0

This equation occurs in a paper I was looking at - a derivation is given, but alas in "electronic supplement" to which i have no access. We are to assume that humans receive two cues x and y which take numerical values. There are exactly two states of the environment 1 and 2. The interest is in the probabilities $P(1|x,y)$ which is the probability that the environment is in state 1 given the human receives cues x and y. It is asserted that Bayes rule gives: \begin{equation}\tag{1} P(1 | x,y) = \dfrac{P(x|1)\cdot P(y|1)}{P(x|1)\cdot P(y|1) + P(x|2)\cdot P(y|2)} \end{equation} I assumed that this should emerge from some simple manipulation like: \begin{split}P(1 | x,y) &= \dfrac{P(x,y|1)\cdot P(1)}{P(x,y)}\\ &=\dfrac{P(x,y|1)\cdot P(1)}{P(x,y|1)\cdot P(1) + P(x,y|2)\cdot P(2))}\end{split} or that given in in this question but if it does I can't see it.

So is formula (1) a version of Bayes Theorem?

peter2108
  • 188
  • 1
  • 8

1 Answers1

1

I learn that it is an assumption of conditional independence that \begin{equation}\tag{1} P(x,y |1) = P(x|1)\cdot P(y|1) \end{equation} and that this is a Bayesian assumption. The paper has the word "Bayesian" in its title so I think I can reckon on that assumption. So I can continue: \begin{split}P(1 | x,y) &= \dfrac{P(x,y|1)\cdot P(1)}{P(x,y)}\\[1ex] &=\dfrac{P(x,y|1)\cdot P(1)}{P(x,y|1)\cdot P(1) + P(x,y|2)\cdot P(2))}\\[2ex] &= \dfrac{P(x|1)\cdot P(y|1)\cdot P(1)}{P(x|1)\cdot P(y|1)\cdot P(1) + P(x|2)\cdot P(y|2)\cdot P(2))} \end{split} There is an earlier remark in the paper "Thus, over the long run, the environment is equally likely to be in each state" from which I conclude that $P(1)=P(2)$ which gives the equation (1).

peter2108
  • 188
  • 1
  • 8