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

Let $X$ be a continuous random variable of the exponential distribution with parameter $\beta$ for some $\beta \in \R_{>0}$.
Let $n$ be a strictly positive integer. 
Then the $n$th raw moment of $X$ is given by: 
:$\expect {X^n} = n! \beta^n$

Proof:
From Moment Generating Function of Exponential Distribution, the moment generating function of $X$ is given by:
:$\map {M_X} t = \dfrac 1 {1 - \beta t}$
By Moment in terms of Moment Generating Function:
:$\expect {X^n} = \map {M^{\paren n}_X} 0$
We have: 
{{begin-eqn}}
{{eqn	| l = \map {M^{\paren n}_X} t
	| r = \frac {\d^n} {\d t^n} \paren {\frac 1 {1 - \beta t} }
}}
{{eqn	| r = -\frac 1 \beta \frac {\d^n} {\d t^n} \paren {\frac 1 {t - \frac 1 \beta} }
}}
{{eqn	| r = -\frac {\paren {-1}^n n!} {\beta \paren {t - \frac 1 \beta}^{n + 1} }
	| c = Nth Derivative of Reciprocal of Mth Power: Corollary, Chain Rule for Derivatives
}}
{{eqn	| r = \frac {\paren {-1}^{n + 1} n!} {\frac \beta {\beta^{n + 1} } \paren {\beta t - 1}^{n + 1} }
}}
{{eqn	| r = \frac {\paren {-1}^{n + 1} n! \beta^n} {\paren {\beta t - 1}^{n + 1} } 
}}
{{end-eqn}}
Setting $t = 0$ gives:
{{begin-eqn}}
{{eqn	| l = \expect {X^n}
	| r = \frac {\paren {-1}^{n + 1} n! \beta^n} {\paren {-1}^{n + 1} }
}}
{{eqn	| r = n! \beta^n
}}
{{end-eqn}}
{{qed}}
Category:Exponential Distribution
Category:Raw Moments
d7ta17w22v5w3z5j8p9y16k6a2xhpwh
Theorem: Central Moment of Exponential Distribution

Let $X$ be a continuous random variable of the exponential distribution with parameter $\beta$ for some $\beta \in \R_{> 0}$
Let $n$ be a strictly positive integer. 
Then the $n$th central moment $\mu_n$ of $X$ is given by: 
:$\ds \mu_n = n! \beta^n \sum_{k \mathop = 0}^n \frac {\paren {-1}^k} {k!}$

Proof:
From definition of central moment we have:
:$\mu_n = \expect {\paren {x - \mu}^n}$
By Expectation of Exponential Distribution we have:
:$\mu = \beta$
So: 
{{begin-eqn}}
{{eqn | l = \mu_2
      | r = \expect {\sum_{k \mathop = 0}^n \binom n k X^{n - k} \paren {-\beta}^k}
      | c = Binomial Theorem
}}
{{eqn | r = \sum_{k \mathop = 0}^n \binom n k \paren {-\beta}^k \expect {X^{n - k} }
      | c = Expectation is Linear
}}
{{eqn | r = \sum_{k \mathop = 0}^n \beta^k \binom n k \paren {-1}^k \paren {\paren {n - k}! \beta^{n - k} }
      | c = Raw Moment of Exponential Distribution
}}
{{eqn | r = \sum_{k \mathop = 0}^n \beta^n \paren {\frac {n!} {k! \paren {n - k}!} } \paren {-1}^k \paren {n - k}!
      | c = {{Defof|Binomial Coefficient}}
}}
{{eqn | r = n! \beta^n \sum_{k \mathop = 0}^n \frac {\paren {-1}^k} {k!}
}}
{{end-eqn}}
{{qed}}
Category:Exponential Distribution
Category:Central Moments
202j9l6940pl1hzy2ym4qi96u11h17w
Theorem: Excess Kurtosis of Bernoulli Distribution

Let $X$ be a discrete random variable with a Bernoulli distribution with parameter $p$.
Then the excess kurtosis $\gamma_2$ of $X$ is given by: 
:$\gamma_2 = \dfrac {1 - 6 p q} {p q}$
where $q = 1 - p$.

