avewright/proofwiki-math huggingface dataset:
Theorem: Subset is Compatible with Ordinal Successor

Let $x$ and $y$ be ordinals and let $x^+$ denote the successor set of $x$.
Let $x \in y$.
Then:
:$x^+ \in y^+$

Proof:
First note that by Successor Set of Ordinal is Ordinal, $x^+$ and $y^+$ are ordinals.
Let $x \in y$.
We wish to show that $x^+ \in y^+$.
By Ordinal Membership is Trichotomy:
:Either $x^+ \in y^+$, $y^+ \in x^+$, or $x^+ = y^+$.
{{AimForCont}} $y^+ = x^+$.
Then $y \in x$ or $y = x$ by the definition of successor set.
If $y = x$ then $x \in x$, contradicting the fact that Ordinal is not Element of Itself.
If $y \in x$ then since an ordinal is transitive, $y \in y$, again contradicting Ordinal is not Element of Itself.
Thus $y^+ ≠ x^+$.
{{AimForCont}} $y^+ \in x^+$.
By definition of successor set, $y^+ \in x$ or $y^+ = x$.
If $y^+ \in x$, then since $y^+$ and $x$ are both ordinals, $y^+ \subsetneqq x$.
Then $y \in x$.
Since $y$ is transitive, $y \in y$, contradicting Ordinal is not Element of Itself.
If instead $y^+ = x$, then $x \in y \in y^+ = x$, so the same contradiction arises because $x$ is transitive.
Thus $y^+ \notin x^+$.
So the only remaining possibility, that $x^+ \in y^+$, must hold.
{{qed}}
Theorem: Subset is Compatible with Ordinal Successor

Let $x$ and $y$ be ordinals and let $x^+$ denote the successor set of $x$.
Let $x \in y$.
Then:
:$x^+ \in y^+$

Proof:
First note that by Successor Set of Ordinal is Ordinal, $x^+$ and $y^+$ are ordinals.
By Ordinal Membership is Trichotomy, one of the following must be true:
{{begin-eqn}}
{{eqn | l = x^+
      | r = y^+
}}
{{eqn | l = y^+
      | o = \in
      | r = x^+
}}
{{eqn | l = x^+
      | o = \in
      | r = y^+
}}
{{end-eqn}}
We will show that the first two are both false, so that the third must hold.
Two preliminary facts:
{{begin-axiom}}
{{axiom | n = 1
        | ml= x
        | mo= \ne
        | mr= y
        | rc= $x \in y$ and Ordinal is not Element of Itself
}}
{{axiom | n = 2
        | ml= y
        | mo= \notin
        | mr= x
        | rc= $x \in y$ and Ordinal Membership is Asymmetric
}}
{{end-axiom}}
By $(1)$ and Equality of Successors:
:$x^+ \ne y^+$
Thus the first of the three possibilities is false.
{{AimForCont}} $y^+ \in x^+$.
Then:
{{begin-eqn}}
{{eqn | l = y
      | o = \in
      | r = y^+
      | c = {{Defof|Successor Set}}
}}
{{eqn | ll= \leadsto
      | l = y
      | o = \in
      | r = x^+
      | c = $y^+ \in x^+$ and $x^+$ is an ordinal and therefore transitive.
}}
{{eqn | ll= \leadsto
      | l = y
      | o = \in
      | r = x
      | c = {{Defof|Successor Set}}
}}
{{eqn | lo= \lor
      | l = y
      | r = x
}}
{{end-eqn}}
But we already know that $y \notin x$ by $(2)$ and $y \ne x$ by $(1)$.
So this is a contradiction, and we conclude that $y^+ \notin x^+$.
Thus we have shown that the second possibility is false.
Thus the third and final one must hold: $x^+ \in y^+$.
{{qed}}
Theorem: Successor is Less than Successor/Sufficient Condition/Proof 1

Let $x$ and $y$ be ordinals and let $x^+$ denote the successor set of $x$.
Let $x^+ \in y^+$.
Then:
: $x \in y$

