avewright/proofwiki-math huggingface dataset:
Theorem: Excess Kurtosis 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 excess kurtosis $\gamma_2$ of $X$ is equal to $-\dfrac 6 5$.

Proof:
From the definition of excess kurtosis, we have: 
:$\gamma_2 = \expect {\paren {\dfrac {X - \mu} \sigma}^4} - 3$
From the definition of the continuous uniform distribution, $X$ has probability density function: 
:$\map {f_X} x = \begin {cases} \dfrac 1 {b - a} & : a \le x \le b \\ 0 & : \text {otherwise} \end {cases}$
From the definition of the expected value of a continuous random variable: 
:$\ds \expect X = \int_{-\infty}^\infty x \map {f_X} x \rd x$
So:
{{begin-eqn}}
{{eqn | l = \expect {\paren {\dfrac {X - \mu} \sigma}^4} - 3
      | r = \int_a^b \paren {\dfrac {x - \mu} \sigma}^4 \paren {\frac 1 {b - a} } dx - 3
}}
{{eqn | r = \frac 1 {\sigma^4 \paren {b - a} } \int_a^b \paren {x - \mu}^4  d \paren {x - \mu} - 3
      | c = Integration by Substitution
}}
{{eqn | r = \frac 1 {\sigma^4 \paren {b - a} } \intlimits {\frac {\paren {x - \mu}^5} 5} a b - 3
      | c = Primitive of Power, Fundamental Theorem of Calculus
}}
{{eqn | r = \frac {12^2} {\paren {b - a}^4 \paren {b - a} } \paren{\frac {\paren {b - \frac {a + b} 2}^5 - \paren {a - \frac {a + b} 2}^5} 5} - 3
      | c = Variance of Continuous Uniform Distribution, Expectation of Continuous Uniform Distribution
}}
{{eqn | r = \frac {144} {5 \paren {b - a}^5} \paren {\paren {\frac {b - a} 2}^5 - \paren {\frac {a - b} 2}^5} - 3
}}
{{eqn | r = \frac {144} {5 \paren {b - a}^5} \paren {2 \paren {\frac {b - a} 2}^5} - 3
}}
{{eqn | r = \frac {144} 5 \paren {2 \paren {\frac 1 2}^5} - 3
}}
{{eqn | r = -\frac 6 5
}}
{{end-eqn}}
{{qed}}
Category:Kurtosis
Category:Continuous Uniform Distribution
4j8ts1haipcn8sjl2x24neyq2fn07tb
Theorem: Square of Standard Normal Random Variable has Chi-Squared Distribution

Let $X \sim \Gaussian 0 1$ where $\Gaussian 0 1$ is the standard normal distribution. 
Then $X^2 \sim \chi^2_1$ where $\chi^2_1$ is the chi-square distribution with $1$ degree of freedom.

Proof:
Let $Y \sim \chi^2_1$. 
We aim to show that: 
:$\map \Pr {Y < x^2} = \map \Pr {-x < X < x}$
for all real $x \ge 0$.
We have: 
{{begin-eqn}}
{{eqn | l = \map \Pr {Y < x^2}
      | r = \frac 1 {\sqrt 2 \map \Gamma {1 / 2} } \int_0^{x^2} t^{\paren {1 / 2} - 1} e^{-t / 2} \rd t
      | c = {{Defof|Chi-Squared Distribution}}
}}
{{eqn | r = \frac 1 {\sqrt {2 \pi} } \int_0^{x^2} \frac {e^{-t / 2} } {\sqrt t} \rd t
      | c = Gamma Function of One Half
}}
{{eqn | r = \frac 2 {\sqrt {2 \pi} } \int_0^x u \frac {e^{-u^2 / 2} } {\sqrt {u^2} } \rd u
      | c = substituting $t = u^2$
}}
{{eqn | r = \frac 2 {\sqrt {2 \pi} } \int_0^x e^{-u^2 / 2} \rd u
}}
{{eqn | r = \frac 1 {\sqrt {2 \pi} } \int_{-x}^x e^{-u^2 / 2} \rd u
      | c = Definite Integral of Even Function
}}
{{eqn | r = \map \Pr {-x < X < x}
      | c = {{Defof|Normal Distribution}}
}}
{{end-eqn}}
{{qed}}
Category:Normal Distribution
Category:Chi-Squared Distribution
rl4ztdawo70r9fbjdzuhqdhgl0jmdje
Theorem: Sum of Chi-Squared Random Variables

