nyuuzyou/wikis huggingface dataset:
Sizes of Tetrahedra of Same Height are as Bases

== Theorem ==
== Proof ==
:
Let there be two tetrahedra of the same height whose bases are $ABC, DEF$ and whose apices are $G$ and $H$.
It is to be demonstrated that the ratio of $\triangle ABC$ to $\triangle DEF$ equals the ratio of tetrahedron $ABCG$ to tetrahedron $DEFH$.
Suppose the ratio of $ABCG$ to $DEFH$ is not equal to the ratio of $\triangle ABC$ to $\triangle DEF$.
Then the ratio of $\triangle ABC$ to $\triangle DEF$ equals the ratio of $ABCG$ to some solid figure $W$ which is either smaller than $DEFH$ or larger than $DEFH$.
First suppose that $W$ is smaller than $DEFH$.
From :
:let $DEFH$ be divided into two equal tetrahedra which are similar to $DEFH$ and two equal prisms.
Then the two prisms are greater than half of $DEFH$.
Again, let those tetrahedra be similarly divided.
:
:Let this be done continually until there are left over from $DEFH$ some tetrahedra which are less than the excess by which $DEFH$ is greater than $W$.
Let these, for the sake of argument, be $DQRS$ and $STUH$.
Therefore their remainders, the prisms in $DEFH$, are greater than $W$.
Let $ABCG$ be divided similarly and the same number of times as $DEFH$.
From :
:the ratio of $\triangle ABC$ to $\triangle DEF$ equals the ratio of the prisms in $ABCG$ to the prisms in $DEFH$.
But:
:$\triangle ABC : \triangle DEF = ABCG : W$
So from :
:the ratio of $ABCG$ to $W$ equals the ratio of the prisms in $ABCG$ to the prisms in $DEFH$.
So from :
:the ratio of $ABCG$ to the prisms in it is equal to the ratio of $W$ to the prisms in $DEFH$.
But $ABCG$ is greater than the prisms in it.
Therefore $W > DEFG$.
But by hypothesis $W < DEFG$.
Therefore it is not the case that the ratio of $\triangle ABC$ to $\triangle DEF$ equals the ratio of $ABCG$ to some solid figure smaller than $DEFH$.
Now suppose that the ratio of $\triangle ABC$ to $\triangle DEF$ equals the ratio of $ABCG$ to some solid figure $W$ greater than $DEFH$.
Thus:
:$\triangle DEF : \triangle ABC = W : ABCG$
But using the same technique as :
:$W : ABCG = DEFH : X$
where $X$ is some solid figure less than $ABCG$.
Therefore from :
:$\triangle DEF : \triangle ABC = DEFG : X$
But this has been demonstrated to be absurd.
Therefore it is not the case that the ratio of $\triangle ABC$ to $\triangle DEF$ equals the ratio of $ABCG$ to some solid figure greater than $DEFH$.
Hence the result.
== Sources ==
* : Book $\text{XII}$. Propositions
Category:Tetrahedra
Skew Lines Cannot Exist in Euclidean Plane

== Theorem ==
Let $\PP$ be a Euclidean plane.
Let $\LL_1$ and $\LL_2$ be straight lines embedded in $\PP$.
Then it is not possible for $\LL_1$ and $\LL_2$ to be skew
== Proof ==
The proof then follows directly from :
That is, if $\LL_1$ and $\LL_2$ do not intersect, they are parallel.
== Sources ==
* 
* 
Category:Skew Lines
Skewness in terms of Non-Central Moments

== Theorem ==
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 ==
Category:Skewness
Skewness of Bernoulli Distribution

== Theorem ==
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:
Category:Bernoulli Distribution
Category:Skewness
Skewness of Beta Distribution

