avewright/proofwiki-math huggingface dataset:
Theorem: Conditional Probability Defines Probability Space

Let $\struct {\Omega, \Sigma, \Pr}$ be a measure space.
Let $B \in \Sigma$ such that $\map \Pr B > 0$.
Let $Q: \Sigma \to \R$ be the real-valued function defined as:
:$\map Q A = \condprob A B$
where:
:$\condprob A B = \dfrac {\map \Pr {A \cap B} }{\map \Pr B}$
is the conditional probability of $A$ given $B$.
Then $\struct {\Omega, \Sigma, Q}$ is a probability space.

Proof:
It is to be shown that $Q$ is a probability measure on $\struct {\Omega, \Sigma}$.
As $\Pr$ is a measure, we have that:
:$\forall A \in \Omega: \map Q A \ge 0$
Also, we have that:
{{begin-eqn}}
{{eqn | l = \map Q \Omega
      | r = \condprob \Omega B
      | c =
}}
{{eqn | r = \frac {\map \Pr {\Omega \cap B} } {\map \Pr B}
      | c =
}}
{{eqn | r = \frac {\map \Pr B} {\map \Pr B}
      | c = Intersection with Universal Set
}}
{{eqn | r = 1
      | c = as $\map \Pr B > 0$
}}
{{end-eqn}}
Now, suppose that $A_1, A_2, \ldots$ are disjoint events in $\Sigma$.
Then:
{{begin-eqn}}
{{eqn | l = \map Q {\bigcup_{i \mathop = 1}^\infty A_i}
      | r = \frac 1 {\map \Pr B} \map \Pr {\paren {\bigcup_{i \mathop = 1}^\infty A_i} \cap B}
      | c =
}}
{{eqn | r = \frac 1 {\map \Pr B} \map \Pr {\bigcup_{i \mathop = 1}^\infty \paren {A_i \cap B} }
      | c = Intersection Distributes over Union
}}
{{eqn | r = \sum_{i \mathop = 1}^\infty \frac {\map \Pr {A_i \cap B} } {\map \Pr B}
      | c = as $\Pr$ is a measure
}}
{{eqn | r = \sum_{i \mathop = 1}^\infty \map Q {A_i}
}}
{{end-eqn}}
{{qed}}
Theorem: Bayes' Theorem

Let $\Pr$ be a probability measure on a probability space $\struct {\Omega, \Sigma, \Pr}$.
Let $\condprob A B$ denote the conditional probability of $A$ given $B$.
Let $\map \Pr A > 0$ and $\map \Pr B > 0$.
Then:
:$\condprob B A = \dfrac {\condprob A B \, \map \Pr B} {\map \Pr A}$

Proof:
From the definition of conditional probabilities, we have:
:$\condprob A B = \dfrac {\map \Pr {A \cap B} } {\map \Pr B}$
:$\condprob B A = \dfrac {\map \Pr {A \cap B} } {\map \Pr A}$
After some algebra:
:$\condprob A B \, \map \Pr B = \map \Pr {A \cap B} = \condprob B A \, \map \Pr A$
Dividing both sides by $\map \Pr A$ (we are told that it is non-zero), the result follows:
:$\condprob B A = \dfrac {\condprob A B \, \map \Pr B} {\map \Pr A}$
{{Qed}}
Theorem: Pairwise Independence does not imply Independence

Just because all the events in a family of events in a probability space are pairwise independent, it does not mean that the family is independent.