Let $n_1, n_2, \ldots, n_k$ be strictly positive integers which sum to $N$.
Let $X_i \sim {\chi^2}_{n_i}$ for $1 \le i \le k$, where ${\chi^2}_{n_i}$ is the chi-squared distribution with $n_i$ degrees of freedom. 
Then: 
:$\ds X = \sum_{i \mathop = 1}^k X_i \sim {\chi^2}_N$

Proof:
Let $Y \sim {\chi^2}_N$. 
By Moment Generating Function of Chi-Squared Distribution, the moment generating function of $X_i$ is given by: 
:$\map {M_{X_i} } t = \paren {1 - 2 t}^{-n_i / 2}$
Similarly, the moment generating function of $Y$ is given by: 
:$\map {M_Y} t = \paren {1 - 2 t}^{-N / 2}$
By Moment Generating Function of Linear Combination of Independent Random Variables, the moment generating function of $X$ is given by: 
:$\ds \map {M_X} t = \prod_{i \mathop = 1}^k \map {M_{X_i} } t$
We aim to show that: 
:$\map {M_X} t = \map {M_Y} t$
By Moment Generating Function is Unique, this ensures $X = Y$.
We have:
{{begin-eqn}}
{{eqn	| l = \map {M_X} t
	| r = \prod_{i \mathop = 1}^k \paren {1 - 2 t}^{-n_i / 2}
}}
{{eqn	| r = \paren {1 - 2 t}^{-\paren {n_1 + n_2 + \ldots + n_k} / 2}
}}
{{eqn	| r = \paren {1 - 2 t}^{-N / 2}
}}
{{eqn	| r = \map {M_Y} t
}}
{{end-eqn}}
{{qed}}
Category:Chi-Squared Distribution
4ibw7t4edkstupabflwavzo6afwpyqq
Theorem: Moment Generating Function 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 moment generating function of $X$, $M_X$, is given by: 
:$\map {M_X} t = \begin{cases} \paren {1 - 2 t}^{-n / 2} & : t < \dfrac 1 2 \\ \text{does not exist} & : t \ge \dfrac 1 2 \end{cases}$

