nyuuzyou/wikis huggingface dataset:
Exists Rational Cut Between two Cuts

== Theorem ==
Let $\alpha$ and $\beta$ be cuts.
Let $\alpha < \beta$, where $<$ denotes the strict ordering on cuts.
Then there exists a rational cut $r^*$ associated with the rational number $r$ such that:
:$\alpha < r^* < \beta$
== Proof ==
Let $\alpha < \beta$.
Then there exists a rational number $p$ such that $p \in \beta$ and $p \notin \alpha$.
Let $r \in \Q$ such that $r > p$ and $r \in \beta$.
Because $r \in \beta$ and $r \notin r^*$, we have that $r^* < \beta$.
Because $p \in r^*$ and $p \notin \alpha$, we have that $\alpha < r^*$.
== Sources ==
* : Chapter $1$: The Real and Complex Number Systems: Dedekind Cuts: $1.29$. Theorem
Category:Cuts
Exists Subset which is not Element/Proof 1

== Theorem ==
== Proof ==
Let $S$ be a set.
Let $T$ be the set of all elements of $S$ which do not contain $S$ as elements.
Then by the corollary to Russell's paradox $T$ itself cannot be an element of $S$.
This $T$ is the required subset.
== Sources ==
* : Chapter $1$: General Background: $\S 9$ Zermelo set theory
Category:Exists Subset which is not Element
Exists Subset which is not Element/Proof 2

== Theorem ==
== Proof ==
Consider the power set $\powerset S$ of $S$.
From Cantor's Theorem, there is no surjection $f: S \to \powerset S$.
That is, there are more subsets of $S$ than there are elements of $S$.
So there must be at least one subset of $S$ which is not an element of $S$.
== Sources ==
* : Chapter $1$: General Background: $\S 9$ Zermelo set theory
Category:Exists Subset which is not Element
Exp (-x^2) is Schwartz Test Function

== Theorem ==
$\map \exp {-x^2}$ is a Schwartz test function.
== Proof ==
Let:
:$\dfrac {\d^n}{\d x^n} \map \exp {-x^2} = \map {p_n} x \map \exp {-x^2}$
We have that $\map {p_0} x = 1$ and $\map {p_1} x = -2 x$.
Then:
where 
:$\map {p_{n + 1}} x := \map {p_n'} x - 2 x \map {p_n} x$
Note that each subsequent $\map {p_n} x$ is constructed from derivatives, products and differences of polynomials.
Hence, $\map {p_n} x$ is a polynomial.
Consider the Maclaurin series of $\map \exp {x^2}$ .
Suppose $\size x \ge 1$.
Then:
Suppose $\size x \le 1$.
We have that $x^k \map \exp {-x^2}$ is a continuous function.
Hence, it is bounded:
:$\forall x \in \closedint {-1} 1 \exists M \in \R : \size {x^k \map \exp {-x^2}} \le M$
Altogether:
:$\sup \size {x^k \map \exp {-x^2}} \le \map \max {M, k!} < \infty$
Thus:
:$\map \exp {-x^2} \in \map \SS \R$.
== Sources ==
* : Chapter $\S 6.5$: A glimpse of distribution theory. Fourier transform of (tempered) distributions
Category:Examples of Schwartz Test Functions
Expanded Number/Examples/163

== Example of Expanded Number ==
The number expressed as $163$ in decimal notation is expressed in expanded form as:
:$1 \times 10^2 + 6 \times 10^1 + 3 \times 10^0$
or more compactly as:
:$1 \times 10^2 + 6 \times 10 + 3$
== Sources ==
* 
* 
Category:Examples of Expanded Numbers
Expansion Theorem for Determinants/Corollary

