hcju/proofwikiqa huggingface dataset:
Proof by induction: In the below, we assume that $A_1, A_2, \ldots$ are all elements of $\mathcal A$. For all $n \in \N_{>0}$, let $P \left({n}\right)$ be the proposition: :$\displaystyle f \left({\bigcup_{i \mathop = 1}^n A_i}\right) \le \sum_{i \mathop = 1}^n f \left({A_i}\right)$ $P(1)$ is trivially true, as this just says $f \left({A_1}\right) \le f \left({A_1}\right)$. === Basis for the Induction === $P(2)$ is the case $f \left({A_1 \cup A_2}\right) \le f \left({A_1}\right) + f \left({A_2}\right)$, which comes from the definition of a subadditive function. 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: :$\displaystyle f \left({\bigcup_{i \mathop = 1}^k A_i}\right) \le \sum_{i \mathop = 1}^k f \left({A_i}\right)$ Then we need to show: :$\displaystyle f \left({\bigcup_{i \mathop = 1}^{k+1} A_i}\right) \le \sum_{i \mathop = 1}^{k+1} f \left({A_i}\right)$ === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l = f \left({\bigcup_{i \mathop = 1}^{k+1} A_i}\right)       | r = f \left({\bigcup_{i \mathop = 1}^k A_i \cup A_{k+1} }\right)       | c =  }} {{eqn | o = \le       | r = f \left({\bigcup_{i \mathop = 1}^k A_i}\right) + f \left({A_{k+1} }\right)       | c = from the base case }} {{eqn | o = \le       | r = \sum_{i \mathop = 1}^k f \left({A_i}\right) + f \left({A_{k+1} }\right)       | c = from the induction hypothesis }} {{eqn | r = \sum_{i \mathop = 1}^{k+1} f \left({A_i}\right)       | 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: :$\displaystyle \forall n \in \N_{>0}: \left({\bigcup_{i \mathop = 1}^n A_i}\right) \le \sum_{i \mathop = 1}^n f \left({A_i}\right)$ {{qed}} Category:Set Systems lp90h89jj0uhx97boj1slj6j9algrpj
First we check that the statement of the theorem makes sense. And we have that from Properties of Algebras of Sets we have that $\varnothing \in \mathcal A$. Now let $X \in \mathcal A$. From Intersection with Empty Set we have that $X \cap \varnothing = \varnothing$, and so $X$ and $\varnothing$ are disjoint. As $f$ is additive, it follows by definition that: :$f \left({X \cup \varnothing}\right) = f \left({X}\right) + f \left({\varnothing}\right)$ But $X \cup \varnothing = X$ from Union with Empty Set and so: :$f \left({X}\right) = f \left({X}\right) + f \left({\varnothing}\right)$ Hence the result: :$f \left({\varnothing}\right) = 0$ {{qed}} Category:Set Systems sdrw0pnbj942ytb03176sfb0thwva4w
From Set Difference and Intersection form Partition, we have that: :$A$ is the union of the two disjoint sets $A \setminus B$ and $A \cap B$ :$B$ is the union of the two disjoint sets $B \setminus A$ and $A \cap B$. So, by the definition of additive function: :$\map f A = \map f {A \setminus B} + \map f {A \cap B}$ :$\map f B = \map f {B \setminus A} + \map f {A \cap B}$ We also have from Set Difference is Disjoint with Reverse that: :$\paren {A \setminus B} \cap \paren {B \setminus A} = \O$ Hence: {{begin-eqn}} {{eqn | l = \map f A + \map f B       | r = \map f {A \setminus B} + 2 \map f {A \cap B} + \map f {B \setminus A} }} {{eqn | r = \map f {\paren {A \setminus B} \cup \paren {A \cap B} \cup \paren {B \setminus A} } + \map f {A \cap B}       | c = Set Difference and Intersection form Partition: Corollary 1 }} {{eqn | r = \map f {A \cup B} + \map f {A \cap B} }} {{end-eqn}} Hence the result. {{qed}} Category:Set Systems 7s6ygdc8s3z1zmduj3vix5scn9l4k1l