== Theorem == 
Let $X \sim \BetaDist \alpha \beta$ for some $\alpha, \beta > 0$, where $\operatorname {Beta}$ denotes the Beta distribution. 
Then the skewness $\gamma_1$ of $X$ is given by: 
:$\gamma_1 = \dfrac {2 \paren {\beta - \alpha} \sqrt {\alpha + \beta + 1} } {\paren {\alpha + \beta + 2} \sqrt {\alpha \beta} }$
== 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 expectation of $X$. 
:$\sigma$ is the standard deviation of $X$.
We have, by Expectation of Beta Distribution:
:$\expect X = \dfrac {\alpha} {\alpha + \beta}$
By Variance of Beta Distribution:
:$\var X = \sigma^2 = \dfrac {\alpha \beta} {\paren {\alpha + \beta}^2 \paren {\alpha + \beta + 1} }$
so:
:$\sigma = \dfrac {\sqrt {\alpha \beta} } {\paren {\alpha + \beta} \paren {\sqrt {\alpha + \beta + 1 } } }$
From Raw Moment of Beta Distribution, we have:
So..
Category:Skewness
Category:Beta Distribution
Skewness of Binomial Distribution

== Theorem ==
Let $X$ be a discrete random variable with a Binomial distribution with parameter $n$ and $p$ for some $n \in \N$ and $0 \le p \le 1$.
Then the skewness $\gamma_1$ of $X$ is given by: 
:$\gamma_1 = \dfrac {1 - 2 p} {\sqrt {n p q} }$
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 Binomial Distribution:
:$\mu = n p$
By Variance of Binomial Distribution, we also have: 
:$\var X = \sigma^2 = n p \paren {1 - p}$
so: 
:$\sigma = \sqrt {n p \paren {1 - p} }$
To 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 Binomial Distribution, this is given by: 
:$\map {M_X} t = \paren {1 - p + p e^t}^n$
From Moment in terms of Moment Generating Function:
:$\expect {X^3} = \map {M_X} 0$
So:
Setting $t = 0$:
Plugging this result back into our equation above:
Category:Binomial Distribution
Category:Skewness
Skewness of Chi-Squared Distribution

== Theorem == 
Let $n$ be a strictly positive integer. 
Let $X \sim \chi^2_n$ where $\chi^2_n$ is the chi-squared distribution with $n$ degrees of freedom. 
Then the skewness $\gamma_1$ of $X$ is given by: 
:$\gamma_1 = \sqrt{\dfrac 8 n}$
== Proof ==
From Skewness in terms of Non-Central Moments, we have: 
:$\gamma_1 = \dfrac {\expect {X^3} - 3 \mu \sigma^2 - \mu^3} {\sigma^3}$
where: 
:$\mu$ is the expectation of $X$. 
:$\sigma$ is the standard deviation of $X$.
By Expectation of Chi-Squared Distribution we have: 
:$\mu = n$
By Variance of Chi-Squared Distribution we have: 
:$\sigma = \sqrt {2 n}$
We also have: 
So: 
Category:Skewness
Category:Chi-Squared Distribution
Skewness of Continuous Uniform Distribution

== Theorem ==
Let $X$ be a continuous random variable which is uniformly distributed on a closed real interval $\closedint a b$. 
Then the 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: 
We have: 
:$\paren {-u}^3 = -u^3$
so we can see the integrand is odd. 
So, by Definite Integral of Odd Function: 
:$\ds \gamma_1 = \frac 1 {\paren {b - a} \sigma^3} \int_{-\frac {b - a} 2}^{\frac {b - a} 2} u^3 \rd u = 0$
Category:Skewness
Category:Continuous Uniform Distribution
Skewness of Erlang Distribution

== Theorem ==
Let $k$ be a strictly positive integer. 
Let $\lambda$ be a strictly positive real number. 
Let $X$ be a continuous random variable with an Erlang distribution with parameters $k$ and $\lambda$.
Then the skewness $\gamma_1$ of $X$ is given by: 
:$\gamma_1 = \dfrac 2 {\sqrt k}$
== Proof ==
From Skewness in terms of Non-Central Moments, we have: 
:$\gamma_1 = \dfrac {\expect {X^3} - 3 \mu \sigma^2 - \mu^3} {\sigma^3}$
where: 
:$\mu$ is the expectation of $X$. 
:$\sigma$ is the standard deviation of $X$.
By Expectation of Erlang Distribution we have: 
:$\mu = \dfrac k \lambda$
By Variance of Erlang Distribution we have: 
:$\sigma = \dfrac {\sqrt k} \lambda$
We also have:
So: 
Category:Skewness
Category:Erlang Distribution
Skewness of Exponential Distribution