== Corollary to Expansion Theorem for Determinants ==
Let $\mathbf A = \sqbrk a_n$ be a square matrix of order $n$.
Let $D = \map \det {\mathbf A}$ be the determinant of $\mathbf A$.
Let $a_{p q}$ be an element of $\mathbf A$.
Let $A_{p q}$ be the cofactor of $a_{p q}$ in $D$.
Let $\delta_{rs}$ be the Kronecker delta.
Then:
:$(1): \quad \ds \forall r \in \closedint 1 n: \sum_{k \mathop = 1}^n a_{r k} A_{s k} = \delta_{r s} D$
:$(2): \quad \ds \forall r \in \closedint 1 n: \sum_{k \mathop = 1}^n a_{k r} A_{k s} = \delta_{r s} D$
That is, if you multiply each element of a row or column by the cofactor of another row or column, the sum of those products is zero.
== Proof ==
Let $D'$ be the determinant obtained by replacing row $s$ with row $r$.
Let $a'_{ij}$ be an element of $D'$.
Then:
:$a'_{ij} = \begin {cases} a_{ij} & : i \ne s \\ a_{rj} & : i = s \end {cases}$
Let the cofactor of $a'_{ij}$ in $D'$ be denoted by $A'_{i j}$.
Then:
:$\forall k \in \closedint 1 n: A'_{s k} = A_{s k}$
So by the Expansion Theorem for Determinants:
:$\ds D' = \sum_{k \mathop = 1}^n a'_{s k} A'_{s k}$
But the $r$th and $s$th rows are identical.
So by Square Matrix with Duplicate Rows has Zero Determinant:
:$D' = 0$
Hence the result.
The result for columns follows from Determinant of Transpose. 
Category:Determinants
Expansion Theorem for Determinants/Examples/Arbitrary Order 3 Example 1

== Example of Use of Expansion Theorem for Determinants ==
Consider the order $3$ determinant:
:$\begin {vmatrix} 1 & 2 & 3 \\ 4 & 1 & 2 \\ 6 & 5 & 4 \end {vmatrix}$
We have:
== Sources ==
* 
* 
Category:Expansion Theorem for Determinants
Expansion Theorem for Determinants/Mistake

== Source Work ==
:
:cofactor
== Mistake ==
:(i) The expression $a_{i1} A_{i1} + a_{i2} + \cdots + a_{in} A_{in}$ has the same value for any $i$, and is the definition of $\det \mathbf A$, the determinant of $\mathbf A$.
== Correction ==
The expression should read:
:$a_{i1} A_{i1} + a_{i2} A_{i2} + \cdots + a_{in} A_{in}$
In the $6$th edition, this section has been removed.
== Sources ==
* 
Category:Mistakes/The Concise Oxford Dictionary of Mathematics
Expansion of Associated Reduced Quadratic Irrational

== Lemma ==
Let $\alpha$ be a reduced quadratic irrational which is associated to $n$.
Let $\alpha$ be transformed into its integer part and fractional part via:
:$\alpha = \left \lfloor {\alpha} \right \rfloor + \dfrac 1{\alpha'}$
Then the resulting quadratic irrational $\alpha'$ is also reduced and associated to $n$.
== Proof ==
Let $\alpha = \dfrac{\sqrt n + P} Q$ and $X = \left \lfloor {\alpha} \right \rfloor$
Then we have:
:$\dfrac 1 {\alpha'} = \dfrac {\sqrt n - \left({Q X - P}\right)} Q$
Because $\sqrt n$ is irrational, we must have:
:$\dfrac 1 {\alpha'} > 0$
Because $\dfrac 1 {\alpha'}$ is the fractional part:
:$0 < \dfrac 1 {\alpha'} < 1 \implies \alpha' > 1$
Transforming:
we have:
:$P' = Q X - P$
and:
:$Q' = \dfrac 1 Q \left({n - \left({Q X - P}\right)^2}\right)$
To show $Q' \in \Z$:
We have that:
:$n - \left({Q X - P}\right)^2 \equiv n - P^2 \pmod Q$
Because $\alpha$ is associated to $n$, $Q$ must divide this quantity.
Hence $Q'$ is an integer as defined.
Because $X = \left \lfloor{\dfrac{\sqrt n + P} Q}\right \rfloor$ is an integer and $\alpha$ is irrational:
:$X < \dfrac {\sqrt n + P} Q$
Hence:
:$P' = QX - P < \sqrt n $
forcing:
:$\tilde{\alpha}' < 0$
Because $\alpha > 1$:
:$X \ge 1 \iff 0 \le X - 1$
Thus:
Hence $\tilde \alpha' > -1$ and $\alpha'$ is reduced.
Because $Q' = \dfrac 1 Q \left({n - \left({P'}\right)^2}\right)$:
:$n - \left({P'}\right)^2 \equiv Q Q' \equiv 0 \pmod {Q'}$
Hence $\alpha'$ is associated to $n$.
Thus $\alpha'$ is both reduced and associated to $n$.
Category:Continued Fractions
Category:Quadratic Irrationals
Expansion of Characteristic Polynomial of Matrix