We assume that $S, T \subseteq \mathbb U$ where $\mathbb U$ is the universe. Then we can use the definition of Set Difference as Intersection with Complement. {{begin-eqn}} {{eqn | o =       | r = \paren {S \setminus T} \cap \paren {T \setminus S}       | c =  }} {{eqn | r = \paren {S \cap \map \complement T} \cap \paren {T \cap \map \complement S}       | c = Set Difference as Intersection with Complement }} {{eqn | r = \paren {S \cap \map \complement S} \cap \paren {T \cap \map \complement T}       | c = Intersection is Associative and Intersection is Commutative }} {{eqn | r = \O \cap \O       | c = Intersection with Complement }} {{eqn | r = \O       | c = Empty Set Disjoint with Itself }} {{end-eqn}} {{qed}} Category:Set Difference Category:Set Intersection im14fc2uli4oodxcfdlxk33odzab1ed
A direct corollary of Non-Negative Additive Function is Monotone. {{qed}}
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}}
As by definition a probability measure is a measure, we can directly use the result Measure is Monotone. {{qed}}
We have that $f$ is defined as countably additive {{iff}}: :$\displaystyle f \left({\bigcup_{i \mathop \ge 1} A_i}\right) = \sum_{i \mathop \ge 1} f \left({A_i}\right)$ where $\left \langle {A_i} \right \rangle$ is any sequence of pairwise disjoint elements of $\mathcal A$. We need to show that: :$\displaystyle \forall n \in \N: f \left({\bigcup_{i \mathop = 1}^n A_i}\right) = \sum_{i \mathop = 1}^n f \left({A_i}\right)$ Let $n \in \N$ be any arbitrary natural number. Let $\left \langle {B_i}\right \rangle$ be the sequence of pairwise disjoint elements of $\mathcal A$ defined as: :$B_i = \begin{cases} A_i & : i \le n \\ \varnothing & : i > n \end{cases}$ It follows that: :$\displaystyle \bigcup_{i \mathop \ge 1} B_i = \bigcup_{i \mathop = 1}^n A_i$ Thus: {{begin-eqn}} {{eqn | l = f \left({\bigcup_{i \mathop = 1}^n A_i}\right)       | r = f \left({\bigcup_{i \mathop \ge 1} B_i}\right) }} {{eqn | r = \sum_{i \mathop \ge 1} f \left({B_i}\right) }} {{eqn | r = \sum_{i \mathop = 1}^n f \left({A_i}\right) + \sum_{i \mathop > n} f \left({\varnothing}\right) }} {{eqn | r = \sum_{i \mathop = 1}^n f \left({A_i}\right)       | c = from Countably Additive Function of Null Set }} {{end-eqn}} Hence the result. {{qed}} Category:Set Systems Category:Measure Theory t5bk2aansp9swbgmjlcj7c6kgg1wcux
Suppose that $A \in \mathcal A$ such that $f \left({A}\right)$ is a finite number. So, let $f \left({A}\right) = x$. Consider the sequence $\left \langle {S_i}\right \rangle \subseteq \mathcal A$ defined as: :$\forall i \in \N: S_i = \begin{cases} A & : i = 0 \\ \varnothing & : i > 0 \end{cases}$ Then: :$\displaystyle \bigcup_{i \mathop \ge 0} S_i = A$ Hence: {{begin-eqn}} {{eqn | l = f \left({A}\right)       | r = x       | c =  }} {{eqn | ll= \implies       | l = f \left({\bigcup_{i \mathop \ge 0} S_i}\right)       | r = x       | c =  }} {{eqn | ll= \implies       | l = \sum_{i \mathop \ge 0} f \left({S_i}\right)       | r = x       | c =  }} {{eqn | ll= \implies       | l = f \left({A}\right) + \sum _{i \mathop > 0} f \left({\varnothing}\right)       | r = x       | c =  }} {{eqn | ll= \implies       | l = \sum_{i \mathop > 0} f \left({\varnothing}\right)       | r = 0       | c =  }} {{end-eqn}} It follows directly that: : $f \left({\varnothing}\right) = 0$ {{qed}} Category:Set Systems ifdz9s9c4xyi2s3e5hxub3g5tseqae4
