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I chanced on this article on wikipedia on Bayesian search. In the mathematics section, it states how the posteriors are estimated. While I understand how $p^{'}$ is calculated, I can't seem to figure out how the $r^{'}$ are updated. How does the posterior on other cells on the grid change given a no-find on a given grid cell? If someone could give any pointers it would be helpful. Thanks

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I hope I can help you. First, remember the most simple statement of Bayes' theorem: $$ P(A|B)=\frac{P(B|A)\cdot P(A)}{P(B)} $$ Let's assume that we have only two grid cells to search. Denote the prior probability that the wreck is in grid cell 1 as $P(W_{1})=p$, the prior probability that the wreck is in grid cell 2 as $P(W_{2})=r$, the probability of finding the wreck in a certain grid cell if it is really there as $P(F|W)=q$. Let's say that grid cell 1 has been searched but no wreck has been found. The probability that the wreck is in grid cell 1 although it hasn't been found there is given by: $$ p'=P(W_{1}| \overline{ F}_{1})=\frac{P(\overline{ F}_{1}|W_{1})\cdot P(W_{1})}{P(\overline{F}_{1})}=\frac{P(\overline{ F}_{1}|W_{1})\cdot P(W_{1})}{P(\overline{F}_{1}|W_{1}) \cdot P(W_{1}) +P(\overline{F}_{1}|\overline{W}_{1}) \cdot P(\overline{W}_{1})} = \frac{(1-q)\cdot p}{(1-q)\cdot p + 1\cdot (1-p)}= \frac{(1-q)\cdot p}{1-pq} $$ For the other grid cell 2, we can proceed in the same manner: $$ r'=P(W_{2}| \overline{ F}_{1})=\frac{P(\overline{ F}_{1}|W_{2})\cdot P(W_{2})}{P(\overline{F}_{1})}=\frac{1\cdot r}{(1-q)\cdot p + 1\cdot (1-p)}=\frac{1\cdot r}{1-pq} $$ Note that the denominator is the same for both situations. Also the probability of not finding the wreck in grid cell 1 when it is in fact in grid 2 (or in any other grid cell apart from cell 1) is assumed to be 1 (i.e. $P(\overline{ F}_{1}|W_{2})=1$).

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From what I'm reading on that page, I would set the new value of every other cell to be r'. r' would be calculated for each square based upon what the current probability of the square is plugged into that equation as r. Then just plug p and q in.

Does that make sense or am I missing something?

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  • $\begingroup$ Not quite. Lets say there is another cell with its prior being set to $r$, then its posterior is $r\frac{1}{1 - pq}$. How is this arrived at? I understand why it would change and go up, but don't follow how it is calculated to that expression $\endgroup$
    – broccoli
    Jul 31 '13 at 0:41

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