nyuuzyou/wikis huggingface dataset:
Expectation of Exponential Distribution/Proof 1

== Theorem ==
== Proof == 
The expectation of a continuous random variable $X$ with sample space $\Omega_X$ is given by:
:$\ds \expect X := \int_{x \mathop \in \Omega_X} x \map {f_X} x \rd x$
where $f_X$ is the probability density function of $X$.
For the exponential distribution:
:$\Omega_X = \hointr 0 \infty$
From Probability Density Function of Exponential Distribution:
:$\ds \map {f_X} x = \frac 1 \beta \map \exp {- \frac x \beta}$
So:
:$\ds \expect X = \int_0^\infty x \frac 1 \beta \map \exp {- \frac x \beta} \rd x$
where $\exp$ is the exponential function. 
Substituting $u = \dfrac x \beta$, we have:
:$\ds \expect X = \beta \int_0^\infty u \map \exp {-u} \rd u$
The integral evaluates to:
:$\ds \expect X = \bigintlimits {-\beta \paren {u + 1} \map \exp {-u} } 0 \infty$
So:
:$\ds \expect X = \beta - \beta \lim_{u \mathop \to \infty} \frac {u + 1} {\exp u}$
By Limit at Infinity of Polynomial over Complex Exponential, it follows that this limit is zero, so that:
:$\expect X = \beta$
Category:Expectation of Exponential Distribution
Expectation of Exponential Distribution/Proof 2

== Theorem ==
== Proof ==
From Moment Generating Function of Exponential Distribution, the moment generating function $M_X$ of $X$, is given by:
:$\map {M_X} t = \dfrac 1 {1 - \beta t}$
By Moment in terms of Moment Generating Function:
:$\expect X = \map {M_X'} 0$
We have: 
 
Setting $t = 0$ gives:
== Sources ==
* 
* 
Category:Expectation of Exponential Distribution
Expectation of Function of Continuous Random Variable

== Theorem ==
Let $X$ be a continuous random variable.
Let $\expect X$ be the expectation of $X$.
Let $g: \R \to \R$ be a real function.
Then:
:$\ds \expect {g \sqbrk X} = \int_{-\infty}^\infty \map g x \map f x $
== Proof ==
== Sources ==
* 
* 
Category:Expectation
Expectation of Function of Discrete Random Variable

== Theorem ==
Let $X$ be a discrete random variable.
Let $\expect X$ be the expectation of $X$.
Let $g: \R \to \R$ be a real function.
Then:
:$\ds \expect {g \sqbrk X} = \sum_{x \mathop \in \Omega_X} \map g x \, \map \Pr {X = x}$
whenever the sum is absolutely convergent.
== Proof ==
Let $\Omega_X = \Img X = I$.
Let $Y = g \sqbrk X$.
Thus:
:$\Omega_Y = \Img Y = g \sqbrk I$
So:
From the definition of expectation, this last sum applies only when the last sum is absolutely convergent.
== Sources ==
* : $\S 2.4$: Expectation: Theorem $2 \ \text{B}$
* 
* 
Category:Expectation
Expectation of Function of Joint Probability Mass Distribution

== Theorem ==
Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $X$ and $Y$ be discrete random variables on $\struct {\Omega, \Sigma, \Pr}$.
Let $\expect X$ be the expectation of $X$.
Let $g: \R^2 \to \R$ be a real-valued function
Let $p_{X, Y}$ be the joint probability mass function of $X$ and $Y$.
:$\ds \expect {\map g {X, Y} } = \sum_{x \mathop \in \Omega_X} \sum_{y \mathop \in \Omega_Y} \map g {x, y} \map {p_{X, Y} } {x, y}$
whenever the sum is absolutely convergent.
== Proof ==
Let $\Omega_X = \Img X = I_X$ and $\Omega_Y = \Img Y = I_Y$.
Let $Z = \map g {X, Y}$.
Thus $\Omega_Z = \Img Z = g \sqbrk {I_X, I_Y}$.
So:
From the definition of expectation, this last sum applies only when the last sum is absolutely convergent.
== Sources ==
* : $\S 3.2$: Expectation in the multivariate case: Theorem $3 \ \text{A}$
Category:Expectation
Expectation of Gamma Distribution

