A confusion on the proof of Yao's theorem (Yao 82)
He's doing a pretty poor job of expressing a very simple idea here, which is that if there exists a distinguisher $D$ for which $Pr\lbrack D(H^{i-1})=1\rbrack$ > $Pr\lbrack D(H^{i})=1\rbrack$ (which means the advantage is negative before taking the absolute value), there also exists a distinguisher $\overline{D}$ for which $Pr\lbrack D(H^{i})=1\rbrack$ > $Pr\lbrack D(H^{i-1})=1\rbrack$ (i.e. the advantage is positive before taking the absolute value) by just defining $\overline{D}(H^*) = 1-D(H^*)$ for $H^* \in \{H^i,H^{i-1}\}$. Informally, this distinguisher runs $D$ on its input and just flips whatever $D$ outputs. 

Thus, w.l.o.g. we can assume the distinguisher has a higher probability of outputting a 1 given a sample from $H^i$ than from $H^{i-1}$, which means we don't need the absolute value.