corrected some pages

In all pages from the section "Probability axioms" (and some others), "P(...)" was replaced by "p(...)" throughout.

Author: JoramSoch

D/prob-ax.md

@@ -39,22 +39,22 @@ username: "JoramSoch"
 ---
 
 
-**Definition:** Let there be a [sample space](/D/samp-spc) $\Omega$, an [event space](/D/eve-spc) $\mathcal{E}$ and a [probability measure](/D/prob-meas) $P$, such that $P(E)$ is the [probability](/D/prob) of some [event](/D/reve) $E \in \mathcal{E}$. Then, we introduce three axioms of probability:
+**Definition:** Let there be a [sample space](/D/samp-spc) $\Omega$, an [event space](/D/eve-spc) $\mathcal{E}$ and a [probability measure](/D/prob-meas) $P$, such that $p(E)$ is the [probability](/D/prob) of some [event](/D/reve) $E \in \mathcal{E}$. Then, we introduce three axioms of probability:
 
 * First axiom: The probability of an event is a non-negative real number:
 
 $$ \label{eq:prob-ax1}
-P(E) \in \mathbb{R}, \; P(E) \geq 0, \; \text{for all } E \in \mathcal{E} \; .
+p(E) \in \mathbb{R}, \; p(E) \geq 0, \; \text{for all } E \in \mathcal{E} \; .
 $$
 
 * Second axiom: The probability that at least one elementary event in the sample space will occur is one:
 
 $$ \label{eq:prob-ax2}
-P(\Omega) = 1 \; .
+p(\Omega) = 1 \; .
 $$
 
 * Third axiom: The probability of any countable sequence of disjoint (i.e. [mutually exclusive](/D/exc)) events $E_1, E_2, E_3, \ldots$ is equal to the sum of the probabilities of the individual events:
 
 $$ \label{eq:prob-ax3}
-P\left( \bigcup_{i=1}^\infty E_i \right) = \sum_{i=1}^\infty P(E_i) \; .
+p\left( \bigcup_{i=1}^\infty E_i \right) = \sum_{i=1}^\infty p(E_i) \; .
 $$

P/bonf-ineq.md

@@ -36,30 +36,30 @@ username: "JoramSoch"
 **Theorem:** The [probability](/D/prob) of the intersection of $A$ and $B$ is larger than or equal to the sum of the probabilities of $A$ and $B$ minus one:
 
 $$ \label{eq:bool-ineq}
-P(A \cap B) \geq P(A) + P(B) - 1 \; .
+p(A \cap B) \geq p(A) + p(B) - 1 \; .
 $$
 
 
 **Proof:** The [addition law of probability](/P/prob-add) states that, for two [events](/D/reve) $A$ and $B$, it holds true that
 
 $$ \label{eq:prob-add}
-P(A \cup B) = P(A) + P(B) - P(A \cap B) \; .
+p(A \cup B) = p(A) + p(B) - p(A \cap B) \; .
 $$
 
-Rearranging for $P(A \cap B)$, we have:
+Rearranging for $p(A \cap B)$, we have:
 
 $$ \label{eq:bool-ineq-s1}
-P(A \cap B) = P(A) + P(B) - P(A \cup B) \; .
+p(A \cap B) = p(A) + p(B) - p(A \cup B) \; .
 $$
 
 The [range of probability](/P/prob-range) is
 
 $$ \label{eq:prob-range}
-0 \leq P(E) \leq 1 \; .
+0 \leq p(E) \leq 1 \; .
 $$
 
-Thus, $P(A \cup B)$ is at most one. (This is the case, if $A$ and $B$ are [collectively exhaustive](/P/prob-exh) and their union is thus [equal to the sample space](/D/prob-ax) $\Omega$.) With this, we are able to derive a lower bound for $P(A \cap B)$, as given by the theorem:
+Thus, $p(A \cup B)$ is at most one. (This is the case, if $A$ and $B$ are [collectively exhaustive](/P/prob-exh) and their union is thus [equal to the sample space](/D/prob-ax) $\Omega$.) With this, we are able to derive a lower bound for $p(A \cap B)$, as given by the theorem:
 
 $$ \label{eq:bool-ineq-qed}
-P(A \cap B) \geq P(A) + P(B) - 1 \; .
+p(A \cap B) \geq p(A) + p(B) - 1 \; .
 $$

