INGENIA

PRB-02

Bayes

Invert the conditional.

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ProbabilityBayes

Governing equation

P(A∣B)=P(B∣A)P(A)/P(B)P(A|B)=P(B|A)P(A)/P(B)

P(A)∈[0.001,0.5]P(A) \in \left[0.001, 0.5\right]

where

P(B|A)
P(B|A) (—)
P(A)
P(A) (—)
P(B)
P(B) (—)
P(A|B)
P(A|B) (—)

The figure, part by part

Each numbered marker wears the color of its card. Hover a card to light the marker. Tap a marker to jump to that card. The slider walks a red point along the curve; Animate drawing traces it like a pen.

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①

Governing law — red

P(A∣B)=P(B∣A)P(A)/P(B)P(A|B)=P(B|A)P(A)/P(B)

P(A)∈[0.001,0.5]P(A) \in \left[0.001, 0.5\right]

P(A)=0.001P(A) = 0.001P(A∣B)=0.018P(A|B) = 0.018

②

Domain of the variable — blue

P(A)min⁡=0.001,P(A)max⁡=0.5P(A)_{\min} = 0.001,\quad P(A)_{\max} = 0.5

P(A)∈[0.001,0.5]P(A) \in \left[0.001, 0.5\right]

P(A)=0.001P(A) = 0.001P(A)=0.001P(A) = 0.001

③

Response curve — green

P(A∣B)=P(B∣A)P(A)/P(B)P(A|B)=P(B|A)P(A)/P(B)

P(A)∈[0.001,0.5]P(A) \in \left[0.001, 0.5\right]

P(A)=0.001P(A) = 0.001P(A∣B)=0.018P(A|B) = 0.018

④

Rate of change — amber

dP(A∣B)dP(A)\frac{\mathrm{d}P(A|B)}{\mathrm{d}P(A)}

P(A)∈[0.001,0.5]P(A) \in \left[0.001, 0.5\right]

P(A)=0.004P(A) = 0.004P(A∣B)′=18P(A|B)^{\prime} = 18

Lecture brief

Historical brief

Bernoulli trials, Poisson, Gauss, Bayes and Shannon entropy turned chance into a calculus of belief and noise. The lab computes likelihood, interval and information in SI-free counts. This sheet (PRB-02 — Bayes) is the form associated with Bayes. Working symbols: P(B∣A)P(B|A), P(A)P(A), P(B)P(B) →\rightarrow P(A∣B)P(A|B). Invert the conditional.

Purpose

Purpose: compute P(A∣B)P(A|B) from P(B∣A)P(B|A), P(A)P(A), P(B)P(B) in Probability & statistics via P(A∣B)=P(B∣A)P(A)/P(B)P(A|B)=P(B|A)P(A)/P(B) Invert the conditional. Use it when a real probability & statistics question must be answered in SI before a code check.

Live realistic example

In symbols

Live case. Given P(B∣A)=0.900 —P(B|A) = 0.900\,\mathrm{—}, P(A)=0.010 —P(A) = 0.010\,\mathrm{—}, P(B)=0.050 —P(B) = 0.050\,\mathrm{—}, the governing relation P(A∣B)=P(B∣A)P(A)/P(B)P(A|B)=P(B|A)P(A)/P(B) yields P(A∣B)=0.1800 —P(A|B) = 0.1800\,\mathrm{—}. Invert the conditional. Move a slider: the numbers are this situation, not a canned story.

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Inputs

Outputs

  • P(A|B) P(A|B)0.1800 —
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Narration of this film

Invert the conditional.

Invert the conditional.

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