Proof:
Suppose $y^+ \in x^+$.
By the definition of successor, $y^+ \in x \lor y^+ = x$.
Suppose $y^+ = x$.
By Ordinal is Less than Successor, $y \in x$.
Suppose $y^+ \in x$.
By Ordinal is Less than Successor, $y \in y^+$.
By Ordinal is Transitive, $y \in x$.
Theorem: Successor is Less than Successor/Sufficient Condition/Proof 2

Let $x$ and $y$ be ordinals and let $x^+$ denote the successor set of $x$.
Let $x^+ \in y^+$.
Then:
: $x \in y$

Proof:
First note that by Successor Set of Ordinal is Ordinal, $x^+$ and $y^+$ are ordinals.
Let $x^+ \in y^+$.
Then since $y^+$ is transitive, $x^+ \subseteq y^+$.
Thus $x \in y$ or $x = y$.
If $x = y$ then $x^+ \in x^+$, contradicting Ordinal is not Element of Itself.
Thus $x \in y$.
Theorem: Successor is Less than Successor/Sufficient Condition

== Proof 1 ==
{{:Successor is Less than Successor/Sufficient Condition/Proof 1}}

Proof:
Suppose $y^+ \in x^+$.
By the definition of successor, $y^+ \in x \lor y^+ = x$.
Suppose $y^+ = x$.
By Ordinal is Less than Successor, $y \in x$.
Suppose $y^+ \in x$.
By Ordinal is Less than Successor, $y \in y^+$.
By Ordinal is Transitive, $y \in x$.
Theorem: Successor is Less than Successor/Sufficient Condition

== Proof 1 ==
{{:Successor is Less than Successor/Sufficient Condition/Proof 1}}

Proof:
First note that by Successor Set of Ordinal is Ordinal, $x^+$ and $y^+$ are ordinals.
Let $x^+ \in y^+$.
Then since $y^+$ is transitive, $x^+ \subseteq y^+$.
Thus $x \in y$ or $x = y$.
If $x = y$ then $x^+ \in x^+$, contradicting Ordinal is not Element of Itself.
Thus $x \in y$.
Theorem: Probability of Event not Occurring

:$\forall A \in \Sigma: \map \Pr {\Omega \setminus A} = 1 - \map \Pr A$

Proof:
From the conditions for $\Pr$ to be a probability measure, we have:
:$(1): \quad \forall A \in \Sigma: 0 \le \map \Pr A$
:$(2): \quad \map \Pr \Omega = 1$
:$(3): \quad \ds \map \Pr {\bigcup_{i \mathop \ge 1} A_i} = \sum_{i \mathop \ge 1} \map \Pr {A_i}$ where all $A_i$ are pairwise disjoint.
Let $A \in \Sigma$ be an event.
Then $\paren {\Omega \setminus A} \in \Sigma$ by definition of Event Space: Axiom $(ES \ 2)$.
From Intersection with Relative Complement is Empty:
:$A \cap \paren {\Omega \setminus A} = \O$
From Union with Relative Complement:
:$A \cup \paren {\Omega \setminus A} = \Omega$
So:
:$\map \Pr A + \map \Pr {\Omega \setminus A} = 1$
from above, and so:
:$\map \Pr {\Omega \setminus A} = 1 - \map \Pr A$
{{qed}}
Theorem: Delta-Algebra is Sigma-Algebra

Every $\delta$-algebra is a $\sigma$-algebra.

Proof:
Let $\DD$ be a $\delta$-algebra whose unit is $\mathbb U$.
Let $A_1, A_2, \ldots$ be a countably infinite collection of elements of $\DD$.
Then:
{{begin-eqn}}
{{eqn | q = \forall i
      | l = \mathbb U \setminus A_i
      | o = \in
      | r = \DD
      | c = $\DD$ is closed under relative complement with $\mathbb U$
}}
{{eqn | ll= \leadsto
      | l = \bigcap_{i \mathop = 1}^\infty \paren {\mathbb U \setminus A_i}
      | o = \in
      | r = \DD
      | c = $\DD$ is closed under countable intersections
}}
{{eqn | ll= \leadsto
      | l = \mathbb U \setminus \bigcup_{i \mathop = 1}^\infty A_i
      | o = \in
      | r = \DD
      | c = De Morgan's Laws: Complement of Union
}}
{{eqn | ll= \leadsto
      | l = \bigcup_{i \mathop = 1}^\infty A_i
      | o = \in
      | r = \DD
      | c = $\DD$ is closed under relative complement with $\mathbb U$
}}
{{end-eqn}}
Thus $\DD$ is also a $\sigma$-algebra.
{{qed}}
Theorem: Countable Intersection of Events is Event

