nyuuzyou/wikis huggingface dataset:
Variance of Hat-Check Distribution

== Theorem ==
Let $X$ be a discrete random variable with the Hat-Check distribution with parameter $n$.
Then the variance of $X$ is given by:
:$\var X = 1$
== Proof ==
From the definition of Variance as Expectation of Square minus Square of Expectation:
:$\var X = \expect {X^2} - \paren {\expect X}^2$
From Expectation of Function of Discrete Random Variable:
:$\ds \expect {X^2} = \sum_{x \mathop \in \Omega_X} x^2 \, \map \Pr {X = x}$
So:
Then:
== Also see ==
* Expectation of Hat-Check Distribution
Category:Variance
Category:Hat-Check Distribution
Category:Hat-Check Problem
Variance of Linear Combination of Random Variables

== Theorem ==
Let $X$ and $Y$ be random variables. 
Let the variances of $X$ and $Y$ be finite. 
Let $a$ and $b$ be real numbers. 
Then the variance of $a X + b Y$ is given by: 
:$\var {a X + b Y} = a^2 \, \var X + b^2 \, \var Y + 2 a b \, \cov {X, Y}$
where $\cov {X, Y}$ is the covariance of $X$ and $Y$. 
=== Corollary ===
== Proof == 
Category:Variance
Variance of Linear Combination of Random Variables/Corollary

== Theorem ==
Let $X$ and $Y$ be independent random variables. 
 
Let the variances of $X$ and $Y$ be finite. 
Let $a$ and $b$ be real numbers. 
Then the variance of $a X + b Y$ is given by: 
:$\var {a X + b Y} = a^2 \, \var X + b^2 \, \var Y$
== Proof ==
From Variance of Linear Combination of Random Variables, we have: 
:$\var {a X + b Y} = a^2 \, \var X + b^2 \, \var Y + 2 a b \, \cov {X, Y}$
where $\cov {X, Y}$ is the covariance of $X$ and $Y$. 
From Covariance of Independent Random Variables is Zero: 
:$2 a b \, \cov {X, Y} = 0$
The result follows.
Category:Variance
Variance of Linear Function of Observations of Stationary Process

== Theorem ==
Let $S$ be a strictly stationary stochastic process giving rise to a time series $T$.
Let $\sequence {s_n}$ be a sequence of $n$ successive values of $T$:
:$\sequence {s_n} = \tuple {z_1, z_2, \dotsb, z_n}$
Let $L_t$ be a linear function of $\sequence {s_n}$:
:$L_t = l_1 z_t + l_2 z_{t - 1} + \dotsb + l_n z_{t - n + 1}$
Then the variance of $L_t$ is given by:
:$\var {L_t} = \ds \sum_{i \mathop = 1}^n \sum_{j \mathop = 1}^n l_i l_j \gamma {\size {j - i} }$
where $\gamma_k$ is the autocovariance of $S$ at lag $k$.
== Proof ==
== Sources ==
* :
::Part $\text {I}$: Stochastic Models and their Forecasting:
:::$2$: Autocorrelation Function and Spectrum of Stationary Processes:
::::$2.1$ Autocorrelation Properties of Stationary Models:
:::::$2.1.3$ Positive Definiteness and the Autocovariance Matrix
Category:Stationary Stochastic Processes
Variance 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 variance of $X$ is given by: 
:$\var X = \map \exp {2 \mu + \sigma^2} \paren {\map \exp {\sigma^2} - 1}$
== Proof ==
By Variance as Expectation of Square minus Square of Expectation:
:$\var X = \expect {X^2} - \paren {\expect X}^2$
By Expectation of Log Normal Distribution:
:$\expect X = \map \exp {\mu + \dfrac {\sigma^2} 2}$
From Raw Moment of Log Normal Distribution:
The $n$th raw moment $\expect {X^n}$ of $X$ is given by: 
:$\expect {X^n} = \map \exp {n \mu + \dfrac {\sigma^2 n^2} 2}$
Therefore, for $n = 2$ we have:
:$\expect {X^2} = \map \exp {2 \mu + 2 \sigma^2}$
So:
Category:Variance
Category:Log Normal Distribution
Variance of Logistic Distribution