Proof:
Consider throwing a fair four-sided die.
This gives us an event space $\Omega = \set {1, 2, 3, 4}$, with each $\omega \in \Omega$ equally likely to occur:
:$\forall \omega \in \Omega: \map \Pr \omega = \dfrac 1 4$
Consider the set of events:
:$\SS = \set {A, B, C}$
where:
:$A = \set {1, 2}, B = \set {1, 3}, C = \set {1, 4}$
We have that:
:$\map \Pr A = \map \Pr B = \map \Pr C = \dfrac 1 2$
We also have that:
:$\map \Pr {A \cap B} = \map \Pr {A \cap C} = \map \Pr {B \cap C} = \map \Pr {\set 1} = \dfrac 1 4$
Thus:
:$\map \Pr A \map \Pr B = \map \Pr {A \cap B}$
:$\map \Pr A \map \Pr C = \map \Pr {A \cap C}$
:$\map \Pr B \map \Pr C = \map \Pr {B \cap C}$
Thus the events $A, B, C$ are pairwise independent.
Now, consider:
:$\map \Pr {A \cap B \cap C} = \map \Pr {\set 1} = \dfrac 1 4$
But:
:$\map \Pr A \map \Pr B \map \Pr C = \dfrac 1 8 \ne \map \Pr {A \cap B \cap C}$
So, although $\SS$ is pairwise independent, it is not independent.
{{qed}}
Theorem: Event Independence is Symmetric

Let $A$ and $B$ be events in a probability space.
Let $A$ be independent of $B$.
Then $B$ is independent of $A$.
That is, '''is independent of''' is a symmetric relation.

Proof:
We assume throughout that $\map \Pr A > 0$ and $\map \Pr B > 0$.
Let $A$ be independent of $B$.
Then by definition:
:$\condprob A B = \map \Pr A$
From the definition of conditional probabilities, we have:
:$\condprob A B = \dfrac {\map \Pr {A \cap B} } {\map \Pr B}$
and also:
:$\condprob B A = \dfrac {\map \Pr {A \cap B} } {\map \Pr A}$
So if $\condprob A B = \map \Pr A$ we have:
{{begin-eqn}}
{{eqn | l = \map \Pr A
      | r = \frac {\map \Pr {A \cap B} } {\map \Pr B}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \map \Pr B
      | r = \frac {\map \Pr {A \cap B} } {\map \Pr A}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \map \Pr B
      | r = \condprob B A
      | c = 
}}
{{end-eqn}}
So by definition, $B$ is independent of $A$.
{{qed}}
Theorem: Disjoint Independent Events means One is Void

Let $A$ and $B$ be events in a probability space.
Suppose $A$ and $B$ are:
:disjoint
:independent.
Then either $\map \Pr A = 0$ or $\map \Pr B = 0$.
That is, if two events are disjoint and independent, at least one of them can't happen.

Proof:
For $A$ and $B$ to be independent:
:$\map \Pr {A \cap B} = \map \Pr A \, \map \Pr B$
For $A$ and $B$ to be disjoint:
:$\map \Pr {A \cap B} = 0$
So:
:$\map \Pr A \, \map \Pr B = 0$
Hence the result.
{{qed}}
Theorem: Set Difference and Intersection form Partition

Let $S$ and $T$ be sets such that:
:$S \setminus T \ne \O$
:$S \cap T \ne \O$
where $S \setminus T$ denotes set difference and $S \cap T$ denotes set intersection.
Then $S \setminus T$ and $S \cap T$ form a partition of $S$.

Proof:
From Set Difference and Intersection are Disjoint:
:$S \setminus T$ and $S \cap T$ are disjoint.
Next from Set Difference Union Intersection:
:$S = \paren {S \setminus T} \cup \paren {S \cap T}$
Thus by definition $S \setminus T$ and $S \cap T$ form a partition of $S$.
{{qed}}
Category:Set Difference
Category:Set Intersection
Category:Set Partitions
Category:Set Difference and Intersection form Partition
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Theorem: Independent Events are Independent of Complement

Let $A$ and $B$ be events in a probability space $\struct {\Omega, \Sigma, \Pr}$.
Then $A$ and $B$ are independent {{iff}} $A$ and $\Omega \setminus B$ are independent.