:$\quad A_1, A_2, \ldots \in \Sigma \implies \ds \bigcap_{i \mathop = 1}^\infty A_i \in \Sigma$

Proof:
By definition, a probability space $\struct {\Omega, \Sigma, \Pr}$ is a measure space.
So, again by definition, an event space $\Sigma$ is a $\sigma$-algebra on $\Omega$.
From Sigma-Algebra is Delta-Algebra:
:$\ds A_1, A_2, \ldots \in \Sigma \implies \bigcap_{i \mathop = 1}^\infty A_i \in \Sigma$
by definition of $\delta$-algebra.
{{qed}}
Theorem: Subset Relation is Ordering/General Result

Let $\mathbb S$ be a set of sets.
Then $\subseteq$ is an ordering on $\mathbb S$.
In other words, let $\struct {\mathbb S, \subseteq}$ be the relational structure defined on $\mathbb S$ by the subset relation $\subseteq$.
Then $\struct {\mathbb S, \subseteq}$ is an ordered set.

Proof:
To establish that $\subseteq$ is an ordering, we need to show that it is reflexive, antisymmetric and transitive.
So, checking in turn each of the criteria for an ordering:
Theorem: Noetherian Domain is Factorization Domain

Let $R$ be a noetherian integral domain.
Then $R$ is a factorization domain.

Proof:
Let $\FF$ be the set of ideals of $R$ of the form $x R$, with $x$ not a unit and such that $x$ cannot be decomposed in the form:
:$x = u p_1 \dotsm p_r$
where $u$ is a unit and $p_1, \dotsc, p_r$ irreducible.
We show by contradiction that $\FF = \O$.
{{AimForCont}} $\FF \ne \O$.
Since $R$ is noetherian, we can choose a maximal element $a R \in \FF$.
By construction, $a$ is not irreducible, so we can write:
:$a = b c$
with $b, c$ non-units and not associates.
By the definition of associates in a commutative ring this means that $b R \subsetneq a R$ and $a R \subsetneq b R$.
Since $a R$ is assumed maximal, this means that $b R$ and $c R$ do not belong to $\FF$.
Therefore there exist units $u, v$ and irreducible elements $p_1, \dotsc, p_r, q_1, \dotsc, q_s$ such that:
:$b = u p_1 \dotsm p_r$
and:
:$c = v q_1 \dotsm q_s$
But this implies that:
:$a = b c = \paren {u v} p_1 \cdots p_r \cdot q_1 \dotsm q_s$
This is a contradiction, since {{hypothesis}} $a$ cannot be written in this form.
{{Qed}}
Category:Ring Theory
Category:Factorization
pwdftf4nlll0jvmkbnklkb54pib727a
Theorem: Probability Measure is Monotone

Let $A, B \in \Sigma$ such that $A \subseteq B$.
Then:
:$\map \Pr A \le \map \Pr B$

Proof:
From Set Difference Union Second Set is Union:
:$A \cup B = \paren {B \setminus A} \cup A$
From Set Difference Intersection with Second Set is Empty Set:
:$\paren {B \setminus A} \cap A = \O$
From the Addition Law of Probability:
:$\map \Pr {A \cup B} = \map \Pr {B \setminus A} + \map \Pr A$
From Union with Superset is Superset:
:$A \subseteq B \implies A \cup B = B$
Thus:
:$\map \Pr B = \map \Pr {B \setminus A} + \map \Pr A$
By definition of probability measure:
:$\map \Pr {B \setminus A} \ge 0$
from which it follows that:
:$\map \Pr B \ge \map \Pr A$
{{qed}}
Theorem: Probability Measure is Monotone

Let $A, B \in \Sigma$ such that $A \subseteq B$.
Then:
:$\map \Pr A \le \map \Pr B$

Proof:
As by definition a probability measure is a measure, we can directly use the result Measure is Monotone.
{{qed}}
Theorem: Transfinite Induction/Principle 1/Proof 2

Let $\On$ denote the class of all ordinals.
Let $A$ denote a class.
Suppose that:
:For each element $x$ of $\On$, if $\forall y \in \On: \paren {y < x \implies y \in A}$ then $x$ is an element of $A$.
Then $\On \subseteq A$.

