Bayesian Odds

Bayes' theorem

\[ p(D|C) = \frac{p(C|D)\cdot p(D)}{p(C)}. \]

joins in one formula the a priori probabilities of events \(C\) and \(D\) and the posterior probabilities \(p(C|D)\) of event \(C\) provided \(D\) took place - and \(p(D|C)\), the probability of \(D\) provided that \(C\) is known to take place.

On occasion when there are two events, say \(A\) and \(B\), whose comparative posterior probabilities are of interest, it may be more advantageous to consider the ratios, i.e. the odds:

\[ \frac{p(A|C)}{p(B|C)} = \frac{p(C|A)}{p(C|B)}\cdot \frac{p(A)}{p(B)}. \]

Ward Edwards gives a simple example where the latter formula comes in handy:

There are two bags, one containing \(700\) red and \(300\) blue chips, the other containing \(300\) red and \(700\) blue chips. Flip a fair coin to determine which one of the bags to use. Chips are drawn with replacement. In \(12\) samples, \(8\) red and \(4\) blue chips showed up. What is the probability that it was the predominantly red bag?

Author Edwards writes

Clearly the sought probability is higher than \(0.5\). Please don't continue reading till you have written down your esimate.

Better yet, try to derive your estimate formally. Make use of Bayes' theorem, of course.

Solution

References

  1. W. Edwards, Conservatism in Human Information Processing, in Judegment under uncertainty: Heuristics and biases, D. Kahneman, P. Slovic, A. Tversky (editors), Cambridge University Press, 1982

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Copyright © 1996-2017 Alexander Bogomolny

There are two bags, one containing \(700\) red and \(300\) blue chips, the other containing \(300\) red and \(700\) blue chips. Flip a fair coin to determine which one of the bags to use. Chips are drawn with replacement. In \(12\) samples, \(8\) red and \(4\) blue chips showed up. What is the probability that it was the predominantly red bag?

Solution

Let \(A\) be the event of selecting the first bag. Let \(B\) be the event of selecting the second bag. Finally, let C be the result of the experiment, i.e., drawing \(8\) red and \(4\) blue chips from the selected bag. Clearly,

\begin{align} p(C|A) & = \big(\frac{7}{10}\big)^{8}\big(\frac{3}{10}\big)^{4}{12 \choose 8} \\ p(C|B) & = \big(\frac{7}{10}\big)^{4}\big(\frac{3}{10}\big)^{8}{12 \choose 4} \end{align}

so that \(\displaystyle\frac{p(C|A)}{p(C|B)} = \big(\frac{7}{3}\big)^{4} \approx 29.642\). Now, \(p(A) = p(B) = 0.5\), implying that

\[ \frac{p(A|C)}{p(B|C)} \approx {29.642}. \]

This is the odds of selecting one of the bags relative to the other. From \(p(A|C) + p(B|C) = 1\), it follows that \(p(A|C) \approx \displaystyle\frac{29.642}{30.642} \approx 0.967\).

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Copyright © 1996-2017 Alexander Bogomolny

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