Page 170 in Philip Tetlock's et al. Superforecasting book shows Bayes' theorem in odds form as:

$$\frac{P (H|D)}{P (\neg H|D)} = P (D|H) P (D|\neg H) \frac{ P (H)}{P (\neg H)}$$

Posterior Odds = Likelihood Ratio • Prior Odds

Shouldn't the Likelihood Ratio be $\frac{P (D|H)}{P (D|\neg H)}$, i.e. division instead of multiplication?

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    $\begingroup$ Quite right: this is a typo. $\endgroup$ – Xi'an Nov 26 '15 at 14:02

So this doesn't remain unanswered (except in comments) let's derive the odds ratio form from scratch:

We know $P(A|B)=\frac{P(B|A)\cdot P(A)}{P(B)}$, so:

$$\frac{P(A|B)}{P(\bar{A}|B)}=\frac{\frac{P(B|A)\cdot P(A)}{P(B)}}{\frac{P(B|\bar{A})\cdot P(\bar{A})}{P(B)}}$$

$$=\frac{P(B|A)\cdot P(A)}{{P(B|\bar{A})\cdot P(\bar{A})}}$$

$$=\frac{P(B|A)}{{P(B|\bar{A})}}\cdot \frac{P(A)}{P(\bar{A})}$$

So yes, it's a typo as you suggest and should be divided rather than multiplied.


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