== Theorem ==
Let $R$ be a commutative ring with unity.
Let $R \sqbrk x$ be the polynomial ring in one variable over $R$.
Let $\mathbf A$ be a square matrix over $R$ of order $n > 0$.
Let $\map {p_{\mathbf A} } x$ be the characteristic polynomial of $\mathbf A$.
Then $\map {p_{\mathbf A} } x$ can be expressed as:
:$\map {p_{\mathbf A} } x = x^n + a_{n - 1} x^{n - 1} + \cdots + a_0$
where:
== Proof ==
== Sources ==
* 
Category:Characteristic Polynomial of Matrix
Expansion of Included Set Topology

== Theorem ==
Let $S$ be a set.
Let $A_1 \subseteq S$ and $A_2 \subseteq S$.
Let $T_1 = \struct {S, \tau_{A_1} }$ and $T_2 = \struct {S, \tau_{A_2} }$ be included set spaces on $S$.
Then:
:$(1): \quad T_1 \ge T_2 \iff A_1 \subseteq A_2$
:$(2): \quad T_1 > T_2 \iff A_1 \subsetneq A_2$
where:
:$T_1 \ge T_2$ denotes that $T_1$ is finer than $T_2$
:$T_1 > T_2$ denotes that $T_1$ is strictly finer than $T_2$.
== Proof ==
Let $T_1 = \struct {S, \tau_{A_1} }$ and $T_2 = \struct {S, \tau_{A_2} }$ be included set spaces on $S$.
=== Necessary Condition ===
Let $T_1 \ge T_2$.
We have that $A_2 \in \tau_{A_2}$ by definition of included set topology.
But by definition of finer topology, $A_2 \in \tau_{A_1}$.
So by definition of included set topology, $A_1 \subseteq A_2$.
Now suppose that $T_1 > T_2$.
That is, $T_1 \ge T_2$ and $T_1 \ne T_2$.
It follows that $\exists X \in \tau_{A_1}: X \notin \tau_{A_2}$.
Then $A_1 \subseteq X \subsetneq A_2$ from which it follows that $A_1 \subsetneq A_2$.
=== Sufficient Condition ===
Let $A_1 \subseteq A_2$.
Then:
So $T_1 \ge T_2$ by definition of finer topology.
Now suppose that $A_1 \subsetneq A_2$.
Then $A_2 \nsubseteq A_1$ and so $A_1 \notin \tau_{A_2}$.
Thus $\tau_{A_1} \nsubseteq \tau_{A_2}$ and so $T_1 > T_2$.
Both necessity and sufficiency have been proved, hence the result.
== Sources ==
* : Chapter $\text {I}$: Topological Spaces: $1$. Open Sets and Closed Sets: Exercise $1 \ \text{(a)}$
Category:Included Set Topology
Expectation Preserves Inequality

== Theorem ==
Let $X$, $Y$ be random variables.
Let $\map \Pr {X \ge Y} = 1$.
Then: 
:$\expect X \ge \expect Y$
== Proof ==
Note that we have: 
:$\map \Pr {X - Y \ge 0} = 1$
From Expectation of Non-Negative Random Variable is Non-Negative, we then have: 
:$\expect {X - Y} \ge 0$
From Sum of Expectations of Independent Trials, we have: 
:$\expect X + \expect {-Y} \ge 0$
From Expectation of Linear Transformation of Random Variable, we have:
:$\expect X - \expect Y \ge 0$
That is: 
:$\expect X \ge \expect Y$
Category:Expectation
Expectation and Variance of Poisson Distribution equal its Parameter

== Theorem ==
Let $X$ be a discrete random variable with the Poisson distribution with parameter $\lambda$.
Then the expectation of $X$ equals the variance of $X$, that is, $\lambda$ itself.
== Proof ==
From Expectation of Poisson Distribution:
:$\expect X = \lambda$
From Variance of Poisson Distribution:
:$\var X = \lambda$
== Sources ==
* : $\S 4.3$: Moments: Exercise $5$
* 
* 
Category:Poisson Distribution
Expectation is Linear

== Theorem ==
Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $X$ and $Y$ be integrable random variables on $\struct {\Omega, \Sigma, \Pr}$.
Let $\expect X$ denote the expectation of $X$.
Then:
:$\forall \alpha, \beta \in \R: \expect {\alpha X + \beta Y} = \alpha \, \expect X + \beta \, \expect Y$
== Proof ==
=== Discrete Random Variable ===
 
=== Continuous Random Variable ===
=== General Case ===
 
== Sources ==
* : $\S 3.2$: Expectation in the multivariate case: $(7)$
Category:Expectation is Linear
Category:Expectation
Expectation is Linear/Continuous

