nyuuzyou/wikis huggingface dataset:
Probability Generating Function 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 p.g.f. of $X$ is:
:$\ds \map {\Pi_X} s = \paren {\frac {p s} {1 - q s} }^n$
where $q = 1 - p$.
== Proof ==
From the definition of p.g.f:
:$\ds \map {\Pi_X} s = \sum_{k \mathop \ge 0} \map {p_X} k s^k$
From the definition of the type $1$ negative binomial distribution:
:$\map {p_X} k = \dbinom {k - 1} {n - 1} p^n q^{k - n}$
where $q = 1 - p$.
So:
Hence the result.
== Also see ==
* Probability Generating Function of Negative Binomial Distribution (Type 2)
== Sources ==
* : $\S 4.2$: Integer-valued random variables: $(13)$
Category:Probability Generating Function of Negative Binomial Distribution
Category:Negative Binomial Distribution (Type 1)
Category:Probability Generating Functions
Probability Generating Function of Negative Binomial Distribution/Type 2

== Theorem ==
Let $X$ be a discrete random variable with the type $2$ negative binomial distribution with parameters $n$ and $p$.
Then the p.g.f. of $X$ is:
:$\map {\Pi_X} s = \paren {\dfrac q {1 - p s} }^n$
where $q = 1 - p$.
== Proof ==
From the definition of p.g.f:
:$\ds \map {\Pi_X} s = \sum_{k \mathop \ge 0} \map {p_X} k s^k$
From the definition of the type $2$ negative binomial distribution:
:$\map {p_X} k = \dbinom {n + k - 1} {n - 1} p^k q^n$
where $q = 1 - p$.
So:
For the third equality, the equation in the second line is rewritten in terms of binomial series.
Hence the result.
== Also see ==
* Probability Generating Function of Negative Binomial Distribution (Type 1)
Category:Probability Generating Function of Negative Binomial Distribution
Category:Negative Binomial Distribution (Type 2)
Category:Probability Generating Functions
Probability Generating Function of One

== Theorem ==
Let $X$ be a discrete random variable whose codomain, $\Omega_X$, is a subset of the natural numbers $\N$.
Let $p_X$ be the probability mass function for $X$.
Let $\map {\Pi_X} s$ be the probability generating function for $X$.
Then:
:$\map {\Pi_X} 1 = 1$
== Proof ==
== Sources ==
* : $\S 4.2$: Integer-valued random variables: $(6)$
Category:Probability Generating Functions
Probability Generating Function of Poisson Distribution

== Theorem ==
Let $X$ be a discrete random variable with the Poisson distribution with parameter $\lambda$.
Then the p.g.f. of $X$ is:
:$\map {\Pi_X} s = e^{-\lambda \paren {1 - s} }$
== Proof ==
From the definition of p.g.f:
:$\ds \map {\Pi_X} s = \sum_{x \mathop \ge 0} \map {p_X} x s^x$
From the definition of the Poisson distribution:
:$\ds \forall k \in \N, k \ge 0: \map {p_X} k = \frac {e^{-\lambda} \lambda^k} {k!}$
So:
Hence the result.
== Sources ==
* : $\S 4.2$: Integer-valued random variables: $(12)$
Category:Probability Generating Functions
Category:Poisson Distribution
Probability Generating Function of Scalar Multiple of Random Variable

== Theorem ==
Let $X$ be a discrete random variable whose probability generating function is $\map {\Pi_X} s$.
Let $k \in \Z_{\ge 0}$ be a positive integer.
Let $Y$ be a discrete random variable such that $Y = m X$.
Then
:$\map {\Pi_Y} s = \map {\Pi_X} {s^m}$
where $\map {\Pi_Y} s$ is the probability generating function of $Y$.
== Proof ==
From the definition of p.g.f:
:$\ds \map {\Pi_X} s = \sum_{k \mathop \ge 0} \map \Pr {X = k} s^k$
We have :
:$\map \Pr {Y = m k} = \map \Pr {X = k}$
Thus:
:$\ds \map {\Pi_Y} s = \sum_{m k \mathop \ge 0} \map \Pr {X = k} s^{m k}$
From the definition of a probability generating function:
:$\map {\Pi_Y} s = \map {\Pi_X} {s^m}$
== Sources ==
* : $\S 4.2$: Integer-valued random variables: Exercise $3$
Category:Probability Generating Functions
Probability Generating Function of Shifted Geometric Distribution