Proof:
<onlyinclude>
{{AimForCont}} $\neg \On \subseteq A$.
Then:
:$\paren {\On \setminus A} \ne \O$
From Set Difference is Subset, $\On \setminus A$ is a subclass of the ordinals.
By Class of All Ordinals is Well-Ordered by Subset Relation, $\On \setminus A$ must have a smallest element $y$.
Then every strict predecessor of $y$ must lie in $A$, so by the premise, $y$ must also be an element of $A$.
This contradicts the fact that $y$ is an element of $\On \setminus A$.
Therefore $\On \subseteq A$.
{{qed}}
{{LEM|Reductio ad Absurdum|4}}
Category:Transfinite Induction
7g74p8tq95kw7avuvsggqard03hcaq3
Theorem: Immediate Successor is Unique in Toset

Let $(S, \preceq)$ be a totally ordered set.
Let $x, y \in S$.
Suppose that $y$ is an immediate successor of $x$.
Then $y$ is the ''only'' immediate successor of $x$.

Proof:
{{AimForCont}} that $x$ has another immediate successor, $z ≠ y$.
Then by definition:
:$x \prec y$
:$x \prec z$
Since $\preceq$ is a total ordering and $y ≠ z$:
:$y \prec z$ or $z \prec y$.
If $y \prec z$ then $x \prec y \prec z$, contradicting the fact that $z$ is an immediate successor of $x$.
If $z \prec y$ then $x \prec z \prec y$, contradicting the fact that $y$ is an immediate successor of $x$.
Thus we conclude that there can be no such $z$.
{{qed}}
Theorem: Mapping from Set to Class of All Ordinals is Bounded Above

Let $x$ be a set.
Let $\On$ be the class of all ordinals.
Let $f: x \to \On$ be a mapping.
Then $f$ has an upper bound.

Proof:
Let $I$ be the image of $f$.
By {{Corollary|Union of Set of Ordinals is Ordinal}}, $\bigcup I$ is an ordinal.
But by Union is Smallest Superset, each element of $I$ is a subset of $\bigcup I$.
Thus $\bigcup I$ is an upper bound of $f$.
{{qed}}
Category:Class of All Ordinals
Category:Class Mappings
dn066mvr3jb0qbefutzuphpvq5nrvdl
Theorem: Equivalence of Definitions of Independent Events

Let $\EE$ be an experiment with probability space $\struct {\Omega, \Sigma, \Pr}$.
Let $A, B \in \Sigma$ be events of $\EE$ such that $\map \Pr A > 0$ and $\map \Pr B > 0$.
{{TFAE|def = Independent Events}}

Proof:
{{begin-eqn}}
{{eqn | l = \condprob A B
      | r = \map \Pr A
      | c = 
}}
{{eqn | ll= \leadstoandfrom
      | l = \dfrac {\map \Pr {A \cap B} } {\map \Pr B}
      | r = \map \Pr A
      | c = {{Defof|Conditional Probability}}
}}
{{eqn | ll= \leadstoandfrom
      | l = \map \Pr {A \cap B}
      | r = \map \Pr A \, \map \Pr B
      | c = which is valid, as $\map \Pr B > 0$
}}
{{end-eqn}}
{{qed}}
Theorem: Independent Events are Independent of Complement/General Result

Let $A_1, A_2, \ldots, A_m$ be events in a probability space $\struct {\Omega, \Sigma, \Pr}$.
Then $A_1, A_2, \ldots, A_m$ are independent {{iff}} $\Omega \setminus A_1, \Omega \setminus A_2, \ldots, \Omega \setminus A_m$ are also independent.