== Theorem ==
Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $X$ and $Y$ be random variables on $\struct {\Omega, \Sigma, \Pr}$.
Let $E$ denote the expectation function.
Then:
:$\forall \alpha, \beta \in \R: \expect {\alpha X + \beta Y} = \alpha \expect X + \beta \expect Y$
== Proof == 
Let $\map \supp X$ and $\map \supp Y$ be the supports of $X$ and $Y$ respectively. 
Let $f_{X, Y} : \map \supp X \times \map \supp Y \to \R$ be the joint probability density function of $X$ and $Y$. 
Let $f_X$ and $f_Y$ be the marginal probability density functions of $X$ and $Y$. 
Then:
 
Category:Expectation is Linear
Expectation is Linear/Discrete

== Theorem ==
Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $X$ and $Y$ be random variables on $\struct {\Omega, \Sigma, \Pr}$.
Let $\expect X$ denote the expectation of $X$.
Then:
:$\forall \alpha, \beta \in \R: \expect {\alpha X + \beta Y} = \alpha \, \expect X + \beta \, \expect Y$
== Proof == 
Follows directly from Expectation of Function of Joint Probability Mass Distribution, thus:
Category:Expectation is Linear
Expectation is Linear/General Case

== Theorem ==
== Proof ==
From Integral of Integrable Function is Homogeneous, we have: 
:$\alpha X$ and $\beta Y$ are $\Pr$-integrable.
From Integral of Integrable Function is Additive, we have: 
:$\alpha X + \beta Y$ is $\Pr$-integrable.
From Linear Combination of Real-Valued Random Variables is Real-Valued Random Variable, we have: 
:$\alpha X + \beta Y$ is a real-valued random variable.
Then: 
Category:Expectation is Linear
Expectation is Monotone

== Theorem ==
Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $X$ and $Y$ be integrable random variables such that:
:$\forall \omega \in \Omega: \map X \omega \le \map Y \omega$
Then:
:$\expect X \le \expect Y$
== Proof ==
From the definition of expectation we have:
:$\ds \expect X = \int X \rd \Pr$
and:
:$\ds \expect Y = \int Y \rd \Pr$
The result follows directly from Integral of Integrable Function is Monotone.
Category:Expectation
Expectation of Almost Surely Constant Random Variable

== Theorem ==
Let $X$ be an almost surely constant random variable.
That is, there exists some $c \in \R$ such that: 
:$\map \Pr {X = c} = 1$
Then: 
:$\expect X = c$
== Proof ==
Note that since $\map \Pr {X = c} = 1$, we have $\map \Pr {X \ne c} = 0$ from Probability of Event not Occurring.
Therefore: 
:$\map {\mathrm {supp} } X = \set c$
We therefore have:
Category:Expectation
Expectation of Bernoulli Distribution

== Theorem ==
Let $X$ be a discrete random variable with a Bernoulli distribution with parameter $p$.
Then the expectation of $X$ is given by:
:$\expect X = p$
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Proof 4 ==
Category:Expectation
Category:Bernoulli Distribution
Category:Expectation of Bernoulli Distribution
Expectation of Bernoulli 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 Bernoulli distribution:
:$\expect X = 1 \times p + 0 \times \paren {1 - p}$
Hence the result.
Category:Expectation of Bernoulli Distribution
Expectation of Bernoulli Distribution/Proof 3

== Theorem ==
== Proof ==
From the Probability Generating Function of Bernoulli Distribution, we have:
:$\map {\Pi_X} s = q + p s$
where $q = 1 - p$.
From Expectation of Discrete Random Variable from PGF, we have:
:$\expect X = \map { {\Pi_X}'} 1$
From Derivatives of PGF of Bernoulli Distribution:
:$\map { {\Pi_X}'} s = p$
Hence the result.
Category:Expectation of Bernoulli Distribution
Expectation of Bernoulli Distribution/Proof 4

== Theorem ==
== Proof ==
From Moment Generating Function of Bernoulli Distribution, the moment generating function of $X$, $M_X$, is given by: 
:$\map {M_X} t = q + p e^t$
where $q = 1 - p$. 
By Moment in terms of Moment Generating Function:
:$\expect X = \map {M_X'} 0$
We have: 
Setting $t = 0$ gives: 
Category:Expectation of Bernoulli Distribution
Expectation of Beta Distribution