== Theorem ==
Let $X$ be a continuous random variable of the exponential distribution with parameter $\beta$ for some $\beta \in \R_{> 0}$
Then the skewness $\gamma_1$ of $X$ is equal to $2$.
== Proof ==
From Skewness in terms of Non-Central Moments, we have: 
:$\gamma_1 = \dfrac {\expect {X^3} - 3 \mu \sigma^2 - \mu^3} {\sigma^3}$
where: 
:$\mu$ is the expectation of $X$. 
:$\sigma$ is the standard deviation of $X$.
By Expectation of Exponential Distribution we have: 
:$\mu = \beta$ 
By Variance of Exponential Distribution we have: 
:$\sigma = \beta$ 
By Raw Moment of Exponential Distribution we also have: 
:$\expect {X^3} = 3! \beta^3 = 6 \beta^3$
So: 
Category:Skewness
Category:Exponential Distribution
Skewness of Gamma Distribution

== Theorem == 
Let $X \sim \map \Gamma {\alpha, \beta}$ for some $\alpha, \beta > 0$, where $\Gamma$ is the Gamma distribution.
Then the skewness $\gamma_1$ of $X$ is given by:
:$\gamma_1 = \dfrac 2 {\sqrt \alpha}$
== Proof 1 == 
== Proof 2 == 
Category:Gamma Distribution
Category:Skewness
Category:Skewness of Gamma Distribution
Skewness of Gamma Distribution/Proof 1

== Theorem == 
== Proof == 
From Skewness in terms of Non-Central Moments, we have: 
:$\gamma_1 = \dfrac {\expect {X^3} - 3 \mu \sigma^2 - \mu^3} {\sigma^3}$
where: 
:$\mu$ is the expectation of $X$. 
:$\sigma$ is the standard deviation of $X$.
By Expectation of Gamma Distribution, we have: 
:$\mu = \dfrac \alpha \beta$
By Variance of Gamma Distribution, we have: 
:$\sigma = \dfrac {\sqrt \alpha} \beta$ 
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 Gamma Distribution: Third Moment: 
:$\map { {M_X}} t = \dfrac {\beta^\alpha \alpha \paren {\alpha + 1} \paren {\alpha + 2} } {\paren {\beta - t}^{\alpha + 3} }$
Setting $t = 0$:
So:
Category:Skewness of Gamma Distribution
Skewness of Gamma Distribution/Proof 2

== Theorem == 
== Proof == 
From Expectation of Power of Gamma Distribution‎, we have:
:$\expect {X^n} = \dfrac {\alpha^{\overline n} } {\beta^n}$
where $\alpha^{\overline n}$ denotes the $n$th rising factorial of $\alpha$.
Hence:
Category:Skewness of Gamma Distribution
Skewness of Geometric Distribution/Formulation 1

== Theorem ==
Let $X$ be a discrete random variable with the geometric distribution with parameter $p$ for some $0 < p < 1$.
:$\map X \Omega = \set {0, 1, 2, \ldots} = \N$
:$\map \Pr {X = k} = \paren {1 - p} p^k$
Then the skewness of $X$ is given by:
:$\gamma_1 = \dfrac {1 + p} {\sqrt p}$
== 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 1, we have: 
:$\mu = \dfrac p {1 - p}$
By Variance of Geometric Distribution: Formulation 1, we have: 
:$\sigma = \dfrac {\sqrt p} {1 - p}$ 
So: 
To calculate $\gamma_1$, 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: 
:$\map { {M_X}} t = p \paren {1 - p} e^t \paren {\dfrac {1 + p e^t} {\paren {1 - p e^t}^3} }$
From Moment Generating Function of Geometric Distribution: Third Moment: 
:$\map { {M_X}} t = p \paren {1 - p } e^t \paren {\dfrac {1 + 4p e^t + p^2 e^{2 t} } {\paren {1 - p e^t}^4 } }$
Setting $t = 0$ and from Exponential of Zero, we have:
:$\map {M_X} 0 = \dfrac {p \paren {1 + p} } {\paren {1 - p}^2}$
:$\map {M_X} 0 = \dfrac {p \paren {1 + 4 p + p^2} } {\paren {1 - p}^3}$
So:
Category:Skewness
Category:Skewness of Geometric Distribution
Skewness of Geometric Distribution/Formulation 2