Proof:
From the definition of excess kurtosis, we have: 
:$\gamma_2 = \expect {\paren {\dfrac {X - \mu} \sigma}^4} - 3$
where: 
:$\mu$ is the expectation of $X$. 
:$\sigma$ is the standard deviation of $X$.
By Expectation of Bernoulli Distribution, we have: 
:$\mu = p$
By Variance of Bernoulli Distribution, we have: 
:$\sigma = \sqrt {p q}$ 
So: 
{{begin-eqn}}
{{eqn	| l = \gamma_2
	| r = \dfrac {\expect {X^4} - 4 \mu \expect {X^3} + 6 \mu^2 \expect {X^2} - 3 \mu^4} {\sigma^4} - 3
	| c = Kurtosis in terms of Non-Central Moments
}}
{{eqn	| r = \frac {p - 4p^2 + 6p^3 - 3 p^4} {\paren {p q}^2} - 3
	| c = Raw Moment of Bernoulli Distribution
}}
{{eqn	| r = \frac {p \paren {1 - 4p + 6p^2 - 3p^3} } {\paren {p q}^2} - 3
}}
{{eqn	| r = \frac {p \paren {1 - p} \paren {3p^2 - 3p + 1} } {\paren {p q}^2} - 3
}}
{{eqn	| r = \frac {p q \paren {1 - 3 p \paren {1 - p} } - 3 \paren {p q}^2} {\paren {p q}^2}
}}
{{eqn	| r = \frac {1 - 3 p q - 3 p q} {p q}
}}
{{eqn	| r = \frac {1 - 6 p q} {p q}
}}
{{end-eqn}}
{{qed}}
Category:Kurtosis
Category:Bernoulli Distribution
bnkjl2n26i0xtcdgcppt74sgu83ax6a
Theorem: Product of Limits of Real Sequences (1 + x over n)^n and (1 - x over n)^n equals 1

Let $\sequence {a_n}$ be the sequence defined as:
:$a_n = \paren {1 + \dfrac x n}^n$
Let $\sequence {b_n}$ be the sequence defined as:
:$b_n = \paren {1 - \dfrac x n}^n$
Then the product of the limits of $\sequence {a_n}$ and $\sequence {b_n}$ equals $1$

Proof:
From Real Sequence $\paren {1 + \dfrac x n}^n$ is Convergent, $\sequence {a_n}$ is convergent.
Setting $x \to -x$, it follows that $\sequence {\paren {1 + \dfrac {\paren {-x} } n}^n} = \sequence {b_n}$ is also convergent.
Then:
{{begin-eqn}}
{{eqn | l = \paren {1 - \dfrac x n}^n \paren {1 - \dfrac x n}^n
      | r = \paren {\paren {1 - \dfrac x n} \paren {1 - \dfrac x n} }^n
      | c = 
}}
{{eqn | r = \paren {1 - \dfrac {x^2} {n^2} }^n
      | c = Difference of Two Squares
}}
{{eqn | r = \paren {\paren {1 - \dfrac {x^2} {n^2} }^{n^2} }^{1/n}
      | c = 
}}
{{end-eqn}}
From Real Sequence $\paren {1 + \dfrac x n}^n$ is Convergent, $\sequence {\paren {1 - \dfrac {x^2} {n^2} }^{n^2} }$ is convergent to a positive limit.
It follows that:
:$\exists a, b \in \R_{>0}: a < \paren {1 - \dfrac {x^2} {n^2} }^{n^2} < b$
for sufficiently large $n$.
Hence:
:$a^{1/n} < \paren {1 - \dfrac x n}^n \paren {1 - \dfrac x n}^n < b^{1/n}$
The result follows from Limit of Root of Positive Real Number and the Squeeze Theorem for Real Sequences.
{{qed}}
Theorem: Farey Sequence is not Convergent

Consider the Farey sequence:
:$F = \dfrac 1 2, \dfrac 1 3, \dfrac 2 3, \dfrac 1 4, \dfrac 2 4, \dfrac 3 4, \dfrac 1 5, \dfrac 2 5, \dfrac 3 5, \dfrac 4 5, \dfrac 1 6, \ldots$
$F$ is not convergent.