== Theorem == 
Let $X$ be a continuous random variable which satisfies the logistic distribution:
:$X \sim \map {\operatorname {Logistic} } {\mu, s}$
The variance of $X$ is given by:
:$\var X = \dfrac {s^2 \pi^2} 3$
=== Lemma 1 ===
=== Lemma 2 ===
=== Lemma 3 ===
=== Lemma 4 ===
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
Category:Variance
Category:Logistic Distribution
Category:Variance of Logistic Distribution
Variance of Logistic Distribution/Lemma 3

== Lemma for Variance of Logistic Distribution ==
:$\ds \int_{\to 0}^{\to 1} \map \ln u \map \ln {1 - u} \rd u = 2 - \dfrac {\pi^2} 6$
== Proof ==
From Corollary to Power Series Expansion for Logarithm of 1 + x we have:
:$\ds \ln \paren {1 - x} = -\sum_{n \mathop = 1}^\infty \dfrac {x^n} n$
Therefore: 
Category:Variance of Logistic Distribution
Variance of Logistic Distribution/Proof 1

== Theorem == 
== Proof ==
From the definition of the logistic distribution, $X$ has probability density function:
:$\map {f_X} x = \dfrac {\map \exp {-\dfrac {\paren {x - \mu} } s} } {s \paren {1 + \map \exp {-\dfrac {\paren {x - \mu} } s} }^2}$
From Variance as Expectation of Square minus Square of Expectation: 
:$\ds \var X = \int_{-\infty}^\infty x^2 \, \map {f_X} x \rd x - \paren {\expect X}^2$
So:
:$\ds \var X = \frac 1 s \int_{-\infty}^\infty \dfrac {x^2 \map \exp {-\dfrac {\paren {x - \mu} } s} } {\paren {1 + \map \exp {-\dfrac {\paren {x - \mu} } s} }^2} \rd x - \mu^2$
let:
and also:
Then:
 
Category:Variance of Logistic Distribution
Variance of Logistic Distribution/Proof 2

== Theorem == 
== Proof == 
By Moment Generating Function of Logistic Distribution, the moment generating function of $X$ is given by: 
:$\ds \map {M_X} t = \map \exp {\mu t} \int_{\to 0}^{\to 1} \paren {\dfrac {1 - u} u}^{-s t} \rd u$
for $\size t < \dfrac 1 s$. 
From Variance as Expectation of Square minus Square of Expectation: 
:$\ds \var X = \expect {X^2} - \paren {\expect X}^2$
From Expectation of Logistic Distribution, we have:
:$\expect X = \mu$
From Moment Generating Function of Logistic Distribution: Second Moment: 
:$\ds \map { {M_X}} t = \map \exp {\mu t} \paren {\mu^2 \int_{\to 0}^{\to 1} \paren {\dfrac {1 - u} u}^{-s t} \rd u - 2 s \mu \int_{\to 0}^{\to 1} \map \ln {\dfrac {1 - u} u} \paren {\dfrac {1 - u} u}^{-s t} \rd u + s^2 \int_{\to 0}^{\to 1} \map {\ln^2} {\dfrac {1 - u} u} \paren {\dfrac {1 - u} u}^{-s t} \rd u}$
Hence setting $t = 0$:
So: 
Category:Variance of Logistic Distribution
Variance of Logistic Distribution/Proof 3

== Theorem == 
== Proof ==
From the definition of the logistic distribution, $X$ has probability density function:
:$\map {f_X} x = \dfrac {\map \exp {-\dfrac {\paren {x - \mu} } s} } {s \paren {1 + \map \exp {-\dfrac {\paren {x - \mu} } s} }^2}$
From Variance as Expectation of Square minus Square of Expectation: 
:$\ds \var X = \int_{-\infty}^\infty x^2 \, \map {f_X} x \rd x - \paren {\expect X}^2$
So:
From Sum of Infinite Geometric Sequence, we have:
:$\ds \sum_{n \mathop = 0}^\infty \paren {-1}^n \map \exp {-z}^n = \dfrac 1 {1 + \map \exp {-z} }$
Taking the derivative of both sides, we have:
Therefore:
 