== Theorem == 
Let $X \sim \map \Gamma {\alpha, \beta}$ for some $\alpha, \beta > 0$, where $\Gamma$ is the Gamma distribution.
The expectation of $X$ is given by:
:$\expect X = \dfrac \alpha \beta$
== Proof 1 == 
 
== Proof 2 ==
== Sources ==
* : Appendix $13$: Probability distributions
* : Appendix $15$: Probability distributions
Category:Expectation of Gamma Distribution
Category:Gamma Distribution
Category:Expectation
Expectation 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 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:
Category:Expectation of Gamma Distribution
Expectation 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 Moment in terms of Moment Generating Function: 
:$\expect X = \map { {M_X}'} 0$
From Moment Generating Function of Gamma Distribution: First Moment: 
:$\map { {M_X}'} t = \dfrac {\beta^\alpha \alpha} {\paren {\beta - t}^{\alpha + 1} }$
Hence setting $t = 0$:
Category:Expectation of Gamma Distribution
Expectation 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 expectation of $X$ is given by:
:$\expect X = \dfrac p {1 - p}$
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
Category:Expectation of Geometric Distribution
Expectation of Geometric Distribution/Formulation 1/Proof 1

== Theorem ==
== Proof ==
From the definition of expectation:
:$\ds \expect X = \sum_{x \mathop \in \Omega_X} x \map \Pr {X = x}$
By definition of geometric distribution:
:$\ds \expect X = \sum_{k \mathop \in \Omega_X} k p^k \paren {1 - p}$
Let $q = 1 - p$:
Category:Expectation of Geometric Distribution
Expectation of Geometric Distribution/Formulation 1/Proof 2

== Theorem ==
== Proof ==
From the Probability Generating Function of Geometric Distribution:
:$\map {\Pi_X} s = \dfrac q {1 - p s}$
where $q = 1 - p$.
From Expectation of Discrete Random Variable from PGF:
:$\expect X = \map {\Pi'_X} 1$
We have:
Plugging in $s = 1$:
Hence the result.
Category:Expectation of Geometric Distribution
Expectation of Geometric Distribution/Formulation 1/Proof 3

== Theorem ==
== Proof ==
From the definition of expectation:
:$\ds \expect X = \sum_{x \mathop \in \Omega_X} x \map \Pr {X = x}$
Then
By the Ratio Test, both $\sum_{k \mathop \ge 1} k p^k$ and $\sum_{k \mathop \ge 1} k p^{k+1}$ converge absolutely.
From Absolutely Convergent Real Series is Convergent, both series converge.
Category:Expectation of Geometric Distribution
Expectation 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 expectation of $X$ is given by:
:$\map E X = \dfrac {1-p} p$
== Proof 1 ==
== Proof 2 ==
== Also presented as ==
== Sources ==
* 
Category:Expectation of Geometric Distribution
Expectation of Geometric Distribution/Formulation 2/Also presented as

== Expectation of Geometric Distribution: Formulation 2: Also presented as == 
The Expectation of Geometric Distribution is also presented in the form:
:$\var X = \dfrac q p$
where $q$ has been defined conventionally as $q = 1 - p$.
== Sources ==
* 
Category:Expectation of Geometric Distribution
Expectation of Geometric Distribution/Formulation 2/Proof 1

== Theorem ==
== Proof ==
From the definition of expectation:
:$\ds \expect X = \sum_{x \mathop \in \Omega_X} x \map \Pr {X = x}$
By definition of geometric distribution:
:$\ds \expect X = \sum_{k \mathop \in \Omega_X} k p \paren {1 - p}^k$
Let $q = 1 - p$:
Category:Expectation of Geometric Distribution
Expectation 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 Moment in terms of Moment Generating Function: 
:$\expect X = \map { {M_X}'} 0$
From Moment Generating Function of Geometric Distribution: First Moment: 
:$\map { {M_X}'} t = \dfrac {p \paren {1 - p} e^t } {\paren {1 - \paren {1 - p} e^t}^2 }$
Hence setting $t = 0$:
Category:Expectation of Geometric Distribution
Expectation of Geometric Distribution/Proof 1