== Theorem == 
Let $X \sim \BetaDist \alpha \beta$ for some $\alpha, \beta > 0$, where $\operatorname{Beta}$ denotes the beta distribution. 
Then:
:$\expect X = \dfrac \alpha {\alpha + \beta}$
== Proof 1 == 
 
== Proof 2 ==
Category:Expectation
Category:Beta Distribution
Category:Expectation of Beta Distribution
Expectation 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 the definition of the expected value of a continuous random variable: 
:$\ds \expect X = \int_0^1 x \map {f_X} x \rd x$
So:
Category:Expectation of Beta Distribution
Expectation of Binomial Distribution

== Theorem ==
Let $X$ be a discrete random variable with the binomial distribution with parameters $n$ and $p$ for some $n \in \N$ and $0 \le p \le 1$.
Then the expectation of $X$ is given by:
:$\expect X = n p$
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Proof 4 ==
== Sources ==
* : $\S 2.4$: Expectation: Exercise $9$
* 
* 
* 
* : Appendix $13$: Probability distributions
* : Appendix $15$: Probability distributions
Category:Expectation of Binomial Distribution
Category:Expectation
Category:Binomial Distribution
Expectation of Binomial Distribution/Proof 2

== Theorem ==
== Proof ==
From Bernoulli Process as Binomial Distribution, we see that $X$ as defined here is a sum of discrete random variables $Y_i$ that model the Bernoulli distribution:
:$\ds X = \sum_{i \mathop = 1}^n Y_i$
Each of the Bernoulli trials is independent of each other, by definition of a Bernoulli process.
It follows that:
Category:Expectation of Binomial Distribution
Expectation of Binomial Distribution/Proof 3

== Theorem ==
== Proof ==
From the Probability Generating Function of Binomial Distribution, we have:
:$\map {\Pi_X} s = \paren {q + p s}^n$
where $q = 1 - p$.
From Expectation of Discrete Random Variable from PGF, we have:
:$\expect X = \map {\Pi'_X} 1$
We have:
 
Plugging in $s = 1$:
:$\map {\Pi'_X} 1 = n p \paren {q + p}$
Hence the result, as $q + p = 1$.
Category:Expectation of Binomial Distribution
Expectation of Binomial Distribution/Proof 4

== Theorem ==
== Proof ==
From Moment Generating Function of Binomial Distribution, the moment generating function of $X$, $M_X$, is given by:
:$\ds \map {M_X} t = \paren {1 - p + p e^t}^n$
By Moment in terms of Moment Generating Function:
:$\ds \expect X = \map {M_X'} 0$
We have: 
Setting $t = 0$ gives:
Category:Expectation of Binomial Distribution
Expectation of Bounded Random Variable

== Theorem ==
Let $X$ be a random variable. 
Let $a$ and $b$ be real numbers with $b \ge a$.
Let: 
:$\map \Pr {a \le X \le b} = 1$
Then:
:$a \le \expect X \le b$
where $\expect X$ denotes the expectation of $X$. 
== Proof ==
From:
:$\map \Pr {a \le X \le b} = 1$
it follows that: 
:$\map \Pr {X \ge a} = 1$
That is: 
:$\map \Pr {X - a \ge 0} = 1$
From Expectation of Non-Negative Random Variable is Non-Negative, we therefore have that: 
:$\expect {X - a} \ge 0$
From Expectation of Linear Transformation of Random Variable, we have: 
:$\expect X - a \ge 0$
That is:
:$\expect X \ge a$
Note that we also have: 
:$\map \Pr {X \le b} = 1$
That is: 
:$\map \Pr {b - X \ge 0} = 1$
Again applying Expectation of Non-Negative Random Variable is Non-Negative, we have: 
:$\expect {b - X} \ge 0$
From Expectation of Linear Transformation of Random Variable, we have: 
:$b - \expect X \ge 0$
giving: 
:$\expect X \le b$
Putting these inequalities together, we have: 
:$a \le \expect X \le b$
Category:Expectation
Expectation of Chi-Squared Distribution