Proof:
For $A$ and $B$ to be independent:
:$\map \Pr {A \cap B} = \map \Pr A \map \Pr B$
We need to show that:
:$\map \Pr {A \cap \paren {\Omega \setminus B} } = \map \Pr A \map \Pr {\Omega \setminus B}$
First note that $\Omega \setminus B \equiv \relcomp \Omega B$ where $\complement_\Omega$ denotes the relative complement.
From Set Difference as Intersection with Relative Complement, we have then that $A \cap \paren {\Omega \setminus B} = A \setminus B$.
From Set Difference and Intersection form Partition, we have that:
:$\paren {A \setminus B} \cup \paren {A \cap B} = A$
:$\paren {A \setminus B} \cap \paren {A \cap B} = \O$
So from the Kolmogorov axioms, we have that:
:$\map \Pr A = \map \Pr {A \setminus B} + \map \Pr {A \cap B}$
Hence:
{{begin-eqn}}
{{eqn | l = \map \Pr {A \setminus B}
      | r = \map \Pr A - \map \Pr {A \cap B}
      | c = 
}}
{{eqn | r = \map \Pr A - \map \Pr A \map \Pr B
      | c = as $A$ and $B$ are independent
}}
{{eqn | r = \map \Pr A \paren {1 - \map \Pr B}
      | c = 
}}
{{eqn | r = \map \Pr A \map \Pr {\Omega \setminus B}
      | c = Elementary Properties of Probability Measure
}}
{{end-eqn}}
But as $A \setminus B = A \cap \paren {\Omega \setminus B}$ we have:
:$\map \Pr {A \cap \paren {\Omega \setminus B} } = \map \Pr A \map \Pr {\Omega \setminus B}$
which is what we wanted to show.
Now, suppose $A$ and $\Omega \setminus B$ are independent.
From the above, we have that $A$ and $\Omega \setminus \paren {\Omega \setminus B}$ are independent.
But $\Omega \setminus \paren {\Omega \setminus B} = B$ from Relative Complement of Relative Complement hence the result.
{{qed}}
Theorem: Probability of Independent Events Not Happening

Let $\EE = \struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $A_1, A_2, \ldots, A_m \in \Sigma$ be independent events in the event space of $\EE$.
Then the probability of none of $A_1$ to $A_m$ occurring is:
:$\ds \prod_{i \mathop = 1}^m \paren {1 - \map \Pr {A_i} }$

Proof:
Let $A_1, A_2, \ldots, A_m \in \Sigma$ be independent events.
From Independent Events are Independent of Complement, we have that $\Omega \setminus A_1, \Omega \setminus A_2, \ldots, \Omega \setminus A_m \in \Sigma$ are also independent.
From the definition of occurrence, if $A$ does not happen then $\Omega \setminus A$ ''does'' happen.
So for none of $A_1, A_2, \ldots, A_m$ to happen, ''all'' of $\Omega \setminus A_1, \Omega \setminus A_2, \ldots, \Omega \setminus A_m$ must happen.
From Elementary Properties of Probability Measure:
:$\forall A \in \Omega: \map \Pr {\Omega \setminus A} = 1 - \map \Pr A$
So the probability of none of $A_1$ to $A_m$ occurring is:
:$\ds \prod_{i \mathop = 1}^m \paren {1 - \map \Pr {A_i} }$
{{qed}}
Theorem: Probability of Occurrence of At Least One Independent Event

Let $\EE = \struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $A_1, A_2, \ldots, A_m \in \Sigma$ be independent events in the event space of $\EE$.
Then the probability of at least one of $A_1$ to $A_m$ occurring is:
:$\ds 1 - \prod_{i \mathop = 1}^m \paren {1 - \map \Pr {A_i} }$