== Theorem ==
== Proof ==
From the definition of expectation:
:$\ds \expect X = \sum_{x \mathop \in \Omega_X} x \map \Pr {X = x}$
By definition of geometric distribution:
:$\ds \expect X = \sum_{k \mathop \in \Omega_X} k p^k \paren {1 - p}$
Let $q = 1 - p$:
Category:Expectation of Geometric Distribution
Expectation of Geometric Distribution/Proof 2

== Theorem ==
== Proof ==
From the Probability Generating Function of Geometric Distribution:
:$\map {\Pi_X} s = \dfrac q {1 - p s}$
where $q = 1 - p$.
From Expectation of Discrete Random Variable from PGF:
:$\expect X = \map {\Pi'_X} 1$
We have:
Plugging in $s = 1$:
Hence the result.
Category:Expectation of Geometric Distribution
Expectation of Hat-Check Distribution

== Theorem ==
Let $X$ be a discrete random variable with the Hat-Check distribution with parameter $n$.
Then the expectation of $X$ is given by:
:$\expect X = n - 1$
== Proof ==
From the definition of expectation:
:$\ds \expect X = \sum_{x \mathop \in \Img X} x \map \Pr {X = x}$
By definition of hat-check distribution:
:$\ds \expect X = \sum_{k \mathop = 0}^n k \dfrac 1 {\paren {n - k }!} \sum_{s \mathop = 0}^k \dfrac {\paren {-1}^s} {s!}$
Then:
== Also see ==
* Variance of Hat-Check Distribution
Category:Expectation
Category:Hat-Check Distribution
Category:Hat-Check Problem
Expectation of Linear Transformation of Random Variable

== Theorem ==
Let $X$ be a random variable. 
Let $a, b$ be real numbers. 
Let $\expect X$ denote the expectation of $X$. 
Then we have: 
:$\expect {a X + b} = a \expect X + b$
if that expectation exists.
== Proof ==
=== Discrete Random Variable ===
=== Continuous Random Variable ===
Category:Expectation
Category:Expectation of Linear Transformation of Random Variable
Expectation of Linear Transformation of Random Variable/Continuous

== Theorem ==
Let $X$ be a continuous random variable. 
Let $a, b$ be real numbers. 
Then we have: 
:$\expect {a X + b} = a \expect X + b$
where $\expect X$ denotes the expectation of $X$. 
== Proof ==
Let $\map \supp X$ be the support of $X$. 
Let $f_X : \map \supp X \to \R$ be the probability density function of $X$.
Then:
Category:Expectation of Linear Transformation of Random Variable
Expectation of Linear Transformation of Random Variable/Discrete

== Theorem ==
Let $X$ be a discrete random variable. 
Let $a, b$ be real numbers. 
Then we have: 
:$\expect {a X + b} = a \expect X + b$
where $\expect X$ denotes the expectation of $X$. 
== Proof ==
We have: 
Category:Expectation of Linear Transformation of Random Variable
Expectation of Log Normal Distribution

== Theorem == 
Let $X$ be a continuous random variable with the Log Normal distribution with $\mu \in \R, \sigma \in \R_{> 0}$.
The expectation of $X$ is given by:
:$\expect X = \exp {\paren {\mu + \dfrac {\sigma^2 } 2 } }$
== Proof == 
From Raw Moment of Log Normal Distribution, we have:
The $n$th raw moment $\expect {X^n}$ of $X$ is given by: 
:$\expect {X^n} = \exp {\paren {n\mu + \dfrac {\sigma^2 n^2 } 2 } }$
Therefore, for $n = 1$ we have:
:$\expect X = \exp {\paren {\mu + \dfrac {\sigma^2 } 2 } }$
Hence the result.
Category:Log Normal Distribution
Category:Expectation
Expectation of Logistic Distribution

