Stochastic Modeling

Author: marbetschar

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     - title: My Areas
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projects/masters-degree-in-data-science/stochastic-modeling.md

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+# Stochastic Modeling
+
+:::{note} Historical Context
+**1654**: Blaise Pascal and Pierre de Fermat  
+Chevalier de Méré wrote a letter to Pascal, which became the origin of probability theory stemming from gambling problems.
+
+**1932**: Kolmogorov strictly defined probability theory.
+:::
+
+## Review of Probability Theory
+
+Modeling of a random experiment involves three key components:
+
+- **Sample space** $\Omega$: the set of possible outcomes
+- **Events** $A \subseteq \Omega$: where $\mathcal{A}$ is the set of all measurable events
+- **Probability Measure** $\mathbb{P}$: see definition below
+
+### Definition of a Probability Measure
+
+:::{prf:definition} Probability Measure
+:label: def-prob-measure
+
+A function $\mathbb{P}: \mathcal{A} \rightarrow \mathbb{R}$ with the following three properties:
+
+**A1)** $0 \le \mathbb{P}(E) \le 1$ for any $E \in \mathcal{A}$
+
+**A2)** $\mathbb{P}(\Omega) = 1$
+
+**A3)** If $E_1, E_2, \dots$ are pairwise disjoint events (i.e., $E_i \cap E_j = \emptyset$ if $i \neq j$), then
+
+$$
+\mathbb{P}\left(\bigcup_{i=1}^{\infty} E_i\right) = \sum_{i=1}^{\infty} \mathbb{P}(E_i)
+$$ (eq:countable-additivity)
+
+Any function satisfying A1, A2, A3 is called a **Probability Function**. These axioms (A1–A3) are known as the **Kolmogorov Axioms**.
+:::
+
+:::{note} Interpretation
+- $\mathbb{P}(E) = 0$: almost impossible
+- $\mathbb{P}(E) = 1$: almost certain
+:::
+
+### Example: Fair Die Tossing
+
+:::{prf:example} Fair Die
+:label: ex-fair-die
+
+Consider a fair six-sided die:
+
+- Sample space: $\Omega = \{1, 2, 3, 4, 5, 6\}$
+- Event space: $\mathcal{A} = \mathcal{P}(\Omega)$ (power set with $|\mathcal{A}| = |\mathcal{P}(\Omega)| = 2^6$)
+- Probability: $\mathbb{P}(\{i\}) = \frac{1}{6}$ for $i=1, \dots, 6$
+
+Let $A = \{2, 4, 6\}$ be the event "roll an even number". Then:
+
+$$
+\mathbb{P}(A) \stackrel{\text{A3}}{=} \mathbb{P}(\{2\}) + \mathbb{P}(\{4\}) + \mathbb{P}(\{6\}) = \frac{3}{6} = \frac{1}{2}
+$$
+
+This follows the classical probability formula:
+
+$$
+\mathbb{P}(A) = \frac{\text{# of favorable outcomes}}{\text{# of all outcomes}}
+$$
+:::
+
+### Properties of a Probability Measure
+
+:::{prf:proposition} Basic Properties
+:label: prop-prob-properties
+
+For any probability measure $\mathbb{P}$ and events $A, B \in \mathcal{A}$:
+
+a) $\mathbb{P}(\emptyset) = 0$
+
+b) $\mathbb{P}(A^c) = 1 - \mathbb{P}(A)$
+
+c) $\mathbb{P}(A \setminus B) = \mathbb{P}(A) - \mathbb{P}(A \cap B)$
+
+d) $\mathbb{P}(A \cup B) = \mathbb{P}(A) + \mathbb{P}(B) - \mathbb{P}(A \cap B)$
