(New page: If we let O = event object is present and N = event object not present we can set up the following <br> <math>\textrm{Pr}[D_k | O] =\tbinom nk (p_2)^k(1-p_2)^k</math><br> <math>\textrm{Pr...) |
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| − | If we let O = event object is present | + | If we let <math>O</math> = event object is present, <math>N</math> = event object not present, and <math>D_k</math> be k detections (of n echoes) we can set up the following:<br> |
<math>\textrm{Pr}[D_k | O] =\tbinom nk (p_2)^k(1-p_2)^k</math><br> | <math>\textrm{Pr}[D_k | O] =\tbinom nk (p_2)^k(1-p_2)^k</math><br> | ||
Latest revision as of 11:00, 11 November 2008
If we let $ O $ = event object is present, $ N $ = event object not present, and $ D_k $ be k detections (of n echoes) we can set up the following:
$ \textrm{Pr}[D_k | O] =\tbinom nk (p_2)^k(1-p_2)^k $
$ \textrm{Pr}[D_k | N] =\tbinom nk (p_1)^k(1-p_1)^k $
and what we're looking for is to find the value of k that means that it is more likely that an ojbect is there rather than not there. In other words:
$ \frac{\textrm{Pr}[D_k | O]} {\textrm{Pr}[D_k | N]} > 1 $
If you set up the equations appropriately, you can take the log of both sides to find a value of k. It gets pretty messy, but I got the following:
$ k > \frac{-n \cdot ln(\frac{1-p_2}{1-p_1})}{ln(\frac{p_2(1-p_1)}{p_1(1-p_2)})} $