== Theorem == 
Let $X$ be a continuous random variable which satisfies the logistic distribution:
:$X \sim \map {\operatorname {Logistic} } {\mu, s}$
The expectation of $X$ is given by:
:$\expect X = \mu$
=== Lemma 1 ===
=== Lemma 2 ===
=== Lemma 3 ===
== Proof 1 == 
 
== Proof 2 ==
Category:Logistic Distribution
Category:Expectation
Category:Expectation of Logistic Distribution
Expectation of Logistic Distribution/Lemma 3

== Lemma for Expectation of Logistic Distribution ==
:$\forall k \in \N$:
:$\ds \int_{\to 0}^{\to 1} \map {\ln^{2k + 1} } {\dfrac {1 - u} u} \rd u = 0$
== Proof ==
let:
Then:
Category:Expectation of Logistic Distribution
Expectation 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 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:
:$\ds \expect X = \frac 1 s \int_{-\infty}^\infty \dfrac {x \map \exp {-\dfrac {\paren {x - \mu} } s} } {\paren {1 + \map \exp {-\dfrac {\paren {x - \mu} } s} }^2} \rd x$
let:
and also:
Then:
Category:Expectation of Logistic Distribution
Expectation 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 Moment in terms of Moment Generating Function: 
:$\expect X = \map { {M_X}'} 0$
From Moment Generating Function of Logistic Distribution: First Moment: 
:$\ds \map { {M_X}'} t = \map \exp {\mu t} \paren {\mu \int_{\to 0}^{\to 1} \paren {\dfrac {1 - u} u}^{-s t} \rd u - s \int_{\to 0}^{\to 1} \map \ln {\dfrac {1 - u} u} \paren {\dfrac {1 - u} u}^{-s t} \rd u}$
Hence setting $t = 0$:
Category:Expectation of Logistic Distribution
Expectation of Negative Binomial Distribution/Type 1

== Theorem ==
Let $X$ be a discrete random variable with the type 1 negative binomial distribution with parameters $n$ and $p$.
Then the expectation of $X$ is given by:
:$\expect X = \dfrac n p$
== Proof ==
From Probability Generating Function of Negative Binomial Distribution (Type 1), we have:
:$\map {\Pi_X} s = \paren {\dfrac {p s} {1 - q s} }^n$
where $q = 1 - p$.
From Expectation of Discrete Random Variable from PGF, we have:
:$\expect X = \map {\Pi'_X} 1$
We have:
Plugging in $s = 1$:
== Sources ==
* : $\S 4.3$: Moments: Exercise $6$
Category:Expectation
Category:Negative Binomial Distribution (Type 1)
Expectation 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 expectation of $X$ is given by:
:$\expect X = \dfrac {r p} q$
where $q = 1 - p$.
== Proof ==
From Probability Generating Function of Negative Binomial Distribution (Type 2):
:$\map {\Pi_X} s = \paren {\dfrac q {1 - p s} }^r$
From Expectation of Discrete Random Variable from PGF:
:$\expect X = \map {\Pi'_X} 1$
We have:
Plugging in $s = 1$:
== Also defined as ==
== Sources ==
* : Appendix $13$: Probability distributions
* : Appendix $15$: Probability distributions
Category:Expectation of Negative Binomial Distribution
Category:Negative Binomial Distribution (Type 2)
Category:Expectation
Expectation of Negative Binomial Distribution (Type 2)/Also defined as

== Theorem ==
Let $X$ be a discrete random variable with the type $2$ negative binomial distribution with parameters $r$ and $p$.
Let $X$ use the definition as the number of failures before $r$ successes have occurred.
Then the expectation of $X$ is given by:
:$\expect X = \dfrac {r q} p$
where $q = 1 - p$.
== Proof ==
Let $Y$ be a discrete random variable with the type $2$ negative binomial distribution with parameters $r$ and $p$.
Let $Y$ use the definition as the number of successes before $r$ failures have occurred.
From Expectation of Negative Binomial Distribution (Type 2):
:$\expect X = \dfrac {r p} q$
By definition:
:$p$ is the probability of success.
:$q$ is the probability of failure.
Hence the result by exchanging $q$ and $p$.
== Sources ==
* 
* 
Category:Expectation of Negative Binomial Distribution
Category:Negative Binomial Distribution (Type 2)
Category:Expectation
Expectation of Non-Negative Random Variable is Non-Negative

