nyuuzyou/wikis huggingface dataset:
Modus Tollendo Ponens/Proof Rule

== Proof Rule ==
Modus tollendo ponens is a valid argument in types of logic dealing with disjunctions $\lor$ and negation $\neg$.
This includes propositional logic and predicate logic, and in particular natural deduction.
As a proof rule it is expressed in either of the two forms:
:$(1): \quad$ If we can conclude $\phi \lor \psi$, and we can also conclude $\neg \phi$, then we may infer $\psi$.
:$(2): \quad$ If we can conclude $\phi \lor \psi$, and we can also conclude $\neg \psi$, then we may infer $\phi$.
It can be written:
:$\ds {\paren {\phi \lor \psi} \quad \neg \phi \over \psi} \textrm {MTP}_1 \qquad \text{or} \qquad {\paren {\phi \lor \psi} \quad \neg \psi \over \phi} \textrm {MTP}_2$
=== Tableau Form ===
== Explanation ==
== Also known as ==
== Also see ==
* This is a rule of inference of the following proof systems:
** Definition:Natural Deduction
== Linguistic Note ==
== Technical Note ==
== Sources ==
* : $\text{II}$: 'AND', 'OR', 'IF AND ONLY IF': $\S 3$
Category:Rules of Inference
Category:Modus Tollendo Ponens
Modus Tollendo Ponens/Proof Rule/Tableau Form

== Proof Rule ==
Let $\phi \lor \psi$ be a well-formed formula in a tableau proof whose main connective is the disjunction operator.
The Modus Tollendo Ponens is invoked for $\phi \lor \psi$ in either of the two forms:
;Form 1:
;Form 2:
Category:Tableau Proofs
Modus Tollendo Ponens/Proof Rule/Technical Note

== Technical Note on ==
When invoking in a tableau proof, use the template:
:{{ModusTollendoPonens|line|pool|statement|first|second|1 or 2}}
or:
:{{ModusTollendoPonens|line|pool|statement|first|second|1 or 2|comment}}
where:
:line is the number of the line on the tableau proof where is to be invoked
:pool is the pool of assumptions (comma-separated list)
:statement is the statement of logic that is to be displayed in the Formula column, without the $ ... $ delimiters
:first is the first of the two lines of the tableau proof upon which this line directly depends, the one in the form $p \lor q$
:second is the second of the two lines of the tableau proof upon which this line directly depends, the one in the form $\neg p$
:1 or 2 should hold 1 for ModusTollendoPonens_1, and 2 for ModusTollendoPonens_2
:comment is the (optional) comment that is to be displayed in the Notes column.
Category:Technical Notes
Modus Tollendo Ponens/Sequent Form/Case 1

== Theorem ==
== Proof 1 ==
== Proof by Truth Table ==
== Sources ==
* : $3$: The Method of Deduction: $3.1$: Formal Proof of Validity: Rules of Inference: $4.$
* : $\S \text{II}$: The Logic of Statements $(2): \ 1$: Decision procedures and proofs: $5$
* : $\S 1.14$: Exercise $17 \ \text{(iii)}$
* : $\S 1.2.1$: Rules for natural deduction: Exercises $1.5: \ 1 \ \text{(c)}$
Category:Modus Tollendo Ponens
Modus Tollendo Ponens/Sequent Form/Case 1/Proof by Truth Table

== Theorem ==
== Proof ==
We apply the Method of Truth Tables to the proposition.
$\begin{array}{|ccc|cc||c|} \hline
p & \lor & q & \neg & p & q\\
\hline
\F & \F & \F & \T & \F & \F \\
\F & \T & \T & \T & \F & \T \\
\T & \T & \F & \F & \T & \F \\
\T & \T & \T & \F & \T & \T \\
\hline
\end{array}$
As can be seen, when $p \lor q$ is true, and so is $\neg p$, then $q$ is also true.
== Sources ==
* : $2$ Arguments Containing Compound Statements: $2.3$: Argument Forms and Truth Tables
Category:Modus Tollendo Ponens
Category:Truth Table Proofs
Modus Tollendo Ponens/Sequent Form/Case 2