Proof:
Follows as a direct result of Probability of Independent Events Not Happening.
Let $B$ be the event "None of $A_1$ to $A_m$ happen".
From Probability of Independent Events Not Happening:
:$\ds \map \Pr B = \prod_{i \mathop = 1}^m \paren {1 - \map \Pr {A_i} }$
Then $\Omega \setminus B$ is the event "''Not'' none of $A_1$ to $A_m$ happen", or "At least one of $A_1$ to $A_m$ happens".
From Elementary Properties of Probability Measure:
:$\forall A \in \Omega: \map \Pr {\Omega \setminus A} = 1 - \map \Pr A$
Hence the probability that at least one of $A_1$ to $A_m$ happen is:
:$\ds 1 - \map \Pr B = 1 - \prod_{i \mathop = 1}^m \paren {1 - \map \Pr {A_i} }$
Theorem: Total Probability Theorem

Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $\set {B_1, B_2, \ldots}$ be a partition of $\Omega$ such that $\forall i: \map \Pr {B_i} > 0$.
Then:
:$\ds \forall A \in \Sigma: \map \Pr A = \sum_i \condprob A {B_i} \map \Pr {B_i}$

Proof:
{{begin-eqn}}
{{eqn | l = \map \Pr A
      | r = \map \Pr {A \cap \paren {\bigcup_i B_i} }
      | c = Intersection with Subset is Subset: $\ds \bigcup_i B_i = \Omega$ and $A \subseteq \Omega$
}}
{{eqn | r = \map \Pr {\bigcup_i \paren {A \cap B_i} }
      | c = Intersection Distributes over Union
}}
{{eqn | r = \sum_i \map \Pr {A \cap B_i}
      | c = as all the $A \cap B_i$ are disjoint
}}
{{eqn | r = \sum_i \condprob A {B_i} \map \Pr {B_i}
      | c = {{Defof|Conditional Probability}}
}}
{{end-eqn}}
{{qed}}
Theorem: Boole's Inequality

Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $A_1, A_2, \ldots, A_n$ be events in $\Sigma$.
Then:
:$\ds \map \Pr {\bigcup_{i \mathop = 1}^n A_i} \le \sum_{i \mathop = 1}^n \map \Pr {A_i}$

Proof:
A direct consequence of the facts that:
:a Probability Measure is Subadditive
:the result Finite Union of Sets in Subadditive Function which gives:
::$\ds \map f {\bigcup_{i \mathop = 1}^n A_i} \le \sum_{i \mathop = 1}^n \map f {A_i}$
:for a subadditive function $f$.
{{qed}}
Theorem: Additive Nowhere Negative Function is Subadditive

Let $\AA$ be an algebra of sets.
Let $f: \AA \to \overline \R$ be an additive function such that:
:$\forall A \in \AA: \map f A \ge 0$
Then $f$ is subadditive.

Proof:
If $f$ is additive then by Additive Function is Strongly Additive:
:$\forall A, B \in \AA: \map f {A \cup B} = \map f A + \map f B - \map f {A \cap B}$
As $\map f {A \cap B} \ge 0$, the result follows by definition of subadditive:
:$\forall A, B \in \AA: \map f {A \cup B} \le \map f A + \map f B$
{{qed}}
Category:Set Systems
3443akyby5i5sjjxs843w0c9zdsfmep
Theorem: Probability Measure is Subadditive

Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Then $\Pr$ is a subadditive function.

Proof:
By definition, a probability measure is a measure.
The result follows from Measure is Subadditive
{{qed}}
Category:Probability Measures
dfid6kx4vypxcmoju6wzculb0u571x3
Theorem: Inclusion-Exclusion Principle

Let $\SS$ be an algebra of sets.
Let $A_1, A_2, \ldots, A_n$ be finite sets.
Let $f: \SS \to \R$ be an additive function.
Then:
{{begin-eqn}}
{{eqn | l = \map f {\bigcup_{i \mathop = 1}^n A_i}
      | r = \sum_{i \mathop = 1}^n \map f {A_i}
      | c = 
}}
{{eqn | o = 
      | ro= -
      | r = \sum_{1 \mathop \le i \mathop < j \mathop \le n} \map f {A_i \cap A_j}
      | c = 
}}
{{eqn | o = 
      | ro= +
      | r = \sum_{1 \mathop \le i \mathop < j \mathop < k \mathop \le n} \map f {A_i \cap A_j \cap A_k}
      | c = 
}}
{{eqn | o = 
      | ro= \cdots
      | c = 
}}
{{eqn | o = 
      | ro= +
      | r = \paren {-1}^{n - 1} \map f {\bigcap_{i \mathop = 1}^n A_i}
      | c = 
}}
{{end-eqn}}