== Theorem ==
Let $X$ be a random variable. 
Let $\map \Pr {X \ge 0} = 1$. 
Then:
:$\expect X \ge 0$
where $\expect X$ denotes the expectation of $X$. 
== Proof ==
=== Discrete Random Variable ===
=== Continuous Random Variable ===
Category:Expectation
Category:Expectation of Non-Negative Random Variable is Non-Negative
Expectation of Non-Negative Random Variable is Non-Negative/Continuous

== Theorem ==
Let $X$ be a continuous random variable. 
Let $\map \Pr {X \ge 0} = 1$. 
Then:
:$\expect X \ge 0$
where $\expect X$ denotes the expectation of $X$. 
== Proof ==
Category:Expectation of Non-Negative Random Variable is Non-Negative
Expectation of Non-Negative Random Variable is Non-Negative/Discrete

== Theorem ==
Let $X$ be a discrete random variable. 
Let $\map \Pr {X \ge 0} = 1$. 
Then:
:$\expect X \ge 0$
where $\expect X$ denotes the expectation of $X$. 
== Proof ==
Let $\map \supp X$ be the support of $X$. 
Note that since $X$ is discrete, its sample space and hence support is countable. 
Therefore, there exists some sequence $\sequence {x_i}_{i \mathop \in I}$ such that: 
:$\map \supp X = \set {x_i \mid i \in I}$
for some $I \subseteq \N$. 
By the definition of a sample space, we have: 
:$\map \Pr {X = x_i} \ge 0$
for all $i \in I$. 
Note that since $\map \Pr {X \ge 0} = 1$, we have $\map \Pr {X < 0} = 0$. 
So, for any $x < 0$ we necessarily have $\map \Pr {X = x} = 0$, meaning that $x \notin \map \supp X$. 
We therefore have that any elements of $\map \supp X$ are non-negative.
That is: 
:$x_i \ge 0$
for all $i \in I$.
Therefore: 
:$x_i \map \Pr {X = x_i} \ge 0$
for all $i \in I$. 
Summing over all $i \in I$, we have: 
:$\ds \sum_{i \mathop \in I} x_i \map \Pr {X = x_i} \ge 0$
Hence, by the definition of expectation:
:$\expect X \ge 0$
Category:Expectation of Non-Negative Random Variable is Non-Negative
Expectation 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: 
:$\expect X = \mu$
where $\mu$ denotes the expectation of $X$.
== Proof 1 == 
== Proof 2 ==
== Sources ==
* 
* 
* : Appendix $13$: Probability distributions
* : Appendix $15$: Probability distributions
Category:Expectation of Normal Distribution
Category:Normal Distribution
Category:Expectation
Expectation 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 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:
Category:Expectation of Normal Distribution
Expectation 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 Moment in terms of Moment Generating Function: 
:$\expect X = \map {M'_X} 0$
We have: 
Setting $t = 0$: 
Category:Expectation of Normal Distribution
Expectation of Pareto Distribution

== Theorem == 
Let $X$ be a continuous random variable with the Pareto distribution with $a, b \in \R_{> 0}$.
The expectation of $X$ is given by:
:$\expect X = \begin {cases} \dfrac {a b} {a - 1} & 1 < a \\ \text {does not exist} & 1 \ge a \end {cases}$
== Proof == 
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 $n = 1$ we have:
:$\expect X = \begin {cases} \dfrac {a b} {a - 1} & 1 < a \\ \text {does not exist} & 1 \ge a \end {cases}$
Category:Pareto Distribution
Category:Expectation
Expectation of Poisson Distribution

== Theorem ==
Let $X$ be a discrete random variable with the Poisson distribution with parameter $\lambda$.
Then the expectation of $X$ is given by:
:$\expect X = \lambda$
== 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:Expectation of Poisson Distribution
Category:Expectation
Category:Poisson Distribution
Expectation of Poisson Distribution/Proof 1

