nyuuzyou/wikis huggingface dataset:
Vanishing of Quasi-Coherent Sheaf Cohomology of Noetherian Affine Scheme

== Theorem ==
Let $X = \Spec A$ be the spectrum of a noetherian commutative ring $A$.
Let $\FF$ be a quasi-coherent sheaf on $X$.
Then for all $i \in \Z$ with $i > 0$ the $i$-th sheaf cohomology $\map {H^i} {X, \FF} = 0$.
== Proof 1 ==
Let $M := \map\Gamma{X, \FF}$ be the $\OO_X$-module of global sections of $\FF$.
By Category of Modules has Enough Injectives the category of $A$-modules has enough injectives.
By Injective Resolution Exists iff Enough Injectives $M$ has an injective resolution $I$.
Let 
: $\tilde{} : A\text{-}\mathbf{Mod} \to \mathbf{QCoh}(X)$
be the functor from the category of $A$-modules to the category of quasi-coherent $\OO_X$-modules $\mathbf{QCoh}(X)$.
By Equivalence of Categories is Exact and Exact Functor Preserves Exact Sequences $\tilde I$ is an exact cochain complex.
By Equivalence of Modules and Quasi-Coherent Sheaves there is an isomorphism $\FF \cong \tilde M$.
By Sheaf Associated to Injective Module over Noetherian Ring is Flasque, $\tilde I^i$ is flasque for all $i \mathop \in \Z$.
By Flasque Sheaf is Acyclic $\tilde I$ is an acyclic resolution of $\FF$.
By Equivalence of Modules and Quasi-Coherent Sheaves, the cochain complex $\Gamma(X, \tilde I)$ is exact at $i$ for all positive integers $i \in \Z_{\geq 1}$.
By Cochain Complex is Exact iff Cohomology Vanishes $H^i(\Gamma(X, \tilde I)) = 0$ for all positive integers $i \in \Z_{\geq 1}$.
By Right Derived Functor by Acyclic Resolution the cohomology groups $H^i(\Gamma(X, \tilde I))$ are isomorphic to the sheaf cohomology $H^i(X, \FF)$ for all $i \mathop \in \Z$.
Hence $H^i(X, \FF) = 0$ for all positive integers $i \in \Z_{\geq 1}$.
== Proof 2 ==
This is a special case of Vanishing of Quasi-Coherent Sheaf Cohomology of Affine Scheme.
== Sources ==
* Theorem $\text{III}.3.5$
Category:Algebraic Geometry
Category:Schemes
Variance as Expectation of Square minus Square of Expectation

== Theorem ==
Let $X$ be a random variable.
Then the variance of $X$ can be expressed as:
:$\var X = \expect {X^2} - \paren {\expect X}^2$
That is, it is the expectation of the square of $X$ minus the square of the expectation of $X$.
== Proof ==
=== Discrete Random Variable ===
=== Continuous Random Variable ===
== Motivation ==
This is a significantly more convenient way of defining the variance than the first-principles version.
In particular, it is far easier to program a computer to calculate this (you don't need to maintain a record of all the divergences).
Therefore, this is by far the more usually encountered of the definitions for variance.
== Also see ==
* Sum of Squared Deviations from Mean
== Sources ==
* : $\S 2.4$: Expectation: $(22)$
* 
* 
* 
Category:Variance
Category:Expectation
Category:Variance as Expectation of Square minus Square of Expectation
Variance as Expectation of Square minus Square of Expectation/Continuous

== Theorem == 
Let $X$ be a continuous random variable.
Then the variance of $X$ can be expressed as:
:$\var X = \expect {X^2} - \paren {\expect X}^2$
That is, it is the expectation of the square of $X$ minus the square of the expectation of $X$.
== Proof ==
Let $\mu = \expect X$. 
Let $X$ have probability density function $f_X$. 
As $f_X$ is a probability density function: 
:$\ds \int_{-\infty}^\infty \map {f_X} x \rd x = \Pr \paren {-\infty < X < \infty} = 1$
Then:
 
Category:Variance as Expectation of Square minus Square of Expectation
Variance as Expectation of Square minus Square of Expectation/Discrete

== Theorem == 
Let $X$ be a discrete random variable.
Then the variance of $X$ can be expressed as:
:$\var X = \expect {X^2} - \paren {\expect X}^2$
That is, it is the expectation of the square of $X$ minus the square of the expectation of $X$.
== Proof ==
We let $\mu = \expect X$, and take the expression for variance:
:$\var X := \ds \sum_{x \mathop \in \Img X} \paren {x - \mu}^2 \map \Pr {X = x}$
Then:
Hence the result, from $\mu = \expect X$.
== Sources ==
* : $\S 2.4$: Expectation: $(22)$
Category:Variance as Expectation of Square minus Square of Expectation
Variance is Least Mean Square Deviation about Point