== Theorem ==
Let $X$ be a discrete random variable with the shifted geometric distribution with parameter $p$.
Then the p.g.f. of $X$ is:
:$\map {\Pi_X} s = \dfrac {p s} {1 - q s}$
where $q = 1 - p$.
== Proof ==
From the definition of p.g.f:
:$\ds \map {\Pi_X} s = \sum_{x \mathop \ge 0} \map {p_X} x s^x$
From the definition of the shifted geometric distribution:
:$\forall k \in \N, k \ge 1: \map {p_X} k = p q^{k - 1}$
So:
Hence the result.
== Sources ==
* : $\S 4.2$: Integer-valued random variables: Example $9$
Category:Probability Generating Functions
Category:Geometric Distribution
Probability Generating Function of Shifted Random Variable

== Theorem ==
Let $X$ be a discrete random variable whose probability generating function is $\map {\Pi_X} s$.
Let $k \in \Z_{\ge 0}$ be a positive integer.
Let $Y$ be a discrete random variable such that $Y = X + m$.
Then
:$\map {\Pi_Y} s = s^m \map {\Pi_X} s$
where $\map {\Pi_Y} s$ is the probability generating function of $Y$.
== Proof ==
From the definition of p.g.f:
:$\ds \map {\Pi_X} s = \sum_{k \mathop \ge 0} \map \Pr {X = k} s^k$
By hypothesis:
:$\map \Pr {Y = k + m} = \map \Pr {X = k}$
Thus:
From the definition of a probability generating function:
:$\map {\Pi_Y} s = s^m \map {\Pi_X} s$
== Sources ==
* : $\S 4.2$: Integer-valued random variables: Exercise $3$
Category:Probability Generating Functions
Probability Generating Function of Zero

== Theorem ==
Let $X$ be a discrete random variable whose codomain, $\Omega_X$, is a subset of the natural numbers $\N$.
Let $p_X$ be the probability mass function for $X$.
Let $\map {\Pi_X} s$ be the probability generating function for $X$.
Then:
:$\map {\Pi_X} 0 = \map {p_X} 0$
== Proof ==
== Sources ==
* : $\S 4.2$: Integer-valued random variables: $(6)$
Category:Probability Generating Functions
Probability Mass Function/Examples/Arbitrary Example 1

== Example of Probability Mass Function ==
Consider a population consisting of children the state of whose teeth is being monitored.
The following table consists of a count of the number of teeth with dental caries in a group of $100$ schoolchildren:
$\quad \begin {array} {|l|l|}
\hline
\text {Number of Teeth} & \text {Number of Children} \\
\hline
0 & 53 \\
1 & 29 \\
2 & 14 \\
3 & 1 \\
4 & 3 \\
\hline
\end {array}$
The values of the probability mass function, in this case better referred to as a relative frequency function, are:
== Sources ==
* 
* 
Category:Examples of Probability Mass Functions
Probability Mass Function of Binomial Distribution