== Theorem ==
Let $X$ be a discrete random variable with the geometric distribution with parameter $p$ for some $0 < p < 1$.
:$\map X \Omega = \set {0, 1, 2, \ldots} = \N$
:$\map \Pr {X = k} = p \paren {1 - p}^k$
Then the 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: 
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:
Category:Skewness
Category:Skewness of Geometric Distribution
Skewness of Hat-Check Distribution

== Theorem ==
Let $X$ be a discrete random variable with a Hat-Check distribution with parameter $n$. ($n \gt 2$)
Then the skewness $\gamma_1$ of $X$ is given by: 
:$\gamma_1 = -1$
== 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 expectation of $X$. 
:$\sigma$ is the standard deviation of $X$.
We have, by Expectation of Hat-Check Distribution:
:$\expect X = n - 1$
By Variance of Hat-Check Distribution:
:$\var X = \sigma^2 = 1$
so:
:$\sigma = \sqrt 1 = 1$
To now calculate $\gamma_1$, we must calculate $\expect {X^3}$.
To help complete the sum above, recall that:
Therefore:
So:
Category:Skewness
Category:Hat-Check Distribution
Skewness of Log Normal Distribution

== Theorem ==
Let $X$ be a continuous random variable with the Log Normal distribution with $\mu \in \R, \sigma \in \R_{> 0}$.
Then the skewness $\gamma_1$ of $X$ is given by:
:$\gamma_1 = \paren {\map \exp {\sigma^2} + 2} \sqrt {\paren {\map \exp {\sigma^2} - 1} }$
== Proof ==
From Skewness in terms of Non-Central Moments, we have: 
:$\gamma_1 = \dfrac {\expect {X^3} - 3 \mu \sigma^2 - \mu^3} {\sigma^3}$
where: 
:$\mu$ is the expectation of $X$. 
:$\sigma$ is the standard deviation of $X$.
By Expectation of Log Normal Distribution:
:$\mu = \map \exp {\mu + \dfrac {\sigma^2} 2}$ 
By Variance of Log Normal Distribution:
:$\sigma = \map \exp {\mu + \dfrac {\sigma^2} 2} \sqrt {\paren {\map \exp {\sigma^2} - 1} }$ 
From Raw Moment of Log Normal Distribution:
:$\expect {X^3} = \map \exp {3 \mu + \dfrac {3^2 \sigma^2} 2}$
So:
Category:Skewness
Category:Log Normal Distribution
Skewness of Logistic Distribution

== Theorem ==
Let $X$ be a continuous random variable which satisfies the logistic distribution:
:$X \sim \map {\operatorname {Logistic} } {\mu, s}$
Then the skewness $\gamma_1$ of $X$ is equal to $0$.
== Proof ==
From Skewness in terms of Non-Central Moments, we have: 
:$\gamma_1 = \dfrac {\expect {X^3} - 3 \mu \sigma^2 - \mu^3} {\sigma^3}$
where: 
:$\mu$ is the expectation of $X$. 
:$\sigma$ is the standard deviation of $X$.
By Expectation of Logistic Distribution we have: 
:$\mu = \mu$ 
By Variance of Logistic Distribution we have: 
:$\sigma = \dfrac {s \pi} {\sqrt 3}$ 
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 Derivatives of Moment Generating Function of Logistic Distribution: 
:$\ds \map { {M_X}} 0 = \mu^3 \int_{\to 0}^{\to 1} \rd u - 3 \mu^2 s \int_{\to 0}^{\to 1} \map \ln {\dfrac {1 - u} u} \rd u + 3 \mu s^2 \int_{\to 0}^{\to 1} \map {\ln^2} {\dfrac {1 - u} u} \rd u - s^3 \int_{\to 0}^{\to 1} \map {\ln^3} {\dfrac {1 - u} u} \rd u$
Therefore:
So: 
Category:Skewness
Category:Logistic Distribution
Skewness of Normal Distribution

== Theorem ==
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 1 ==
 
== Proof 2 ==
Category:Skewness of Normal Distribution
Category:Normal Distribution
Category:Skewness
Skewness of Normal Distribution/Proof 1

== Theorem ==
== 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: 
:$\gamma_1 = 0$ 
Category:Skewness of Normal Distribution
Skewness of Normal Distribution/Proof 2

== Theorem ==
== 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: 
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 $t = 0$:
So:
Category:Skewness of Normal Distribution
Skewness of Pareto Distribution

