hcju/proofwikiqa huggingface dataset:
To show that $V$ is a discrete random variable on $\left({\Omega, \Sigma, \Pr}\right)$, we need to show that: :$(1): \quad $The image of $U$ and $V$ are countable subsets of $\R$; :$(2): \quad \forall x \in \R: \left\{{\omega \in \Omega: V \left({\omega}\right) = x}\right\} \in \Sigma$. First we consider any $V_V = \left\{{\omega \in \Omega: V \left({\omega}\right) = v}\right\}$ such that $V_v \ne \varnothing$. We have that: : $V_v = \left\{{\omega \in \Omega: X \left({\omega}\right) Y \left({\omega}\right) = v}\right\}$ Consider any $\omega \in V_v$. If $v = 0$ the result follows immediately, so we assume that $v \ne 0$. Then: :$\omega \in X_x \cap Y_x$ where: : $X_x = \left\{{\omega \in \Omega: X \left({\omega}\right) = x}\right\}, Y_x = \left\{{\omega \in \Omega: Y \left({\omega}\right) = \dfrac v x}\right\}$ Both $X_x \in \Sigma$ and $Y_x \in \Sigma$ (as $X$ and $Y$ are discrete random variables. As $\left({\Omega, \Sigma, \Pr}\right)$ is a probability space then $X_x \cap Y_x \in \Sigma$. Now note that $\displaystyle V_v = \bigcup_{x \mathop \in \R} \left({X_x \cap Y_x}\right)$. That is, it is the union of all such intersections of sets whose discrete random variables multiply together to make $v$. As $X_x$ is a countable set it follows that $V_v$ is a countable union of countable sets. From Countable Union of Countable Sets is Countable it follows that $X_x$ is a countable set. And, by dint of $\left({\Omega, \Sigma, \Pr}\right)$ being a probability space, $V_v \in \Sigma$. Thus $V$ is a discrete random variables on $\left({\Omega, \Sigma, \Pr}\right)$. {{qed}} {{explain|Might need to be more rigorous on the boundary instances.}} {{proofread}}
Follows directly from the definition of shifted geometric distribution. Let $Y$ be the discrete random variable defined as the number of trials for the first success to be achieved. Thus the last trial (and the last trial only) will be a success, and the others will be failures. The probability that $k-1$ failures are followed by a success is: :$\Pr \left({Y = k}\right) = \left({1 - p}\right)^{k-1} p$ Hence the result. {{Qed}}
By definition: * $\Omega \left({Y}\right) = \N_{>0} = \left\{{1, 2, 3, \ldots}\right\}$ * $\Pr \left({Y = k}\right) = p \left({1 - p}\right)^{k-1}$ Then: {{begin-eqn}} {{eqn | l=\Pr \left({\Omega}\right)       | r=\sum_{k \mathop \ge 1} p \left({1 - p}\right)^{k-1}       | c=by definition of shifted geometric distribution }} {{eqn | r=\sum_{j \mathop \ge 0} p \left({1 - p}\right)^j       | c=putting $j = k-1$ }} {{eqn | r=1       | c=Geometric Distribution Gives Rise to Probability Mass Function }} {{end-eqn}} So $Y$ satisfies $\Pr \left({\Omega}\right) = 1$, and hence the result. {{Qed}}
By definition: * $\operatorname{Im} \left({X}\right) = \left\{{0, 1, 2, \ldots}\right\}$ * $\displaystyle \Pr \left({X = k}\right) = \binom {n + k - 1} {n - 1} p^k \left({1-p}\right)^n$ Then: {{begin-eqn}} {{eqn | l=\Pr \left({\Omega}\right)       | r=\sum_{k \mathop \ge n} \binom {n + k - 1} {n - 1} p^k \left({1-p}\right)^n       | c= }} {{eqn | r=\left({1-p}\right)^n \sum_{k \mathop \ge n} \binom {n + k - 1} k p^k       | c=Symmetry Rule for Binomial Coefficients }} {{eqn | r=\left({1-p}\right)^n \sum_{j \mathop \ge 0} \binom {-n} k p^k       | c=Negated Upper Index of Binomial Coefficient }} {{eqn | r=\left({1-p}\right)^n p^{-n}       | c=Binomial Theorem }} {{eqn | r=1       | c= }} {{end-eqn}} So $X$ satisfies $\Pr \left({\Omega}\right) = 1$, and hence the result. {{Qed}} Category:Negative Binomial Distribution 94mz54htjumappbdxccj4njghtkr9vr
