You initially place probability p< 5, on this occuring. Certainly one might think that your expected gain from this bet is (1-p)10-p10or(1-2p)10>0
You initially place probability p.5, on this occuring. Certainly one might think that your expected gain from this bet is (1-p)10-p10 or (1-2p)100
Your opponent would offer the bet only if (2p-1)10>0, so if p>5>p, it seems like you should have a horse race But this can't actually happen To see this let's put some structure on learning Remeber bayes Rule P(B|4)=P(B∩AP(A) This is always the key to understanding belief formations So more information structure The unconditional probability of a lemon squirting is p If the lemon is squirting, then every person receives a positive signal with probability Z If the lemon is not squirting then every person recieves a positive signal with
Your opponent would offer the bet only if (2p’-1)100, so if p’.5p, it seems like you should have a horse race. But this can’t actually happen. To see this, let’s put some structure on learning. Remeber Bayes’ Rule PB|A PB A/PA This is always the key to understanding belief formations. So more information structure. The unconditional probability of a lemon squirting is p. If the lemon is squirting, then every person receives a positive signal with probability z. If the lemon is not squirting then every person recieves a positive signal with
probability 1-z The probability of lemon squirting conditional upon recieving a positive signal is zp/(zp+(1-z(1-p)) There are actually four possible states of the world Signal, Lemon Signal, No Lemon No Signal, lemon No signal, no lemon (1) signal, no lemon, (2)signal, lemon, (3)no signal, no lemon, 4)no signal, lemon The probabilties of each of these states are(1)(1-z)(1-p),(2)Zp,(③3)z(1-p),(4) (1-z)p
probability 1-z. The probability of lemon squirting conditional upon recieving a positive signal is zp/(zp(1-z)(1-p)). There are actually four possible states of the world: Signal, Lemon Signal, No Lemon No Signal, Lemon No Signal, No Lemon (1) signal, no lemon, (2) signal, lemon, (3) no signal, no lemon, (4) no signal, lemon. The probabilties of each of these states are (1) (1-z)(1-p), (2) zp, (3) z(1-p), (4) (1-z)p
If you havent received a signal-your probability assessment is(1-z)p/ (z+p-2pz If you have recieved a signal, your probability assessment is zp/(1-z-p+2pz) If there are two people, the situation is even harder- there are actually eight states of the world based on the signal that each person has recieved
If you haven’t received a signal– your probability assessment is (1-z)p/(zp-2pz). If you have recieved a signal, your probability assessment is zp/(1-z-p2pz). If there are two people, the situation is even harder– there are actually eight states of the world based on the signal that each person has recieved
In this gambling example we need to deal with two people So we have the following states of the world 2(1-p)(1-z)2 pz(1-2)(1-p)z(1-z) pz(1-)(1-p)z(1-z) p(1-2)2(1-p) If the other person has bet- what should you assume? So conditional upon you not getting a
In this gambling example we need to deal with two people: So we have the following states of the world: pz2 1 p1 z2 pz1 z 1 pz1 z pz1 z 1 pz1 z p1 z2 1 pz2 If the other person has bet– what should you assume? So conditional upon you not getting a