== Theorem ==
Let $S$ be an object upon which:
:the mean square deviation
:the expectation
:the variance
is defined.
For each point $x$ in $S$, let $\map M x$ denote the mean square deviation of $S$ about $x$.
Then the value of $x$ for which $\map M x$ is the minimum is the variance of $S$.
That is, the expectation $\bar x$ of $S$ is the value of $S$ for which the mean square deviation is smallest.
== Proof ==
== Sources ==
* 
* 
Category:Mean Square Deviation
Category:Variance
Variance of Bernoulli Distribution

== Theorem ==
Let $X$ be a discrete random variable with the Bernoulli distribution with parameter $p$:
:$X \sim \Bernoulli p$
Then the variance of $X$ is given by:
:$\var X = p \paren {1 - p}$
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Proof 4 ==
== Proof 5 ==
Category:Variance
Category:Bernoulli Distribution
Category:Variance of Bernoulli Distribution
Variance of Bernoulli Distribution/Proof 2

== Theorem ==
== Proof ==
From 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:
:$\ds \expect {X^2} = \sum_{x \mathop \in \Img X} x^2 \, \map \Pr {X = x}$
So:
Then:
Category:Variance of Bernoulli Distribution
Variance of Bernoulli Distribution/Proof 4

== 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 Bernoulli Distribution, we have:
:$\map {\Pi_X} s = q + p s$
where $q = 1 - p$.
From Expectation of Bernoulli Distribution, we have $\mu = p$.
We have $\map { {\Pi_X}} s = 0$ from Derivatives of PGF of Bernoulli Distribution.
Hence:
:$\var X = 0 - \mu - \mu^2 = p - p^2 = p \paren {1 - p}$
Category:Variance of Bernoulli Distribution
Variance of Bernoulli Distribution/Proof 5

== Theorem ==
== Proof ==
From Moment Generating Function of Bernoulli Distribution, the moment generating function $M_X$ of $X$ is given by: 
:$\map {M_X} t = q + p e^t$
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:
:$\expect {X^2} = \map {M_X} 0$
We have: 
Setting $t = 0$ gives: 
In Expectation of Bernoulli Distribution, it is shown that: 
:$\expect X = p$
So:
Variance of Beta Distribution

== Theorem ==
Let $X \sim \map \Beta {\alpha, \beta}$ for some $\alpha, \beta > 0$, where $\Beta$ is the Beta distribution. 
Then:
:$\var X = \dfrac {\alpha \beta} {\paren {\alpha + \beta}^2 \paren {\alpha + \beta + 1} }$
== Proof 1 ==
 
== Proof 2 ==
 
Category:Variance
Category:Beta Distribution
Category:Variance of Beta Distribution
Variance of Beta Distribution/Proof 1

== Theorem ==
== Proof ==
From the definition of the Beta distribution, $X$ has probability density function:
:$\map {f_X} x = \dfrac {x^{\alpha - 1} \paren {1 - x}^{\beta - 1} } {\map \Beta {\alpha, \beta} }$
From Variance as Expectation of Square minus Square of Expectation: 
:$\ds \var X = \int_0^1 x^2 \map {f_X} X \rd x - \paren {\expect X}^2$
So: 
Category:Variance of Beta Distribution
Variance of Beta Distribution/Proof 2

== 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 Beta Distribution:
:$\expect X = \dfrac \alpha {\alpha + \beta}$
From Raw Moment of Beta Distribution:
:$\ds \expect {X^n} = \prod_{r \mathop = 0}^{n - 1} \frac {\alpha + r} {\alpha + \beta + r}$
So:
Therefore:
Category:Variance of Beta Distribution
Variance of Binomial Distribution

== Theorem ==
Let $X$ be a discrete random variable with the binomial distribution with parameters $n$ and $p$.
Then the variance of $X$ is given by:
:$\var X = n p \paren {1 - p}$
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Sources ==
* : $\S 2.4$: Expectation: Exercise $9$
* 
* 
* 
* : Appendix $13$: Probability distributions
* : Appendix $15$: Probability distributions
Category:Variance of Binomial Distribution
Category:Variance
Category:Binomial Distribution
Variance of Binomial 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:
:$\ds \expect {X^2} = \sum_{x \mathop \in \Img X} x^2 \Pr \paren {X = x}$
To simplify the algebra a bit, let $q = 1 - p$, so $p + q = 1$.
So:
Then:
as required.
Category:Variance of Binomial Distribution
Variance of Binomial Distribution/Proof 2