Consider the experiment $\mathcal E$ as described. By Bernoulli Process as a Negative Binomial Distribution: First Form, $\mathcal E$ is modelled by a negative binomial distribution of the first form with parameters $n$ and $p$: : $\displaystyle \forall k \in \Z, k \ge 0: \Pr \left({X = k}\right) = \binom {n + k - 1} {n - 1} p^k q^n $ where $q = 1 - p$. Now consider the experiment $\mathcal E'$ as described. By Bernoulli Process as Geometric Distribution, $\mathcal E'$ is modelled by a geometric distribution with parameter $p$: : $\forall k \in \Z, k \ge 0: \Pr \left({X = k}\right) = p^k q$ where $q = 1 - p$. {{proof wanted|Requires further results}} Category:Negative Binomial Distribution Category:Geometric Distribution fwx2ltasw3i8boetz38c6ulwd41ndyl
The number of Bernoulli trials may be as few as $0$, so the image is correct: :$\Img X = \set {0, 1, 2, \ldots}$ If $X$ takes the value $x$, then there must have been $n + x$ trials altogether. So, after $n + x - 1$ trials, there must have been $n - 1$ failures, as (from the description of the experiment) the last trial is a failure. So the probability of the occurrence of the event $\sqbrk {X = x}$ is given by the binomial distribution, as follows: :$\displaystyle \map {p_X} x = \binom {n + x - 1} {n - 1} p^x \paren {1 - p}^n$ where $x \in \set {0, 1, 2, \ldots}$ Hence the result, by definition of first form of the negative binomial distribution. {{qed}} Category:Negative Binomial Distribution Category:Bernoulli Distribution r399641lr1kow0v4slips2yb0dzthep
First note that the number of Bernoulli trials has to be at least $n$, so the image is correct: $\Img X = \set {n, n + 1, n + 2, \ldots}$. Now, note that if $X$ takes the value $x$, then in the first $x - 1$ trials there must have been $n - 1$ successes. Hence there must have been $x - n$ failures, and so a success happens at trial number $x$. So the probability of the occurrence of the event $\sqbrk {X = x}$ is given by the binomial distribution, as follows: :$\map {p_X} x = \dbinom {x - 1} {n - 1} \paren {1 - p}^{x - n} p^n$ where $x \in \set {k, k + 1, k + 2, \ldots}$ Hence the result, by definition of second form of the negative binomial distribution. {{qed}}
Consider the experiment $\FF$ as described. By Bernoulli Process as a Negative Binomial Distribution: Second Form, $\FF$ is modelled by a negative binomial distribution of the second form with parameters $n$ and $p$: :$\ds \forall k \in \Z, k \ge n: \map \Pr {Y = k} = \binom {k - 1} {n - 1} q^{k - n} p^n$ where $q = 1 - p$. Now consider the experiment $\FF'$ as described. By Bernoulli Process as Shifted Geometric Distribution, $\FF'$ is modelled by a geometric distribution with parameter $p$: :$\forall k \in \Z, k \ge 1: \map \Pr {Y = k} = q^{k - 1} p$ where $q = 1 - p$. {{proof wanted|Requires further results, which are on their way.}} Category:Negative Binomial Distribution Category:Geometric Distribution 78o0joww7g5rz1fgq0y5nyfz1df6c3r