Proof:
Proof by induction:
For all $n \in \N: n \ge 2$, let $\map P n$ be the proposition:
:$A_1, A_2, \ldots, A_n$ are independent {{iff}} $\Omega \setminus A_1, \Omega \setminus A_2, \ldots, \Omega \setminus A_n$ are independent.
Theorem: Probability of Independent Events Not Happening/Corollary

Let $A$ be an event in an event space of an experiment $\EE$ whose probability space is $\struct {\Omega, \Sigma, \Pr}$.
Let $\map \Pr A = p$.
Suppose that the nature of $\EE$ is that its outcome is independent of previous trials of $\EE$.
Then the probability that $A$ does not occur during the course of $m$ trials of $\EE$ is $\paren {1 - p}^m$.

Proof:
This is an instance of Probability of Independent Events Not Happening with all of $A_1, A_2, \ldots, A_m$ being instances of $A$.
The result follows directly.
{{qed}}
Category:Probability Theory
jfor173v5hwwv46csmqgmacah73w0e4
Theorem: Probability of Limit of Sequence of Events/Increasing

Let $\sequence {A_n}_{n \mathop \in \N}$ be an increasing sequence of events.
Let $\ds A = \bigcup_{i \mathop \in \N} A_i$ be the limit of $\sequence {A_n}_{n \mathop \in \N}$.
Then:
:$\ds \map \Pr A = \lim_{n \mathop \to \infty} \map \Pr {A_n}$

Proof:
Let $\ds B_i = A_i \setminus A_{i - 1}$ for $i \in \N: i > 0$.
Then:
:$A = A_0 \cup B_1 \cup B_2 \cup \cdots$
is the union of disjoint events in $\Sigma$.
By definition of probability measure:
{{begin-eqn}}
{{eqn | l = \map \Pr A
      | r = \map \Pr {A_0} + \map \Pr {B_1} + \map \Pr {B_2} + \cdots
      | c = 
}}
{{eqn | r = \map \Pr {A_0} + \lim_{n \mathop \to \infty} \sum_{k \mathop = 1}^n \map \Pr {B_k}
      | c = 
}}
{{end-eqn}}
But we have:
:$\map \Pr {B_i} = \map \Pr {A_i} - \map \Pr {A_{i - 1} }$ for $i \in \N: i > 0$.
So:
:$\ds \map \Pr A = \map \Pr {A_0} + \lim_{n \mathop \to \infty} \sum_{k \mathop = 1}^n \paren {\map \Pr {A_i} - \map \Pr {A_{i - 1} } }$
The last sum telescopes.
Hence the result:
:$\ds \map \Pr A = \lim_{n \mathop \to \infty} \map \Pr {A_n}$
{{qed}}
Theorem: Probability of Limit of Sequence of Events/Decreasing

Let $\sequence {B_n}_{n \mathop \in \N}$ be a decreasing sequence of events.
Let $\ds B = \bigcap_{i \mathop \in \N} B_i$ be the limit of $\sequence {B_n}_{n \mathop \in \N}$.
Then:
:$\ds \map \Pr B = \lim_{n \mathop \to \infty} \map \Pr {B_n}$

Proof:
{{expand|The below needs to be done properly.}}
Set $A_i = \Omega \setminus B_i$ and then apply De Morgan's laws and the result for an increasing sequence of events.
{{qed}}
Theorem: Sum of Discrete Random Variables

Let $U: \Omega \to \R$ be defined as:
:$\forall \omega \in \Omega: \map U \omega = \map X \omega + \map Y \omega$
Then $U$ is also a discrete random variable on $\struct {\Omega, \Sigma, \Pr}$.