== Theorem ==
== 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 Binomial Distribution:
:$\map {\Pi_X} s = \paren {q + p s}^n$
where $q = 1 - p$.
From Expectation of Binomial Distribution:
:$\mu = n p$
From Derivatives of PGF of Binomial Distribution:
:$\map {\Pi' '_X} s = n \paren {n - 1} p^2 \paren {q + p s}^{n - 2}$
Setting $s = 1$ and using the formula $\map {\Pi' '_X} 1 + \mu - \mu^2$:
:$\var X = n \paren {n - 1} p^2 + n p - n^2 p^2$
Hence the result.
== Sources ==
* 
* 
Category:Variance of Binomial Distribution
Variance of Binomial Distribution/Proof 3

== Theorem ==
== Proof ==
From Bernoulli Process as Binomial Distribution, we see that $X$ as defined here is the sum of the discrete random variables that model the Bernoulli distribution.
Each of the Bernoulli trials is independent of each other.
Hence we can use Sum of Variances of Independent Trials.
The Variance of Bernoulli Distribution is $p \paren {1 - p}$.
Thus the variance of $\Binomial n p$ is $n p \paren {1 - p}$.
Category:Variance of Binomial Distribution
Variance 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 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: 
So: 
== Sources ==
* 
* 
* : Appendix $13$: Probability distributions
* : Appendix $15$: Probability distributions
Category:Variance
Category:Chi-Squared Distribution
Variance of Chi Distribution

== Theorem == 
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 variance of $X$ is given by: 
:$\var X = n - 2 \paren {\dfrac {\map \Gamma {\paren {n + 1} / 2} } {\map \Gamma {n / 2} } }^2$
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 Square of Chi Random Variable has Chi-Squared Distribution: 
:$X^2 \sim \chi^2_n$
So, by Expectation of Chi-Squared Distribution:
:$\expect {X^2} = n$
By Expectation of Chi Distribution:
:$\expect X = \sqrt 2 \dfrac {\map \Gamma {\paren {n + 1} / 2} } {\map \Gamma {n / 2} }$
Hence the result.
Category:Variance
Category:Chi Distribution
Variance of Continuous Uniform Distribution

== Theorem == 
Let $a, b \in \R$ such that $a < b$.
Let $X \sim \ContinuousUniform a b$ be the continuous uniform distribution over $\closedint a b$. 
Then: 
:$\var X = \dfrac {\paren {b - a}^2} {12}$
== Proof ==
From the definition of the continuous uniform distribution, $X$ has probability density function: 
:$\map {f_X} x = \begin{cases} \dfrac 1 {b - a} & a \le x \le b \\ 0 & \text{otherwise} \end{cases}$
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:
== Also see ==
* Variance of Discrete Uniform Distribution
== Sources ==
* : Appendix $13$: Probability distributions
* : Appendix $15$: Probability distributions
Category:Variance
Category:Continuous Uniform Distribution
Variance of Discrete Random Variable from PGF

== Theorem ==
Let $X$ be a discrete random variable whose probability generating function is $\map {\Pi_X} s$.
Then the variance of $X$ can be obtained from the second derivative of $\map {\Pi_X} s$ $s$ at $x = 1$:
:$\var X = \map {\Pi' '_X} 1 + \mu - \mu^2$
where $\mu = \expect X$ is the expectation of $X$.
== Proof ==
From the definition of the probability generating function:
:$\ds \map {\Pi_X} s = \sum_{x \mathop \ge 2} \map p x s^x$
From Derivatives of Probability Generating Function at One:
:$\ds \map {\Pi' '_X} s = \sum_{x \mathop \ge 2} x \paren {x - 1} \map p x s^{x - 2}$
But it also holds when you include $x = 0$ and $x = 1$ in the sum, as in both cases the term evaluates to zero and therefore vanishes.
So:
:$\ds \map {\Pi' '_X} s = \sum_{x \mathop \ge 0} x \paren {x - 1}\map p x s^{x - 2}$
Plugging in $s = 1$ gives:
The result follows from the definition of variance:
:$\var X = \expect {X^2} - \paren {\expect X}^2$
after a little algebra.
== Motivation ==
== Sources ==
* : $\S 4.3$: Moments: $(20)$
* 
* 
Category:Variance of Discrete Random Variable from PGF
Category:Variance
Category:Probability Generating Functions
Variance of Discrete Random Variable from PGF/Motivation