=== Definition 1 equivalent to Definition 2 === Let $\var X$ be defined as: :$\var X := \expect {\paren {X - \expect X}^2}$ Let $\mu = \expect X$. Let $\map f X = \paren {X - \mu}^2$ be considered as a function of $X$. Then by applying Expectation of Function of Discrete Random Variable: :$\displaystyle \expect {\map f X} = \sum_{x \mathop \in \Omega_X} \map f x \, \map \Pr {X = x}$ which leads to: :$\displaystyle \expect {\paren {X - \mu}^2} = \sum_{x \mathop \in \Omega_X} \paren {\paren {X - \mu}^2} \map \Pr {X = x}$ thus demonstrating the equality of Definition 1 to Definition 2. {{qed|lemma}} === Definition 2 equivalent to Definition 3 === Let $\mu = \expect X$, and take the expression for variance: :$\displaystyle \var X := \sum_{x \mathop \in \Img X} \paren {x - \mu}^2 \map \Pr {X = x}$ Then from Variance as Expectation of Square minus Square of Expectation: :$\var X = \expect {X^2} - \paren {\expect X}^2$ {{qed}}
We have from Power Rule for Derivatives that: :$\displaystyle \frac {\d} {\d x} \sum_{n \mathop \ge 1} \paren {n + 1} x^n = \sum_{n \mathop \ge 1} n \paren {n + 1} x^{n - 1}$ But from Sum of Infinite Geometric Sequence: {{begin-eqn}} {{eqn | l = \sum_{n \mathop \ge 1} \paren {n + 1} x^n       | r = \sum_{m \mathop \ge 2} m x^{m - 1}       | c =  }} {{eqn | r = \sum_{m \mathop \ge 1} m x^{m - 1} - 1       | c = taking into account the first term }} {{eqn | r = \frac 1 {\paren {1 - x}^2} - 1       | c = from main result above }} {{end-eqn}} The result follows by Power Rule for Derivatives and the Chain Rule for Derivatives applied to $\dfrac 1 {\paren {1 - x}^2}$. {{Qed}} Category:Analysis Category:Differential Calculus Category:Geometric Sequences drbl8sx80mkdsg4thpnflm3zofrmrx9
=== Definition 1 implies Definition 2 === Let $S$ be a countable set by Definition 1. Then there is an injection $f: S \to \N$. By Law of Excluded Middle, $S$ is finite or infinite. If $S$ is finite, then it trivially satisfies Definition 2. {{explain|Implicitly we take $S \cong f[S]$ here}} If $S$ is infinite, then it is countably infinite by Infinite Set of Natural Numbers is Countably Infinite, and thus satisfies Definition 2. {{qed|lemma}} === Definition 2 implies Definition 1 === Let $S$ be a countable set by Definition 2. Then $S$ is finite or countably infinite. If $S$ is countably infinite, then there is a bijection $f: S \to \N$, which is injective by definition. If $S$ is finite, then for some natural number $n$ there is a bijection $f: S \to \N_n$, where $\N_n$ is the initial segment of $\N$ determined by $n$. Let $f': S \to \N$ be the extension of $f$ to $\N$. {{explain}} Then $f'$ is an injection. Thus in either case $S$ is countable by Definition 1. {{qed}} {{questionable|Does it really need LEM for this? LEM states that "if $A$ is not false, then it is true", not "if $A$ is not true, then it is not-true". As it defined here, a set is infinite if it is not finite. That should not require LEM to accept.}} {{LEM}} {{finish|Include the proof for Definition:Countable Set/Definition 3}}
By Infinite Set has Countably Infinite Subset, we have an injection $g: \N \to S$ But by Cantor-Bernstein-Schröder Theorem/Lemma this produces a bijection $f: \N \to S$ {{qed}}
{{begin-eqn}} {{eqn | o =        | r = 1 + 2 z + 4 z^2 + \cdots       | c =  }} {{eqn | r = \sum_{n \mathop \ge 0} \paren {2 z}^n       | c =  }} {{eqn | r = \frac 1 {1 - 2 z}       | c = Sum of Infinite Geometric Sequence }} {{end-eqn}} This is valid for: :$\size {2 z} < 1$ from which: :$\size z < \dfrac 1 2$ follows directly by division by $2$. The result follows from the definition of a generating function. {{qed}}
