Oct 16-20 - Bayes’ Rule
 
1. Conditional Probabilities – PIL Ch 5
 
• Consider a chance set-up with dependent outcomes: specifically, a coin is tossed to select between two packs of cards, and then a card is drawn. We don’t suppose either that the coin is fair or that the packs are ordinary ones.
    Let H = ‘the coin comes up heads’
    Let T = ‘the coin comes up tails’
    Let AS = ‘the card drawn is the ace of spades’
• Suppose that N trials are performed – with the following numbers for the different outcomes:
   
AS - a times
   
¬AS - b times
H - n times
n + m = N
a + b = n
T - m times
  
c + d = m
   
AS - c times
   
¬AS - d times
• Let Fr(X) be the (relative) frequency with which X occurs in the trials:
    Fr(H) = n/N
    Fr(AS) = ?
(*)    Fr(H ∧ AS) = a/N
    Fr(T ∧ ¬AS) = ?
• Let Fr(X/Y) be the frequency with which X occurs in the trials in which Y occurs – i.e., given that Y occurs:
(**)    Fr(AS/H) = a/n
    Fr(¬AS/T) = ?
• From (*) and (**) we see:
    Fr(AS/H) = Fr(AS ∧ H) ÷ Fr(H)
• In the ‘long run’ probabilities should agree with frequencies – thus we have:
    1. (v)Pr(AS/H) = Pr(AS ∧ H) ÷ Pr(H)
 
2. Bayes’ Rule – PIL Ch 7
 
• Pr(X) = Pr(X ∧ Y) + Pr (X ∧ ¬Y) – why?
• Bayes’ Rule:
Pr(A/B) =
Pr(B/A) × Pr(A)
 
P(B)
   
=
Pr(B/A) × Pr(A)
 
Pr(B/A) × Pr(A) + Pr(B/¬A) × Pr(¬A)
• E.g., let the coin be fair, let the heads pack be an ordinary 52 card deck, and let the tails pack contain nothing but the ace of spades – fill in the probabilities.
    1. (a)Suppose someone performs a trial without your seeing – what is the probability that they tossed heads?
    2. (b)Suppose they show you that the card drawn was the ace of spades – what is the probability that they tossed heads?
    3. (c)Suppose that they return the card, shuffle and draw a second ace of spades – what is the probability that they tossed heads?
    Notice how the evidence keeps effects the probability as it mounts up.
• Answer Q2-3 (p.56)
• Suppose your probability of having a certain type of cancer is Pr(C). Suppose further that a test is 99% reliable – if you have cancer, it yields a positive result 99% of the time, so Pr(+/C) = 99/100. Suppose that the chance of a ‘false positive’ is 1/200: P(+/¬C) = 1/200. If you receive a positive result, what is the chance that you have cancer, given:
    1. (a)Pr(C) = 1/10,000?
    2. (b)Pr(C) = 1/1,000?
    3. (c)Pr(C) = 1/10?
    What we see is that the ‘base rate’ is crucial to evaluating such probabilities.
• Q2 (p.77).