Proof:
From the definition of the chi-squared distribution, $X$ has probability density function:
:$\map {f_X} x = \dfrac 1 {2^{n / 2} \map \Gamma {n / 2} } x^{\paren {n / 2} - 1} e^{- x / 2}$
From the definition of a moment generating function:
:$\ds \map {M_X} t = \expect {e^{t X} } = \int_0^\infty e^{t x} \map {f_X} x \rd x$
So: 
{{begin-eqn}}
{{eqn | l = \map {M_X} t
      | r = \frac 1 {2^{n / 2} \map \Gamma {n / 2} } \int_0^\infty x^{\paren {n / 2} - 1} e^{t x - \paren {x / 2} } \rd x
}}
{{eqn | r = \frac 1 {2^{n / 2} \map \Gamma {n / 2} } \int_0^\infty x^{\paren {n / 2} - 1} e^{-\paren {\paren {1 / 2} - t} x } \rd x
}}
{{end-eqn}}
Note that if $\dfrac 1 2 - t < 0$, then $e^{-\paren {\paren {1 / 2} - t} x} \to \infty$ by Exponential Tends to Zero and Infinity, so the integral diverges in this case and the expectation fails to exist.
If $\dfrac 1 2 - t = 0$, then $e^{-\paren {\paren {1 / 2} - t} x} = 1$ by Exponential of Zero, so the integral is equal to: 
:$\ds \int_0^\infty x^{\paren {n / 2} - 1} \rd x$
However, as $n$ is a strictly positive integer, we have $\dfrac n 2 - 1 > -1$. 
So, the integral diverges in this case by Integral to Infinity of Reciprocal of Power of x, and hence the expectation also fails to exist in this case.
For the integral to converge, we must have $\dfrac 1 2 - t > 0$. That is, when $t < \dfrac 1 2$. 
In this case: 
{{begin-eqn}}
{{eqn | l = \frac 1 {2^{n / 2} \map \Gamma {n / 2} } \int_0^\infty x^{\paren {n / 2} - 1} e^{-\paren {\paren {1 / 2} - t} x } \rd x
      | r = \frac 1 {2^{n / 2} \map \Gamma {n / 2} \paren {\frac 1 2 - t} } \int_0^\infty \paren {\frac u {\frac 1 2 - t} }^{n / 2 - 1} e^{-u} \rd u
      | c = Integration by Substitution: $u = \paren {\frac 1 2 - t} x$
}}
{{eqn | r = \frac 1 {2^{n / 2} \map \Gamma {n / 2} \paren {\frac 1 2 - t}^{n / 2} } \int_0^\infty u^{n / 2 - 1} e^{-u} \rd u
}}
{{eqn | r = \frac {\map \Gamma {n / 2} } {\map \Gamma {n / 2} } \paren {1 - 2 t}^{-n / 2}
      | c = {{Defof|Gamma Function}}
}}
{{eqn | r = \paren {1 - 2 t}^{-n / 2}
}}
{{end-eqn}}
{{qed}}
Theorem: Moment Generating Function of Linear Combination of Independent Random Variables

Let $X_1, X_2, \ldots, X_n$ be independent random variables. 
Let $k_1, k_2, \ldots, k_n$ be real numbers.
Let: 
:$\ds X = \sum_{i \mathop = 1}^n k_i X_i$
Let $M_{X_i}$ be the moment generating function of $X_i$ for $1 \le i \le n$.
Then: 
:$\ds \map {M_X} t = \prod_{i \mathop = 1}^n \map {M_{X_i}} {k_i t}$
for all $t$ such that $M_{X_i}$ exists for all $1 \le i \le n$.

Proof:
{{begin-eqn}}
{{eqn	| l = \map {M_X} t
	| r = \expect {\map \exp {t X} }
	| c = {{Defof|Moment Generating Function}}
}}
{{eqn	| r = \expect {\map \exp {t \sum_{i \mathop = 1}^n k_i X_i} }
}}
{{eqn	| r = \expect {\prod_{i \mathop = 1}^n \map \exp {t k_i X_i} }
	| c = Exponential of Sum
}}
{{eqn	| r = \prod_{i \mathop = 1}^n \expect {\map \exp {t k_i X_i} }
	| c = Condition for Independence from Product of Expectations
}}
{{eqn	| r = \prod_{i \mathop = 1}^n \map {M_{X_i} } {k_i t}
	| c = {{Defof|Moment Generating Function}}
}}
{{end-eqn}}
{{qed}}
Category:Moment Generating Functions
5sz5czlxm95jwxm0tpgng2g98u9cxh9
Theorem: Sum of Squares of Standard Normal Random Variables has Chi-Squared Distribution

Let $X_1, X_2, \ldots, X_n$ be independent random variables.
Let $X_i \sim \Gaussian 0 1$ for $1 \le i \le n$ where $\Gaussian 0 1$ is the standard normal distribution.
Then: 
:$\ds \sum_{i \mathop = 1}^n X^2_i \sim \chi^2_n$
where $\chi^2_n$ is the chi-squared distribution with $n$ degrees of freedom.