== Theorem ==
Let $X$ be a continuous random variable with the Pareto distribution with $a, b \in \R_{> 0}$.
Then the skewness $\gamma_1$ of $X$ is given by:
:$\gamma_1 = \begin {cases} \paren {\sqrt {\dfrac {\paren {a - 2} } a } } \paren { \dfrac {2 \paren {a + 1} } {a - 3} } & 3 < a \\ \text {does not exist} & 3 \ge a \end {cases}$
== Proof ==
From Skewness in terms of Non-Central Moments, we have: 
:$\gamma_1 = \dfrac {\expect {X^3} - 3 \mu \sigma^2 - \mu^3} {\sigma^3}$
where: 
:$\mu$ is the expectation of $X$. 
:$\sigma$ is the standard deviation of $X$.
By Expectation of Pareto Distribution we have: 
:$\mu = \dfrac {a b } {\paren {a - 1} }$ 
By Variance of Pareto Distribution we have: 
:$\sigma = \dfrac {\sqrt a b } {\sqrt {\paren {a - 2} } \paren {a - 1} }$ 
From Raw Moment of Pareto Distribution, we have:
:$\expect {X^3} = \begin {cases} \dfrac {a b^3} {a - 3} & 3 < a \\ \text {does not exist} & 3 \ge a \end {cases}$
So: 
Category:Skewness
Category:Pareto Distribution
Skewness of Poisson Distribution

== Theorem ==
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:
Setting $t = 0$:
So:
Category:Skewness
Category:Poisson Distribution
Skewness of Snedecor's F-Distribution

== Theorem ==
Let $n, m$ be strictly positive integers. 
Let $X \sim F_{n, m}$ where $F_{n, m}$ is Snedecor's $F$-distribution with $\tuple {n, m}$ degrees of freedom.
Then the skewness $\gamma_1$ of $X$ is given by: 
:$\gamma_1 = \dfrac {2 \paren {m + 2 n - 2} } {m - 6} \sqrt {\dfrac {2 \paren {m - 4} } {n \paren {m + n - 2} } }$
for $m > 6$, and does not exist otherwise. 
== Proof ==
Category:Snedecor's F-Distribution
Category:Skewness
Skewness of Student's t-Distribution

== Theorem ==
Let $k$ be a strictly positive integer. 
Let $X \sim t_k$ where $t_k$ is the $t$-distribution with $k$ degrees of freedom.
Then the skewness $\gamma_1$ of $X$ is equal to $0$ for $k > 3$, and does not exist otherwise. 
== Proof ==
Category:Skewness
Category:Student's t-Distribution
Skewness of Weibull Distribution

== Theorem ==
Let $X$ be a continuous random variable with the Weibull distribution with $\alpha, \beta \in \R_{> 0}$.
Then the skewness $\gamma_1$ of $X$ is given by:
:$\gamma_1 = \dfrac {\map \Gamma {1 + \dfrac 3 \alpha} - 3 \map \Gamma {1 + \dfrac 1 \alpha} \map \Gamma {1 + \dfrac 2 \alpha} + 2 \map \Gamma {1 + \dfrac 1 \alpha}^3} {\paren {\map \Gamma {1 + \dfrac 2 \alpha} - \paren {\map \Gamma {1 + \dfrac 1 \alpha} }^2}^{\frac 3 2 } }$
where $\Gamma$ is the Gamma function.
== Proof ==
From Skewness in terms of Non-Central Moments, we have: 
:$\gamma_1 = \dfrac {\expect {X^3} - 3 \mu \sigma^2 - \mu^3} {\sigma^3}$
where: 
:$\mu$ is the expectation of $X$. 
:$\sigma$ is the standard deviation of $X$.
By Expectation of Weibull Distribution we have: 
:$\mu = \beta \, \map \Gamma {1 + \dfrac 1 \alpha}$ 
By Variance of Weibull Distribution we have: 
:$\sigma = \beta \, \paren {\map \Gamma {1 + \dfrac 2 \alpha} - \paren {\map \Gamma {1 + \dfrac 1 \alpha} }^2}^{\frac 1 2}$ 
From Raw Moment of Weibull Distribution, we have:
:$\expect {X^3} = \beta^3 \map \Gamma {1 + \dfrac 3 \alpha}$
So: 
Category:Skewness
Category:Weibull Distribution