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2 changes: 1 addition & 1 deletion book/dt.tex
Original file line number Diff line number Diff line change
Expand Up @@ -501,7 +501,7 @@ \section{Formalizing the Learning Problem}
$\hat \vx$ to corresponding prediction $\hat y$. The key property
that $f$ should obey is that it should do well (as measured by $\ell$)
on future examples that are \emph{also} drawn from $\cD$. Formally,
it's \concept{expected loss} $\ep$ over $\cD$ with repsect to $\ell$
its \concept{expected loss} $\ep$ over $\cD$ with repsect to $\ell$
should be as small as possible:
\begin{align} \label{eq:expectederror}
\ep
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4 changes: 2 additions & 2 deletions book/ens.tex
Original file line number Diff line number Diff line change
Expand Up @@ -235,7 +235,7 @@ \section{Boosting Weak Learners}
$\vx$; otherwise $f(\vx) = -1$ for all $\vx$. To make the problem
moderately interesting, suppose that in the original training set,
there are $80$ positive examples and $20$ negative examples. In this
case, $f\oth(\vx)=+1$. It's weighted error rate will be $\hat\ep\oth
case, $f\oth(\vx)=+1$. Its weighted error rate will be $\hat\ep\oth
= 0.2$ because it gets every negative example wrong. Computing, we
get $\al\oth = \frac12\log 4$. Before normalization, we get the new
weight for each positive (correct) example to be $1 \exp[-\frac12\log
Expand Down Expand Up @@ -271,7 +271,7 @@ \section{Boosting Weak Learners}

In fact, a very popular weak learner is a decision \concept{decision
stump}: a decision tree that can only ask \emph{one} question. This
may seem like a silly model (and, in fact, it is on it's own), but
may seem like a silly model (and, in fact, it is on its own), but
when combined with boosting, it becomes very effective. To understand
why, suppose for a moment that our data consists only of binary
features, so that any question that a decision tree might ask is of
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