CSE555: Introduction to Pattern Recognition Midterm ... - CEDAR

More documents

Recommendations

Info

We solve this equation and findwhich can be rewritten asThe final solution is then(1 − ˆP(wn∑i )) z ik = ˆP(wn∑i ) (1 − z ik )k=1k=1n∑z ik = ˆP(wn∑i ) z ik + n ˆP(w i ) − ˆP(wn∑i ) z ikk=1k=1k=1ˆP(w i ) = 1 n∑z iknk=1(c) (3pts) Interpret the meaning of your result in words.Answer:In this question, we apply the maximum-likelihood method <strong>to</strong> estimate the priorprobability. From the result in part (b), it can be observed that the estimate ofthe probability of category w i is merely the probability of obtaining its indica<strong>to</strong>ryvalue in the training data, just as we would expect.5. (20pts) Consider an HMM with an explicit absorber state w 0 and unique null visiblesymbol v 0 with the following transition probabilities a ij and symbol probabilities b jk(where the matrix indexes begin at 0):a ij =⎛⎜⎝1 0 00.2 0.3 0.50.4 0.5 0.1⎞⎟⎠ b jk =⎛⎜⎝1 0 00 0.7 0.30 0.4 0.6⎞⎟⎠(a) (7pts) Give a graph representation of this Hidden Markov Model.Answer:0.3ω0.210.3 0.7v2v110.50.5ω01v00.40.4v 1ω20.6v20.16
(b) (10pts) Suppose the initial hidden state at t = 0 is w 1 . Starting from t = 1, whatis the probability it generates the particular sequence V 3 = {v 2 ,v 1 ,v 0 }?Answer:The probability of observing the sequence V 3 is 0.03678. See the figure below forthe details.vv2 10vW0W1W20 0.036781*.3*.3 *.3*.70 0.30.09 0.1239*.5*.6*.5*.7 *.5*.4*.4*1*.1*.40.03*.2*1t=0 1 2 3(c) (3pts) Given the above sequence V 3 , what is the most probable sequence of hiddenstates?Answer:From the figure above and by using the decoding algorithm, one can observe thatthe most probable sequence of hidden states is {w 1 ,w 2 ,w 1 ,w 0 }.7
Page 1 and 2: CSE555: Introduction to Pattern Rec
Page 3 and 4: Thus, we first compute the sample m
Page 5: 4. (20pts) Let samples be drawn by

CSE555: Introduction to Pattern Recognition Midterm ... - CEDAR

Create successful ePaper yourself

Delete template?

Save as template?