== Theorem ==
The probability mass function (pmf) of a binomially distributed random variable $X$ is equal to:
:$\map \Pr {X = x} = \dbinom n x p^x \paren {1 - p}^{n - x}$
where $n$ is the number of trials and $p$ is the probability of success.
== Proof ==
Let $B_i: i = 1, 2, \ldots, \dbinom n x$ be events such that:
:$(1): \quad B_i$ is the $i$th possible way to see $x$ successes in $n$ Bernoulli trials
:$(2): \quad \forall i \ne j: B_i \cap B_j = \O$
We can see that:
:$\forall i: \map \Pr {B_i} = p^x \paren {1 - p}^{n - x}$
This is true since there will be $x$ successes, each with probability $p$ of occurring, and $n - x$ failures each with probability $1 - p$ of occurring.
Furthermore we can assume independent trials and thus the result follows.
See Bernoulli Process as Binomial Distribution for further analysis of this.
Now our task becomes finding:
:$\ds \map \Pr {X = x} = \map \Pr {\bigcup_{i \mathop = 1}^{\binom n x} B_i}$
which is the probability of one of the $\dbinom n x$ outcomes occurring.
Then by the Inclusion-Exclusion Principle considered as an extension of the Addition Law of Probability we have that for any countable union of events:
:$\ds \map \Pr {\bigcup_{i \mathop = 1}^n A_i} = \sum_{i \mathop = 1}^n \map \Pr {A_i} - \sum_{i \mathop \ne j: i, j \mathop = 1}^n \map \Pr {A_i \cap A_j} - \map \Pr {\bigcap_{i \mathop = 1}^n A_i}$
Fortunately in this case the above reduces to:
:$\ds \map \Pr {\bigcup_{i \mathop = 1}^n A_i} = \sum_{i \mathop = 1}^n \map \Pr {A_i}$
since the events are pairwise disjoint and $\map \Pr \O = 0$.
Thus:
== Sources ==
* 
* 
* : Appendix $13$: Probability distributions
* : Appendix $15$: Probability distributions
Category:Binomial Distribution
Category:Examples of Probability Mass Functions
Probability Mass Function of Function of Discrete Random Variable

== Theorem ==
Let $X$ be a discrete random variable.
Let $Y = \map g X$, where $g: \R \to \R$ is a real function.
Then the probability mass function of $Y$ is given by:
:$\ds \map {p_Y} y = \sum_{x \mathop \in \inv g y} \map \Pr {X = x}$
== Proof ==
By Function of Discrete Random Variable‎ we have that $Y$ is itself a discrete random variable.
Thus:
== Sources ==
* : $\S 2.3$: Functions of discrete random variables: $(18)$
Category:Probability Mass Functions
Probability Mass Function of Hypergeometric Distribution

== Theorem ==
The probability mass function of a hypergeometric distribution on a discrete random variable $X$ is given by:
:$\map \Pr {X = r} = \dfrac {\dbinom M r \dbinom N {n - r} } {\dbinom {M + N} r}$
where:
:$n$ is the number of trials
:$M + N$ is the total population, consisting of $M$ satisfactory units and $N$ unsatisfactory units
:$r$ is the number of satisfactory units drawn from the population without replacement.
== Proof ==
== Sources ==
* 
* 
Category:Hypergeometric Distribution
Category:Examples of Probability Mass Functions
Probability Mass Function of Negative Binomial Distribution/Type 1

== Theorem ==
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 ==
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.
=== Also presented as ===
== Also defined as ==
== Also see ==
* Negative Binomial Distribution (Type 1) as Generalized Geometric Distribution
* Negative Binomial Distribution (Type 1) Gives Rise to Probability Mass Function
* Probability Mass Function of Negative Binomial Distribution (Type 2)
== Sources ==
* : $\S 2.2$: Examples: $(10)$
* 
* : Appendix $13$: Probability distributions
* : Appendix $15$: Probability distributions
Category:Probability Mass Function of Negative Binomial Distribution
Category:Negative Binomial Distribution (Type 1)
Category:Probability Mass Functions
Probability Mass Function of Negative Binomial Distribution/Type 2

== Theorem ==
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 ==
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.
=== Also presented as ===
== Also defined as ==
== Also see ==
* Negative Binomial Distribution (Type 2) as Generalized Geometric Distribution‎
* Negative Binomial Distribution (Type 2) Gives Rise to Probability Mass Function
* Probability Mass Function of Negative Binomial Distribution (Type 1)
== Technical Note ==
Category:Probability Mass Function of Negative Binomial Distribution
Category:Negative Binomial Distribution (Type 2)
Category:Probability Mass Functions
Probability Mass Function of Negative Binomial Distribution (Type 1)/Also defined as