Proof:
We have the following subsequences of $F$ which are all convergent to a different limit:
:$\dfrac 1 2, \dfrac 2 4, \dfrac 3 6, \dfrac 4 8 \to \dfrac 1 2$ as $n \to \infty$
:$\dfrac 1 2, \dfrac 1 3, \dfrac 1 4, \dfrac 1 5 \to 0$ as $n \to \infty$
:$\dfrac 1 2, \dfrac 2 3, \dfrac 3 4, \dfrac 4 5 \to 1$ as $n \to \infty$
It follows from Limit of Subsequence equals Limit of Real Sequence that $F$ cannot be convergent.
{{qed}}
Theorem: Skewness of Gamma Distribution

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:
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 $t = 0$:
{{begin-eqn}}
{{eqn	| l = \expect {X^3}
	| r = \dfrac {\beta^\alpha \alpha \paren {\alpha + 1} \paren {\alpha + 2} } {\paren {\beta - 0}^{\alpha + 3} }
}}
{{eqn	| r = \dfrac {\alpha \paren {\alpha + 1} \paren {\alpha + 2} } {\beta^3}
	| c = 
}}
{{end-eqn}}
So:
{{begin-eqn}}
{{eqn	| l = \gamma_1
	| r = \frac {\expect {X^3} - 3 \paren {\dfrac \alpha \beta} \paren {\dfrac \alpha {\beta^2} } - \paren {\dfrac \alpha \beta}^3} {\paren {\dfrac {\alpha \sqrt \alpha} {\beta^3} } } 
}}
{{eqn	| r = \frac {\dfrac {\alpha \paren {\alpha + 1} \paren {\alpha + 2} } {\beta^3} - 3 \paren {\dfrac \alpha \beta} \paren {\dfrac \alpha {\beta^2} } - \paren {\dfrac \alpha \beta}^3} {\paren {\dfrac {\alpha \sqrt \alpha} {\beta^3} } } 
}}
{{eqn	| r = \frac {\alpha^3 + 3 \alpha^2 + 2 \alpha - 3 \alpha^2 - \alpha^3} {\paren {\alpha \sqrt \alpha } } 
        | c = $\beta^3$ cancels
}}
{{eqn	| r = \dfrac 2 {\sqrt \alpha}
}}
{{end-eqn}}
{{qed}}
Theorem: Skewness of Gamma Distribution

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:
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:
{{begin-eqn}}
{{eqn | l = \gamma_1
      | r = \expect {\paren {\dfrac {X - \mu} \sigma}^3}
      | c = {{Defof|Skewness}}
}}
{{eqn | 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 \expect X \expect {X^2} + 3 \paren {\expect X}^3 - \paren {\expect X}^3} {\paren {\sqrt {\var X} }^3}
      | c = Definition of $\mu$ and $\sigma$
}}
{{eqn | r = \paren {\expect {X^3} - 3 \expect X \expect {X^2} + 2 \paren {\expect X}^3} \paren {\sqrt {\dfrac {\beta^2} \alpha} }^3
      | c = simplification, Variance of Gamma Distribution
}}
{{eqn | r = \dfrac {\alpha^{\overline 3} } {\beta^3} - 3 \dfrac {\alpha^{\overline 1} } {\beta^1} \dfrac {\alpha^{\overline 2} } {\beta^2} + 2 \paren {\dfrac {\alpha^{\overline 1} } {\beta^1} }^3 \dfrac {\beta^3} {\alpha \sqrt \alpha}
      | c = Expectation of Power of Gamma Distribution
}}
{{eqn | r = \paren {\dfrac {\alpha \paren {\alpha + 1} \paren {\alpha + 2} } {\beta^3} - 3 \dfrac {\alpha} \beta \dfrac {\alpha \paren {\alpha + 1} } {\beta^2} + 2 \paren {\dfrac \alpha \beta}^3} \dfrac {\beta^3} {\alpha \sqrt \alpha}
      | c = {{Defof|Rising Factorial}}
}}
{{eqn | r = \dfrac {\alpha^3 + 3 \alpha^2 + 2 \alpha - 3 \alpha^3 - 3 \alpha^2 + 2 \alpha^3} {\beta^3} \dfrac {\beta^3} {\alpha \sqrt \alpha}
      | c = multiplying out
}}
{{eqn | r = \dfrac 2 {\sqrt \alpha}
      | c = simplification
}}
{{end-eqn}}
{{qed}}
Theorem: Excess Kurtosis of Gamma Distribution

