-periodic element of $f$.
Combining Antiperiodic Element is Multiple of Antiperiod, Divides is Reflexive, and Common Divisor Divides Difference yields: 
:$A \divides \paren {A + L} \land A \divides A \implies A \divides L$
Hence the result.
{{qed}}
Category:Antiperiodic Functions
hywff4b2rkdxz178xdjt5efld55qywm
Theorem: Skewness of Continuous Uniform Distribution

Let $X$ be a continuous random variable which is uniformly distributed on a closed real interval $\closedint a b$. 
Then the skewness $\gamma_1$ of $X$ is equal to $0$.

Proof:
From the definition of skewness: 
:$\gamma_1 = \expect {\paren {\dfrac {X - \mu} \sigma}^3}$
where: 
:$\mu$ is the mean of $X$
:$\sigma$ is the standard deviation of $X$. 
From the definition of the continuous uniform distribution, $X$ has probability density function:
:$\map {f_X} x = \dfrac 1 {b - a}$
So, from Expectation of Function of Continuous Random Variable:
:$\ds \gamma_1 = \expect {\paren {\dfrac {X - \mu} \sigma}^3} = \frac 1 {\paren {b - a} \sigma^3} \int_a^b \paren {x - \mu}^3 \rd x$
Then: 
{{begin-eqn}}
{{eqn	| l = \frac 1 {\paren {b - a} \sigma^3} \int_a^b \paren {x - \mu}^3 \rd x
	| r = \frac 1 {\paren {b - a} \sigma^3} \int_a^b \paren {x - \frac {a + b} 2}^3 \rd x
	| c = Expectation of Continuous Uniform Distribution
}}
{{eqn	| r = \frac 1 {\paren {b - a} \sigma^3} \int_{a - \frac {a + b} 2}^{b - \frac {a + b} 2} u^3 \rd u
	| c = substituting $u = x - \dfrac {a + b} 2$
}}
{{eqn	| r = \frac 1 {\paren {b - a} \sigma^3} \int_{-\frac {b - a} 2}^{\frac {b - a} 2} u^3 \rd u
}}
{{end-eqn}}
We have: 
:$\paren {-u}^3 = -u^3$
so we can see the integrand is odd. 
So, by Definite Integral of Odd Function: 
:$\ds \gamma_1 = \frac 1 {\paren {b - a} \sigma^3} \int_{-\frac {b - a} 2}^{\frac {b - a} 2} u^3 \rd u = 0$
{{qed}}
Category:Skewness
Category:Continuous Uniform Distribution
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