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update grammar #276

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4 changes: 2 additions & 2 deletions book/ens.tex
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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
Expand Down