Proof:
Proof by induction:
For all $n \in \N_{>0}$, let $\map P n$ be the proposition:
{{begin-eqn}}
{{eqn | l = \map f {\bigcup_{i \mathop = 1}^n A_i}
      | r = \sum_{i \mathop = 1}^n \map f {A_i}
      | c = 
}}
{{eqn | o = 
      | ro= -
      | r = \sum_{1 \mathop \le i \mathop < j \mathop \le n} \map f {A_i \cap A_j}
      | c = 
}}
{{eqn | o = 
      | ro= +
      | r = \sum_{1 \mathop \le i \mathop < j \mathop < k \mathop \le n} \map f {A_i \cap A_j \cap A_k}
      | c = 
}}
{{eqn | o = 
      | ro= \cdots
      | c = 
}}
{{eqn | o = 
      | ro= +
      | r = \paren {-1}^{n - 1} \map f {\bigcap_{i \mathop = 1}^n A_i}
      | c = 
}}
{{end-eqn}}
$\map P 1$ is true, as this just says $\map f {A_1} = \map f {A_1}$.
Theorem: Existence of Probability Space and Discrete Random Variable

Let $I$ be an arbitrary countable indexing set.
Let $S = \set {s_i: i \in I} \subset \R$ be a countable set of real numbers.
Let $\set {\pi_i: i \in I} \subset \R$ be a countable set of real numbers which satisfies:
:$\ds \forall i \in I: \pi_i \ge 0, \sum_{i \mathop \in I} \pi_i = 1$
Then there exists:
:a probability space $\struct {\Omega, \Sigma, \Pr}$
and:
:a discrete random variable $X$ on $\struct {\Omega, \Sigma, \Pr}$
such that the probability mass function $p_X$ of $X$ is given by:
{{begin-eqn}}
{{eqn | l = \map {p_X} {s_i}
      | r = \pi_i
      | c = if $i \in I$
}}
{{eqn | l = \map {p_X} s
      | r = 0
      | c = if $s \notin S$
}}
{{end-eqn}}

Proof:
Take $\Omega = S$ and $\Sigma = \powerset S$ (the power set of $S$).
Then let:
:$\ds \map \Pr A = \sum_{i: s_i \mathop \in A} \pi_i$
for all $A \in \Sigma$.
Then we can define $X: \Omega \to \R$ by:
:$\forall \omega \in \Omega: \map X \omega = \omega$
This suits the conditions of the assertion well enough.
{{qed}}
Theorem: Countable Function on Power Set of Sample Space is Discrete Random Variable

Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space such that $\Sigma$ is the power set of $\Omega$.
Let $f: \Omega \to \R$ be a function such that $\Img f$ is countable.
Then $f$ is a discrete random variable on $\struct {\Omega, \Sigma, \Pr}$.

Proof:
By definition, $\map {f^{-1} } x \subseteq \Omega$.
But then $\map {f^{-1} } x \in \powerset \Omega$.
Hence the result.
{{qed}}
Theorem: Characteristic Function on Event is Discrete Random Variable

Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $E \in \Sigma$ be any event of $\struct {\Omega, \Sigma, \Pr}$.
Let $\chi_E: \Omega \to \set {0, 1}$ be the characteristic function of $E$.
Then $\chi_E$ is a discrete random variable on $\struct {\Omega, \Sigma, \Pr}$.