Category:Variance of Logistic Distribution
Variance of Negative Binomial Distribution/Type 1

== Theorem ==
Let $X$ be a discrete random variable with the type $1$ negative binomial distribution with parameters $r$ and $p$.
Then the variance of $X$ is given by:
:$\var X = \dfrac {r q} {p^2}$
where $q = 1 - p$.
== Proof ==
From Variance of Discrete Random Variable from PGF:
:$\var X = \map {\Pi' '_X} 1 + \mu - \mu^2$
where $\mu = \expect X$ is the expectation of $X$.
From the Probability Generating Function of Negative Binomial Distribution (Type 1):
:$\map {\Pi_X} s = \dfrac {p s} {1 - q s}$
From Expectation of Negative Binomial Distribution (Type 1):
:$\mu = \dfrac r p$
From Second Derivative of PGF of Negative Binomial Distribution (Type 1):
:$\dfrac {\d^2} {\d s^2} \map {\Pi_X} s = \paren {\dfrac {p s} {1 - q s} }^{r + 2} \paren {\dfrac {r \paren {r - 1} + 2 r q s} {\paren {p s^2}^2} }$
Putting $s = 1$ and using the formula $\map {\Pi' '_X} 1 + \mu - \mu^2$:
== Sources ==
* : $\S 4.3$: Moments: Exercise $6$
* : Appendix $13$: Probability distributions
* : Appendix $15$: Probability distributions
Category:Variance of Negative Binomial Distribution
Category:Negative Binomial Distribution (Type 1)
Category:Variance
Variance of Negative Binomial Distribution/Type 2

== Theorem ==
Let $X$ be a discrete random variable with the type $2$ negative binomial distribution with parameters $r$ and $p$.
Then the variance of $X$ is given by:
:$\var X = \dfrac {r q} {p^2}$
where $q = 1 - p$.
== Proof ==
== Sources ==
* 
* 
Category:Variance of Negative Binomial Distribution
Category:Negative Binomial Distribution (Type 2)
Category:Variance
Variance of Normal Distribution

== Theorem ==
Let $X \sim N \paren {\mu, \sigma^2}$ for some $\mu \in \R, \sigma \in \R_{> 0}$, where $N$ is the normal distribution.
Then: 
:$\var X = \sigma^2$
== Proof 1 ==
 
== Proof 2 ==
== Sources ==
* 
* 
* : Appendix $13$: Probability distributions
* : Appendix $15$: Probability distributions
Category:Variance of Normal Distribution
Category:Variance
Category:Normal Distribution
Variance of Normal Distribution/Proof 1

== Theorem ==
== Proof ==
From the definition of the normal distribution, $X$ has probability density function:
:$\map {f_X} x = \dfrac 1 {\sigma \sqrt {2 \pi} } \map \exp {-\dfrac {\paren {x - \mu}^2} {2 \sigma^2} }$
From Variance as Expectation of Square minus Square of Expectation: 
:$\ds \var X = \int_{-\infty}^\infty x^2 \map {f_X} x \rd x - \paren {\expect X}^2$
So:
Category:Variance of Normal Distribution
Variance of Normal Distribution/Proof 2

== Theorem == 
== Proof == 
By Moment Generating Function of Normal Distribution, the moment generating function of $X$ is given by: 
:$\map {M_X} t = \map \exp {\mu t + \dfrac 1 2 \sigma^2 t^2}$
From Variance as Expectation of Square minus Square of Expectation: 
:$\var X = \expect {X^2} - \paren {\expect X}^2$
From Moment Generating Function of Normal Distribution: Second Moment: 
:$\map { {M_X}' '} t = \paren {\sigma^2 + \paren {\mu + \sigma^2 t}^2 } \map \exp {\mu t + \dfrac 1 2 \sigma^2 t^2}$
From Moment in terms of Moment Generating Function, we also have: 
:$\expect {X^2} = \map { {M_X}' '} 0$
Setting $t = 0$, we obtain the second moment: 
So: 
Category:Variance of Normal Distribution
Variance of Pareto Distribution