Proof:
To show that $U$ is discrete random variable on $\struct {\Omega, \Sigma, \Pr}$, we need to show that:
:$(1): \quad$ The image of $U$ is a countable subset of $\R$;
:$(2): \quad \forall x \in \R: \set {\omega \in \Omega: \map U \omega = x} \in \Sigma$.
First we consider any $U_u = \set {\omega \in \Omega: \map U \omega = u}$ such that $U_u \ne \O$.
We have that $U_u = \set {\omega \in \Omega: \map X \omega + \map Y \omega = u}$.
Consider any $\omega \in U_u$.
Then:
:$\omega \in X_x \cap Y_x$
where:
:$X_x = \set{\omega \in \Omega: \map X \omega = x}, Y_x = \set {\omega \in \Omega: \map Y \omega = u - x}$
Because $X$ and $Y$ are discrete random variables, both $X_x \in \Sigma$ and $Y_x \in \Sigma$.
As $\struct {\Omega, \Sigma, \Pr}$ is a probability space, then $X_x \cap Y_x \in \Sigma$.
Now note that:
:$\ds U_u = \bigcup_{x \mathop \in \R} \paren {X_x \cap Y_x}$
That is, it is the union of all such intersections of sets whose discrete random variables add up to $u$.
As $X_x$ is a countable set it follows that $U_u$ 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 $\struct {\Omega, \Sigma, \Pr}$ being a probability space, $U_u \in \Sigma$.
Thus $U$ is a discrete random variables on $\struct {\Omega, \Sigma, \Pr}$.
{{qed}}
{{explain|Might need to be more rigorous on the boundary instances.}}
{{proofread}}
Theorem: Product of Discrete Random Variables

Let $V: \Omega \to \R$ be defined as:
:$\forall \omega \in \Omega: \map V \omega = \map X \omega \map Y \omega$
Then $V$ is also a discrete random variable on $\struct {\Omega, \Sigma, \Pr}$.

Proof:
To show that $V$ is a discrete random variable on $\struct {\Omega, \Sigma, \Pr}$, we need to show that:
:$(1): \quad $The image of $U$ and $V$ are countable subsets of $\R$;
:$(2): \quad \forall x \in \R: \set {\omega \in \Omega: \map V \omega = x} \in \Sigma$.
First we consider any $V_V = \set {\omega \in \Omega: \map V \omega = v}$ such that $V_v \ne \O$.
We have that:
: $V_v = \set {\omega \in \Omega: \map X \omega \map Y \omega = v}$
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:
: $X_x = \set {\omega \in \Omega: \map X \omega = x}, Y_x = \set {\omega \in \Omega: \map Y \omega = \dfrac v x}$
Both $X_x \in \Sigma$ and $Y_x \in \Sigma$ (as $X$ and $Y$ are discrete random variables.
As $\struct {\Omega, \Sigma, \Pr}$ is a probability space then $X_x \cap Y_x \in \Sigma$.
Now note that $\ds V_v = \bigcup_{x \mathop \in \R} \paren {X_x \cap Y_x}$.
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 $\struct {\Omega, \Sigma, \Pr}$ being a probability space, $V_v \in \Sigma$.
Thus $V$ is a discrete random variables on $\struct {\Omega, \Sigma, \Pr}$.
{{qed}}
{{explain|Might need to be more rigorous on the boundary instances.}}
{{proofread}}
Theorem: Bernoulli Process as Geometric Distribution/Shifted

Let $\sequence {Y_i}$ be a binomial experiment with parameter $p$.
Let $\EE$ be the experiment which consists of performing the Bernoulli trial $Y_i$ as many times as it takes to achieve a success, and then stop.
Let $k$ be the number of Bernoulli trials to achieve a success.
Then $k$ is modelled by a shifted geometric distribution with parameter $p$.

Proof:
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:
:$\map \Pr {Y = k} = \paren {1 - p}^{k - 1} p$
Hence the result.
{{Qed}}
Theorem: Probability Mass Function of Negative Binomial Distribution/Type 2

Let $X$ be a discrete random variable on a probability space $\struct {\Omega, \Sigma, \Pr}$.
Let $X$ have the '''type $2$ negative binomial distribution with parameters $r$ and $p$'''.
Then the probability mass function of $X$ is given by:
:$\map \Pr {X = k} = \dbinom {k + r - 1} {r - 1} p^k \paren {1 - p}^r$
where:
:$\Img X = \set {0, 1, 2, \ldots}$
:$0 < p < 1$.