Let $X \sim \map \Gamma {\alpha, \beta}$ for some $\alpha, \beta > 0$, where $\Gamma$ is the Gamma distribution.
Then the excess kurtosis $\gamma_2$ of $X$ is given by: 
:$\gamma_2 = \dfrac 6 \alpha$

Proof:
From the definition of excess kurtosis, we have: 
:$\gamma_2 = \expect {\paren {\dfrac {X - \mu} \sigma}^4} - 3$
where: 
:$\mu = \expect X$ is the expectation of $X$
:$\sigma = \sqrt {\var X}$ is the standard deviation of $X$.
By Expectation of Gamma Distribution, we have: 
:$\mu = \dfrac \alpha \beta$
By Variance of Gamma Distribution, we have: 
:$\sigma^2 = \dfrac \alpha {\beta^2}$ 
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:
{{begin-eqn}}
{{eqn	| l = \gamma_2
	| r = \dfrac {\expect {X^4} - 4 \mu \expect {X^3} + 6 \mu^2 \expect {X^2} - 3 \mu^4} {\sigma^4} - 3
	| c = Kurtosis in terms of Non-Central Moments
}}
{{eqn	| r = \dfrac {\dfrac {\alpha^{\overline 4} } {\beta^4} - 4 \paren {\dfrac \alpha \beta} \dfrac {\alpha^{\overline 3} } {\beta^3} + 6 \paren {\dfrac \alpha \beta}^2 \dfrac {\alpha^{\overline 2} } {\beta^2} - 3 \paren {\dfrac \alpha \beta}^4} {\dfrac {\alpha^2} {\beta^4} } - 3
	| c = Expectation of Gamma Distribution, Variance of Gamma Distribution, Expectation of Power of Gamma Distribution
}}
{{eqn	| r = \paren {\dfrac {\alpha \paren {\alpha + 1} \paren {\alpha + 2} \paren {\alpha + 3} } {\beta^4} - 4 \paren {\dfrac \alpha \beta} \dfrac {\alpha \paren {\alpha + 1} \paren {\alpha + 2} } {\beta^3} + 6 \paren {\dfrac \alpha \beta}^2 \dfrac {\alpha \paren {\alpha + 1} } {\beta^2} - 3 \paren {\dfrac \alpha \beta}^4} \dfrac {\beta^4} {\alpha^2} - 3
	| c = {{Defof|Rising Factorial}}
}}
{{eqn	| r = \dfrac {\alpha \paren {\alpha + 1} \paren {\alpha + 2} \paren {\alpha + 3} - 4 \alpha^2 \paren {\alpha + 1} \paren {\alpha + 2} + 6 \alpha^3 \paren {\alpha + 1} - 3 \alpha^4} {\alpha^2} - 3
	| c = simplifying
}}
{{eqn	| r = \dfrac {\paren {\alpha^4 + 6 \alpha^3 + 11 \alpha^2 + 6 \alpha} - 4 \paren {\alpha^4 + 3 \alpha^3 + 2 \alpha^2} + 6 \paren {\alpha^4 + \alpha^3 } - 3 \alpha^4} {\alpha^2} - 3
	| c = multiplying out
}}
{{eqn	| r = \dfrac {\paren {1 - 4 + 6 - 3} \alpha^4 + \paren {6 - 12 + 6} \alpha^3 + \paren {11 - 8} \alpha^2 + 6 \alpha} {\alpha^2} - 3
	| c = gathering terms
}}
{{eqn	| r = \dfrac {6 \alpha + 3 \alpha^2} {\alpha^2} - 3
	| c = simplifying
}}
{{eqn	| r = \dfrac 6 \alpha
	| c = simplifying
}}
{{end-eqn}}
{{qed}}
Theorem: Skewness 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 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: 
{{begin-eqn}}
{{eqn | l = \gamma_1
      | r = \frac {6 \beta^3 - 3\beta^3 - \beta^3} {\beta^3}
}}
{{eqn | r = \frac {2 \beta^3} {\beta^3}
}}
{{eqn | r = 2
}}
{{end-eqn}}
{{qed}}
Theorem: Excess Kurtosis 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 excess kurtosis $\gamma_2$ of $X$ is equal to $6$.