== Theorem ==
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 $p$, defined as number of number of Bernoulli trials before the $r$th failure has occurred.
Then the probability mass function of $X$ is given by:
:$\map \Pr {X = k} = \dbinom {k + r - 1} {r - 1} p^r \paren {1 - p}^k$
where:
:$\Img X = \set {0, 1, 2, \ldots}$
:$0 < p < 1$.
Then the probability mass function of $X$ is given by:
:$\map \Pr {X = k} = \dbinom {k - 1} {r - 1} p^{k - r} \paren {1 - p}^r$
where:
:$0 < p < 1$
:$\Img X = \set {r, r + 1, r + 2, \dotsc}$
== Proof ==
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 $r - 1$ failures.
Hence there must have been $k - r$ successes, and so a failure 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:
:$\map {p_X} k = \dbinom {k - 1} {r - 1} p^{k - r} \paren {1 - p}^r$
where $k \in \set {r, r + 1, r + 2, \ldots}$
Hence the result, by definition of probability mass function.
== Technical Note ==
Category:Probability Mass Function of Negative Binomial Distribution
Category:Negative Binomial Distribution (Type 1)
Category:Probability Mass Functions
Probability Mass Function of Negative Binomial Distribution (Type 1)/Also presented as

== Probability Mass Function of Negative Binomial Distribution (Type 1): Also presented as == 
The probability mass function of the type $1$ negative binomial distribution with parameters $r$ and $p$ can also be presented as:
:$\map \Pr {X = k} = \dbinom {k - 1} {k - r} p^k \paren {1 - p}^r$
It can also be seen in the form:
:$\map \Pr {X = k} = \dbinom {k - 1} {r - 1} p^r q^{k - r}$
or:
:$\map \Pr {X = k} = \dbinom {k - 1} {k - r} p^r q^{k - r}$
where $q = 1 - p$.
== Proof ==
We have that:
:$\paren {k - 1} - \paren {r - 1} = k - r$
Hence from Symmetry Rule for Binomial Coefficients:
:$\dbinom {k - 1} {k - r} = \dbinom {k - 1} {r - 1}$
Category:Probability Mass Function of Negative Binomial Distribution
Probability Mass Function of Negative Binomial Distribution (Type 2)/Also defined as

== Theorem ==
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 $p$, defined as the number of failures before $r$ successes have occurred.
Then the probability mass function of $X$ is given by:
:$\map \Pr {X = k} = \dbinom {k + r - 1} {r - 1} p^r \paren {1 - p}^k$
where:
:$\Img X = \set {0, 1, 2, \ldots}$
:$0 < p < 1$.
== Proof ==
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 $r - 1$ successes, as (from the description of the experiment) the last trial is a success.
So the probability of the occurrence of the event $\sqbrk {X = k}$ is given by the binomial distribution, as follows:
:$\map {p_X} k = \dbinom {k + r - 1} {r - 1} p^r \paren {1 - p}^k$
where $k \in \set {0, 1, 2, \ldots}$
Hence the result, by definition of probability mass function.
== Also see ==
* Negative Binomial Distribution (Type 2) as Generalized Geometric Distribution‎
* Negative Binomial Distribution (Type 2) Gives Rise to Probability Mass Function
* Probability Mass Function of Negative Binomial Distribution (Type 1)
== Sources ==
* 
* 
Category:Probability Mass Function of Negative Binomial Distribution
Category:Negative Binomial Distribution (Type 2)
Category:Probability Mass Functions
Probability Mass Function of Negative Binomial Distribution (Type 2)/Also presented as

== Probability Mass Function of Negative Binomial Distribution (Type 2): Also presented as == 
The probability mass function of the type $2$ negative binomial distribution with parameters $r$ and $p$ can also be presented as:
:$\map \Pr {X = k} = \dbinom {k + r - 1} k p^k \paren {1 - p}^r$
It can also be seen in the form:
:$\map \Pr {X = k} = \dbinom {k + r - 1} {r - 1} p^k q^r$
or:
:$\map \Pr {X = k} = \dbinom {k + r - 1} k p^k q^r$
where $q = 1 - p$.
== Proof ==
We have that:
:$\paren {k + r - 1} - \paren {r - 1} = k$
Hence from Symmetry Rule for Binomial Coefficients:
:$\dbinom {k + r - 1} k = \dbinom {k + r - 1} {r - 1}$
Category:Probability Mass Function of Negative Binomial Distribution
Probability Measure is Monotone