Take the sequence: :$S_n = 1, 1, 1, \ldots$ From Generating Function for Constant Sequence, this has the generating function: :$\displaystyle G \paren z = \sum_{n \mathop = 1}^\infty z^n = \frac 1 {1 - z}$ By Derivative of Generating Function: :$\displaystyle \dfrac \d {\d z} G \paren z = 0 + 1 + 2 z + 3 z^2 \cdots = \sum_{n \mathop \ge 0} \paren {n + 1} z^n$ which is the generating function for the sequence $\sequence {a_n}$. But: : $G \paren z = \dfrac 1 {1 - z}$ and so by Power Rule for Derivatives and the Chain Rule for Derivatives: :$\dfrac \d {\d z} G \paren z = \dfrac 1 {\paren {1 - z}^2}$ The result follows from the definition of a generating function. {{qed}}
Since $S$ is non-empty, we can choose an element $x \in S$. Since $f$ is an injection, for each $t \in \Img f$ there exists a unique $s \in S$ such that: :$\map f s = t$ Thus by Law of Excluded Middle there exists a well-defined mapping $T \to S$ given by: :$\map g t = \begin {cases} s & : \paren {t \in \Img f} \land \paren {\map f s = t} \\ x & : t \notin \Img f \end {cases}$ By construction, for any given $s \in S$, the element $\map f s$ maps to $s$ under $g$. Therefore $g: T \to S$ is a surjection. {{qed}}
By Injection iff Left Inverse, $f$ has a left inverse $g: T \to S$. By Left Inverse Mapping is Surjection, $g$ is a surjection. {{qed}}
Since $T$ is not empty, it has an element $t_0$. Define a mapping $g: S \setminus \Preimg {\mathcal R}$ by letting $\map g x = t_0$ for all $x \in S$. Let $f = \mathcal R \cup g$ be a relation on $S \times T$. By Union of Many-to-One Relations with Disjoint Domains is Many-to-One, $f$ is a many-to-one relation. By Union with Relative Complement, $f$ is total. Thus $f$ is a mapping. {{qed}} {{LEM|Union with Relative Complement}} Category:Relation Theory Category:Mapping Theory ffdavrb9jiakr6m6tolfarfqxxjoglv
{{begin-eqn}} {{eqn | l = \left({1 + z}\right)^m       | r = \sum_{n \mathop = 0}^m \binom m n z^n       | c = Binomial Theorem }} {{end-eqn}} The result follows from the definition of a generating function. {{qed}}
{{begin-eqn}} {{eqn | l=\Pi_X \left({1}\right)       | r=p_X \left({0}\right) + 1^1.p_X \left({1}\right) + 1^2.p_X \left({2}\right) + \cdots       | c=Definition of probability generating function }} {{eqn | r=p_X \left({0}\right) + p_X \left({1}\right) + p_X \left({2}\right) + \cdots       | c=as $\forall n \in \N_{>0}: s^n = 1$ }} {{eqn | r=\sum_{n \mathop = 0}^\infty p_X \left({x}\right)       | c= }} {{eqn | r=\sum_{x \mathop \ge 0} p_X \left({x}\right)       | c=by definition of $\Omega_X$ }} {{eqn | r=1       | c=Definition of Probability Mass Function }} {{end-eqn}} {{qed}}
By definition of probability generating function: :$\displaystyle \Pi_X \left({s}\right) := \sum_{n \mathop = 0}^\infty p_X \left({n}\right) s^n = p_X \left({0}\right) + p_X \left({1}\right) s +  p_X \left({2}\right) s^2 + \cdots$ where $p_X$ is the probability mass function for $X$. For any real function $g: \R \to \R$, by Expectation of Function of Discrete Random Variable: :$\displaystyle E \left({g \left({X}\right)}\right) = \sum_{x \mathop \in \Omega_X} g \left({x}\right) \Pr \left({X = x}\right)$ whenever the sum is absolutely convergent. In this case, $g \left({X}\right) = s^X$. Thus: :$\displaystyle E \left({s^X}\right) = \sum_{x \mathop \in \Omega_X} s^x \Pr \left({X = x}\right)$ or, using $p_X \left({x}\right) := \Pr \left({X = x}\right)$: :$\displaystyle E \left({s^X}\right) = \sum_{x \mathop \in \Omega_X} p_X \left({x}\right) s^x$ By definition of $X$, this can then be expressed as: :$\displaystyle E \left({s^X}\right) = \sum_{n \mathop = 0}^\infty p_X \left({n}\right) s^n$ Thus whenever $\displaystyle \sum_{n \mathop = 0}^\infty p_X \left({n}\right) s^n$ is absolutely convergent: :$\Pi_X \left({s}\right) = E \left({s^X}\right)$ by definition of probability mass function. As $p_X$ is a probability mass function for $X$: :$\displaystyle \sum_{k \mathop = 0}^\infty p_X \left({k}\right) = 1$ Thus the condition on absolute convergence is satisfied by all $s$ such that $\left|{s}\right| < 1$ by the observation: :$\displaystyle \sum_{k \mathop = 0}^\infty \left|{p_X \left({k}\right) s^k}\right| \le \sum_{k \mathop = 0}^\infty p_X \left({k}\right) = 1$ {{qed}}
By the definition of PGF, it follows that if: :$\forall k \in \N: \map \Pr {X = k} = \map \Pr {Y = k}$ then: :$\forall s \in \closedint {-1} 1: \map {\Pi_X} s = \map {\Pi_Y} s$ Suppose that $\map {\Pi_X} s = \map {\Pi_Y} s$ for all $s \in \closedint {-1} 1$. From Probability Generating Function as Expectation the radius of convergence of both $\map {\Pi_X} s$ and $\Pi_Y \left({s}\right)$ is at least $1$. Therefore they have unique power series expansions about $s = 0$: :$\ds \map {\Pi_X} s = \sum_{n \mathop = 0}^\infty s^n \map \Pr {X = k}$ :$\ds \map {\Pi_Y} s = \sum_{n \mathop = 0}^\infty s^n \map \Pr {Y = k}$ As $\map {\Pi_X} s = \map {\Pi_Y} s$, these two power series have identical coefficients. {{handwaving|It needs to be demonstrated that those power series expansions actually are unique. This has not been done.}} Hence the result. {{qed}}
From the definition of p.g.f: :$\displaystyle \Pi_X \left({s}\right) = \sum_{k \mathop \ge 0} p_X \left({k}\right) s^k$ From the definition of the negative binomial distribution (first form): :$\displaystyle p_X \left({k}\right) = \binom {n + k - 1} {n - 1} p^k q^n$ where $q = 1 - p$. So: {{begin-eqn}} {{eqn | l=\Pi_X \left({s}\right)       | r=\sum_{k \mathop \ge 0} \binom {n + k - 1} {n - 1} p^k q^n s^k       | c= }} {{eqn | r=q^n \sum_{k \mathop \ge 0} \binom {n + k - 1} {n - 1} \left({p s}\right)^k       | c= }} {{eqn | r=\left({\frac q {1 - ps} }\right)^n       | c= }} {{end-eqn}} For the third equality, the equation in the second line is rewritten in terms of binomial series. {{explain|Yes all very well, but it's still not trivial.}} Hence the result. {{qed}} Category:Negative Binomial Distribution 3zi186690bgsli7m9f052bhjfk7zdh0
From the definition of p.g.f: :$\displaystyle \Pi_X \left({s}\right) = \sum_{k \mathop \ge 0} p_X \left({k}\right) s^k$ From the definition of the negative binomial distribution (second form): :$\displaystyle p_X \left({k}\right) = \binom {k-1} {n-1} p^n q^{k-n}$ where $q = 1 - p$. So: {{begin-eqn}} {{eqn | l=\Pi_X \left({s}\right)       | r=\sum_{k \mathop \ge n} \binom {k-1} {n-1} p^n q^{k-n} s^k       | c= }} {{eqn | r=\frac {p^n}{q^n} \sum_{k \mathop \ge n} \binom {k-1} {n-1} \left({q s}\right)^k       | c= }} {{eqn | r=\left({\frac {ps} {1 - qs} }\right)^n       | c= }} {{end-eqn}} Hence the result. {{qed}}
By the chain rule: :$\dfrac \d {\d t} \map F {\map \gamma t} = \map {F'} {\map \gamma t} \map {\gamma'} t = \map f {\map \gamma t} \map {\gamma'} t$ Thus: {{begin-eqn}} {{eqn | l = \int_\gamma \map f z \rd z       | r = \int_a^b \map f {\map \gamma t} \map {\gamma'} t \rd t       | c = {{Defof|Complex Primitive|Antiderivative}} }} {{eqn | r = \int_a^b \frac \d {\d t} \map F {\map \gamma t} \rd t       | c =  }} {{eqn | r = \map F {\map \gamma b} - \map F {\map \gamma a}       | c = Fundamental Theorem of Calculus }} {{end-eqn}} {{qed}} Category:Complex Analysis Category:Contour Integration Category:Fundamental Theorems r7gs9g2amaaio8wbkmbby6dbd948e08