Proof:
From the definition of excess kurtosis, we have: 
:$\gamma_2 = \expect {\paren {\dfrac {X - \mu} \sigma}^4} - 3$
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: 
:$\sigma = \beta$
So: 
{{begin-eqn}}
{{eqn	| l = \gamma_2
	| r = \dfrac {\expect {X^4} - 4 \mu \expect {X^3} + 6 \mu^2 \expect {X^2} - 3 \mu^4} {\sigma^4} - 3
	| c = Kurtosis in terms of Non-Central Moments
}}
{{eqn	| r = \frac {4! \beta^4 - 4 \beta \paren {3! \beta^3} + 6 \beta^2 \paren {2! \beta^2} - 3 \beta^4} {\beta^4} - 3
	| c = Raw Moment of Exponential Distribution
}}
{{eqn	| r = \frac {24 \beta^4 - 24 \beta^4 + 12 \beta^4 - 3 \beta^4} {\beta^4} - 3
}}
{{eqn	| r = \frac {9 \beta^4} {\beta^4} - 3
}}
{{eqn	| r = 6
}}
{{end-eqn}}
{{qed}}
Category:Kurtosis
Category:Exponential Distribution
pjq2zj0543fo4q2kw4kb2a9d60kc8jr
Theorem: Farey Sequence has Convergent Subsequences for all x in Closed Unit Interval

Consider the Farey sequence:
:$\sequence {a_n} = \dfrac 1 2, \dfrac 1 3, \dfrac 2 3, \dfrac 1 4, \dfrac 2 4, \dfrac 3 4, \dfrac 1 5, \dfrac 2 5, \dfrac 3 5, \dfrac 4 5, \dfrac 1 6, \ldots$
Every element of the closed real interval $\closedint 0 1$ is the limit of a subsequence of $\sequence {a_n}$.

Proof:
We have that every rational number $\dfrac p q$ between $0$ and $1$ occurs infinitely often in $\sequence {a_n}$:
:$\dfrac p q, \dfrac {2 p} {2 q}, \dfrac {3 p} {3 q}, \ldots$
Let $x \in \closedint 0 1$.
From Between two Real Numbers exists Rational Number, a term $a_{n_1}$ of $\sequence {a_n}$ can be found such that:
:$x - 1 < a_{n_1} < x + 1$
Then a term $a_{n_2}$ of $\sequence {a_n}$ such that $n_2 > n_1$ can be found such that:
:$x - \dfrac 1 2 < a_{n_2} < x + \dfrac 1 2$
Hence we can create a subsequence $\sequence {a_{n_k} }$ such that:
:$x - \dfrac 1 k < a_{n_k} < x + \dfrac 1 k$
and so by the Squeeze Theorem for Real Sequences:
:$r_{n_k} \to x$ as $k \to \infty$
{{qed}}
Theorem: Expectation of Chi-Squared Distribution

Let $n$ be a strictly positive integer. 
Let $X \sim \chi_n^2$ where $\chi_n^2$ is the chi-squared distribution with $n$ degrees of freedom. 
Then the expectation of $X$ is given by: 
:$\expect X = n$

Proof:
{{begin-eqn}}
{{eqn | l = \expect X
      | r = \prod_{k \mathop = 0}^0 \paren {n + 2 k}
      | c = Raw Moment of Chi-Squared Distribution
}}
{{eqn | r = n
}}
{{end-eqn}}
{{qed}}
Theorem: Variance of Chi-Squared Distribution

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 variance of $X$ is given by: 
:$\var X = 2 n$

