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@ -395,7 +395,7 @@ Logistic regression has the following properties:
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\bottomrule
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\end{tabular}
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\end{center}
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A possible way to sample $\bar{x}^{(1)}$ is (w.r.t. the indexes of the examples in $x$) $[2, 3, 3, 2, 4, 6, 2, 4, 1, 9, 1]$.
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A possible way to sample $\bar{x}^{(1)}$ is (w.r.t. the indexes of the examples in $x$) $[2, 3, 3, 2, 4, 6, 2, 4, 1, 9]$.
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\end{example}
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\end{description}
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@ -190,7 +190,7 @@
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This is equivalent to computing the probability of the whole sentence, which expanded using the chain rule becomes:
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\[
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\begin{split}
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\prob{w_1, \dots, w_{i-1} w_i} &= \prob{w_1} \prob{w_2 | w_1} \prob{w_3 | w_{1..2}} \dots \prob{w_n | w_{1..n-1}} \\
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\prob{w_1, \dots, w_{i-1}, w_i} &= \prob{w_1} \prob{w_2 | w_1} \prob{w_3 | w_{1..2}} \dots \prob{w_n | w_{1..n-1}} \\
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&= \prod_{i=1}^{n} \prob{w_i | w_{1..i-1}}
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\end{split}
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\]
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@ -224,7 +224,7 @@
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\begin{description}
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\item[Estimating $\mathbf{N}$-gram probabilities]
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Consider the bigram case, the probability that a token $w_i$ follows $w_{i-1}$ can be determined through counting:
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\[ \prob{w_i | w_{i-1}} = \frac{\texttt{count}(w_{i-1} w_i)}{\texttt{count}(w_{i-1})} \]
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\[ \prob{w_i | w_{i-1}} = \frac{\texttt{count}(w_{i-1}, w_i)}{\texttt{count}(w_{i-1})} \]
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\end{description}
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\begin{remark}
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