== Theorem ==
== Proof ==
From the definition of expectation:
:$\ds \expect X = \sum_{x \mathop \in \Img X} x \map \Pr {X = x}$
By definition of Poisson distribution:
:$\ds \expect X = \sum_{k \mathop \ge 0} k \frac 1 {k!} \lambda^k e^{-\lambda}$
Then:
Category:Expectation of Poisson Distribution
Expectation of Poisson Distribution/Proof 2

== Theorem ==
== Proof ==
From Probability Generating Function of Poisson Distribution:
:$\map {\Pi_X} s = e^{-\lambda \paren {1 - s} }$
From Expectation of Discrete Random Variable from PGF:
:$\expect X = \map {\Pi'_X} 1$
We have:
Plugging in $s = 1$:
:$\map {\Pi'_X} 1 = \lambda e^{- \lambda \paren {1 - 1} } = \lambda e^0$
Hence the result from Exponential of Zero:
:$e^0 = 1$
== Sources ==
* : $\S 4.3$: Moments: Exercise $5$
Category:Expectation of Poisson Distribution
Expectation 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: 
:$\map {M_X} t = e^{\lambda \paren {e^t - 1} }$
By Moment in terms of Moment Generating Function:
:$\expect X = \map {M_X'} 0$
We have: 
Setting $t = 0$ gives: 
Category:Expectation of Poisson Distribution
Expectation of Power of Gamma Distribution

== Theorem ==
Let $X \sim \map \Gamma {\alpha, \beta}$ for some $\alpha, \beta > 0$, where $\Gamma$ is the Gamma distribution.
Then:
:$\expect {X^n} = \dfrac {\alpha^{\overline n} } {\beta^n}$
where:
:$\expect {X^n}$ denotes the expectation of $X^n$
:$\alpha^{\overline n}$ denotes the $n$th rising factorial of $\alpha$.
== Proof ==
From Moment in terms of Moment Generating Function:
:$\expect {X^n} = \map { {M_X}^{\paren n} } 0$
where ${M_X}^{\paren n}$ denotes the $n$th derivative of $M_X$.
Then:
Category:Gamma Distribution
Expectation of Product of Independent Random Variables is Product of Expectations/Corollary

== Corollary to Expectation of Product of Independent Random Variables is Product of Expectations ==
Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space. 
Let $X$ and $Y$ be integrable random variables that are independent.
Then:
:$\expect {X Y} = \expect X \expect Y$
== Proof ==
We have: 
Category:Expectation of Product of Independent Random Variables is Product of Expectations
Expectation of Random Variable as Integral with respect to Probability Distribution

== Theorem ==
Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $X$ be an integrable real-valued random variable on $\struct {\Omega, \Sigma, \Pr}$. 
Let $P_X$ be the probability distribution of $X$.
Then: 
:$\ds \expect X = \int_\R x \map {\rd P_X} x$
where $\expect X$ is the expected value of $X$. 
== Proof ==
From the definition of expectation: 
:$\ds \expect X = \int_\Omega X \rd \Pr$
We can write: 
:$\ds \int_\Omega X \rd \Pr = \int_\Omega I_\R \circ X \rd \Pr$
where $I_\R$ is the identity map for $\R$. 
From the definition of probability distribution, we have: 
:$P_X = X_* \Pr$
where $X_* \Pr$ is the pushforward of $\Pr$ on $\tuple {\R, \map \BB \R}$, where $\map \BB \R$ denotes the Borel $\sigma$-algebra on $\R$.
So, from Integral with respect to Pushforward Measure: Corollary, we have: 
:$I_\R$ is $P_X$-integrable
and:
:$\ds \int_\Omega I_\R \circ X \rd \Pr = \int_\R I_\R \rd P_X$
That is: 
:$\ds \int_\Omega X \rd \Pr = \int_\R x \map {\rd P_X} x$
== Sources ==
* : $10$: Probability: $10.1$: Basics
Category:Expectation