Bifalgorithm post · Brisk · Chapter 1

What rules must a valid probability measure satisfy?

What rules must a valid probability measure satisfy? (Brisk · Chapter 1) — "In order for a function to be a valid probability measure, it must satisfy several…

Ideas in this flowchart

  • Reglas para que una función sea una medida de probabilidad válida
  • Tres reglas fundamentales
  • Pr(Ω) = 1: el conjunto de todos los resultados posibles tiene probabilidad 1
  • Pr(A) ≥ 0 para todo A ⊆ Ω: no existen probabilidades negativas
  • Pr(A ∪ B) = Pr(A) + Pr(B) − Pr(A ∩ B): propiedad de aditividad
  • La unión representa un «o» inclusivo, por lo que ambos eventos pueden ocurrir
  • Se resta Pr(A ∩ B) porque la intersección se contó dos veces
  • Ejemplo U y R: 0.4 + 0.7 − 0.2 = 0.9
  • Ejemplo U y F: como U ∩ F = U, 0.4 + 0.5 − 0.4 = 0.5
  • Si dos eventos son mutuamente excluyentes, su intersección es vacía y no hay nada que restar
  • Ejemplo U y C: 0.4 + 0.1 − 0 = 0.5
  • Reglas que se siguen lógicamente
  • Pr(∅) = 0
  • Pr(complemento de A) = 1 − Pr(A)
  • Si A ⊆ B, entonces Pr(A) ≤ Pr(B)
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What rules must a valid probability measure satisfy?

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Brisk

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"In order for a function to be a valid probability measure, it must satisfy several rules: 1. Pr(Ω) = 1. 2. Pr(A) ≥ 0 for all A ⊆ Ω. 3. Pr(A ∪ B) = Pr(A) + Pr(B) − Pr(A ∩ B). Rule 1 means that something has to happen: if an event includes every possible outcome, then there is a probability of 1 that the event will occur. Rule 2 says there are no negative probabilities. Rule 3 is called the additivity property. In computer science, “or” will almost always mean an inclusive or unless explicitly noted otherwise, so Pr(U ∪ R) asks for the probability that either an underage winner or a rock star wins, including the possibility that both descriptions apply. We add Pr(U) and Pr(R), but then subtract Pr(U ∩ R), because the intersection was double-counted. In the All-time Idol example, Pr(U) = .4 and Pr(R) = .7, and Kelly’s probability of .2 is counted twice. The probability is therefore .4 + .7 − .2 = .9. For an underage or female winner, Pr(U) = .4 and Pr(F) = .5, while U ∩ F is U, so the result is .4 + .5 − .4 = .5. If two events are mutually exclusive, their intersection is empty and there is nothing to subtract. For example, if C is the event that a country singer wins, then C contains only Carrie, so U and C are mutually exclusive: Pr(U ∪ C) = .4 + .1 − 0 = .5. Other rules that follow logically are: 4. Pr(∅) = 0. 5. Pr(the complement of A) = 1 − Pr(A). 6. Pr(A) ≤ Pr(B) if A ⊆ B."

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