Adding pieces of evidence

Imagine now the following variant of the previous toy experiment. After the white ball is observed, you put it again in the box, shake well and make a second extraction. You get white the second time too. Calling $ W_1$ and $ W_2$ the two observations, we have now:11
$\displaystyle \frac{P(B_1\,\vert\,W_1,W_2,I_0)}{P(B_2\,\vert\,W_1,W_2,I_0)}$ $\displaystyle =$ $\displaystyle \frac{P(W_1,W_2\,\vert\,B_1,I_0)}{P(W1,W_2\,\vert\,B_2,I_0)} \times
\frac{P(B_1\,\vert\,I_0)}{P(B_2\,\vert\,I_0)}$ (11)
  $\displaystyle =$ $\displaystyle \frac{P(W_2\,\vert\,W_1,B_1,I_0)\cdot P(W_1\,\vert\,B_1,I_0)}
...ot P(W_1\,\vert\,B_2,I_0)}
\frac{P(B_1\,\vert\,I_0)}{P(B_2\,\vert\,I_0)}$ (12)
  $\displaystyle =$ $\displaystyle \frac{P(W_2\,\vert\,B_1,I_0)}{P(W_2\,\vert\,B_2,I_0)}
\times \fra...
\times \frac{P(B_1\,\vert\,I_0)}{P(B_2\,\vert\,I_0)}$ (13)
  $\displaystyle =$ $\displaystyle \frac{P(W_2\,\vert\,B_1,I_0)}{P(W_2\,\vert\,B_2,I_0)}
\times \frac{P(B_1\,\vert\,W_1,I_0)}{P(B_2\,\vert\,W_1,I_0)}\,,$ (14)

that, using the compact notation introduced above, we can rewrite in the following enlighting forms. The first is [Eq. (14)]
$\displaystyle O_{1,2}(W_1,W_2,I)$ $\displaystyle =$ $\displaystyle \tilde O_{1,2}(W_2,I) \times O_{1,2}(W_1,I)\,,$ (15)

that is, the final odds after the first inference become the initial odds of the second inference (and so on, if there are several pieces of evidence). Therefore, beginning from a situation in which $ B_1$ was thirteen times more credible than $ B_2$ is exactly equivalent to having started from unitary odds updated by a factor 13 due to a piece of evidence.

The second form comes from Eq. (13):

$\displaystyle O_{1,2}(W_1,W_2,I)$ $\displaystyle =$ $\displaystyle \tilde O_{1,2}(W_1,I) \times \tilde O_{1,2}(W_2,I)
\times O_{1,2}(I)$ (16)
  $\displaystyle =$ $\displaystyle \tilde O_{1,2}(W_1,W_2,I) \times O_{1,2}(I)\,,$ (17)

$\displaystyle \tilde O_{1,2}(W_1,W_2,I)$ $\displaystyle =$ $\displaystyle \tilde O_{1,2}(W_1,I) \times
\tilde O_{1,2}(W_2,I)\,:$ (18)

Bayes factors due to independent13pieces of evidence multiply. That is, two independent pieces of evidence ($ W_1$ and $ W_2$) are equivalent to a single piece of evidence (` $ W_1\cap W_2$'), whose Bayes factor is the product of the individual ones. In our case $ \tilde O_{1,2}(W_1\cap W_2,I)=169$.

In general, if we have several hypotheses $ H_i$ and several independent pieces of evidence, $ E_1$, $ E_2$, ..., $ E_n$, indicated all together as $ {\mbox{\boldmath $E$}}$, then Eq. (4) becomes

$\displaystyle O_{i,j}({\mbox{\boldmath$E$}},I)$ $\displaystyle =$ $\displaystyle \left[\,\prod_{k=1}^n
\tilde O_{i,j}(E_k,I)
\times O_{i,j}(I)\,,$ (19)

$\displaystyle \tilde O_{i,j}({\mbox{\boldmath$E$}},I)$ $\displaystyle =$ $\displaystyle \prod_{k=1}^n
\tilde O_{i,j}(E_k,I)\,,$ (20)

where $ \prod$ stand for `product' (analogous to $ \sum$ for sums).

Giulio D'Agostini 2010-09-30