Let $A \subseteq B$, and let $f$ be an additive function. From Set Difference Union Intersection: :$B = \paren {B \setminus A} \cup \paren {A \cap B}$ From Intersection with Subset is Subset we have that: :$A \subseteq B \implies A \cap B = A$ So: :$A \subseteq B \implies B = \paren {B \setminus A} \cup A$ From Set Difference Intersection with Second Set is Empty Set: :$\paren {B \setminus A} \cap A = \O$ So $B$ is the union of the two disjoint sets $B \setminus A$ and $A \cap B$. So, by the definition of additive function: :$\map f B = \map f {B \setminus A} + \map f A$ But as, by hypothesis, $\map f {B \setminus A} \ge 0$: :$\map f B \ge \map f A$ hence the result. {{qed}} Category:Set Systems cyak66q6ypcr20vkj1auoelzpzlbpap
Let $S$ be a set, and let $\powerset S$ be its power set. We have that a power set is an algebra of sets, and so: :$(1): \quad \forall A, B \in \powerset S: A \cup B \in \powerset S$ :$(2): \quad \relcomp S A \in \powerset S$ Let $\sequence {A_i}$ be a countably infinite sequence of sets in $\powerset S$. Then from Power Set is Closed under Countable Unions: :$\ds \bigcup_{i \mathop \in \N} A_i \in \powerset S$ So, by definition, $\powerset S$ is a sigma-algebra. {{qed}}
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: :$\map \Pr A = \dfrac {\card A} n$ {{qed}} Category:Probability Theory qsq264ohyabwaxnluz7vp40wwwcdsgs
It is to be shown that $Q$ is a probability measure on $\left({\Omega, \Sigma}\right)$. As $\Pr$ is a measure, we have that: :$\forall A \in \Omega: Q \left({A}\right) \ge 0$ Also, we have that: {{begin-eqn}} {{eqn | l=Q \left({\Omega}\right)       | r=\Pr \left({\Omega \mid B}\right)       | c= }} {{eqn | r=\frac {\Pr \left({\Omega \cap B}\right)} {\Pr \left({B}\right)}       | c= }} {{eqn | r=\frac {\Pr \left({B}\right)} {\Pr \left({B}\right)}       | c=Intersection with Universe }} {{eqn | r=1       | c=as $\Pr \left({B}\right) > 0$ }} {{end-eqn}} Now, suppose that $A_1, A_2, \ldots$ are disjoint events in $\Sigma$. Then: {{begin-eqn}} {{eqn | l=Q \left({\bigcup_{i \mathop = 1}^\infty A_i}\right)       | r=\frac 1 {\Pr \left({B}\right)} \Pr \left({\left({\bigcup_{i \mathop = 1}^\infty A_i}\right) \cap B}\right)       | c= }} {{eqn | r=\frac 1 {\Pr \left({B}\right)} \Pr \left({\bigcup_{i \mathop = 1}^\infty \left({A_i \cap B}\right)}\right)       | c=Intersection Distributes over Union }} {{eqn | r=\sum_{i \mathop = 1}^\infty \frac {\Pr \left({A_i \cap B}\right)} {\Pr \left({B}\right)}       | c=as $\Pr$ is a measure }} {{eqn | r=\sum_{i \mathop = 1}^\infty Q \left({A_i}\right) }} {{end-eqn}} {{qed}}
From the definition of conditional probabilities, we have: :$\map \Pr {A \mid B} = \dfrac {\map \Pr {A \cap B} } {\map \Pr B}$ :$\map \Pr {B \mid A} = \dfrac {\map \Pr {A \cap B} } {\map \Pr A}$ After some algebra: :$\map \Pr {A \mid B} \, \map \Pr B = \map \Pr {A \cap B} = \map \Pr {B \mid A} \, \map \Pr A$ Dividing both sides by $\map \Pr A$ (we are told that it is non-zero), the result follows: :$\map \Pr {B \mid A} = \dfrac {\map \Pr {A \mid B} \, \map \Pr B} {\map \Pr A}$ {{Qed}}
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}}
We assume throughout that $\map \Pr A > 0$ and $\map \Pr B > 0$. Let $A$ be independent of $B$. Then by definition: :$\map \Pr {A \mid B} = \map \Pr A$ From the definition of conditional probabilities, we have: :$\map \Pr {A \mid B} = \dfrac {\map \Pr {A \cap B} } {\map \Pr B}$ and also: :$\map \Pr {B \mid A} = \dfrac {\map \Pr {A \cap B} } {\map \Pr A}$ So if $\map \Pr {A \mid B} = \map \Pr A$ 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 = \map \Pr {B \mid A}       | c =  }} {{end-eqn}} So by definition, $B$ is independent of $A$. {{qed}}
