# More about Beliefs. We saw how articulating a Symmetry Belief could benefit us in both constructing and sensibly interpreting the results of a hypothesis test.   Could an even better articulation of our beliefs benefit us even more?  Yes!!

# Recall from last Wednesday (May 3rd, "RMay3.txt"), that we could test whether data came for an American or French Roulette table.  We was the set up:

#Parameters 

p0=18/38 
p1=18/37
q0=1-p0
q1=1-p1

#We learned how to do a cost benefit analysis and BEFORE could sensibly fix:

#N=Sample Size
#alpha = P(pd=p1| p=p0) = P(pThat>pstar | p=p0) 
#beta = P(pd=p0 | p=p1) = P(pThat<pstar | p=p1) 
#pstar=critical parameter

#AFTER the test should report:

#pThat
#pd
#P=P(phat>pThat | p=p0)=1-pnorm((pThat-p0)/(sqrt((p0*q0)/N)),0,1)
#P1=P(phat<pThat | p=p1)


# We ran this test:

N=10000
pstar=p0+((p1-p0)/(sqrt(p0*q0)+sqrt(p1*q1)))*sqrt(p0*q0)
pstar  
pT=p1
Successes=rbinom(1,N,pT)
pThat=Successes/N
pThat
P=1-pnorm((pThat-p0)/(sqrt((p0*q0)/N)),0,1)
P
P1=pnorm((pThat-p1)/(sqrt((p1*q1)/N)),0,1)
P1

#and ended with this question:

##################################

# Suppose before the test we felt that the wheel was much more likely to be American then French how does this change our interpretation of the results? 

####################
#Will answer this by attempting to find
 
#P(H0 is true | our data)

# and hence 

# P(H1 is true | our data) = 1-P(H0 is true | our data)

#Notice in general this is what we'd really like to know, though we can express it as an apologetic cagey  paragraph (as in the text) or mathematically.  In general doing this requires the Calculus (and is done in MAth 50), but it's good to experience it at least once! 

####################

#A Solution.  We have our prior beliefs: 

#Prior0 = P(p=p0 )
#Prior1= (p=p1). 


# We'd like to find our Posteriors beliefs:   
#Post0=P(p=p0 | phat= pThat) 
#Post 1= P(p=p1 | phat = pThat). 

#  Baye's Thereom tells us Post0= P(p=p0 | phat =  pThat) = P(phat = pThat | p=p0) (P(p=p0)/(P(phat = pThat)). Notice P(phat = pThat) =  P(phat = pThat | p=p0) *P(p=p0)+P(phat = pThat | p=p1) *P(p=p1) so 

#Post0 =  (P(phat = pThat | p=p0)/(P(phat = pThat | p=p0) *P(p=p0)+P(phat = pThat | p=p1) *P(p=p1))

# Given 

Prior0 = .5
 Prior1=.5

# what at are the posteriors after our Roulette experiment?


Post0=(dbinom(Successes,N,p0)*Prior0)/(dbinom(Successes,N,p0)*Prior0+dbinom(Successes,N,p1)*Prior1)
Post1=1-Post0
Post0
Post1

##### A comment on priors:

# In practice, the prior probabilities can be subtle to estimate. In scientific experiments they often correspond to an expert's "degree of belief".  One might worry that degrees of beliefs don't satisfy the rules of probability well enough to be sensibly utilized in this manner and this it is fun and interesting to consider this possibility.  Among non-experts this often debated (i.e. the  "Linda Paradox"  and the conjunction fallacy.) However,  among experts one might hope  that if you ask an expert "If E true?" the and they said it is true with a 60% probability that if you were look at all experts who have answered such a question that roughly 60% of them were in fact correct.  This attaches something akin to book's meaning to these belief probabilities but in such a way that would be hard to actually measure (one might try it out on weather reports and in cooked up examples one still finds examples of the conjunction fallacy).