Proof:
By definition of characteristic function:
:$\forall \omega \in \Omega: \chi_E = \begin{cases}
1 & : \omega \in E \\
0 & : \omega \notin E \\
\end{cases}$
Then clearly:
:$\forall x \in \R: \map { {\chi_E}^{-1} } x = \begin{cases}
E & : x = 1 \\
\Omega \setminus E & : x = 0 \\
\O & : x \notin \set {0, 1}
\end{cases}$
So whatever the value of $x \in \R$, its preimage is in $\Sigma$.
Hence the result.
{{qed}}
Theorem: Poisson Distribution Gives Rise to Probability Mass Function

Let $X$ be a discrete random variable on a probability space $\struct {\Omega, \Sigma, \Pr}$.
Let $X$ have the poisson distribution with parameter $\lambda$ (where $\lambda > 0$).
Then $X$ gives rise to a probability mass function.

Proof:
By definition:
:$\Img X = \N$
:$\map \Pr {X = k} = \dfrac 1 {k!} \lambda^k e^{-\lambda}$
Then:
{{begin-eqn}}
{{eqn | l = \map \Pr \Omega
      | r = \sum_{k \mathop \in \N} \frac 1 {k!} \lambda^k e^{-\lambda}
      | c = 
}}
{{eqn | r = e^{-\lambda} \sum_{k \mathop \in \N} \frac 1 {k!} \lambda^k
      | c = 
}}
{{eqn | r = e^{-\lambda} e^{\lambda}
      | c = Taylor Series Expansion for Exponential Function
}}
{{eqn | r = 1
      | c = 
}}
{{end-eqn}}
So $X$ satisfies $\map \Pr \Omega = 1$, and hence the result.
{{Qed}}
Theorem: Geometric Distribution Gives Rise to Probability Mass Function

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

Proof:
By definition:
:$\map \Omega X = \N = \set {0, 1, 2, \ldots}$
:$\map \Pr {X = k} = p^k \paren {1 - p}$
Then:
{{begin-eqn}}
{{eqn | l = \map \Pr \Omega
      | r = \sum_{k \mathop \ge 0} p^k \paren {1 - p}
      | c = {{Defof|Geometric Distribution}}
}}
{{eqn | r = \paren {1 - p} \sum_{k \mathop \ge 0} p^k
      | c = 
}}
{{eqn | r = \paren {1 - p} \frac 1 {1 - p}
      | c = Sum of Infinite Geometric Sequence
}}
{{eqn | r = 1
      | c = 
}}
{{end-eqn}}
The above result is valid, because $0 < p < 1$.
So $X$ satisfies $\map \Pr \Omega = 1$, and hence the result.
{{Qed}}
Category:Geometric Distribution
hxbb08zw9l0ggho03b26pc2zwg838uu
Theorem: Negative Binomial Distribution Gives Rise to Probability Mass Function/Type 1

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

Proof:
By definition:
:$\Img X = \set {n, n + 1, n + 2, \ldots}$
:$\map \Pr {X = k} = \dbinom {k - 1} {n - 1} p^n \paren {1 - p}^{k - n}$
Then:
{{begin-eqn}}
{{eqn | l = \map \Pr \Omega
      | r = \sum_{k \mathop \ge n} \binom {k - 1} {n - 1} p^n \paren {1 - p}^{k - n}
      | c = 
}}
{{eqn | r = p^n \sum_{j \mathop \ge 0} \binom {n + j - 1} j \paren {1 - p}^j
      | c = substituting $j = k - n$
}}
{{eqn | r = p^n \sum_{j \mathop \ge 0} \binom {-n} {j} \paren {p - 1}^j
      | c = Negated Upper Index of Binomial Coefficient
}}
{{eqn | r = p^n \paren {1 - \paren {p - 1} }^{-n}
      | c = Binomial Theorem
}}
{{eqn | r = 1
      | c = 
}}
{{end-eqn}}
So $X$ satisfies $\map \Pr \Omega = 1$, and hence the result.
{{Qed}}
Theorem: Coin-Tossing Modeled by Bernoulli Trial

The game of coin-tossing can be modeled as a Bernoulli trial.
This applies whether the coin is fair or biased.