P/bool-ineq.md

@@ -30,60 +30,60 @@ username: "JoramSoch"
 **Theorem:** The [probability](/D/prob) of any countable sequence of [events](/D/reve) $A_1, A_2, A_3, \ldots$ is smaller than or equal to the sum of the [probabilities of the individual events](/D/prob-ax):
 
 $$ \label{eq:bool-ineq}
-P\left( \bigcup_{i=1}^\infty A_i \right) \leq \sum_{i=1}^\infty P(A_i) \; .
+p\left( \bigcup_{i=1}^\infty A_i \right) \leq \sum_{i=1}^\infty p(A_i) \; .
 $$
 
 
 **Proof:** The [addition law of probability](/P/prob-add) states that, for two [events](/D/reve) $A$ and $B$, we have:
 
 $$ \label{eq:prob-add}
-P(A \cup B) = P(A) + P(B) - P(A \cap B) \; .
+p(A \cup B) = p(A) + p(B) - p(A \cap B) \; .
 $$
 
 We will prove the statement by induction, i.e. observe that it holds for $n=1$ event and show that, if it holds for $n$ events, it also holds for $n+1$ events. For $n=1$, it is obviously true that:
 
 $$ \label{eq:bi-1}
-P\left( A_1 \right) \leq P(A_1) \; .
+p\left( A_1 \right) \leq p(A_1) \; .
 $$
 
 We shall suppose that statement \eqref{eq:bool-ineq} holds for $A_1, \ldots, A_n$:
 
 $$ \label{eq:bi-n}
-P\left( \bigcup_{i=1}^n A_i \right) \leq \sum_{i=1}^n P(A_i) \; .
+p\left( \bigcup_{i=1}^n A_i \right) \leq \sum_{i=1}^n p(A_i) \; .
 $$
 
 Using \eqref{eq:prob-add}, for $A_1, \ldots, A_n, A_{n+1}$, we then have:
 
 $$ \label{eq:bi-n+1-s1}
-  P\left( \bigcup_{i=1}^{n+1} A_i \right)
-= P\left( \bigcup_{i=1}^{n} A_i \cup A_{n+1} \right)
-= P\left( \bigcup_{i=1}^{n} A_i \right) + P(A_{n+1}) - P\left( \bigcup_{i=1}^{n} A_i \cap A_{n+1} \right) \; .
+  p\left( \bigcup_{i=1}^{n+1} A_i \right)
+= p\left( \bigcup_{i=1}^{n} A_i \cup A_{n+1} \right)
+= p\left( \bigcup_{i=1}^{n} A_i \right) + p(A_{n+1}) - p\left( \bigcup_{i=1}^{n} A_i \cap A_{n+1} \right) \; .
 $$
 
 Since, by the [first axiom of probability](/D/prob-ax), any probability is positive
 
 $$ \label{eq:prob-pos}
 \begin{split}
-P\left( \bigcup_{i=1}^{n} A_i \cap A_{n+1} \right) \geq 0 \; ,
+p\left( \bigcup_{i=1}^{n} A_i \cap A_{n+1} \right) \geq 0 \; ,
 \end{split}
 $$
 
 it follows from \eqref{eq:bi-n+1-s1} that
 
 $$ \label{eq:bi-n+1-s2}
-P\left( \bigcup_{i=1}^{n+1} A_i \right) \leq P\left( \bigcup_{i=1}^{n} A_i \right) + P(A_{n+1})
+p\left( \bigcup_{i=1}^{n+1} A_i \right) \leq p\left( \bigcup_{i=1}^{n} A_i \right) + p(A_{n+1})
 $$
 
 and using \eqref{eq:bi-n}, we get
 
 $$ \label{eq:bi-n+1-s3}
-P\left( \bigcup_{i=1}^{n+1} A_i \right) \leq \sum_{i=1}^n P(A_i) + P(A_{n+1}) \; ,
+p\left( \bigcup_{i=1}^{n+1} A_i \right) \leq \sum_{i=1}^n p(A_i) + p(A_{n+1}) \; ,
 $$
 
 such that
 
 $$ \label{eq:bi-n+1-s4}
-P\left( \bigcup_{i=1}^{n+1} A_i \right) \leq \sum_{i=1}^{n+1} P(A_i) \; .
+p\left( \bigcup_{i=1}^{n+1} A_i \right) \leq \sum_{i=1}^{n+1} p(A_i) \; .
 $$
 
 Together, \eqref{eq:bi-1} and the transition from \eqref{eq:bi-n} to \eqref{eq:bi-n+1-s4}, prove the theorem.