Proof:
By Variance as Expectation of Square minus Square of Expectation, we have: 
:$\var X = \expect {X^2} - \paren {\expect X}^2$
By Expectation of Chi-Squared Distribution, we have: 
:$\expect X = n$
We also have: 
{{begin-eqn}}
{{eqn | l = \expect {X^2}
      | r = \prod_{k \mathop = 0}^1 \paren {n + 2 k}
      | c = Raw Moment of Chi-Squared Distribution
}}
{{eqn | r = n \paren {n + 2}
}}
{{eqn | r = n^2 + 2 n
}}
{{end-eqn}}
So: 
{{begin-eqn}}
{{eqn | l = \var X
      | r = n^2 + 2 n - n^2
}}
{{eqn | r = 2 n
}}
{{end-eqn}}
{{qed}}
Theorem: Skewness of Chi-Squared Distribution

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: 
{{begin-eqn}}
{{eqn	| l = \expect {X^3}
	| r = \prod_{k \mathop = 0}^2 \paren {n + 2 k}
	| c = Raw Moment of Chi-Squared Distribution
}}
{{eqn	| r = n \paren {n + 2} \paren {n + 4}
}}
{{eqn	| r = n^3 + 6 n^2 + 8 n
}}
{{end-eqn}}
So: 
{{begin-eqn}}
{{eqn	| l = \gamma_1
	| r = \frac {n^3 + 6 n^2 + 8 n - 6 n^2 - n^3} {\paren {\sqrt {2 n} }^3}
}}
{{eqn	| r = \frac {\sqrt {\paren {8 n}^2} } {\sqrt {\paren {2 n}^3} }
}}
{{eqn	| r = \frac {\sqrt {64 n^2} } {\sqrt {8 n^3} } 
}}
{{eqn	| r = \sqrt {\frac 8 n}
}}
{{end-eqn}}
{{qed}}
Category:Skewness
Category:Chi-Squared Distribution
15ar3wso9nvwv3wsx9sst0s4vp1k21a
Theorem: Excess Kurtosis of Chi-Squared Distribution

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 excess kurtosis $\gamma_2$ of $X$ is given by: 
:$\gamma_2 = \dfrac {12} n$

Proof:
From the definition of excess kurtosis, we have: 
:$\gamma_2 = \expect {\paren {\dfrac {X - \mu} \sigma}^4} - 3$
where: 
:$\mu$ is the expectation of $X$. 
:$\sigma$ is the standard deviation of $X$.
By Expectation of Chi-Squared Distribution:
:$\mu = n$
By Variance of Chi-Squared Distribution:
:$\sigma = \sqrt {2 n}$
As is shown in Skewness of Chi-Squared Distribution, we have: 
:$\expect {X^3} = n^3 + 6 n^2 + 8 n$
We also have: 
{{begin-eqn}}
{{eqn | l = \expect {X^4}
      | r = \prod_{k \mathop = 0}^3 \paren {n + 2 k}
      | c = Raw Moment of Chi-Squared Distribution
}}
{{eqn | r = n \paren {n + 2} \paren {n + 4} \paren {n + 6}
}}
{{eqn | r = n \paren {n^2 + 6 n + 8} \paren {n + 6}
}}
{{eqn | r = n^4 + 12 n^3 + 44 n^2 + 48 n
}}
{{end-eqn}}
So:
{{begin-eqn}}
{{eqn | l = \gamma_2
      | r = \frac {\expect {X^4 - 4 X^3 n + 6 X^2 n^2 - 4 X n^3 + n^4} } {\paren {\sqrt {2 n} }^4} - 3
      | c = Fourth Power of Sum, Expectation is Linear
}}
{{eqn | r = \frac {\expect {X^4} - 4 n \expect {X^3} + 6 n^2 \expect {X^2} - 4 n^3 \expect X + n^4} {4 n^2} - 3
      | c = Expectation is Linear
}}
{{eqn | r = \frac {n^4 + 12 n^3 + 44 n^2 + 48 n - 4 n \paren {n^3 + 6 n^2 + 8 n} + 6 n^2 \paren {n^2 + 2 n} - 4 n^4 + n^4} {4 n^2} - 3
      | c = Variance as Expectation of Square minus Square of Expectation
}}
{{eqn | r = \frac {n^4 + 12 n^3 + 44 n^2 + 48 n - 4n^4 - 24 n^3 - 32 n^2 + 6 n^4 + 12 n^3 - 4 n^4 + n^4 - 12 n^2} {4 n^2}
}}
{{eqn | r = \frac {48 n} {4 n^2}
}}
{{eqn | r = \frac {12} n
}}
{{end-eqn}}
{{qed}}
Category:Kurtosis
Category:Chi-Squared Distribution
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Theorem: Raw Moment of Chi-Squared Distribution