Proof:
When a coin is tossed, it can land either with the head side up, or the tail side up.
Not taking into account tricks and teases in which the coin is given due consideration for landing on its edge, these are the only two possible outcomes.
Let the probability that it lands heads be $p$.
Let $\EE$ be the experiment of tossing a coin up into the air and seeing which side lands uppermost.
Let $\EE$ be modeled as the probability space $\struct {\Omega, \Sigma, \Pr}$.
We have that:
:$\Omega = \set {\mathrm H, \T}$
where $\mathrm H$ and $\T$ are the elementary events '''heads land uppermost''' and '''tails land uppermost''' respectively.
We define $\Pr: \Omega \to \R$ as follows:
:<nowiki>$\forall \omega \in \Omega: \map \Pr\omega = \begin{cases}
p & : \omega = \mathrm H \\
1 - p & : \omega = \T
\end{cases}$</nowiki>
in accordance with the definition of the experiment.
Finally, we let $\Sigma = \powerset \Omega$.
This is, by definition, clearly a Bernoulli trial.
{{qed}}
Category:Coin-Tossing
Category:Bernoulli Trials
sfs0uj46k2moop9ydrf4vj3w8dcauo2
Theorem: Binomial Experiment has Binomial Distribution

Let $\sequence {X_i}$ be a finite binomial experiment of length $n$ such that each of the $X_i$ in the sequence is a Bernoulli trial with parameter $p$.
Then the number of successes in $\sequence {X_i}$ is modelled by a binomial distribution with parameters $n$ and $p$.

Proof:
Consider the sample space $\Omega$ of all sequences $\sequence {X_i}$ of length $n$.
The $i$th entry of any such sequence is the result of the $i$th trial.
We have that $\Omega$ is finite.
Let us take the event space $\Sigma$ to be the power set of $\Omega$.
As the elements of $\Omega$ are independent, by definition of the binomial experiment, we have that:
:$\forall \omega \in \Omega: \map \Pr \omega = p^{\map s \omega} \paren {1 - p}^{n - \map s \omega}$
where $\map s \omega$ is the number of successes in $\omega$.
In the same way:
:$\ds \forall A \in \Sigma: \map \Pr A = \sum_{\omega \mathop \in A} \map \Pr \omega$
Now, let us define the discrete random variable $Y_i$ as follows:
:<nowiki>$\map {Y_i} \omega = \begin{cases}
1 & : \text {$\omega_i$ is a success} \\
0 & : \text {$\omega_i$ is a failure}
\end{cases}$</nowiki>
where $\omega_i$ is the $i$th element of $\omega$.
Thus, each $Y_i$ has image $\set {0, 1}$ and a probability mass function:
:$\map \Pr {Y_i = 0} = \map \Pr {\set {\omega \in \Omega: \text {$\omega_i$ is a success} } }$
Thus we have:
{{begin-eqn}}
{{eqn | l = \map \Pr {Y_i = 1}
      | r = \sum_{\omega: \text {$\omega_i$ success} } p^{\map s \omega} \paren {1 - p}^{n - \map s \omega}
}}
{{eqn | r = \sum_{r \mathop = 1}^n \sum_{\substack {\omega: \text {$\omega_i$ success} \\ \map s \omega = r} } p^r \paren {1 - p}^{n - r}
}}
{{eqn | r = \sum_{r \mathop = 1}^n \binom {n - 1} {r - 1} p^r \paren {1 - p}^{n - r}
      | c = as we already know the position of one success (namely $i$)
}}
{{eqn | r = p \sum_{r \mathop = 0}^{n - 1} \binom {n - 1} r p^r \paren {1 - p}^{\paren {n - 1} - r}
      | c = switching summation index
}}
{{eqn | r = p \paren {p + \paren {1 - p} }^{n - 1}
      | c = Binomial Theorem
}}
{{eqn | r = p
}}
{{end-eqn}}
Then:
:$\map \Pr {Y_i = 0} = 1 - \map \Pr {Y_i = 1} = 1 - p$
So (by a roundabout route) we have confirmed that $Y_i$ has the Bernoulli distribution with parameter $p$.
Now, let us define the random variable:
:$\ds \map {S_n} \omega = \sum_{i \mathop = 1}^n \map {Y_i} \omega$
By definition:
:$\map {S_n} \omega$ is the number of successes in $\omega$
:$S_n$ takes values in $\set {0, 1, 2, \ldots, n}$ (as each $Y_i$ can be $0$ or $1$).
Also, we have that:
{{begin-eqn}}
{{eqn | l = \map \Pr {S_n = k}
      | r = \map \Pr {\set {\omega \in \Omega: \map s \omega = k} }
}}
{{eqn | r = \sum_{\omega: \map s \omega \mathop = k} \map \Pr \omega
}}
{{eqn | r = \sum_{\omega: \map s \omega \mathop = k} p^k \paren {1 - p}^{n - k}
}}
{{eqn | r = \binom n k p^k \paren {1 - p}^{n - k}
}}
{{end-eqn}}
Hence the result.
{{qed}}
Theorem: Binomial Distribution Approximated by Poisson Distribution