== Variance of Discrete Random Variable from PGF: Also known as == 
Variance of Discrete Random Variable from PGF shows how to find the variance of a discrete random variable without the need to go through the tedious process of what might be a complicated and fiddly summation.
All you need to do is differentiate its PGF twice, and plug in $1$.
Assuming, of course, you know what its PGF is.
Category:Variance of Discrete Random Variable from PGF
Variance of Discrete Uniform Distribution

== Theorem ==
Let $X$ be a discrete random variable with the discrete uniform distribution with parameter $n$.
Then the variance of $X$ is given by:
:$\var X = \dfrac {n^2 - 1} {12}$
== 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
Category:Discrete Uniform Distribution
Variance 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 variance of $X$ is given by: 
:$\var X = \dfrac k {\lambda^2}$
== 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 Erlang Distribution, we have: 
:$\expect X = \dfrac k \lambda$
We also have: 
So:
Category:Variance
Category:Erlang Distribution
Variance of Exponential Distribution

== Theorem ==
Let $X$ be a continuous random variable with the exponential distribution with parameter $\beta$.
Then the variance of $X$ is:
:$\var X = \beta^2$
 
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Expectation of Exponential Distribution
== Sources ==
* 
* 
* : Appendix $13$: Probability distributions
* : Appendix $15$: Probability distributions
Category:Variance of Exponential Distribution
Category:Exponential Distribution
Category:Variance
Variance of Exponential Distribution/Proof 1

== Theorem ==
== Proof ==
From Variance as Expectation of Square minus Square of Expectation:
:$\var X = \expect {X^2} - \paren {\expect X}^2$
From Expectation of Exponential Distribution:
:$\expect X = \beta$
The expectation of $X^2$ is:
Thus the variance of $X$ is:
 
Category:Variance of Exponential Distribution
Variance of Exponential Distribution/Proof 2

== Theorem ==
== Proof ==
By Moment Generating Function of Exponential Distribution, the moment generating function $M_X$ of $X$ is given by: 
:$\map {M_X} t = \dfrac 1 {1 - \beta t}$ 
From Variance as Expectation of Square minus Square of Expectation:
:$\var X = \expect {X^2} - \paren {\expect X}^2$ 
From Moment in terms of Moment Generating Function, we also have: 
:$\expect {X^2} = \map {M_X' '} 0$
In Expectation of Exponential Distribution: Proof 2, it is shown that: 
:$\map {M_X'} t = \dfrac \beta {\paren {1 - \beta t}^2}$ 
We have: 
 
Setting $t = 0$ gives: 
By Expectation of Exponential Distribution, we have: 
:$\expect X = \beta$
So: 
== Sources ==
* 
* 
Category:Variance of Exponential Distribution
Variance of Gamma Distribution

== Theorem == 
Let $X \sim \map \Gamma {\alpha, \beta}$ for some $\alpha, \beta > 0$, where $\Gamma$ is the Gamma distribution.
The variance of $X$ is given by:
:$\var X = \dfrac \alpha {\beta^2}$
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Sources ==
* : Appendix $13$: Probability distributions
* : Appendix $15$: Probability distributions
Category:Variance of Gamma Distribution
Category:Variance
Category:Gamma Distribution
Variance of Gamma Distribution/Proof 1

== Theorem == 
== Proof ==
From the definition of the Gamma distribution, $X$ has probability density function:
:$\map {f_X} x = \dfrac {\beta^\alpha x^{\alpha - 1} e^{-\beta x} } {\map \Gamma \alpha}$
From Variance as Expectation of Square minus Square of Expectation: 
:$\ds \var X = \int_0^\infty x^2 \map {f_X} x \rd x - \paren {\expect X}^2$
So:
 
Category:Variance of Gamma Distribution
Variance of Gamma Distribution/Proof 2

== Theorem == 
== Proof == 
By Moment Generating Function of Gamma Distribution, the moment generating function of $X$ is given by: 
:$\map {M_X} t = \paren {1 - \dfrac t \beta}^{-\alpha}$
for $t < \beta$. 
From Variance as Expectation of Square minus Square of Expectation: 
:$\var X = \expect {X^2} - \paren {\expect X}^2$
From Expectation of Gamma Distribution:
:$\expect X = \dfrac \alpha \beta$
From Moment Generating Function of Gamma Distribution: Second Moment: 
:$\map { {M_X}} t = \dfrac {\beta^\alpha \alpha \paren {\alpha + 1} } {\paren {\beta - t}^{\alpha + 2} }$
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 Gamma Distribution
Variance of Gamma Distribution/Proof 3