== Theorem == 
Let $n$ be a strictly positive integer. 
Let $X \sim \chi_n^2$ where $\chi_n^2$ is the chi-squared distribution with $n$ degrees of freedom. 
Then the expectation of $X$ is given by: 
:$\expect X = n$
== Proof ==
== Sources ==
* 
* 
* : Appendix $13$: Probability distributions
* : Appendix $15$: Probability distributions
Category:Expectation
Category:Chi-Squared Distribution
Expectation 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 expectation of $X$ is given by: 
:$\expect X = \sqrt 2 \dfrac {\map \Gamma {\paren {n + 1} / 2} } {\map \Gamma {n / 2} }$
where $\Gamma$ is the gamma function.
== Proof == 
From the definition of the chi distribution, $X$ has probability density function:
:$\map {f_X} x = \dfrac 1 {2^{\paren {n / 2} - 1} \map \Gamma {n / 2} } x^{n - 1} e^{- x^2 / 2}$
From the definition of the expected value of a continuous random variable:
:$\ds \expect X = \int_0^\infty x \map {f_X} x \rd x$
So:
Category:Expectation
Category:Chi Distribution
Expectation of Conditional Expectation

== Theorem ==
Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $\GG \subseteq \Sigma$ be a sub-$\sigma$-algebra.
Let $X$ be a integrable random variable.
Let $\expect {X \mid \GG}$ be a version of the conditional expectation of $X$ given $\GG$.
Then: 
:$\expect {\expect {X \mid \GG} } = \expect X$ 
 
== Proof ==
We have: 
== Sources ==
* : $9.7$: Properties of conditional expectation: a list
Category:Conditional Expectation
Expectation 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: 
:$\expect X = \dfrac {a + b} 2$
== 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 the expected value of a continuous random variable: 
:$\ds \expect X = \int_{-\infty}^\infty x \map {f_X} x \rd x$
So:
== Also see ==
* Variance of Continuous Uniform Distribution
* Expectation of Discrete Uniform Distribution
== Sources ==
* : Appendix $13$: Probability distributions
* : Appendix $15$: Probability distributions
Category:Expectation
Category:Continuous Uniform Distribution
Expectation 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 expectation of $X$ is the value of the first derivative of $\map {\Pi_X} s$ $s$ at $s = 1$.
That is:
:$\expect X = \map {\Pi'_X} 1$
== Proof ==
For ease of notation, write $\map p x$ for $\map \Pr {X = x}$.
From the definition of the probability generating function:
:$\ds \map {\Pi_X} s = \sum_{x \mathop \ge 0} \map p x s^x = \map p 0 + \map p 1 s + \map p 2 s^2 + \map p 3 s^3 + \cdots$
Differentiate this $s$:
Plugging in $s = 1$ gives:
:$\ds \map {\Pi'_X} 1 = \sum_{k \mathop = 0}^\infty k \map \Pr {X = k} 1^{k - 1} = \map p 1 + 2 \map p 2 + 3 \map p 3 + \cdots$
But:
:$\ds \sum_{k \mathop = 0}^\infty k \map \Pr {X = k} 1^{k - 1} = \sum_{k \mathop = 0}^\infty k \map \Pr {X = k}$
is precisely the definition of the expectation.
== Motivation ==
== Sources ==
* : $\S 4.3$: Moments: $(17)$
* 
* 
Category:Expectation of Discrete Random Variable from PGF
Category:Expectation
Category:Probability Generating Functions
Expectation of Discrete Random Variable from PGF/Motivation

== Expectation of Discrete Random Variable from PGF: Also known as == 
Expectation of Discrete Random Variable from PGF shows how to find the expectation 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 the PGF and plug in $1$.
Assuming, of course, you know what the PGF is.
Category:Expectation of Discrete Random Variable from PGF
Expectation of Discrete Uniform Distribution

== Theorem ==
Let $X$ be a discrete random variable with the discrete uniform distribution with parameter $n$.
Then the expectation of $X$ is given by:
:$\expect X = \dfrac {n + 1} 2$
== Proof ==
From the definition of expectation:
:$\ds \expect X = \sum_{x \mathop \in \Omega_X} x \map \Pr {X = x}$
Thus:
Category:Expectation
Category:Discrete Uniform Distribution
Expectation 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 expectation of $X$ is given by:
:$\expect X = \dfrac k \lambda$
== Proof ==
Category:Expectation
Category:Erlang Distribution
Expectation of Exponential Distribution

== Theorem ==
Let $X$ be a continuous random variable of the exponential distribution with parameter $\beta$ for some $\beta \in \R_{> 0}$
Then the expectation of $X$ is given by:
:$\expect X = \beta$
== Proof 1 ==
 
== Proof 2 ==
== Also see ==
* Variance of Exponential Distribution
== Sources ==
* 
* 
* : Appendix $13$: Probability distributions
* : Appendix $15$: Probability distributions
Category:Expectation of Exponential Distribution
Category:Expectation
Category:Exponential Distribution