Let $X$ be a discrete random variable which has the binomial distribution with parameters $n$ and $p$.
Then for $\lambda = n p$, $X$ can be approximated by a Poisson distribution with parameter $\lambda$:
:$\ds \lim_{n \mathop \to \infty} \binom n k p^k \paren {1 - p}^{n - k} = \frac {\lambda^k} {k!}  e^{-\lambda}$

Proof:
Let $X$ be as described.
Let $k \in \Z_{\ge 0}$ be fixed.
We write $p = \dfrac \lambda n$ and suppose that $n$ is large.
Then:
{{begin-eqn}}
{{eqn | l = \lim_{n \mathop \to \infty} \binom n k p^k \paren {1 - p}^{n - k}
      | r = \lim_{n \mathop \to \infty} \binom n k \paren {\frac \lambda n}^k \paren {1 - \frac \lambda n}^n \paren {1 - \frac \lambda n}^{-k}
      | c = 
}}
{{eqn | r = \lim_{n \mathop \to \infty} \frac {n^k} {k!} \paren {\frac \lambda n}^k \paren {1 - \frac \lambda n}^n \paren {1 - \frac \lambda n}^{-k}
      | c = Limit to Infinity of Binomial Coefficient over Power
}}
{{eqn | r = \lim_{n \mathop \to \infty} \frac 1 {k!} \lambda^k \paren {1 + \frac {-\lambda} n}^n \paren {1 - \frac \lambda n}^{-k}
      | c = dividing numerator and denominator by $n^k$
}}
{{eqn | r = \dfrac {\lambda^k} {k!} \lim_{n \mathop \to \infty} \paren {1 + \frac {-\lambda} n}^n \paren {1 - \frac \lambda n}^{-k}
      | c = moving $\dfrac {\lambda^k} {k!}$ outside of the limit
}}
{{eqn | r = \dfrac {\lambda^k} {k!} \lim_{n \mathop \to \infty} \paren {1 + \frac {-\lambda} n}^n \lim_{n \mathop \to \infty} \paren {1 - \frac \lambda n}^{-k}
      | c = Product Rule for Limits of Functions
}}
{{eqn | r = \frac {\lambda^k} {k!} e^{-\lambda} \paren {1 }
      | c = {{Defof|Exponential Function/Real|subdef = Limit of Sequence|Exponential Function}}
}}
{{eqn | r = \frac {\lambda^k} {k!} e^{-\lambda}
      | c = 
}}
{{end-eqn}}
Hence the result.
{{qed}}