== Theorem ==
== Proof ==
== Sources ==
* : Chapter $2$: The Propositional Calculus $2$: $2$: Theorems and Derived Rules: Theorem $53$
* : $\S 1.2.1$: Rules for natural deduction: Exercises $1.5: \ 2 \ \text{(c)}$
Category:Modus Tollendo Ponens
Modus Tollendo Ponens/Variant/Formulation 1

== Theorem ==
:$p \lor q \dashv \vdash \neg p \implies q$
This can be expressed as two separate theorems:
=== Forward Implication ===
=== Reverse Implication ===
Note that the latter proof requires Law of Excluded Middle.
Therefore it is not valid in intuitionistic logic.
== Proof by Truth Table ==
== Sources ==
* : $\S 2.4$: Relations between Truth-Functions
* : Chapter $2$: The Propositional Calculus $2$: $2$: Theorems and Derived Rules: Exercise $5 \ \text{(f)}$
* : $\S 2.3.3$
Category:Modus Tollendo Ponens
Modus Tollendo Ponens/Variant/Formulation 1/Proof by Truth Table

== Theorem ==
== Proof ==
We apply the Method of Truth Tables.
As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations.
$\begin{array}{|ccc||cccc|} \hline
p & \lor & q & \neg & p & \implies & q \\
\hline
\F & \F & \F & \T & \F & \F & \F \\
\F & \T & \T & \T & \F & \T & \T \\
\T & \T & \F & \F & \T & \T & \F \\
\T & \T & \T & \F & \T & \T & \T \\
\hline
\end{array}$
== Sources ==
* : $\S 1.4.1$: Exercise $1.8: \ 1$
Category:Modus Tollendo Ponens
Category:Truth Table Proofs
Modus Tollendo Ponens/Variant/Formulation 1/Reverse Implication

== Theorem ==
:$\neg p \implies q \vdash p \lor q$
== Proof ==
== Sources ==
* : $\text{II}$: 'AND', 'OR', 'IF AND ONLY IF': $\S 5$
* : Chapter $1$: The Propositional Calculus $1$: $5$ Further Proofs: Résumé of Rules: Exercise $1 \ \text {(j)}$
Category:Modus Tollendo Ponens
Modus Tollendo Ponens/Variant/Formulation 2

== Theorem ==
:$\vdash \paren {p \lor q} \iff \paren {\neg p \implies q}$
== Proof 1 ==
== Proof by Truth Table ==
== Sources ==
* : $\text{II}$: 'AND', 'OR', 'IF AND ONLY IF': $\S 5$: Theorem $\text{T45}$
* : $\S \text{II}$: The Logic of Statements $(2):$: The remaining rules of inference: $16$
* : $\S 1$: Informal statement calculus: $\S 1.2$: Truth functions and truth tables: Exercise $5 \ \text{(b)}$
Category:Modus Tollendo Ponens
Modus Tollendo Ponens/Variant/Formulation 2/Proof by Truth Table

== Theorem ==
== Proof ==
We apply the Method of Truth Tables.
As can be seen by inspection, the truth values under the main connective are true for all boolean interpretations.
$\begin{array}{|ccc|c|cccc|} \hline
p & \lor & q & \iff & \neg & p & \implies & q \\
\hline
\F & \F & \F & \T & \T & \F & \F & \F \\
\F & \T & \T & \T & \T & \F & \T & \T \\
\T & \T & \F & \T & \F & \T & \T & \F \\
\T & \T & \T & \T & \F & \T & \T & \T \\
\hline
\end{array}$
== Sources ==
* : $\S \text{I}: 13$: Logical Equivalences: Exercise $\text{(A)} \ 5$
Category:Modus Tollendo Ponens
Category:Truth Table Proofs
Modus Tollendo Tollens

== Proof Rule ==
Modus tollendo tollens is a valid argument in types of logic dealing with conditionals $\implies$ and negation $\neg$.
This includes propositional logic and predicate logic, and in particular natural deduction.
=== Proof Rule ===
=== Sequent Form ===
== Explanation ==
== Also known as ==
== Also see ==
* Definition:Conditional
The following are related argument forms:
* Modus Ponendo Ponens
* Modus Ponendo Tollens
* Modus Tollendo Ponens
The Rule of Transposition is conceptually similar, and can be derived from the MTT by a simple application of the Rule of Implication.
These are classic fallacies:
* Affirming the Consequent
* Denying the Antecedent
== Linguistic Note ==
== Sources ==
* : $2$ Arguments Containing Compound Statements: $2.3$: Argument Forms and Truth Tables
* 
* 
Category:Modus Tollendo Tollens
Category:Conditional
Category:Logical Negation
Modus Tollendo Tollens/Also known as