== Theorem ==
Let $X$ be a continuous random variable with the Pareto distribution with $a, b \in \R_{> 0}$.
Then the variance of $X$ is given by: 
:$\var X = \begin {cases} \dfrac {a b^2 } {\paren {a - 2} \paren {a - 1}^2 } & 2 < a \\ \text {does not exist} & 2 \ge a \end {cases}$
== 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 Pareto Distribution, we have: 
:$\expect X = \begin {cases} \dfrac {a b} {a - 1} & 1 < a \\ \text {does not exist} & 1 \ge a \end {cases}$
From Raw Moment of Pareto Distribution, we have:
The $n$th raw moment $\expect {X^n}$ of $X$ is given by: 
:$\expect {X^n} = \begin {cases} \dfrac {a b^n} {a - n} & n < a \\ \text {does not exist} & n \ge a \end {cases}$
Therefore, for $n = 2$ we have:
:$\expect {X^2} = \begin {cases} \dfrac {a b^2} {a - 2} & 2 < a \\ \text {does not exist} & 2 \ge a \end {cases}$
So:
Category:Variance
Category:Pareto Distribution
Variance of Poisson Distribution

== Theorem ==
Let $X$ be a discrete random variable with the Poisson distribution with parameter $\lambda$.
Then the variance of $X$ is given by:
:$\var X = \lambda$
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Also see ==
* Expectation and Variance of Poisson Distribution equal its Parameter
== Sources ==
* : $\S 2.4$: Expectation: Exercise $10$
* : Appendix $13$: Probability distributions
* : Appendix $15$: Probability distributions
Category:Variance of Poisson Distribution
Category:Variance
Category:Poisson Distribution
Variance of Poisson Distribution/Proof 1

== Theorem ==
== Proof ==
From the definition of Variance as Expectation of Square minus Square of Expectation:
:$\var X = \expect {X^2} - \paren {\expect X}^2$
From Expectation of Function of Discrete Random Variable:
:$\ds \expect {X^2} = \sum_{x \mathop \in \Omega_X} x^2 \, \map \Pr {X = x}$
So:
Then:
Category:Variance of Poisson Distribution
Variance of Poisson Distribution/Proof 2

== Theorem ==
== Proof 2==
From Variance of Discrete Random Variable from PGF, we have:
:$\var X = \map {\Pi_X} 1 + \mu - \mu^2$
where $\mu = \expect X$ is the expectation of $X$.
From the Probability Generating Function of Poisson Distribution, we have:
:$\map {\Pi_X} s = e^{-\lambda \paren {1 - s} }$
From Expectation of Poisson Distribution, we have:
:$\mu = \lambda$
From Derivatives of PGF of Poisson Distribution, we have:
:$\map {\Pi_X} s = \lambda^2 e^{-\lambda \paren {1 - s} }$
Putting $s = 1$ using the formula $\map {\Pi_X} 1 + \mu - \mu^2$:
:$\var X = \lambda^2 e^{-\lambda \paren {1 - 1} } + \lambda - \lambda^2$
and hence the result.
== Sources ==
* : $\S 4.3$: Moments: Exercise $5$
Category:Variance of Poisson Distribution
Variance of Poisson Distribution/Proof 3

== Theorem == 
== Proof ==
From Moment Generating Function of Poisson Distribution, the moment generating function of $X$, $M_X$, is given by: 
:$\map {M_X} t = e^{\lambda \paren {e^t - 1} }$
From Variance as Expectation of Square minus Square of Expectation, we have: 
:$\var X = \expect {X^2} - \paren {\expect X}^2$
From Moment in terms of Moment Generating Function:
:$\expect {X^2} = \map {M_X} 0$
In Expectation of Poisson Distribution, it is shown that: 
:$\map {M_X'} t = \lambda e^t e^{\lambda \paren {e^t - 1} }$
Then:
Setting $t = 0$: 
From Expectation of Poisson Distribution: 
:$\expect X = \lambda$
So: 
:$\var X = \lambda^2 + \lambda - \lambda^2 = \lambda$
Category:Variance of Poisson Distribution
Variance of Random Sample from Normal Distribution has Chi-Squared Distribution

