https://www.wordclouds.com/
Month: February 2018
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Hello
 Cross entropy
 H(p,q) = D(pq) + H(p)
 H(p) is some inherent randomness in p
 D(pq) is what we care about. we can try to get D(pq) by calculating cross entropy.
 Conclusion: a model is good is that it assign good approximation to the observed data. So we need to find some good q
 H(p,q) = D(pq) + H(p)
 Main points:
 Example: She loves her. (It’s a correct string, but English is not like this. It should be “She loves herself.”)
 We need a meaning pair.
 Two orthogonal dimensions:
 probability for the strings.

Units Prob String {a^{n}b^{n}n>=1} P(w1, w2,…, wn) Structure A tree structure PCFG  L1 = L2: Language 1 is equal to Language 2
 Weak equivalence
 Sense of string are the same.
 Strong equivalence
 Structure of language 1 and 2 are the same.
 G1 = G2 iff {x G1 generates string x} = {xG2 generates string x} (all and only the same structures)

G1 G2 S>a s s>s a s>e s>e  G1 and G2 are weak equivalent (they generate the same strings) but not strong equivalent
 Weak equivalence
 Example: Jon loves mary
 Questions:
 How to measure equivalence?
 binary judgements?
 EM
 Question: How to find a good model? Expectation maximization (EM)
 The structure of model is given, we need to find the parameters for the model.
 Coin: H H H H T T T T T T
 MLE: argmax [p(xmu)]
 Solve:
 Result: p = k/n
 HMM <A,B, pi>
 pi: initial probabilities
 N states
 V words
 recipe: http://www.umiacs.umd.edu/~resnik/ling773_sp2011/readings/em_recipe.v2.only_hmm.pdf
 Three fundamental questions for EM:
 What is P(Omu)
 Best hidden events given O, mu?
 What’s the best model I can have? argmax_mu P(Omu)
Lecture 3: HMMs and Expectation Maximization
Lecture3: Information Theory
 Today’s class is about:
 Hypothesis testing
 collocations
 Info theory
 Hypothesis Testing
 Last lecture covered the methodology.
 Collocation
 “religion war”
 PMI, PPMI
 PMI = pointwise mutual information
 PMI = log2(P(x,y)/(P(x)P(y))) = I(x,y)
 PPMI = positive PMI = max(0, PMI)
 Example: Hong Kong, the frequency of “Hong” and “Kong” are low, but the frequency for “Hong Kong” is high.
 Nonstatistical measure but very useful: Not a classic hypothesis testing.
 Note: If the frequency of the word is less than 5, then even if the PMI is high, the bigram is meaningless.
 PMI = pointwise mutual information
 RFR
 rfr(w) = (freq(w in DP)/(#DP))/(freq (w in background corpus)/(#background corpus))
 DP is an child class in background corpus.
 Chisquare test
 Draw the table is important.
 Ttest
 Nonstatistical measure but very useful.
 Not a classic hypothesis testing.
 sliding window scan the sentence to find the bigram.
 found >1
 not found >0
 A fox saw a new company’s emerge…..
 A fox >0
 fox saw > 0
 saw a >0
 a new > 0
 new company > 1
 company emerge > 0
 Nonstatistical measure but very useful.
 Info theory
 Entropy: degree of certainty of X; information content of X; pointwise information of X; (ling) surprisal, perplexity
 X~P, H(X) = E[log2(p(x))] = sum(p(x)log2(p(x)))
 H(X,Y) = H(X) + H(Y) – I(X, Y)
 H(X, Y) is the cost of communicating between X and Y;
 H(X) is the cost of communicating X;
 H(Y) is the cost of communicating Y;
 I(X, Y) is the saving because X and Y can communicate with each other.
 Example: 8 horses run, the probability of winning is:
 horse: 1 . 2 . 3 . 4 . 5 . 6 . 7 . 8
 Prob: 1/2, 1/4 , 1/8, 1/16, 1/32, 1/64, 1/128, 1/256
 Code1: 000, 001, 010, 011, 100, 101, 110, 111
 H(X) = 3;
 Code2:0, 10, 110, 1110, 111100, 1111101, 111110, 111111
 Huffman code.
 H(X) = 2;
 Structure
 [W > encoder] > X> [ ]> Y > [decoder > W^]
 Source channel (P(yx))
 Anything before Y is the noisy channel model.
 P(X,Y) is the joint model, P(XY) discriminative model.