== Also known as ==
Modus Tollendo Tollens is also known as:
* Modus tollens, abbreviated M.T.
* Denying the consequent.
== Sources ==
* : $3$: The Method of Deduction: $3.1$: Formal Proof of Validity: Rules of Inference: $2.$
* 
* 
Category:Modus Tollendo Tollens
Modus Tollendo Tollens/Proof Rule

== Proof Rule ==
Modus tollendo tollens is a valid argument in types of logic dealing with conditionals $\implies$ and negation $\neg$.
This includes propositional logic and predicate logic, and in particular natural deduction.
As a proof rule it is expressed in the form:
:If we can conclude $\phi \implies \psi$, and we can also conclude $\neg \psi$, then we may infer $\neg \phi$.
It can be written:
:$\ds {\phi \implies \psi \quad \neg \psi \over \neg \phi} \text{MTT}$
=== Tableau Form ===
== Explanation ==
== Also known as ==
== Also see ==
* This is a rule of inference of the following proof systems:
** Definition:Natural Deduction
== Linguistic Note ==
== Technical Note ==
== Sources ==
* : $\text{I}$: 'NOT' and 'IF': $\S 3$
* : Chapter $1$: The Propositional Calculus $1$: $2$ Conditionals and Negation
* : $\S \text{II}$: The Logic of Statements $(2): \ 1$: Decision procedures and proofs: $2$
* : $\S 1.2.1$: Rules for natural deduction
Category:Rules of Inference
Category:Modus Tollendo Tollens
Modus Tollendo Tollens/Proof Rule/Tableau Form

== Proof Rule ==
Let $\phi \implies \psi$ be a well-formed formula in a tableau proof whose main connective is the implication operator.
The Modus Tollendo Tollens is invoked for $\phi \implies \psi$ and $\neg \psi$ as follows:
== Sources ==
* : Chapter $1$: The Propositional Calculus $1$: $2$ Conditionals and Negation
Category:Tableau Proofs
Modus Tollendo Tollens/Proof Rule/Technical Note

== Technical Note on ==
When invoking in a tableau proof, use the template:
:{{ModusTollens|line|pool|statement|first|second}}
or:
:{{ModusTollens|line|pool|statement|first|second|comment}}
where:
:line is the number of the line on the tableau proof where is to be invoked
:pool is the pool of assumptions (comma-separated list)
:statement is the statement of logic that is to be displayed in the Formula column, without the $ ... $ delimiters
:first is the first of the two lines of the tableau proof upon which this line directly depends, the one in the form $\phi \implies \psi$
:second is the second of the two lines of the tableau proof upon which this line directly depends, the one in the form $\phi$
:comment is the (optional) comment that is to be displayed in the Notes column.
Category:Technical Notes
Modus Tollendo Tollens/Sequent Form

== Proof Rule ==
The Modus Tollendo Tollens can be symbolised by the sequent:
== Proof 1 ==
== Proof 2 ==
== Proof by Truth Table ==
== Sources ==
* : $3$: The Method of Deduction: $3.1$: Formal Proof of Validity: Rules of Inference: $2.$
* : $\S 1.14$: Exercise $17 \ \text{(i)}$
Category:Modus Tollendo Tollens
Modus Tollendo Tollens/Sequent Form/Proof by Truth Table

== Theorem ==
== Proof ==
We apply the Method of Truth Tables to the proposition.
As can be seen for all boolean interpretations by inspection, where the truth value under the main connective on the is $\T$, that under the one on the is also $\T$:
$\begin{array}{|cccccc||cc|} \hline
(p & \implies & q) & \land & \neg & q & \neg & p \\
\hline
\F & \T & \F & \T & \T & \F & \T & \F \\
\F & \T & \T & \F & \F & \T & \T & \F \\
\T & \F & \F & \F & \T & \F & \F & \T \\
\T & \T & \T & \F & \F & \T & \F & \T \\
\hline
\end{array}$
Hence the result.
Note that the two formulas are not equivalent, as the relevant columns do not match exactly.
== Sources ==
* : $2$ Arguments Containing Compound Statements: $2.3$: Argument Forms and Truth Tables
Category:Truth Table Proofs
Category:Modus Tollendo Tollens
Mole/Examples/Water