== Theorem ==
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 ==
Category:Normal Distribution
Category:Chi-Squared Distribution
Variance of Sample Distribution Function

== Theorem ==
Let $S$ be a sample of ordinal data.
Let $\map {S_n} x$ denote the sample distribution function on $S$.
Then:
:$\var {\map {S_n} x} = \dfrac {\map F x \paren {1 - \map F x} } n$
where:
:$\var {\map {S_n} x}$ denotes the variance of $\map {S_n} x$
:$\map F x$ denotes the cumulative distribution function of the population from which $S$ was taken.
== Proof ==
== Sources ==
* 
Category:Sample Distribution Functions
Category:Variance
Variance of Sample Mean

== Theorem ==
Let $X_1, X_2, \ldots, X_n$ form a random sample from a population with mean $\mu$ and variance $\sigma^2$.
Let: 
:$\ds \overline X = \frac 1 n \sum_{i \mathop = 1}^n X_i$
Then: 
:$\var {\overline X} = \dfrac {\sigma^2} n$
== Proof ==
Category:Variance
Category:Inductive Statistics
Variance of Shifted Geometric Distribution

== Theorem ==
Let $X$ be a discrete random variable with the shifted geometric distribution with parameter $p$.
Then the variance of $X$ is given by:
:$\var X = \dfrac {1 - p} {p^2}$
== Proof 1 ==
== Proof 2 ==
== Also presented as ==
== Sources ==
* 
* : Appendix $13$: Probability distributions
* : Appendix $15$: Probability distributions
Category:Variance of Shifted Geometric Distribution
Category:Variance
Category:Geometric Distribution
Variance of Shifted Geometric Distribution/Proof 1

== Theorem ==
== Proof ==
From the definition of Variance as Expectation of Square minus Square of Expectation:
:$\var X = \expect {X^2} - \paren {\expect X}^2$
From Expectation of Function of Discrete Random Variable:
:$\ds \expect {X^2} = \sum_{x \mathop \in \Img X} x^2 \map \Pr {X = x}$
To simplify the algebra a bit, let $q = 1 - p$, so $p + q = 1$.
Thus:
Then:
== Sources ==
* : $\S 2.4$: Expectation: Example $24$
Category:Variance of Shifted Geometric Distribution
Variance of Shifted Geometric Distribution/Proof 2

== Theorem ==
== Proof ==
From Variance of Discrete Random Variable from PGF, we have:
:$\var X = \map {\Pi' '_X} 1 + \mu - \mu^2$
where $\mu = \expect X$ is the expectation of $X$.
From the Probability Generating Function of Shifted Geometric Distribution, we have:
:$\map {\Pi_X} s = \dfrac {p s} {1 - q s}$
where $q = 1 - p$.
From Expectation of Shifted Geometric Distribution, we have:
:$\mu = \dfrac 1 p$
From Derivatives of PGF of Shifted Geometric Distribution, we have:
:$\map {\Pi' '_X} s = \dfrac {p q} {\paren {1 - q s}^3}$
Putting $s = 1$ using the formula $\map {\Pi' '_X} 1 + \mu - \mu^2$:
:$\var X = \dfrac {p q} {\paren {1 - q}^3} + \dfrac 1 p - \paren {\dfrac 1 p}^2$
and hence the result, after some algebra.
== Sources ==
* : $\S 4.3$: Moments: Example $21$
Category:Variance of Shifted Geometric Distribution
Variance 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 variance of $X$ is given by: 
:$\var X = \dfrac {2 m^2 \paren {m + n - 2} } {n \paren {m - 4} \paren {m - 2}^2}$
for $m > 4$, and does not exist otherwise.
 