From the definition of p.g.f: :$\displaystyle \Pi_X \left({s}\right) = \sum_{k \mathop \ge 0} \Pr \left({X = k}\right) s^k$ By hypothesis: :$\Pr \left({Y = m k}\right) = \Pr \left({X = k}\right)$ Thus: :$\displaystyle \Pi_Y \left({s}\right) = \sum_{m k \mathop \ge 0} \Pr \left({X = k}\right) s^{m k}$ From the definition of a probability generating function: :$\Pi_Y \left({s}\right) = \Pi_X \left({s^m}\right)$ {{qed}}
From the definition of p.g.f: :$\displaystyle \Pi_X \left({s}\right) = \sum_{k \mathop \ge 0} \Pr \left({X = k}\right) s^k$ By hypothesis: :$\Pr \left({Y = k + m}\right) = \Pr \left({X = k}\right)$ Thus: :$\displaystyle \Pi_Y \left({s}\right) = \sum_{k + m \mathop \ge 0} \Pr \left({X = k}\right) s^{k + m}$ {{begin-eqn}} {{eqn | l = \Pi_Y \left({s}\right)       | r = \sum_{k + m \mathop \ge 0} \Pr \left({X = k}\right) s^{k + m}       | c =  }} {{eqn | r = \sum_{k + m \mathop \ge 0} \Pr \left({X = k}\right) s^m s^k       | c =  }} {{eqn | r = s^m \sum_{k \mathop \ge 0} \Pr \left({X = k}\right) s^k       | c = Translation of Index Variable of Summation }} {{end-eqn}} {{finish|The last step needs to be expanded and explained as to why we can just lose the $m$ out of the index.}} From the definition of a probability generating function: :$\Pi_Y \left({s}\right) = s^m \Pi_X \left({s}\right)$ {{qed}}
Proof by induction: For all $n \in \N_{> 0}$, let $P \left({n}\right)$ be the proposition: :$\dfrac {\mathrm d^n} {\mathrm d s^n} \Pi_X \left({1}\right) = E \left({X \left({X - 1}\right) \cdots \left({X - n + 1}\right)}\right)$ === Basis for the Induction === $P \left({1}\right)$ is the case: :$\dfrac {\mathrm d} {\mathrm d s} \Pi_X \left({1}\right) = E \left({X}\right)$ for $n = 1, 2, \ldots$ This is demonstrated in Expectation of Discrete Random Variable from PGF. This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 2$, then it logically follows that $P \left({k+1}\right)$ is true. So this is our induction hypothesis: :$\dfrac {\mathrm d^k} {\mathrm d s^k} \Pi_X \left({1}\right) = E \left({X \left({X - 1}\right) \cdots \left({X - k + 1}\right)}\right)$ Then we need to show: :$\dfrac {\mathrm d^{k+1}} {\mathrm d s^{k+1}} \Pi_X \left({1}\right) = E \left({X \left({X - 1}\right) \cdots \left({X - k}\right)}\right)$ === Induction Step === This is our induction step: {{proof wanted|only the boilerplate in place as yet}} {{begin-eqn}} {{eqn | l=       | o= =       | r=       | c= }} {{eqn | ll=\implies       | l=       | o= =       | r=       | c= }} {{end-eqn}} So $P \left({k}\right) \implies P \left({k+1}\right)$ and the result follows by the Principle of Mathematical Induction. Therefore: :$proposition_n$
From Differentiation of Power Series: :$\displaystyle \frac {\mathrm d^n}{\mathrm dx^n} \sum_{m \mathop \ge 0} a_m \left({x - \xi}\right)^m = \sum_{m \mathop \ge n} a_m m^{\underline n} \left({x - \xi}\right)^{m - n}$ where $m^{\underline n}$ denotes the falling factorial. When $x = \xi$ all the terms in $\left({x - \xi}\right)^{m - n}$ vanish except when $m = n$. When $m = n$, from Nth Derivative of Nth Power we have: : $\displaystyle \frac {\mathrm d^n}{\mathrm dx^n} a_m \left({x - \xi}\right)^m = a_m n!$ But then $m = n$ and the result follows. {{qed}} Category:Differential Calculus Category:Power Series hox2vbo4tplohvtd3ro6c8dcm3iud59