Let $n$ and $m$ be strictly positive integers. 
Let $X \sim \chi^2_n$ where $\chi^2_n$ is the chi-squared distribution with $n$ degrees of freedom. 
Then the $m$th raw moment $\expect {X^m}$ of $X$ is given by: 
:$\ds \expect {X^m} = \prod_{k \mathop = 0}^{m - 1} \paren {n + 2 k}$

Proof:
From the definition of the chi-squared distribution, $X$ has probability density function:
:$\ds \map {f_X} x = \dfrac 1 {2^{n / 2} \map \Gamma {n / 2} } x^{\paren {n / 2} - 1} e^{- x / 2}$
From the definition of the expected value of a continuous random variable: 
:$\ds \expect {X^m} = \int_0^\infty x^m \map {f_X} x \rd x$
So: 
{{begin-eqn}}
{{eqn	| l = \expect {X^m}
	| r = \frac 1 {2^{n / 2} \map \Gamma {n / 2} } \int_0^\infty x^{\paren {n / 2} + m - 1} e^{-x / 2} \rd x
}}
{{eqn	| r = \frac 2 {2^{n / 2} \map \Gamma {n / 2} } \int_0^\infty \paren {2 u}^{\paren {n / 2} + m - 1} e^{-u} \rd u
	| c = substituting $x = 2 u$
}}
{{eqn	| r = \frac {2^{\paren {n / 2} + m - 1 + 1} } {2^{n / 2} \map \Gamma {n / 2} } \int_0^\infty u^{\paren {n / 2} + m - 1} e^{-u} \rd u
}}
{{eqn	| r = \frac {2^m} {\map \Gamma {n / 2} } \map \Gamma {\paren {n / 2} + m}
	| c = {{Defof|Gamma Function}}
}}
{{eqn	| r = \frac {\map \Gamma {n / 2} } {\map \Gamma {n / 2} } 2^m \prod_{k \mathop = 0}^{m - 1} \paren {\frac n 2 + k}
	| c = Gamma Difference Equation
}}
{{eqn	| r = \prod_{k \mathop = 0}^{m - 1} \paren {n + 2 k}
}}
{{end-eqn}}
{{qed}}
Category:Raw Moments
Category:Chi-Squared Distribution
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Theorem: Sequence of Square Roots of Natural Numbers is not Cauchy

Let $\sequence {x_n}_{n \mathop \in \N_{>0} }$ be the sequence in $\R$ defined as:
:$x_n = \sqrt n$
Then, despite the fact that from Difference Between Adjacent Square Roots Converges:
:$\size {\sqrt {n + 1} - \sqrt n} \to 0$ as $n \to \infty$
it is not the case that $\sequence {x_n}$ is a Cauchy sequence.

Proof:
{{AimForCont}} $\sequence {x_n}$ is bounded.
Let $H \in \R$ be an upper bound of $\sequence {x_n}$
By the Axiom of Archimedes:
:$\exists N \in \N: N > H^2$
and so:
:$\exists N \in \N: \sqrt N > H$
But $\sqrt N$ is a term of $\sequence {x_n}$.
So $H$ cannot be an upper bound of $\sequence {x_n}$
Hence by Proof by Contradiction, $\sequence {x_n}$ is not bounded.
From Real Cauchy Sequence is Bounded, it follows by the Rule of Transposition that $\sequence {x_n}$ is not a Cauchy sequence
So, while we have the case that:
:$\size {\sqrt {n + 1} - \sqrt n} \to 0$ as $n \to \infty$
it is not the case that:
:$\forall \epsilon \in \R_{>0}: \exists N \in \N: \forall m, n \in \N: m, n \ge N: \size {\sqrt n - \sqrt m} < \epsilon$
{{qed}}