== 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:Variance of Gamma Distribution
Variance 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 variance of $X$ is given by:
:$\var X = \dfrac p {\paren {1-p}^2}$
== Proof 1 ==
== Proof 2 ==
Category:Variance
Category:Variance of Geometric Distribution
Variance of Geometric Distribution/Formulation 1/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:
:$\ds \expect {X^2} = \sum_{x \mathop \in \Omega_X} x^2 \map \Pr {X = x}$
:Let $q = 1 - p$
Thus:
Then:
Category:Variance of Geometric Distribution
Variance of Geometric Distribution/Formulation 1/Proof 2

== Theorem ==
== Proof ==
From Variance of Discrete Random Variable from PGF, we have:
:$\var X = \map {\Pi' '_X} 1 + \mu - \mu^2$
where $\mu = \map E x$ is the expectation of $X$.
From the Probability Generating Function of Geometric Distribution, we have:
:$\map {\Pi_X} s = \dfrac q {1 - ps}$
where $q = 1 - p$.
From Expectation of Geometric Distribution, we have:
:$\mu = \dfrac p q$
From Derivatives of PGF of Geometric Distribution, we have:
:$\map {\Pi' '_X} s = \dfrac {2 q p^2} {\paren {1 - ps}^3}$
Putting $s = 1$ using the formula $\map {\Pi' '_X} 1 + \mu - \mu^2$:
Category:Variance of Geometric Distribution
Variance 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 variance of $X$ is given by:
:$\var X = \dfrac {1 - p} {p^2}$
== Proof 1 ==
== Proof 2 ==
== Also presented as ==
== Sources ==
* 
Category:Variance
Category:Variance of Geometric Distribution
Variance of Geometric Distribution/Formulation 2/Also presented as

== Variance of Geometric Distribution: Formulation 2: Also presented as == 
The Variance of Geometric Distribution is also presented in the form:
:$\var X = \dfrac q {p^2}$
where $q$ has been defined conventionally as $q = 1 - p$.
== Sources ==
* 
Category:Variance of Geometric Distribution
Variance of Geometric Distribution/Formulation 2/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:
:$\ds \expect {X^2} = \sum_{x \mathop \in \Omega_X} x^2 \map \Pr {X = x}$
:Let $q = 1 - p$
Thus:
Then:
Category:Variance of Geometric Distribution
Variance of Geometric Distribution/Formulation 2/Proof 2

== Theorem == 
== Proof == 
By Moment Generating Function of Geometric Distribution, the moment generating function of $X$ is given by: 
:$\map {M_X} t = \dfrac p {1 - \paren {1 - p} e^t}$
for $t < -\map \ln {1 - p}$, and is undefined otherwise.
From Variance as Expectation of Square minus Square of Expectation: 
:$\ds \var X = \expect {X^2} - \paren {\expect X}^2$
From Expectation of Geometric Distribution/Formulation 2, we have:
:$\expect X = \dfrac {1 - p} p$
From Moment Generating Function of Geometric Distribution: Second Moment: 
:$\map { {M_X}' '} t = \dfrac {p \paren {1 - p} e^t + p \paren {1 - p}^2 e^{2 t} } {\paren {1 - \paren {1 - p} e^t}^3}$
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 Geometric Distribution
Variance of 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:
:$\ds \expect {X^2} = \sum_{x \mathop \in \Omega_X} x^2 \map \Pr {X = x}$
To simplify the algebra a bit, let $q = 1 - p$, so $p + q = 1$.
Thus:
Then:
Category:Variance of Geometric Distribution
Variance of Geometric Distribution/Proof 2

== Theorem ==
== 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 Geometric Distribution:
:$\map {\Pi_X} s = \dfrac q {1 - p s}$
where $q = 1 - p$.
From Expectation of Geometric Distribution:
:$\mu = \dfrac p q$
From Derivatives of PGF of Geometric Distribution:
:$\map {\Pi_X} s = \dfrac {2 q p^2} {\left({1 - ps}\right)^3}$
Putting $s = 1$ using the formula $\map {\Pi_X} 1 + \mu - \mu^2$:
:$\var X = \dfrac {2 q p^2} {\paren {1 - p}^3} + \dfrac p q - \paren {\dfrac p q}^2$
and hence the result, after some algebra.
Category:Variance of Geometric Distribution
Variance of Grouped Data

== Theorem ==
Let $S$ be a set of grouped data.
Let it be assumed that $S$ approximates a normal distribution
The variance of $S$ can be calculated by:
:$(1): \quad$ assuming that the elements of $S$ are all at the mid-interval values
:$(2): \quad$ subtracting Sheppard's correction of $\dfrac {h^2} {12}$, where $h$ is the bin width.
== Proof ==
== Sources ==
* 
* 
Category:Grouped Data