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courses:rg:2014:crf [2014/10/28 14:16]
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courses:rg:2014:crf [2014/10/28 14:22] (current)
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-=====Questions ======+===== Conditional Random Fields - Questions =====
  
-(1Definition of CRF in Section 3 contains a formula with a shortcut notation: <latex>P(Y_v | X, Y_w, w \neq v) = P(Y_v | X, Y_w, w \sim v)</latex>.+1Definition of CRF in Section 3 contains a formula with a shortcut notation: <latex>P(Y_v | X, Y_w, w \neq v) = P(Y_v | X, Y_w, w \sim v)</latex>.
  
 a) Try to rewrite this general formula using some more clear notation (or explain it in your words). a) Try to rewrite this general formula using some more clear notation (or explain it in your words).
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 **Hint:** If you don't understand the shortcut notation, just ignore it and use your intuition (vertices connected by edges are not independent). **Hint:** If you don't understand the shortcut notation, just ignore it and use your intuition (vertices connected by edges are not independent).
      
-(2MEMMs suffer from Label Bias Problem. What about HMMs? Why?+2MEMMs suffer from Label Bias Problem. What about HMMs? Why?
  
-(3Which of the following features are meaningful? Why?+3Which of the following features are meaningful? Why?
  
-a) <latex>X_i</latex> == can +a) <latex>X_i</latex> == "can" 
-b) <latex>X_i</latex> == can && <latex>Y_i</latex> == N +b) <latex>X_i</latex> == "can&& <latex>Y_i</latex> == N 
-c) <latex>X_i</latex> == can && <latex>Y_{i-1}</latex> == N +c) <latex>X_i</latex> == "can&& <latex>Y_{i-1}</latex> == N 
-d) <latex>X_{i-1}</latex>== can && <latex>Y_i</latex> == N && <latex>Y_{i-1}</latex> == V +d) <latex>X_{i-1}</latex>== "can&& <latex>Y_i</latex> == N && <latex>Y_{i-1}</latex> == V 
-e) <latex>X_{i-1}</latex> == can && <latex>Y_i</latex> == N && <latex>Y_{i+1}</latex> == V +e) <latex>X_{i-1}</latex> == "can&& <latex>Y_i</latex> == N && <latex>Y_{i+1}</latex> == V 
-f) <latex>X_{i-2}</latex> == can && <latex>Y_i</latex> == V && <latex>Y_{i-1}</latex> == N +f) <latex>X_{i-2}</latex> == "can&& <latex>Y_i</latex> == V && <latex>Y_{i-1}</latex> == N 
-g) <latex>X_{i+3}</latex> == can && <latex>Y_i</latex> == N && <latex>Y_{i-2}</latex> == V +g) <latex>X_{i+3}</latex> == "can&& <latex>Y_i</latex> == N && <latex>Y_{i-2}</latex> == V 
-h) <latex>X_1</latex> == The && <latex>Y_{i-1}</latex> == N && <latex>Y_i</latex> == N+h) <latex>X_1</latex> == "The&& <latex>Y_{i-1}</latex> == N && <latex>Y_i</latex> == N
 i) <latex>X_i</latex> has more letters than <latex>X_{i-1}</latex> && <latex>Y_i</latex> == N i) <latex>X_i</latex> has more letters than <latex>X_{i-1}</latex> && <latex>Y_i</latex> == N
 j) <latex> X</latex> contains word "dog" && (<latex>Y_i</latex> == N || <latex>Y_i</latex> == V) j) <latex> X</latex> contains word "dog" && (<latex>Y_i</latex> == N || <latex>Y_i</latex> == V)
  
-(4Let's suppose, that we have a CRF for the data "he/N can/V can/V a/N can/N" and these features:+4Let's suppose, that we have a CRF for the data "he/N can/V can/V a/N can/N" and these features:
  
 <latex>f_1</latex>: <latex>X_i</latex> == can && <latex>Y_i</latex> == V && (<latex>Y_{i-1}</latex> == N || <latex>Y_{i-1}</latex> == V) <latex>f_1</latex>: <latex>X_i</latex> == can && <latex>Y_i</latex> == V && (<latex>Y_{i-1}</latex> == N || <latex>Y_{i-1}</latex> == V)
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 **Hint2:** The vertical bar in <latex>y|_v</latex> does not mean conditional probability, see its definition under Formula 1. **Hint2:** The vertical bar in <latex>y|_v</latex> does not mean conditional probability, see its definition under Formula 1.
  
-d) Let's suppose, that +d) (optional) Let's suppose, that 
-   <latex>\lambda_1</latex> = 1 +    <latex>\lambda_1</latex> = 1 
-   <latex>\lambda_2</latex> = 1 +    <latex>\lambda_2</latex> = 1 
-   <latex>\mu_1</latex> = 1+    <latex>\mu_1</latex> = 1
  
 Show, that exp(...) expression in Formula 1 (page 3) and <latex>\prod_{i=1}^{n+1} M_i(y_{i-1}, y_i | x))</latex> in formula on page 4 Show, that exp(...) expression in Formula 1 (page 3) and <latex>\prod_{i=1}^{n+1} M_i(y_{i-1}, y_i | x))</latex> in formula on page 4
 give the same result. give the same result.

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