== Theorem ==
Let $\EE$ be an experiment with probability space $\struct {\Omega, \Sigma, \Pr}$.
Let $A, B \in \Sigma$ such that $A \subseteq B$.
Then:
:$\map \Pr A \le \map \Pr B$
where $\map \Pr A$ denotes the probability of event $A$ occurring.
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Elementary Properties of Probability Measure
Category:Probability Theory
Category:Probability Measure is Monotone
Probability Measure is Monotone/Proof 1

== Theorem ==
== 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:
:$\map \Pr B \ge \map \Pr A$
== Sources ==
* : $1$: Events and probabilities: $1.4$: Probability spaces: $(16)$
Category:Probability Measure is Monotone
Probability Measure on Equiprobable Outcomes

== Theorem ==
Let $\struct {\Omega, \Sigma, \Pr}$ be an equiprobability space.
Let $\card \Omega = n$.
Then:
:$\forall \omega \in \Omega: \map \Pr \omega = \dfrac 1 n$
:$\forall A \subseteq \Omega: \map \Pr A = \dfrac {\card A} n$.
 
== Proof ==
By definition, $\map \Pr {\omega_i} = \map \Pr {\omega_j}$ for all $\omega_i, \omega_j \in \Omega$.
So let $\map \Pr {\omega_i} = p$.
Also, by definition of probability measure, we have:
:$\map \Pr \Omega = 1$
We have that $\set {\omega_i} \cap \set {\omega_j} = \O$ when $i \ne j$ and so, by definition of definition of probability measure:
:$\map \Pr {\set {\omega_i} \cup \set {\omega_j} } = \map \Pr {\set {\omega_i} } + \map \Pr {\set {\omega_j} }$
Using the fact that $\ds \Omega = \bigcup_{i \mathop = 1}^n \set {\omega_i}$:
:$\ds \map \Pr \Omega = \sum_{i \mathop = 1}^n \map \Pr {\set {\omega_i} } = \sum_{i \mathop = 1}^n p = n p$
But we have that $\map \Pr \Omega = 1$, and so $1 = n p$.
Hence:
:$\forall \omega \in \Omega: \map \Pr \omega = \dfrac 1 n$
Now consider $A \subseteq \Omega$.
Let the cardinality of $A$ be $k$, that is $\card A = k$.
Thus:
:$\ds \map \Pr A = \sum_i \map \Pr {\omega_i} \sqbrk {\omega_i \in A}$
where $\sqbrk {\omega_i \in A}$ uses Iverson's convention.
Hence:
:$\map \Pr A = k p$
and so:
:$\map \Pr A = \dfrac {\card A} n$
Category:Probability Theory
Probability Measure on Finite Sample Space

== Theorem ==
Let $\Omega = \set {\omega_1, \omega_2, \ldots, \omega_n}$ be a finite set.
Let $\Sigma$ be a $\sigma$-algebra on $\Omega$.
Let $p_1, p_2, \ldots, p_n$ be non-negative real numbers such that:
:$p_1 + p_2 + \cdots + p_n = 1$
Let $Q: \Sigma \to \R$ be the mapping defined as:
:$\forall A \in \Sigma: \map Q A = \ds \sum_{i: \omega_i \in A} p_i$
Then $\struct {\Omega, \Sigma, Q}$ constitutes a probability space.
That is, $Q$ is a probability measure on $\struct {\Omega, \Sigma}$.
=== Corollary ===
== Proof ==
Recall the Kolmogorov axioms:
First we determine that $\Pr$ as defined is actually a probability measure.
By definition, we have that $\map \Pr A$ is the sum of some subset of $\set {p_1, p_2, \ldots, p_n}$.
Thus $0 \le \map \Pr A \le 1$ and Axiom $(1)$ is fulfilled trivially by definition.
Let $A \in \Sigma$ be such that:
:$A = \set {\omega_{r_1}, \omega_{r_2}, \ldots, \omega_{r_k} }$
We have that:
:$A = \set {\omega_{r_1} } \cup \set {\omega_{r_2} } \cup \cdots \cup \set {\omega_{r_k} }$
From Simple Events are Mutually Exclusive, $\set {\set {\omega_{r_1} }, \set {\omega_{r_2} }, \ldots, \set {\omega_{r_k} } }$ constitutes a set of pairwise disjoint events.
Hence:
:$\map \Pr {\set {\omega_{r_1} } \cup \set {\omega_{r_2} } \cup \cdots \cup \set {\omega_{r_k} } } = \ds \sum_{i \mathop = 1}^k \map \Pr {\omega_{r_1} }$
and it is seen that axiom $(3)$ is fulfilled.
Then we have that:
Hence axiom $(2)$ is satisfied.
== Sources ==
* : $1$: Events and probabilities: $1.3$: Probabilities: Exercise $6$
Probability Measure on Finite Sample Space/Corollary