P/cdf-probexc.md

@@ -50,17 +50,17 @@ Using the [second axiom of probability](/D/prob-ax), we have:
 
 $$ \label{eq:cdf-prob-exc-s1}
 \begin{split}
-P(\Omega) &= 1 \\
-P\left( \left\lbrace X \mid X > x \right\rbrace \cup \left\lbrace X \mid X \leq x \right\rbrace \right) &= 1 \; .
+p(\Omega) &= 1 \\
+p\left( \left\lbrace X \mid X > x \right\rbrace \cup \left\lbrace X \mid X \leq x \right\rbrace \right) &= 1 \; .
 \end{split}
 $$
 
 Using the [third axiom of probability](/D/prob-ax), we get:
 
 $$ \label{eq:cdf-prob-exc-s2}
 \begin{split}
-P\left( \left\lbrace X \mid X > x \right\rbrace \right) + P\left( \left\lbrace X \mid X \leq x \right\rbrace \right) &= 1 \\
-P\left( \left\lbrace X \mid X > x \right\rbrace \right) &= 1 - P\left( \left\lbrace X \mid X \leq x \right\rbrace \right) \\
+p\left( \left\lbrace X \mid X > x \right\rbrace \right) + p\left( \left\lbrace X \mid X \leq x \right\rbrace \right) &= 1 \\
+p\left( \left\lbrace X \mid X > x \right\rbrace \right) &= 1 - p\left( \left\lbrace X \mid X \leq x \right\rbrace \right) \\
 \mathrm{Pr}(X > x) &= 1 - \mathrm{Pr}(X \leq x) \; .
 \end{split}
 $$

P/ind-self.md

@@ -36,23 +36,23 @@ username: "JoramSoch"
 **Theorem:** Let $E$ be a [random event](/D/reve). Then, $E$ is [independent of itself](/D/ind), if and only if its [probability](/D/prob) is zero or one:
 
 $$ \label{eq:ind-self}
-E \text{ self-independent} \quad \Leftrightarrow \quad P(E) = 0 \quad \text{or} \quad P(E) = 1 \; .
+E \text{ self-independent} \quad \Leftrightarrow \quad p(E) = 0 \quad \text{or} \quad p(E) = 1 \; .
 $$
 
 
 **Proof:** According to the definition of [statistical independence](/D/ind), it must hold that:
 
 $$ \label{eq:ind}
 \begin{split}
-P(E,E) &= P(E) \cdot P(E) \\
-P(E)   &= \left( P(E) \right)^2 \; .
+p(E,E) &= p(E) \cdot p(E) \\
+p(E)   &= \left( p(E) \right)^2 \; .
 \end{split}
 $$
 
-For $0 \leq P(E) \leq 1$, this is only fulfilled, if
+For $0 \leq p(E) \leq 1$, this is only fulfilled, if
 
 $$ \label{eq:ind-self-qed}
-P(E) = 0 \quad \text{or} \quad P(E) = 1 \; .
+p(E) = 0 \quad \text{or} \quad p(E) = 1 \; .
 $$
 
 Both is possible, since the [lower bound of probability is zero](/D/prob-ax) and the [upper bound of probability is one](/P/prob-range).