Proof:
By Square of Standard Normal Random Variable has Chi-Squared Distribution, we have: 
:$X^2_i \sim \chi^2_1$
for $1 \le i \le n$. 
So, by Sum of Chi-Squared Random Variables, we have: 
:$\ds \sum_{i \mathop = 1}^n X^2_i \sim \chi^2_{1 + 1 + 1 \ldots} = \chi^2_n$ 
{{qed}}
Category:Chi-Squared Distribution
ngbp12fmopqe1mfo7dgz66kx8w15hs5
Theorem: Variance of Random Sample from Normal Distribution has Chi-Squared Distribution

Let $X_1, X_2, \ldots, X_n$ form a random sample of size $n$ from the normal distribution $\Gaussian \mu {\sigma^2}$ for some $\mu \in \R, \sigma \in \R_{>0}$. 
Let:
:$\ds \bar X = \frac 1 n \sum_{i \mathop = 1}^n X_i$
and:
:$\ds s^2 = \frac 1 {n - 1} \sum_{i \mathop = 1}^n \paren {X_i - \bar X}^2$
Then: 
:$\dfrac {\paren {n - 1} s^2} {\sigma^2} \sim \chi^2_{n - 1}$
where $\chi^2_{n - 1}$ is the chi-squared distribution with $n - 1$ degrees of freedom.

Proof:
{{ProofWanted}}
Category:Normal Distribution
Category:Chi-Squared Distribution
qnd4tnqldx0ch0su52rzvm28sx9mh8u
Theorem: Differential Equations for Shortest Path on 3d Sphere/Cartesian Coordinates

Let $M$ be a $3$-dimensional Euclidean space.
Let $S$ be a sphere embedded in $M$.
Let $\gamma$ be a curve on $S$.
Let the chosen coordinate system be Cartesian.
Let $\gamma$ begin at $\paren {x_0, y_0, z_0}$ and terminate at $\paren {x_1, y_1, z_1}$.
Let $\map y x$, $\map z x$ be real functions.
Let $\gamma$ connecting both endpoints be of minimum length.
Then $\gamma$ satisfies the following equations of motion:
:$2 y \map \lambda x - \dfrac \d {\d x} \dfrac {y'} {\sqrt {1 + y'^2 + z'^2} } = 0$
:$2 z \map \lambda x - \dfrac \d {\d x} \dfrac {z'} {\sqrt {1 + y'^2 + z'^2} } = 0$

Proof:
In $3$-dimensional Euclidean space the length of the curve is:
:$\ds \int_{x_0}^{x_1} \sqrt {1 + y'^2 + z'^2} \rd x$
The sphere satisfies the following equation:
{{begin-eqn}}
{{eqn | l = \map g {x, y, z}
      | r = x^2 + y^2 + z^2 - a^2
}}
{{eqn | r = 0
}}
{{end-eqn}}
Consider its partial derivatives {{WRT|Differentiation}} $y$ and $z$.
:$\dfrac {\partial g} {\partial y} = 2y$
:$\dfrac {\partial g} {\partial z} = 2z$
$g_y$ and $g_z$ vanish for $y = 0$ and $z = 0$ respectively.
Substitution of this into the sphere equation tells us that $x^2 = a^2$.
Therefore, the following analysis should exclude points with $x = \pm a$.
By Simplest Variational Problem with Subsidiary Conditions for Curve on Surface, the length functional is replaced by the following auxiliary functional:
:$\ds \int_{x_0}^{x_1} \sqbrk {\sqrt {1 + y'^2 + z'^2} + \map \lambda x \paren {x^2 + y^2 + z^2} } \rd x$
It follows that:
:$\map {\dfrac {\partial} {\partial y'} } {\sqrt {1 + y'^2 + z'^2} + \map \lambda x \paren {x^2 + y^2 + z^2} } = \dfrac {y'} {\sqrt {1 + y'^2 + z'^2} }$
:$\map {\dfrac {\partial} {\partial y} } {\sqrt {1 + y'^2 + z'^2} + \map \lambda x \paren {x^2 + y^2 + z^2} } = 2 y \lambda$
Analogous relations hold for $\map z x$. 
Then by Euler's Equations the following equations of motion hold:
:$2 y \map \lambda x - \dfrac \d {\d x} \dfrac {y'} {\sqrt{1 + y'^2 + z'^2} } = 0$
:$2 z \map \lambda x - \dfrac \d {\d x} \dfrac {z'} {\sqrt{1 + y'^2 + z'^2} } = 0$
{{qed}}
Theorem: Skewness of Snedecor's F-Distribution