== Example of Mole ==
$1$ mole of water has a mass of (approximately) $18$ grams.
== Proof ==
By definition, a molecule of water is composed of:
:$2$ atoms of hydrogen
:$1$ atom of oxygen.
It follows that $1$ mole of water is composed of:
:$2$ moles of hydrogen
:$1$ mole of oxygen.
The (by far) most plentiful isotopes of these elements are:
:hydrogen-$1$, with an atomic mass of approximately $1$
:oxygen-$16$, with an atomic mass of approximately $16$.
From Mass of Mole of Isotope of Element:
:the mass of $1$ mole of hydrogen-$1$ is $1 \ \mathrm g$
:the mass of $1$ mole of oxygen-$16$ is $16 \ \mathrm g$.
Hence the mass of $1$ mole of water is $2 \times 1 + 16 = 18$.
== Sources ==
* : $1$: Simple models of the atom: $\text {1-3}$: The Electrical Structure of Matter
Category:Examples of Moles
Moment Generating Function of Bernoulli Distribution

== Theorem ==
Let $X$ be a discrete random variable with a Bernoulli distribution with parameter $p$ for some $0 \le p \le 1$. 
Then the moment generating function $M_X$ of $X$ is given by: 
:$\map {M_X} t = q + p e^t$
where $q = 1 - p$. 
== Proof ==
From the definition of the Bernoulli distribution, $X$ has probability mass function:
:$\map \Pr {X = n} = \begin{cases}
q & : n = 0 \\
p & : n = 1 \\
0 & : n \notin \set {0, 1} \\
\end{cases}$
From the definition of a moment generating function:
:$\ds \map {M_X} t = \expect {e^{t X} } = \sum_{n \mathop = 0}^1 \map \Pr {X = n} e^{t n}$
So:
Category:Moment Generating Functions
Category:Bernoulli Distribution
Moment Generating Function of Beta Distribution

== Theorem == 
Let $X \sim \BetaDist \alpha \beta$ denote the Beta distribution fior some $\alpha, \beta > 0$.
Then the moment generating function $M_X$ of $X$ is given by: 
:$\ds \map {M_X} t = 1 + \sum_{k \mathop = 1}^\infty \paren {\prod_{r \mathop = 0}^{k - 1} \frac {\alpha + r} {\alpha + \beta + r} } \frac {t^k} {k!}$
== 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 the definition of a moment generating function:
:$\ds \map {M_X} t = \expect {e^{t X} } = \int_0^1 e^{t x} \map {f_X} x \rd x$
So:
Category:Moment Generating Functions
Category:Beta Distribution
Moment Generating Function of Binomial Distribution

== Theorem ==
Let $X$ be a discrete random variable with a binomial distribution with parameters $n$ and $p$ for some $n \in \N$ and $0 \le p \le 1$:
:$X \sim \Binomial n p$
Then the moment generating function $M_X$ of $X$ is given by: 
:$\map {M_X} t = \paren {1 - p + p e^t}^n$
== Proof ==
From the definition of the Binomial distribution, $X$ has probability mass function:
:$\map \Pr {X = k} = \dbinom n k p^k \paren {1 - p}^{n - k}$
From the definition of a moment generating function:
:$\ds \map {M_X} t = \expect {e^{t X} } = \sum_{k \mathop = 0}^n \map \Pr {X = k} e^{t k}$
So: 
== Also presented as ==
== Sources ==
* : Appendix $13$: Probability distributions
* : Appendix $15$: Probability distributions
Category:Moment Generating Function of Binomial Distribution
Category:Moment Generating Functions
Category:Binomial Distribution
Moment Generating Function of Binomial Distribution/Also presented as