== Proof ==
Since $m > 4 > 2$, we have by Expectation of Snedecor's F-Distribution:
:$\expect X = \dfrac m {m - 2}$
We now aim to compute $\expect {X^2}$ with a view to apply Variance as Expectation of Square minus Square of Expectation.
Let $Y$ and $Z$ be independent random variables.
Let $Y \sim \chi^2_n$ where $\chi^2_n$ is the chi-squared distribution with $n$ degrees of freedom. 
Let $Z \sim \chi^2_m$ where $\chi^2_m$ is the chi-squared distribution with $m$ degrees of freedom.
Then:
:$\dfrac {Y / n} {Z / m} \sim F_{n, m}$
Therefore:
:$\expect {X^2} = \expect {\paren {\dfrac {Y / n} {Z / m} }^2}$
Let $f_Y$ and $f_Z$ be the probability density functions of $Y$ and $Z$ respectively.
Let $f_{Y, Z}$ be the joint probability density function of $Y$ and $Z$. 
From Condition for Independence from Joint Probability Density Function, we have for each $y, z \in \R_{\ge 0}$: 
:$\map {f_{Y, Z} } {y, z} = \map {f_Y} y \map {f_Z} z$
We therefore have: 
Note that the integral: 
:$\ds \int_0^\infty z^{m / 2 - 3} e^{-z / 2} \rd z$
converges :
:$\dfrac m 2 - 3 > -1$
That is: 
:$m > 4$
With that, we have for $m > 4$: 
Note that the integral: 
:$\ds \int_0^\infty y^{n / 2 + 1} e^{-y / 2} \rd z$
converges :
:$\dfrac n 2 + 1 > -1$
That is: 
:$n > -4$
This is ensured by the fact that $n \in \N$. 
With that, we have: 
We therefore have: 
We therefore have: 
Category:Variance
Category:Snedecor's F-Distribution
Variance 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 variance of $X$ is given by: 
:$\var X = \dfrac k {k - 2}$
for $k > 2$, and does not exist otherwise. 
 
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Appendix $15$: Probability distributions
Category:Variance
Category:Student's t-Distribution
Variance of Student's t-Distribution/Proof 1

== Theorem ==
== 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 $k > 2$. 
:$\expect Y = \expect {X^2} = \dfrac k {k - 2}$
We therefore have: 
Category:Student's t-Distribution
Category:Variance
Variance of Student's t-Distribution/Proof 2

== Theorem ==
== 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 Variance as Expectation of Square minus Square of Expectation: 
:$\ds \var X = \int_{-\infty}^\infty x^2 \, \map {f_X} x \rd x - \paren {\expect X}^2$
So:
:$\ds \var X = \dfrac {\map \Gamma {\frac {k + 1} 2} } {\sqrt {\pi k} \map \Gamma {\frac k 2} } \int_{-\infty}^\infty \dfrac {x^2} {\paren {1 + \dfrac {x^2} k}^{\frac {k + 1} 2} } \rd x - \mu^2$
Recall from Expectation of Student's t-Distribution that $k \gt 1$, so for $k \gt 1$:
Let: 
: $u = \dfrac {\dfrac {x^2} k} {\paren {1 + \dfrac {x^2} k} }$
Then by Quotient Rule for Derivatives, we have:
: $\dfrac k 2 \rd u = \dfrac {x \rd x} {\paren {1 + \dfrac {x^2} k}^2 }$
And:
: $\paren {1 - u } = \dfrac 1 {\paren {1 + \dfrac {x^2} k} }$
We can re-write $u$ as:
: $u = \dfrac 1 {\paren {1 + \dfrac k {x^2}} }$
And in this form, we can see that as $x \to 0$, $u \to 0$ and as $x \to \infty$, $u \to 1$
Plugging these results back into our integral above, we have:
Note: from the , the Beta function is only defined for $\map \Re x, \map \Re y > 0$
Therefore:
: $\dfrac {k - 2} 2 \gt 0 \leadsto k \gt 2$
Category:Variance
Category:Student's t-Distribution
Variance of Truncated Distribution