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:
{{ProofWanted}}
Category:Snedecor's F-Distribution
Category:Skewness
5izl4gd0tmfwpnlsqnpt49clukg58ay
Theorem: Excess Kurtosis of Snedecor's F-Distribution

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 excess kurtosis $\gamma_2$ of $X$ is given by: 
:$\gamma_2 = \dfrac {12 \paren {5 m n^2 - 22 n^2 + 5 m^2 n - 32 m n + 44 n + m^3 - 8 m^2 + 20 m - 16} } {n \paren {m - 6} \paren {m - 8} \paren {m + n - 2} }$
for $m > 8$, and does not exist otherwise.

Proof:
{{ProofWanted}}
Category:Kurtosis
Category:Snedecor's F-Distribution
43hnxj7j0bsok0d5pguqmn6gmsrrgbz
Theorem: Expectation of Chi Distribution

Let $n$ be a strictly positive integer. 
Let $X \sim \chi_n$ where $\chi_n$ is the chi distribution with $n$ degrees of freedom. 
Then the expectation of $X$ is given by: 
:$\expect X = \sqrt 2 \dfrac {\map \Gamma {\paren {n + 1} / 2} } {\map \Gamma {n / 2} }$
where $\Gamma$ is the gamma function.

Proof:
From the definition of the chi distribution, $X$ has probability density function:
:$\map {f_X} x = \dfrac 1 {2^{\paren {n / 2} - 1} \map \Gamma {n / 2} } x^{n - 1} e^{- x^2 / 2}$
From the definition of the expected value of a continuous random variable:
:$\ds \expect X = \int_0^\infty x \map {f_X} x \rd x$
So:
{{begin-eqn}}
{{eqn	| l = \expect X
	| r = \frac 1 {2^{\paren {n / 2} - 1} \map \Gamma {n / 2} } \int_0^\infty x^n e^{- x^2 / 2} \rd x
}}
{{eqn	| r = \frac 1 {2^{\paren {n / 2} - 1 + \paren {1 / 2} } \map \Gamma {n / 2} } \int_0^\infty \frac 1 {\sqrt t} \paren {\sqrt {2 t} }^n e^{- t} \rd t
	| c = substituting $x = \sqrt {2 t}$
}}
{{eqn	| r = \frac {2^{\paren {n / 2} } } {2^{\paren {n / 2} - \paren {1 / 2} } \map \Gamma {n / 2} } \int_0^\infty t^{\paren {n - 1} / 2} e^{- t} \rd t 
}}
{{eqn	| r = \sqrt 2 \dfrac {\map \Gamma {\paren {n + 1} / 2} } {\map \Gamma {n / 2} }
	| c = {{Defof|Gamma Function}}
}}
{{end-eqn}}
{{qed}}
Category:Expectation
Category:Chi Distribution
ok8wg69ssu7r8ldsxp7ku0kportd0qy
Theorem: Mean of Random Sample from Chi-Squared Distribution has Gamma Distribution

Let $n$ be a strictly positive integer. 
Let $X_1, X_2, \ldots, X_k$ form a random sample of size $k$ from the chi-squared distribution with $n$ degrees of freedom. 
Then: 
:$\ds \overline X = \frac 1 k \sum_{i \mathop = 1}^k X_i \sim \map \Gamma {\frac {n k} 2, \frac k 2}$
where $\map \Gamma {\dfrac {n k} 2, \dfrac k 2}$ is the gamma distribution with parameters $\dfrac {n k} 2$ and $\dfrac k 2$.