P/prob-add.md

@@ -42,30 +42,30 @@ username: "JoramSoch"
 **Theorem:** The [probability](/D/prob) of the union of $A$ and $B$ is the sum of the probabilities of $A$ and $B$ minus the probability of the intersection of $A$ and $B$:
 
 $$ \label{eq:prob-add}
-P(A \cup B) = P(A) + P(B) - P(A \cap B) \; .
+p(A \cup B) = p(A) + p(B) - p(A \cap B) \; .
 $$
 
 
 **Proof:** Let $E_1 = A$ and $E_2 = B \setminus A$, such that $E_1 \cup E_2 = A \cup B$. Then, by the [third axiom of probability](/D/prob-ax), we have:
 
 $$ \label{eq:pAoB}
 \begin{split}
-P(A \cup B) &= P(A) + P(B \setminus A) \\
-P(A \cup B) &= P(A) + P(B \setminus [A \cap B]) \; .
+p(A \cup B) &= p(A) + p(B \setminus A) \\
+p(A \cup B) &= p(A) + p(B \setminus [A \cap B]) \; .
 \end{split}
 $$
 
 Then, let $E_1 = B \setminus [A \cap B]$ and $E_2 = A \cap B$, such that $E_1 \cup E_2 = B$. Again, from the [third axiom of probability](/D/prob-ax), we obtain:
 
 $$ \label{eq:pB}
 \begin{split}
-P(B) &= P(B \setminus [A \cap B]) + P(A \cap B) \\
-P(B \setminus [A \cap B]) &= P(B) - P(A \cap B) \; .
+p(B) &= p(B \setminus [A \cap B]) + p(A \cap B) \\
+p(B \setminus [A \cap B]) &= p(B) - p(A \cap B) \; .
 \end{split}
 $$
 
 Plugging \eqref{eq:pB} into \eqref{eq:pAoB}, we finally get:
 
 $$ \label{eq:prob-add-qed}
-P(A \cup B) = P(A) + P(B) - P(A \cap B) \; .
+p(A \cup B) = p(A) + p(B) - p(A \cap B) \; .
 $$

P/prob-comp.md

@@ -42,24 +42,24 @@ username: "JoramSoch"
 **Theorem:** The [probability](/D/prob) of a complement of a set is one minus the probability of this set:
 
 $$ \label{eq:prob-comp}
-P(A^\mathrm{c}) = 1 - P(A)
+p(\overline{A}) = 1 - p(A)
 $$
 
-where $A^\mathrm{c} = \Omega \setminus A$ and $\Omega$ is the [sample space](/D/samp-spc).
+where $\overline{A} = \Omega \setminus A$ and $\Omega$ is the [sample space](/D/samp-spc).
 
 
-**Proof:** Since $A$ and $A^\mathrm{c}$ are [mutually exclusive](/D/exc) and $A \cup A^\mathrm{c} = \Omega$, the [third axiom of probability](/D/prob-ax) implies:
+**Proof:** Since $A$ and $\overline{A}$ are [mutually exclusive](/D/exc) and $A \cup \overline{A} = \Omega$, the [third axiom of probability](/D/prob-ax) implies:
 
 $$ \label{eq:pAAc}
 \begin{split}
-P(A \cup A^\mathrm{c}) &= P(A) + P(A^\mathrm{c}) \\
-P(\Omega) &= P(A) + P(A^\mathrm{c}) \\
-P(A^\mathrm{c}) &= P(\Omega) - P(A) \; .
+p(A \cup \overline{A}) &= p(A) + p(\overline{A}) \\
+p(\Omega) &= p(A) + p(\overline{A}) \\
+p(\overline{A}) &= p(\Omega) - p(A) \; .
 \end{split}
 $$
 
-The [second axiom of probability](/D/prob-ax) states that $P(\Omega) =1$, such that we obtain:
+The [second axiom of probability](/D/prob-ax) states that $p(\Omega) =1$, such that we obtain:
 
 $$ \label{eq:prob-comp-qed}
-P(A^\mathrm{c}) = 1 - P(A) \; .
+p(\overline{A}) = 1 - p(A) \; .
 $$