== Moment Generating Function of Binomial Distribution: Also presented as==
The moment generating function $M_X$ of a binomial distribution $X$ can also be presented as: 
:$\map {M_X} t = \paren {1 + p \paren {e^t - 1} }^n$
== Sources ==
* : Appendix $13$: Probability distributions
* : Appendix $15$: Probability distributions
Category:Moment Generating Function of Binomial Distribution
Moment Generating Function 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 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: 
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: 
== Sources ==
* : Appendix $13$: Probability distributions
* : Appendix $15$: Probability distributions
Category:Moment Generating Functions
Category:Chi-Squared Distribution
Moment Generating Function of Continuous Uniform Distribution

== Theorem == 
Let $X \sim \ContinuousUniform a b$ for some $a, b \in \R$ denote the continuous uniform distribution on the interval $\closedint a b$.
Then the moment generating function of $X$ is given by: 
:$\map {M_X} t = \begin {cases} \dfrac {e^{t b} - e^{t a} } {t \paren {b - a} } & t \ne 0 \\ 1 & t = 0 \end{cases}$
== 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 the definition of a moment generating function:
:$\ds \map {M_X} t = \expect {e^{t X} } = \int_{-\infty}^\infty e^{t x} \map {f_X} x \rd x$
where $\expect \cdot$ denotes expectation.
First, consider the case $t \ne 0$. 
Then: 
In the case $t = 0$, we have $\expect {X^0} = \expect 1 = 1$. 
== Sources ==
* : Appendix $13$: Probability distributions
* : Appendix $15$: Probability distributions
Category:Moment Generating Functions
Category:Continuous Uniform Distribution
Moment Generating Function of Discrete Uniform Distribution

== Theorem ==
Let $X$ be a discrete random variable with a discrete uniform distribution with parameter $n$ for some $n \in \N$.
Then the moment generating function $M_X$ of $X$ is given by: 
:$\map {M_X} t = \dfrac {e^t \paren {1 - e^{n t} } } {n \paren {1 - e^t} }$
== Proof ==
From the definition of the discrete uniform distribution, $X$ has probability mass function:
:$\map \Pr {X = N} = \dfrac 1 n$
From the definition of a moment generating function:
:$\ds \map {M_X} t = \expect {e^{t X} } = \sum_{N \mathop = 1}^n \map \Pr {X = N} e^{N t}$
So: 
Category:Moment Generating Functions
Category:Discrete Uniform Distribution
Moment Generating Function of Exponential Distribution

== Theorem ==
Let $X$ be a continuous random variable with an exponential distribution with parameter $\beta$ for some $\beta \in \R_{> 0}$.
Then the moment generating function $M_X$ of $X$ is given by: 
:$\map {M_X} t = \dfrac 1 {1 - \beta t}$
for $t < \dfrac 1 \beta$, and is undefined otherwise.
== Proof ==
From the definition of the exponential distribution, $X$ has probability density function:
:$\map {f_X} x = \dfrac 1 \beta e^{-\frac x \beta}$
From the definition of a moment generating function:
:$\ds \map {M_X} t = \expect {e^{t X} } = \int_0^\infty e^{t x} \map {f_X} x \rd x$
Then: 
Note that if $t > \dfrac 1 \beta$, then $e^{x \paren {-\frac 1 \beta + t} } \to \infty$ as $x \to \infty$ by Exponential Tends to Zero and Infinity, so the integral diverges in this case. 
If $t = \dfrac 1 \beta$ then the integrand is identically $1$, so the integral similarly diverges in this case. 
If $t < \dfrac 1 \beta$, then $e^{x \paren {-\frac 1 \beta + t} } \to 0$ as $x \to \infty$ from Exponential Tends to Zero and Infinity, so the integral converges in this case. 
Therefore, the function is only well defined for $t < \dfrac 1 \beta$.
Proceeding:
== Sources ==
* 
* 
* : Appendix $13$: Probability distributions
* : Appendix $15$: Probability distributions
Category:Moment Generating Functions
Category:Exponential Distribution
Moment Generating Function of Gamma Distribution