Proof:
{{Recall|Negative Binomial Distribution (Type 2)}}
{{:Definition:Negative Binomial Distribution (Type 2)}}
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 $k$, then there must have been $k + r$ trials altogether.
So, after $r + k - 1$ trials, there must have been $r - 1$ failures, as (from the description of the experiment) the last trial is a failure.
So the probability of the occurrence of the event $\sqbrk {X = k}$ is given by the binomial distribution, as follows:
:$\map {p_X} k = \dbinom {k + r - 1} {r - 1} p^k \paren {1 - p}^r$
where $k \in \set {0, 1, 2, \ldots}$
Hence the result, by definition of probability mass function.
{{qed}}
Theorem: Geometric Distribution Gives Rise to Probability Mass Function/Shifted

Let $Y$ be a discrete random variable on a probability space $\struct {\Omega, \Sigma, \Pr}$.
Let $Y$ have the shifted geometric distribution with parameter $p$ (where $0 < p < 1$).
Then $Y$ gives rise to a probability mass function.

Proof:
By definition:
:$\map \Omega Y = \N_{>0} = \set {1, 2, 3, \ldots}$
:$\map \Pr {Y = k} = p \paren {1 - p}^{k - 1}$
Then:
{{begin-eqn}}
{{eqn | l = \map \Pr \Omega
      | r = \sum_{k \mathop \ge 1} p \paren {1 - p}^{k - 1}
      | c = {{Defof|Shifted Geometric Distribution}}
}}
{{eqn | r = \sum_{j \mathop \ge 0} p \paren {1 - p}^j
      | c = putting $j = k - 1$
}}
{{eqn | r = 1
      | c = Geometric Distribution Gives Rise to Probability Mass Function
}}
{{end-eqn}}
So $Y$ satisfies $\map \Pr \Omega = 1$, and hence the result.
{{Qed}}
Theorem: Probability Mass Function of Negative Binomial Distribution/Type 1

Let $X$ be a discrete random variable on a probability space $\struct {\Omega, \Sigma, \Pr}$.
Let $X$ have the '''type $1$ negative binomial distribution with parameters $r$ and $p$'''..
Then the probability mass function of $X$ is given by:
:$\map \Pr {X = k} = \dbinom {k - 1} {r - 1} p^r \paren {1 - p}^{k - r}$
where:
:$0 < p < 1$
:$\Img X = \set {r, r + 1, r + 2, \dotsc}$

Proof:
{{Recall|Negative Binomial Distribution (Type 1)}}
{{:Definition:Negative Binomial Distribution (Type 1)}}
First note that the number of Bernoulli trials has to be at least $r$, so the image is correct: $\Img X = \set {r, r + 1, r + 2, \ldots}$.
Now, note that if $X$ takes the value $k$, then in the first $k - 1$ trials there must have been $r - 1$ successes.
Hence there must have been $k - r$ failures, and so a success happens at trial number $k$.
So the probability of the occurrence of the event $\sqbrk {X = k}$ is given by the binomial distribution, as follows:
:$\map {p_X} k = \dbinom {k - 1} {r - 1} p^r \paren {1 - p}^{k - r}$
where $k \in \set {r, r + 1, r + 2, \ldots}$
Hence the result, by definition of probability mass function.
{{qed}}
Theorem: Negative Binomial Distribution Gives Rise to Probability Mass Function/Type 2

Let $X$ have the type $2$ negative binomial distribution with parameters $n$ and $p$ ($0 < p < 1$).
Then $X$ gives rise to a probability mass function.

Proof:
By definition:
:$\Img X = \set {0, 1, 2, \ldots}$
:$\map \Pr {X = k} = \dbinom {n + k - 1} {n - 1} p^k \paren {1 - p}^n$
Then:
{{begin-eqn}}
{{eqn | l = \map \Pr \Omega
      | r = \sum_{k \mathop \ge n} \binom {n + k - 1} {n - 1} p^k \paren {1 - p}^n
      | c = 
}}
{{eqn | r = \paren {1 - p}^n \sum_{k \mathop \ge n} \binom {n + k - 1} k p^k
      | c = Symmetry Rule for Binomial Coefficients
}}
{{eqn | r = \paren {1 - p}^n \sum_{j \mathop \ge 0} \binom {-n} k p^k
      | c = Negated Upper Index of Binomial Coefficient
}}
{{eqn | r = \paren {1 - p}^n p^{-n}
      | c = Binomial Theorem
}}
{{eqn | r = 1
      | c = 
}}
{{end-eqn}}
So $X$ satisfies $\map \Pr \Omega = 1$, and hence the result.
{{Qed}}
Category:Negative Binomial Distribution (Type 2)
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