Re: Probability of making a choice by a person, on the basis of earlier made choices?
- From: Ray Vickson <RGVickson@xxxxxxx>
- Date: Tue, 27 May 2008 17:06:44 -0700 (PDT)
On May 27, 1:12 pm, Marcin <marcinkrawczyk.m...@xxxxxxxxx> wrote:
We are asking a man a few questions. After every response we can
check, whether he told the truth or deliberately lied. Lets say, at
the beginning, the probability, that the man will lie and the
probability the man will tell the truth are the same (0.5). If the man
answers the first question and lies, the probability that he will also
lie when asking a second question is (I guess) already greater than
0.5. But, if the man will answer first five questions frankly, the
probability, that in the reply to sixth question he will also tell the
truth is already much higher than 0.5. But how high is the propability
exactly, and how do you calculate that?
(sorry for english :)
You need to specify a model for successive lies, etc. One possible
model is that for any given probability of lying, p, the lies form an
independent sequence, each with probability p. However, you may not
KNOW p, so (if you are a Bayesian) you would assign a probability
distribution to p itself. Then, indeed, successive observations of
lies or non-lies could influence your estimates of p (even though p
is, itself, not changing---just our information about the value of p).
One simple model is that you know nothing at all about p and are
equally happy to let p be any value between 0 and 1; that is, the
prior distribution of p is uniform from 0 to 1. Now ask yourself:
before I query the man, what is the probability that he will lie on
his first question? Well, it is the prior expectation of p, which is
1/2! Note: that does not mean that we think p really is 1/2; it just
means that if we asked a single question of a large number of
independent copies or our subject, half of them would be truthful and
half of them would lie. (Nevertheless, the lie probability p of any
individual could, and likely would, differ from 1/2.)
Now, if the man lies on all of the first k questions, what is his
posterior distribution of p? Well, P{k lies|p} = p^k (p to the power
k), and P{k lies} = sum_{p} Prob{p}*P{k lies|p} = sum_{p} p^k*Prob{p};
this becomes an integral int(p^k dp, p=0..1) in the limit of
continuous, uniform p. Thus, P{k lies} = 1/(k+1). The posterior
distribution of p, given k lies is P{p|k lies} = P{k lies|p}*P{p}/P{k
lies} = (k+1)*p^k, for 0 <= p <= 1. (By an abuse of notation, I am
letting P{p} stand for the density function.) Now the probability he
lies on question (k+1) is the posterior expectation of p, which is
int((k+1)*p*p^k, p=0..1) = (k+1)/(k+2). For example, if he lies 5
times in a row on questions 1--5, the probability he will lie on
question 6 is 6/7. Note, however, that all this depends on using a
uniform prior distribution for p, which might not be very realistic.
Another possible model might be that the sequence of lies form a
Markov chain, as Robert Israel has suggested. There, too, you may not
know the transition probability parameters, but might need to estimate
them, perhaps in a Bayesian fashion. That case is much more
complicated than the "independent lies" case.
R.G. Vickson
.
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