Building a recurrence relation for Bayes' Theorems

Metronome

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Jun 12, 2018
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I have worked out a few successive derivations of Bayes' Theorem...

Pr(W)=Pr(W)\Pr(W) = \Pr(W)
Pr(W  C1) = Pr(C1  W)Pr(W)Pr(C1)\Pr(W\ |\ C_1)\ =\ \frac{\Pr(C_1\ |\ W)\Pr(W)}{\Pr(C_1)}
Pr(W  (C1  C2)) = Pr(C2  (W  C1))Pr(W  C1)Pr(C2  C1)\Pr(W\ |\ (C_1\ \cap\ C_2))\ =\ \frac{\Pr(C_2\ |\ (W\ \cap\ C_1))\Pr(W\ |\ C_1)}{\Pr(C_2\ |\ C_1)}
Pr(W  (C1  C2  C3))=Pr(C3  (W  C1  C2))Pr(W  (C1  C2))Pr(C3  (C1  C2))\Pr(W\ |\ (C_1\ \cap\ C_2\ \cap\ C_3)) = \frac{\Pr(C_3\ |\ (W\ \cap\ C_1\ \cap\ C_2))\Pr(W\ |\ (C_1\ \cap\ C_2))}{\Pr(C_3\ |\ (C_1\ \cap\ C_2))}

I think the pattern is Pr(W  i=1n1Ci) = Pr(Cn1  (W  i=1n2Ci))Pr(W  i=1n2Ci)Pr(Cn1  i=1n2Ci)\Pr(W\ |\ \bigcap\limits_{i = 1}^{n - 1} C_i)\ =\ \frac{\Pr(C_{n - 1}\ |\ (W\ \cap\ \bigcap\limits_{i = 1}^{n - 2} C_i))\Pr(W\ |\ \bigcap\limits_{i = 1}^{n - 2} C_i)}{\Pr(C_{n - 1}\ |\ \bigcap\limits_{i = 1}^{n - 2} C_i)}.

I have the intuition that this should be interpretable as a recurrence relation. As far as I can tell, Pr(W  i=1n1Ci)\Pr(W\ |\ \bigcap\limits_{i = 1}^{n - 1} C_i) is playing the role of ana_n, and Pr(W  i=1n2Ci)\Pr(W\ |\ \bigcap\limits_{i = 1}^{n - 2} C_i) is playing the role of an1a_{n - 1}. However, it is less clear what to do with the other factors. Should I leave them as probabilities, yielding an= Pr(Cn1  (W  i=1n2Ci))Pr(Cn1  i=1n2Ci)an1a_n =\ \frac{\Pr(C_{n - 1}\ |\ (W\ \cap\ \bigcap\limits_{i = 1}^{n - 2} C_i))}{\Pr(C_{n - 1}\ |\ \bigcap\limits_{i = 1}^{n - 2} C_i)} a_{n - 1}, or is there some transformation I need to apply to the coefficient as well? Or should I leave everything in probability notation? The coefficient is certainly not constant in this form, yet I believe the solution should be simple (probably related to Bayesean updating formulas). How can I massage this into an analytically solvable recurrence relation?
 
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