P/prob-emp.md

@@ -42,14 +42,14 @@ username: "JoramSoch"
 **Theorem:** The [probability](/D/prob) of the empty set is zero:
 
 $$ \label{eq:prob-emp}
-P(\emptyset) = 0 \; .
+p(\emptyset) = 0 \; .
 $$
 
 
 **Proof:** Let $A$ and $B$ be two events fulfilling $A \subseteq B$. Set $E_1 = A$, $E_2 = B \setminus A$ and $E_i = \emptyset$ for $i \geq 3$. Then, the sets $E_i$ are pairwise disjoint and $E_1 \cup E_2 \cup \ldots = B$. Thus, from the [third axiom of probability](/D/prob-ax), we have:
 
 $$ \label{eq:pB}
-P(B) = P(A) + P(B \setminus A) + \sum_{i=3}^\infty P(E_i) \; .
+p(B) = p(A) + p(B \setminus A) + \sum_{i=3}^\infty p(E_i) \; .
 $$
 
-Assume that the probability of the empty set is not zero, i.e. $P(\emptyset) > 0$. Then, the right-hand side of \eqref{eq:pB} would be infinite. However, by the [first axiom of probability](/D/prob-ax), the left-hand side must be finite. This is a contradiction. Therefore, $P(\emptyset) = 0$.
+Assume that the probability of the empty set is not zero, i.e. $p(\emptyset) > 0$. Then, the right-hand side of \eqref{eq:pB} would be infinite. However, by the [first axiom of probability](/D/prob-ax), the left-hand side must be finite. This is a contradiction. Therefore, $p(\emptyset) = 0$.

P/prob-emp2.md

@@ -30,17 +30,17 @@ username: "JoramSoch"
 **Theorem:** The [probability](/D/prob) of the empty set is zero:
 
 $$ \label{eq:prob-emp}
-P(\emptyset) = 0 \; .
+p(\emptyset) = 0 \; .
 $$
 
 
 **Proof:** Let $E_i = \emptyset$ for $i = 1,2,\ldots$ Then, $E_i \cap E_j = \emptyset \cap \emptyset = \emptyset$ for $i,j \geq 1$ and $\bigcup_{i=1}^\infty E_i = \bigcup_{i=1}^\infty \emptyset = \emptyset$. Thus, $E_1, E_2, \ldots$ is a countable sequence of disjoint events so that, with the [third axiom of probability](/D/prob-ax), it holds that:
 
 $$ \label{eq:prob-emp-qed}
 \begin{split}
-P\left(\bigcup_{i=1}^\infty E_i \right) &= \sum_{i=1}^\infty P(E_i) \\
-P(\emptyset) &= \sum_{i=1}^\infty P(\emptyset) \; .
+p\left(\bigcup_{i=1}^\infty E_i \right) &= \sum_{i=1}^\infty p(E_i) \\
+p(\emptyset) &= \sum_{i=1}^\infty p(\emptyset) \; .
 \end{split}
 $$
 
-Since, by the [first axiom of probability](/D/prob-ax), probabilities are non-negative, i.e. $P(\emptyset) \geq 0$, we are searching for a non-negative number which, when added to itself infinitely, is equal to itself. The only such number is zero, i.e. $P(\emptyset) = 0$.
+Since, by the [first axiom of probability](/D/prob-ax), probabilities are non-negative, i.e. $p(\emptyset) \geq 0$, we are searching for a non-negative number which, when added to itself infinitely, is equal to itself. The only such number is zero, i.e. $p(\emptyset) = 0$.

P/prob-exh.md

@@ -36,7 +36,7 @@ username: "JoramSoch"
 **Theorem:** Let $B_1, \ldots, B_n$ be [mutually exclusive](/D/exc) and collectively exhaustive subsets of a [sample space](/D/samp-spc) $\Omega$. Then, their [total probability](/P/prob-tot) is one:
 
 $$ \label{eq:prob-exh}
-\sum_i P(B_i) = 1 \; .
+\sum_i p(B_i) = 1 \; .
 $$
 
 
@@ -55,11 +55,11 @@ $$
 Thus, the [third axiom of probability](/D/prob-ax) implies that
 
 $$ \label{eq:prob-exh-s1}
-\sum_i P(B_i) = P(\Omega) \; .
+\sum_i p(B_i) = p(\Omega) \; .
 $$
 
 and the [second axiom of probability](/D/prob-ax) implies that
 
 $$ \label{eq:prob-exh-s2}
-\sum_i P(B_i) = 1 \; .
+\sum_i p(B_i) = 1 \; .
 $$