Exercises 3, calculation of MAP via Dirichlet prior

Exercises 3, calculation of MAP via Dirichlet prior

by Bruno Magalhaes -
Number of replies: 3

Quick question on Exercise III, on the question computing the MAP with a Dirichlet prior distribution.


We know that P(Xc) = 2.05/54.45 and P(Xt)=1.05/54.45.


From my calculation:

P(X) = P(Xt)+P(Xc)+25*P(unseen) = 1.05/54.45 + 2.05/54.45 + 25*0.05/54.45 = 4.35/54.45

leading to P(Xc)/P(X) = 2.05/4.35.


However in the solutions we have:

P(X) = 5.35/54.45 (?) 

that leads to P(Xc)/P(X) = 2.05/5.35


Can you help?


Minor: there's a small typo in 'repeRtitions'

 

In reply to Bruno Magalhaes

Re: Exercises 3, calculation of MAP via Dirichlet prior

by Jean-Cédric Chappelier -

First of all, it's not the MAP but the expected value (marginalizing \theta; the expected value is NOT at the maximum).

Then, for X (whitespace) you forgot one seen bigram: Xh :
P(X) = P(Xt)+P(Xc)+P(Xh)+24*P(unseen)
which indeed leads to a numerator of  5.35