== Corollary to Probability Measure on Finite Sample Space ==
Let $\Omega = \set {\omega_1, \omega_2, \ldots, \omega_n}$ be a finite set.
Let $\Sigma$ be the power set of $\Omega$.
Let $p_1, p_2, \ldots, p_n$ be non-negative real numbers such that:
:$p_1 + p_2 + \cdots + p_n = 1$
Let $Q: \Sigma \to \R$ be the mapping defined as:
:$\forall A \in \Sigma: \map Q A = \ds \sum_{i: \omega_i \in A} p_i$
Then $\struct {\Omega, \Sigma, Q}$ constitutes a probability space.
That is, $Q$ is a probability measure on $\struct {\Omega, \Sigma}$.
== Proof ==
We have that Power Set of Sample Space is Event Space.
Hence this is a special case of Probability Measure on Finite Sample Space.
== Sources ==
* : $1$: Events and probabilities: $1.3$: Probabilities: Exercise $6$
Probability Measure on Single-Subset Event Space

== Theorem ==
Let $\EE$ be an experiment whose sample space is $\Omega$.
Let $\O \subsetneqq A \subsetneqq \Omega$.
Let $\Sigma := \set {\O, A, \Omega \setminus A, \Omega}$ be the event space of $\EE$.
Let $\Pr: \Sigma \to \R$ be a probability measure on $\struct {\Omega, \Sigma}$.
Then $\Pr$ has the form:
for some $p \in \R$ satisfying $0 \le p \le 1$.
== Proof ==
From Event Space from Single Subset of Sample Space, we have that $\Sigma$ is an event space.
Recall the Kolmogorov axioms:
First we determine that $\Pr$ as defined is actually a probability measure.
Axiom $(1)$ and axiom $(2)$ are fulfilled trivially by definition.
Then we note that, apart from $\O$, $\set {A, \Omega \setminus A}$ are the only pairwise disjoint events whose union is $\Omega$.
Hence by definition of $\Pr$ we see that axiom $(3)$ degenerates to:
and thus axiom $(3)$ is fulfilled.
Thus $\Pr$ is indeed a probability measure.
Now it is established that a probability measure on $\Sigma$ is of the form of $\Pr$ as defined.
First we note that from Probability of Empty Event is Zero:
:$\map \Pr \O = 0$
Hence $\text {Pr 1}$ is satisfied.
From Axiom $(2)$ we have that $\map \Pr \Omega = 1$
Hence $\text {Pr 4}$ is satisfied.
By definition of probability measure, $\map \Pr A$ is between $0$ and $1$ inclusive.
That is:
:$\map \Pr A = p$
for some $p \in \R$ such that $0 \le p \le 1$.
This is exactly $\text {Pr 2}$, which is hence seen to be satisfied.
Then we note that from Probability of Event not Occurring:
:$\map \Pr {\Omega \setminus A} = 1 - \map \Pr A$
That is:
:$\map \Pr {\Omega \setminus A} = 1 - p$
Hence $\text {Pr 3}$ is satisfied.
Thus it has been proved that a probability measure on such an an event space $\Sigma$ has to be of the form $\Pr$ as defined.
== Proof ==
* : $1$: Events and probabilities: $1.3$: Probabilities: Example $10$
Category:Probability Theory