Proof:
By Sum of Chi-Squared Random Variables, we have: 
:$\ds \sum_{i \mathop = 1}^k X_i \sim \chi^2_{n k}$
By Multiple of Chi-Squared Random Variable has Gamma Distribution, we then have: 
:$\ds \frac 1 k \sum_{i \mathop = 1}^k X_i \sim \map \Gamma {\frac {n k} 2, \frac 1 {\frac 2 k} } = \map \Gamma {\frac {n k} 2, \frac k 2}$
{{qed}}
Category:Chi-Squared Distribution
Category:Gamma Distribution
4dwkwskew3y3bn2ux8vg9e85s19l38z
Theorem: Mode of Normal Distribution

Let $\mu$ be a real number. 
Let $\sigma$ be a strictly positive real number. 
Let $X \sim \Gaussian \mu {\sigma^2}$ where $\Gaussian \mu {\sigma^2}$ is the normal distribution with parameters $\mu$ and $\sigma^2$. 
Then the mode $M$ of $X$ is equal to $\mu$.

Proof:
By the definition of the normal distribution, $X$ has probability density function: 
:$\map {f_X} x = \dfrac 1 {\sigma \sqrt {2 \pi} } \, \map \exp {-\dfrac {\paren {x - \mu}^2} {2 \sigma^2} }$
By the definition of a mode, $f_X$ attains its global maximum at $M$.  
From Exponential is Strictly Increasing, we have that $\exp x$ is strictly increasing for all real $x$.
Therefore, to maximise $f_X$, we must pick to $x$ to maximise: 
:$-\dfrac {\paren {x - \mu}^2} {2 \sigma^2}$
That is, to minimise:
:$\dfrac {\paren {x - \mu}^2} {2 \sigma^2}$
We have: 
:$\dfrac {\paren {x - \mu}^2} {2 \sigma^2} \ge 0$
with equality occurring at $x = \mu$.
Therefore, $\mu$ is a mode of $X$.
As this is the only value of $x$ for which equality occurs, we have that $\mu$ is the unique mode of $X$. 
{{qed}}
Category:Modes
Category:Normal Distribution
af34ybsspxvhssk1klo0f0t018rnayg
Theorem: Equivalence of Definitions of Separated Sets

Let $T = \struct {S, \tau}$ be a topological space.
Let $A, B \subseteq S$.
{{TFAE|def = Separated Sets}}

Proof:
=== Definition 1 implies Definition 2 ===
{{:Equivalence of Definitions of Separated Sets/Definition 1 implies Definition 2}}{{qed|lemma}}
Theorem: Equivalence of Definitions of Separated Sets/Definition 1 implies Definition 2

Let $A, B \subseteq S$ satisfy:
:$A^- \cap B = A \cap B^- = \O$
where $A^-$ denotes the closure of $A$ in $T$, and $\O$ denotes the empty set.

Proof:
From Topological Closure is Closed, $B^-$ is closed in $T$.
Let $U =  S \setminus B^-$ be the relative complement of $B^-$.
By the definition of a closed set, $U$ is open in $T$.
From Empty Intersection iff Subset of Relative Complement:
:$A \subseteq S \setminus B^- = U$
From Relative Complement of Relative Complement:
:$S \setminus U = B^-$
By the definition of the closure of a subset:
:$B \subseteq B^- = S \setminus U$
From Empty Intersection iff Subset of Relative Complement:
:$U \cap B = \O$
Similarly, let $V =  S \setminus A^-$ then $V \in \tau$ with:
:$B \subset V$
and
:$V \cap A = \O$
Theorem: Equivalence of Definitions of Separated Sets/Definition 2 implies Definition 1

Let $A, B \subseteq S$.
Let $U, V \in \tau$ satisfy:
:$A \subset U$ and $U \cap B = \O$
:$B \subset V$ and $V \cap A = \O$

