I tried to rewrite the equation given below. I get stuck getting rid of the $ P(n|z_{1:t})$ on the left side. How can this be done? $$ P(n|z_{1:t}) = \left[1+ \frac{1-P(n|z_{t})}{P(n|z_t)} \frac{1-P(n|z_{1:t-1})}{P(n|z_{1:t-1})} \frac{P(n)}{1-P(n)}\right]^{-1} $$ With: \begin{equation} L(n) = log\left[\frac{P(n)}{1-P(n)}\right] \end{equation}
Can be rewritten to (how do I get to this?): \begin{equation} L(n|z_{1:t}) = L(n|z_{1:t-1}) + L(n|z_t) \end{equation}
From: 'OctoMap: an efficient probabilistic 3D mapping framework based on octrees'
Evaluating the reciprocal operation on the RHS of the first equation, we can express $P(n|z_{1:t})$ as $$P(n|z_{1:t})=\frac{P(n|z_t)P(n|z_{1:t-1})(1-P(n))}{P(n|z_t)P(n|z_{1:t-1})(1-P(n))+(1-P(n|z_t))(1-P(n|z_{1:t-1}))P(n)}$$ Thus $$1-P(n|z_{1:t})=\frac{(1-P(n|z_t))(1-P(n|z_{1:t-1}))P(n)}{P(n|z_t)P(n|z_{1:t-1})(1-P(n))+(1-P(n|z_t))(1-P(n|z_{1:t-1}))P(n)}$$ resulting in $$\frac{P(n|z_{1:t})}{1-P(n|z_{1:t})}=\frac{P(n|z_t)P(n|z_{1:t-1})(1-P(n))}{(1-P(n|z_t))(1-P(n|z_{1:t-1}))P(n)}$$ Taking the logarithm of both sides leads to $$\log\left[\frac{P(n|z_{1:t})}{1-P(n|z_{1:t})}\right]=\log\left[\frac{P(n|z_{1:t-1})}{1-P(n|z_{1:t-1})}\right]+\log\left[\frac{P(n|z_{t})}{1-P(n|z_{t})}\right]-\log\left[\frac{P(n)}{1-P(n)}\right]$$ Using the notation $L(n)$ we have $$ L(n|z_{1:t})=L(n|z_{1:t-1})+L(n|z_{t})-L(n)$$ Examining the paper referenced (page $5$, bottom of first column of text), a uniform prior probability is assumed where $P(n)=0.5$, leading to $L(n)=0$, so that we have the desired update equation:- $$ L(n|z_{1:t})=L(n|z_{1:t-1})+L(n|z_{t})$$