== Theorem ==
Let $X$ be a probability distribution.
Let $x_0$ be a given value of $X$.
Let $Y$ be a truncated distribution formed from $X$ by ignoring the part either to the left or right of $x_0$.
Then the variance of $Y$ is dependent upon $X$ and $x_0$.
== Proof ==
== Sources ==
* 
* 
Category:Truncated Distributions
Category:Expectation
Variance of Weibull Distribution

== Theorem ==
Let $X$ be a continuous random variable with the Weibull distribution with $\alpha, \beta \in \R_{> 0}$.
Then the variance of $X$ is given by: 
:$\var X = \beta^2 \paren {\map \Gamma {1 + \dfrac 2 \alpha} - \paren {\map \Gamma {1 + \dfrac 1 \alpha} }^2}$
where $\Gamma$ is the Gamma function.
== 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 Weibull Distribution, we have: 
:$\expect X = \beta \, \map \Gamma {1 + \dfrac 1 \alpha}$
From Raw Moment of Weibull Distribution, we have:
The $n$th raw moment $\expect {X^n}$ of $X$ is given by: 
:$\expect {X^n} = \beta^n \map \Gamma {1 + \dfrac n \alpha}$
Therefore, for $n = 2$ we have:
:$\expect {X^2} = \beta^2 \map \Gamma {1 + \dfrac 2 \alpha}$
So:
Category:Variance
Category:Weibull Distribution
Variation in Sign of Polynomial/Examples/Arbitrary Example 1

== Examples of Variations in Sign of Polynomial ==
Consider the polynomial equation over real numbers:
:$x^5 + x^4 - 2 x^3 + x^2 - 1 = 0$
This has three variations in sign:
:from $x^4$ to $-2 x^3$, where it goes from positive to negative
:from $-2 x^3$ to $x^2$, where it goes from negative to positive
:from $x^2$ to $-1$, where it goes from positive to negative.
== Sources ==
* 
* 
Category:Examples of Variations in Sign of Polynomial
Variation of Complex Measure is Finite Measure/Lemma

== Lemma ==
Let $\struct {X, \Sigma}$ be a measurable space.
Let $\mu$ be a complex measure on $\struct {X, \Sigma}$. 
Let $\cmod \mu$ be the variation of $\mu$. 
Let $\tuple {\mu_1, \mu_2, \mu_3, \mu_4}$ be the Jordan decomposition of $\mu$. 
Then: 
:$\map {\cmod \mu} A < \infty$ for each $A \in \Sigma$.
== Proof ==
From Measures in Jordan Decomposition of Complex Measure are Finite, we have: 
:$\mu_1$, $\mu_2$, $\mu_3$, $\mu_4$ are finite.
So there exist real numbers $M_1, M_2, M_3, M_4 > 0$ such that: 
:$\map {\mu_i} A \le M_i$
for each $A \in \Sigma$. 
Taking:
:$M_1 + M_2 + M_3 + M_4 = M$
we have, from Bound for Variation of Complex Measure in terms of Jordan Decomposition: 
:$\ds \sum_{i \mathop = 1}^n \cmod {\map \mu {A_i} } \le \map {\mu_1} A + \map {\mu_2} A + \map {\mu_3} A + \map {\mu_4} A \le M < \infty$ 
for each $A \in \Sigma$, with $M > 0$ a real number independent of $A$.
So, for each: 
:$\ds S \in \set {\sum_{j \mathop = 1}^n \cmod {\map \mu {A_j} } : \set {A_1, A_2, \ldots, A_n} \in \map P A}$
we have: 
:$S \le M$ 
for some $M \in \R$. 
So, from the definition of supremum, we have: 
:$\ds \sup \set {\sum_{j \mathop = 1}^n \cmod {\map \mu {A_j} } : \set {A_1, A_2, \ldots, A_n} \in \map P A} \le M$
that is: 
:$\map {\cmod \mu} A \le M < \infty$
So:
:$\map {\cmod \mu} A < \infty$ for each $A \in \Sigma$.
Category:Variation of Complex Measure is Finite Measure