Proof:
Let $U \in \tau$ be an arbitrary open set of $T$ such that $A \subseteq U$ and $U \cap B = \O$.
From Empty Intersection iff Subset of Relative Complement:
:$B \subseteq S \setminus U$
By the definition of a closed set, the relative complement $S \setminus U$ of $U$ is closed in $T$.
From Set Closure is Smallest Closed Set in Topological Space:
:$B^- \subseteq S \setminus U$
From Empty Intersection iff Subset of Relative Complement:
:$B^- \cap U = \O$
Then
{{begin-eqn}}
{{eqn | l = B^- \cap A 
      | r = B^- \cap \paren {U \cap A}
      | c = Intersection with Subset is Subset
}}
{{eqn | r = \paren {B^- \cap U} \cap A
      | c = Intersection is Associative
}}
{{eqn | r = \O \cap A
      | c = as $B^- \cap U = \O$
}}
{{eqn | r = \O
      | c = Intersection with Empty Set
}}
{{end-eqn}}
Similarly:
:$A^- \cap B = \O$
Theorem: Expectation of Student's t-Distribution

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 expectation of $X$ is equal to $0$ for $k > 1$, and does not exist otherwise.

Proof:
From the definition of the Student's t-Distribution, $X$ has probability density function:
:$\map {f_X} x = \dfrac {\map \Gamma {\frac {k + 1} 2} } {\sqrt {\pi k} \map \Gamma {\frac k 2} } \paren {1 + \dfrac {x^2} k}^{-\frac {k + 1} 2}$
with $k$ degrees of freedom for some $k \in \R_{>0}$.
From the definition of the expected value of a continuous random variable: 
:$\ds \expect X = \int_{-\infty}^\infty x \map {f_X} x \rd x$
So for $k > 1$:
{{begin-eqn}}
{{eqn | l = \expect X
      | r = \dfrac {\map \Gamma {\frac {k + 1} 2} } {\sqrt {\pi k} \map \Gamma {\frac k 2} } \int_{-\infty}^\infty \dfrac x {\paren {1 + \dfrac {x^2} k}^{\frac {k + 1} 2} } \rd x
}}
{{eqn | r = \dfrac {\map \Gamma {\frac {k + 1} 2} } {\sqrt {\pi k} \map \Gamma {\frac k 2} } \paren {\int_{-\infty}^0 \dfrac x {\paren {1 + \dfrac {x^2} k}^{\frac {k + 1} 2} } \rd x + \int_0^\infty \dfrac x {\paren {1 + \dfrac {x^2} k}^{\frac {k + 1} 2} } \rd x}
      | c = Sum of Integrals on Adjacent Intervals for Continuous Functions
}}
{{eqn | r = \dfrac {k \map \Gamma {\frac {k + 1} 2} } {2 \sqrt {\pi k} \map \Gamma {\frac k 2} } \paren {\int_\infty^1 \dfrac 1 {\paren u^{\frac {k + 1} 2} } \rd u + \int_1^\infty \dfrac 1 {\paren u^{\frac {k + 1} 2} } \rd u}
      | c =  U-substitution Substituting $u = 1 + \dfrac {x^2} k$ and $\dfrac k 2 \rd u = x \rd x$
}}
{{eqn | r = \dfrac {k \map \Gamma {\frac {k + 1} 2} } {2 \sqrt {\pi k} \map \Gamma {\frac k 2} } \paren {-\int_1^\infty \dfrac 1 {\paren u^{\frac {k + 1} 2} } \rd u + \int_1^\infty \dfrac 1 {\paren u^{\frac {k + 1} 2} } \rd u}
      | c = Reversal of Limits of Definite Integral
}}
{{eqn | r = \dfrac {\map \Gamma {\frac {k + 1} 2} } {2 \sqrt {\pi k} \map \Gamma {\frac k 2} } \paren {\dfrac {-2} {k - 1 } } \paren {-\intlimits {\paren u^{-\frac {k - 1} 2} } 1 \infty + \intlimits {\paren u^{-\frac {k - 1} 2} } 1 \infty}
      | c = Primitive of Power, Fundamental Theorem of Calculus
}}
{{eqn | r = 0
      | c = Definite Integral of Odd Function/Corollary
}}
{{end-eqn}}
{{qed|lemma}}
When $k = 1$, we have the natural log evaluated at infinity where it is undefined.
Hence: 
{{begin-eqn}}
{{eqn | o = 
      | r = \dfrac {\map \Gamma 1} {2 \sqrt \pi \map \Gamma {\frac 1 2} } \paren {-\int_1^\infty \dfrac 1 u \rd u + \int_1^\infty  \dfrac 1 u \rd u}
      | c =  U-substitution Substituting $u = 1 + \dfrac {x^2} k$ and $\dfrac k 2 \rd u = x \rd x$ and Reversal of Limits of Definite Integral
}}
{{eqn | r = \dfrac {\map \Gamma 1} {2 \sqrt \pi \map \Gamma {\frac 1 2} } \paren {-\bigintlimits {\map \ln u} 1 \infty + \bigintlimits {\map \ln u} 1 \infty}
      | c = Fundamental Theorem of Calculus, {{Defof|Natural Logarithm}}, Logarithm Tends to Infinity
}}
{{end-eqn}}
Therefore, the expectation of $X$ does not exist:
{{qed}}
Theorem: Variance of Student's t-Distribution