+
+e) $\mathbb{P}(A) \le \mathbb{P}(B)$ if $A \subseteq B$
+:::
+
+:::{prf:proof}
+We prove properties (a) and (b):
+
+**Part (a):** 
+
+$$
+1 \stackrel{\text{A2}}{=} \mathbb{P}(\Omega) = \mathbb{P}(\Omega \cup \emptyset) \stackrel{\text{A3}}{=} \mathbb{P}(\Omega) + \mathbb{P}(\emptyset) \stackrel{\text{A2}}{=} 1 + \mathbb{P}(\emptyset)
+$$
+
+Therefore, $\mathbb{P}(\emptyset) = 0$.
+
+**Part (b):** Since $\Omega = A \cup A^c$ and the events are disjoint, we have:
+
+$$
+1 = \mathbb{P}(\Omega) = \mathbb{P}(A) + \mathbb{P}(A^c)
+$$
+
+Therefore, $\mathbb{P}(A^c) = 1 - \mathbb{P}(A)$.
+
+The remaining properties follow similarly.
+:::
+
+## Conditional Probabilities
+
+Consider events $E, F \in \mathcal{A}$ with $\mathbb{P}(F) > 0$.
+
+**Question:** What is the probability of $E$ if we know that $F$ occurs?
+
+- Relevant cases: $\omega \in F$
+- Favorable cases: $\omega \in E \cap F$
+
+:::{prf:definition} Conditional Probability
+:label: def-conditional-prob
+
+The **conditional probability** of $E$ given $F$ is defined as:
+
+$$
+\mathbb{P}(E|F) = \frac{\mathbb{P}(E \cap F)}{\mathbb{P}(F)}
+$$ (eq:conditional-prob)
+:::
+
+### Example: Throwing Two Dice
+
+:::{prf:example} Two Dice
+:label: ex-two-dice
+
+Sample space: $\Omega = \{(i, j) : i, j = 1, 2, \dots, 6\}$
+
+Let $E$ be the event "sum equals 6". Then $\mathbb{P}(E) = \frac{5}{36}$.
+
+Now assume we know that $F$ = "first toss equals 2". Then:
+
+- $\mathbb{P}(F) = \frac{1}{6}$
+- $\mathbb{P}(E \cap F) = \mathbb{P}(\{(2,4)\}) = \frac{1}{36}$
+
+Therefore:
+
+$$
+\mathbb{P}(E|F) = \frac{\mathbb{P}(E \cap F)}{\mathbb{P}(F)} = \frac{1/36}{1/6} = \frac{1}{6} \quad \left[ > \frac{5}{36} \right]
+$$
+
+Notice that knowing the first die increases the probability of the sum being 6.
+:::
+
+:::{important} Multiplication Rule
+From the definition of conditional probability, we obtain:
+
+$$
+\mathbb{P}(E \cap F) = \mathbb{P}(F) \cdot \mathbb{P}(E|F)
+$$ (eq:multiplication-rule)
+:::
+
+## Law of Total Probability
+
+:::{prf:theorem} Law of Total Probability (General Form)
+:label: thm-total-prob
+
+Let $\{A_1, A_2, \dots, A_n\}$ be a partition of $\Omega$, i.e., $\Omega = A_1 \cup A_2 \cup \dots \cup A_n$ with pairwise disjoint sets.
+
+For any event $E \in \mathcal{A}$:
+
+$$
+E = E \cap \Omega = E \cap (A_1 \cup A_2 \cup \dots \cup A_n)
+$$
+
+Therefore:
+
+$$
+\mathbb{P}(E) \stackrel{\text{A3}}{=} \sum_{i=1}^{n} \mathbb{P}(E \cap A_i) = \sum_{i=1}^{n} \mathbb{P}(E|A_i) \cdot \mathbb{P}(A_i)
+$$ (eq:total-prob)
+:::
+
+:::{prf:theorem} Simplified Law of Total Probability
+:label: thm-total-prob-simple
+
+For any event $A$ with $\mathbb{P}(A) > 0$, we have the partition $\Omega = A \cup A^c$. Thus:
+
+$$
+\mathbb{P}(E) = \mathbb{P}(E|A) \cdot \mathbb{P}(A) + \mathbb{P}(E|A^c) \cdot \mathbb{P}(A^c)
+$$ (eq:total-prob-simple)
+:::