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}}
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 Union qydkh813m6g8kbvanxqsu83faye2qgb
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}}
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: :$\displaystyle \prod_{i \mathop = 1}^m \paren {1 - \map \Pr {A_i} }$ {{qed}}
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: :$\displaystyle \map \Pr B = \prod_{i \mathop = 1}^m \paren {1 - \map \Pr {A_i} }$ 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: :$\displaystyle 1 - \map \Pr B = 1 - \prod_{i \mathop = 1}^m \paren {1 - \map \Pr {A_i} }$ === Proof of Corollary === It can immediately be seen that this is an instance of the main result with all of $A_1, A_2, \ldots, A_m$ being instances of $A$. The result follows directly. {{qed}}
{{begin-eqn}} {{eqn | l = \map \Pr A       | r = \map \Pr {A \cap \paren {\bigcup_i B_i} }       | c = Intersection with Subset is Subset: $\bigcup_i B_i = \Omega$ 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 \map \Pr {A \mid B_i} \map \Pr {B_i}       | c = {{Defof|Conditional Probability}} }} {{end-eqn}} {{qed}}
A direct consequence of the facts that: :a Probability Measure is Subadditive :the result Finite Union of Sets in Subadditive Function which gives: ::$\displaystyle \map f {\bigcup_{i \mathop = 1}^n A_i} \le \sum_{i \mathop = 1}^n \map f {A_i}$ :for a subadditive function $f$. {{qed}}
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 3hmgj8333co74uz78j1h6f5oakm5ber
By definition, a probability measure is a measure. The result follows from Measure is Subadditive {{qed}} Category:Probability Theory Category:Measure Theory oeg6eg7clag4m6zwcls4tqk9akcuace
By definition of characteristic function, we have: :$\forall \omega \in \Omega: \chi_E = \begin{cases} 1 & : \omega \in E \\ 0 & : \omega \notin E \\ \end{cases}$ Then clearly: :$\forall x \in \R: {\chi_E}^{-1} \left({x}\right) = \begin{cases} E & : x = 1 \\ \Omega \setminus E & : x = 0 \\ \varnothing & : x \notin \left\{{0, 1}\right\} \end{cases}$ So whatever the value of $x \in \R$, its preimage is in $\Sigma$. Hence the result. {{qed}}
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}}
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
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}}
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-up be $p$. Let $\mathcal E$ be the experiment of tossing a coin up into the air and seeing which side lands uppermost. Let $\mathcal E$ be modeled as the probability space $\left({\Omega, \Sigma, \Pr}\right)$. We have that: :$\Omega = \left\{{H, T}\right\}$ where $H$ and $T$ are the elementary events "Heads land uppermost" and "Tails land uppermost" respectively. We define $\Pr: \Omega \to \R$ as follows: :$\forall \omega \in \Omega: \Pr \left({\omega}\right) = \begin{cases} p & : \omega = H \\ 1 - p & : \omega = T \end{cases}$ in accordance with the definition of the experiment. Finally, we let $\Sigma = \mathcal P \left({\Omega}\right)$. This is, by definition, clearly a Bernoulli trial. {{qed}} Category:Bernoulli Distribution 4y4wo7idt0kxonp69g23byamh5144ot