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 variance of $X$ is given by: 
:$\var X = \dfrac k {k - 2}$
for $k > 2$, and does not exist otherwise.

Proof:
By Expectation of Student's t-Distribution, we have that for $k > 1$: 
:$\expect X = 0$
From Square of Random Variable with t-Distribution has Snedecor's F-Distribution, we have: 
:$\expect {X^2} = \expect Y$
with $Y \sim F_{1, k}$, where $F_{1, k}$ is Snedecor's $F$-distribution with $\tuple {1, k}$ degrees of freedom.
By Expectation of Snedecor's F-Distribution we have that $\expect Y$ exists {{iff}} $k > 2$. 
:$\expect Y = \expect {X^2} = \dfrac k {k - 2}$
We therefore have: 
{{begin-eqn}}
{{eqn | l = \var X
      | r = \expect {X^2} - \paren {\expect X}^2
      | c = Variance as Expectation of Square minus Square of Expectation
}}
{{eqn | r = \frac k {k - 2} - 0^2
}}
{{eqn | r = \frac k {k - 2}
}}
{{end-eqn}}
{{qed}}
Theorem: Excess Kurtosis of Student's t-Distribution

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 excess kurtosis $\gamma_2$ of $X$ is given by: 
:$\gamma_2 = \dfrac 6 {k - 4}$
for $k > 4$, and does not exist otherwise.

Proof:
{{ProofWanted}}
Category:Kurtosis
Category:Student's t-Distribution
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Theorem: Connected Set in Subspace

Let $T = \struct {S, \tau}$ be a topological space.
Let $A \subseteq B \subseteq S$.
Let $T_B = \struct {B, \tau_B}$ be the topological space where $\tau_B$ is the subspace topology on $B$.
Then
:$A$ is connected in $T_B$ {{iff}} $A$ is connected in $T$.

Proof:
Let $\tau_A$ be the subspace topology on $A$ induced by $\tau$.
Let $\tau'_A$ be the subspace topology on $A$ induced by $\tau_B$.
By the definition of a connected set, $A$ is connected in $T$ {{iff}} $\struct {A, \tau_A}$ is a connected topological space.
Similarly, $A$ is connected in $T_B$ {{iff}} $\struct {A, \tau'_A}$ is a connected topological space.
From Subspace of Subspace is Subspace, we have $\tau_A = \tau'_A$.
Hence $A$ is connected in $T_B$ {{iff}} $A$ is connected in $T$.
{{qed}}
Category:Connected Topological Spaces
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