== Theorem == 
Let $X \sim \map \Gamma {\alpha, \beta}$ for some $\alpha, \beta > 0$, where $\Gamma$ is the Gamma distribution.
Then the moment generating function of $X$ is given by: 
:$\map {M_X} t = \begin {cases} \paren {1 - \dfrac t \beta}^{-\alpha} & t < \beta \\ \text {does not exist} & t \ge \beta \end {cases}$
== 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 the definition of a moment generating function:
:$\ds \map {M_X} t = \expect {e^{t X} } = \int_0^\infty e^{t x} \map {f_X} x \rd x$
First take $t < \beta$. 
Then:
Now take $t = \beta$. 
Our integral becomes: 
So $\expect {e^{\beta X} }$ does not exist.
Finally take $t > \beta$.
We have that $-\paren {\beta - t}$ is positive.
As a consequence of Exponential Dominates Polynomial, we have:
:$x^{\alpha - 1} < e^{-\paren {\beta - t} x}$
for sufficiently large $x$. 
Therefore, in this case, the integrand increases without bound. 
We conclude that the integral is divergent.
Hence $\expect {e^{t X} }$ does not exist for $t > \beta$. 
== Examples ==
In the below, $t < \beta$ throughout.
== Sources ==
* : Appendix $13$: Probability distributions
* : Appendix $15$: Probability distributions
Category:Moment Generating Function of Gamma Distribution
Category:Moment Generating Functions
Category:Gamma Distribution
Moment Generating Function of Gamma Distribution/Examples

== Examples of Use of Moment Generating Function of Gamma Distribution ==
Let $X \sim \map \Gamma {\alpha, \beta}$ for some $\alpha, \beta > 0$, where $\Gamma$ is the Gamma distribution.
Let $t < \beta$.
=== First Moment ===
=== Second Moment ===
=== Third Moment ===
=== Fourth Moment ===
=== Derivatives of Moment Generating Function of Gamma Distribution‎ ===
Category:Moment Generating Function of Gamma Distribution
Moment Generating Function of Gamma Distribution/Examples/First Moment

== Examples of Use of Moment Generating Function of Gamma Distribution ==
Let $X \sim \map \Gamma {\alpha, \beta}$ for some $\alpha, \beta > 0$, where $\Gamma$ is the Gamma distribution.
Let $t < \beta$.
The first moment generating function of $X$ is given by: 
:$\map { {M_X}'} t = \dfrac {\beta^\alpha \alpha} {\paren {\beta - t}^{\alpha + 1} }$
== Proof ==
We have:
Category:Moment Generating Function of Gamma Distribution
Moment Generating Function of Gamma Distribution/Examples/Fourth Moment

== Examples of Use of Moment Generating Function of Gamma Distribution ==
Let $X \sim \map \Gamma {\alpha, \beta}$ for some $\alpha, \beta > 0$, where $\Gamma$ is the Gamma distribution.
Let $t < \beta$.
The fourth moment generating function of $X$ is given by: 
:$\map { {M_X}^{\paren 4} } t = \dfrac {\beta^\alpha \alpha \paren {\alpha + 1} \paren {\alpha + 2} \paren {\alpha + 3} } {\paren {\beta - t}^{\alpha + 4} }$
== Proof ==
We have:
Category:Moment Generating Function of Gamma Distribution
Moment Generating Function of Gamma Distribution/Examples/Second Moment

== Examples of Use of Moment Generating Function of Gamma Distribution ==
Let $X \sim \map \Gamma {\alpha, \beta}$ for some $\alpha, \beta > 0$, where $\Gamma$ is the Gamma distribution.
Let $t < \beta$.
The second moment generating function of $X$ is given by: 
:$\map { {M_X}} t = \dfrac {\beta^\alpha \alpha \paren {\alpha + 1} } {\paren {\beta - t}^{\alpha + 2} }$
== Proof ==
We have:
Category:Moment Generating Function of Gamma Distribution
Moment Generating Function of Gamma Distribution/Examples/Third Moment

== Examples of Use of Moment Generating Function of Gamma Distribution ==
Let $X \sim \map \Gamma {\alpha, \beta}$ for some $\alpha, \beta > 0$, where $\Gamma$ is the Gamma distribution.
Let $t < \beta$.
The third moment generating function of $X$ is given by: 
:$\map { {M_X}} t = \dfrac {\beta^\alpha \alpha \paren {\alpha + 1} \paren {\alpha + 2} } {\paren {\beta - t}^{\alpha + 3} }$
== Proof ==
We have:
Category:Moment Generating Function of Gamma Distribution