From Derivatives of Probability Generating Function at One: :$\Pi''_X \paren 1 = \expect {X \paren {X - 1} }$ and from Expectation of Discrete Random Variable from PGF: :$\Pi'_X \paren 1 = \expect X$ So: {{begin-eqn}} {{eqn | l = \expect {X^2}       | r = \expect {X \paren {X - 1} + X}       | c = by algebra: $X \paren {X - 1} + X = X^2 - X + X$ }} {{eqn | r = \expect {X \paren {X - 1} } + \expect X       | c = Linearity of Expectation Function }} {{eqn | r = \Pi''_X \paren 1 + \Pi'_X \paren 1       | c = from above }} {{end-eqn}} {{qed}}
{{BeginTableau|p \implies \neg p, \neg p \implies p \vdash \bot}} {{Premise|1|p \implies \neg p}} {{Premise|2|\neg p \implies p}} {{SequentIntro|3|1|\neg p|1|Proof by Contradiction: Variant 3}} {{ModusPonens|4|1, 2|p|2|3}} {{NonContradiction|5|1, 2|4|3}} {{EndTableau}} {{qed}} Category:Contradiction Category:Implication 8oh1tuksbgbdsmsmvv8c4wsanwetqw1
{{BeginTableau|$\vdash \neg \left({p \iff \neg p}\right)$}} {{Assumption|1|p \iff \neg p}} {{BiconditionalElimination|2|1|p \implies \neg p|1|1}} {{BiconditionalElimination|3|1|\neg p \implies p|1|2}} {{SequentIntro|4|1|\bot|2,3|Non-Equivalence of Proposition and Negation: Formulation 1}} {{Contradiction|5||\neg \left({p \iff \neg p}\right)|1|4}} {{EndTableau}} Category:Biconditional Category:Contradiction 6y8o80gajctqmhj9do2tot370fd1goa
{{AimForCont}} there exists such a $B$. Let $i: B \to S$ be an injection. Let $f: B \to \powerset S$ be a surjection. Let $i^\gets: \powerset S \to \powerset B$ be the inverse image mapping of $i$. By Mapping Induced by Inverse of Injection is Surjection, $i^\gets$ is a surjection. Let $f^\to: \powerset B \to \powerset {\powerset S}$ be the direct image mapping of $f$. By Direct Image Mapping of Surjection is Surjection, $f^\to$ is a surjection: We have that $i^\gets: \powerset S \to \powerset B$ and $f^\to: \powerset B \to \powerset {\powerset S}$ are surjective. By Composite of Surjections is Surjection, their composition $f^\to \circ i^\gets: \powerset S \to \powerset {\powerset S}$ is a surjection by Composite of Surjections is Surjection. But this violates Cantor's Theorem, contradicting the assumption that such a $B$ exists. {{qed}}
Suppose for the sake of contradiction that $f: \powerset S \to S$ is an injection. By Injection has Surjective Left Inverse Mapping, there is a surjection $g: S \to \powerset S$. But this contradicts Cantor's Theorem. Thus there can be no such injection.  {{qed}} {{LEM|Injection has Surjective Left Inverse Mapping|3}}
The identity mapping $f: \powerset S \to \powerset S$ is a surjection by Identity Mapping is Surjection. Thus by the lemma, there can be no injection from $\powerset S$ into $S$. {{qed}}
Let $f^\to: \powerset S \to \powerset T$ be the direct image mapping by $f$. Let $X \in \powerset S$. Let $Y = \map {f^\to} X$. By Subset equals Preimage of Image iff Mapping is Injection: :$\map {f^\gets} Y = X$ As such a $Y$ exists for each $X \in \powerset S$, $f^\gets$ is surjective. {{qed}} Category:Injections Category:Surjections Category:Inverse Image Mappings ltk5ardk0aufh7a49pbn3p9w47lqk9x
From Expectation of Poisson Distribution: :$\expect X = \lambda$ From Variance of Poisson Distribution: :$\var X = \lambda$ {{qed}}