 T^ (estimated T)
 =argmax_T (P(TW))
 = argmax_T(P(WT)P(T)/P(W))
 = argmax_T(P(WT)P(T)) (W can be omitted because its’s just a scaler)
 = argmax_T(P(w1t1)P(w2t2)…p(w.nt.n) P(t1)P(t2t1)P(t3t2,t1)…P(tnt1,t2, …, t.n1))
 = argmax_T(P(w1t1)P(w2t2)…p(w.nt.n) P(t1)P(t2t1)P(t3t2)…P(t.n))
 E^ = argmax P(FE)P(E)
 F = French, E = English, translation from English to French.
 Relative entropy: KL divergence
 We have an estimated model q and a true distribution p
 D(pq) = sum_x(p(x).log(p(x))/ log(q(x)))
 Properties:
 D(pq) >= 0, equal is true when p == q
 D(pq) != D(qp)
 D(pq) = E_p[p(x)] – E_p[q(x)] E_p(q(x)) is the cross entropy
 Example:
 Model q: P(X)P(Y) independent
 truth p: Q(X, Y) joint
 D(pq)
 = sum_x,y [p(x,y) log2(p(x,y)/p(x)p(y))]
 = E_x,y [log2(p(x,y)/p(x)p(y))] average mutual info is the expectation of pointwise mutual info
 Cross entropy: (perplexity)
 Question: how good is my language model? How do I measure?
 p(w1, w2, w3, w4) = p(w1) p(w2w1) … p(w4w1 w2 w3)
 p(w4w1 w2 w3) = p(w4w*, w2 w3)
 make it Markov, and combine everything beside the prev and the prevprev.
 D(pm) = sum[p(x.1 … x.n)log(p(x.1 … x.n)/m(x.1 … x.n))]
 but how do you know the truth?
 what if n is very large?
 H(p,m)=H(p) + D(pm) = sum[p(x)log(m(x))]
 H(p,m) can be an approximation of D(pm), then:
 We can sample from p and check H(p,m) by the samples, then:
 H(L,m) = lim_(n>infinity) [ 1/n sum(p(x.1 … x.n)log(m(x.1 … x.n)))], then:
 we need p(x.1 … x.n) be a good sample of L.
 H(L,m) = lim_(n>infinity) [1/n log m(x.1 … x.n)]
 m(x.1 … x.n) = product(m(x.i x.i1))
 If m(*) = 0 > H(p,m) = infinity
 Should associate the cost with something…
 smoothing (lots of ways for smoothing)
 Perplexity:
 2^(H(x.1 … x.n, m)) i.e. 2^ (cross entropy)
 effective vocabulary size
 recover the unit from “bits” to the original one.
 Entropy: degree of certainty of X; information content of X; pointwise information of X; (ling) surprisal, perplexity
After class (M& S book)
 6.2.1 MLE: gives the highest probability to the training corpus.
 C(w1, … ,wn) is the frequency of ngram w1…wn in the training set.
 N is the number of training instances.
 P_MLE(w1, …, wn) = C(w1, … , wn) / N
 P_MLE(wn  w.1, …, w.n1) = C(w1, … , wn)/C(w1, … , w.n1)
 6.2.2 Laplace’s law, Lidstone’s law and the JeffreysPerks law
 Laplace’s law
 original smoothing
 Lidstone’s law
 JeffreysPerks law
 Expected likelihood estimation (ELE)
 Laplace’s law
 6.2.3 Held Out Estimation
 Take further test and see how often bigrams that appeared r times in the training text tend to turn up in the future text.
 6.2.4 Cross Validation
 Cross validation:
 Rather than using some of the training data only for frequency counts and some only for smoothing probability estimates, more efficient schemes are possible where each part of the training data is used both as initial traiining data and as held out data.
 Deleted Estimation: twoway cross validation:
 Cross validation:
The test speed of neural network?
Basically, the time spent on testing depends on:
 the complexity of the neural network
 For example, the fastest network should be the fullyconnected network.
 CNN should be faster than LSTM because LSTM is sequential (sequential = slow)
 Currently, there are many ways to compress deep learning model (remove nodes with lighter weight)
 the complexity of data
 Try to build parallel models.
Analyze of different deep learning models:
 RNN
 CNN
 LSTM
 GRU
 Seq2seq:
 used for NLP
 Autoencoder:
 learn the features and reduce dimension’
Other machine learning models:
 random forest
 used for classification
Machine learning thought: