Comment responded, fig 2 caption improved and minor improvements

paper/sl.tex

@@ -119,7 +119,7 @@ Following the above assumptions, a judge with leniency \leniency = \leniencyValu
 	F(\featuresValue_0) = \int { \indicator{\prob{\outcome = 0| \decision = 1, \features = \featuresValue} > \prob{\outcome = 0| \decision = 1, \features = \featuresValue_0}}	d\prob{\featuresValue}	} 
 \end{equation}
 
-which should be equal to
+which can be written as
 
 \begin{equation}
 	F(\featuresValue_0) = \int {\prob{\featuresValue}  \indicator{\prob{\outcome = 0| \decision = 1, \features = \featuresValue} > \prob{\outcome = 0| \decision = 1, \features = \featuresValue_0}} d\featuresValue}
@@ -128,7 +128,7 @@ which should be equal to
 \note[RL]{
 	Should the inequality be reversed? With some derivations
 	\begin{equation}
-	F(\featuresValue_0) = \int {\prob{\featuresValue}  \indicator{\score{\featuresValue} > \score{\featuresValue_0} } ~ d\featuresValue}
+	F(\featuresValue_0) = \int {\prob{\featuresValue}  \indicator{\score{\featuresValue} < \score{\featuresValue_0} } ~ d\featuresValue}
 \end{equation}
 }
 
@@ -168,7 +168,7 @@ We wish to calculate the probability of undesired outcome (\outcome = 0) at a fi
 & = \sum_\featuresValue \prob{\outcome = 0 | \decision = 1, \features = \featuresValue} \prob{\decision = 1 | \leniency = \leniencyValue, \features = \featuresValue} \prob{\features = \featuresValue}
 \end{align*}
 
-\antti{Here one can drop do even at the first line according to do-calculus rule 2, i.e. $P(Y=0|do(R=r))=P(Y=0|R=r)$. However, do-calculus formulas should be computed by first learning a graphical model and then computing the marginals using the graphical model. This gives more accurate result. Michael's complicated formula essentially does this, including forcing $P(Y=0|T=0,X)=0$ (the model supports context-specific independence $Y \perp X | T=0$.)}
+\antti{Here one can drop do even at the first line according to do-calculus rule 2, i.e. $P(Y=0|do(R=r))=P(Y=0|R=r)$. However, do-calculus formulas should be computed by first learning a graphical model and then computing the marginals using the graphical model. This gives more accurate result. Michael's complicated formula essentially does this, including forcing $P(Y=0|T=0,X)=0$ (the model supports context-specific independence $Y \perp X | T=0$.)}
 
 Expanding the above derivation for model \score{\featuresValue} learned from the data
 \[
@@ -217,20 +217,20 @@ The causal model for this scenario corresponds to that depicted in Figure \ref{f
 For the analysis, we assigned 500 subjects to each of the 100 judges randomly.
 Every judge's leniency rate $\leniency$ was sampled uniformly from a half-open interval $[0.1; 0.9)$. 
 Private features $\features$ were defined as i.i.d standard Gaussian random variables.
-Next, probabilities for negative results $\outcome = 0$ were modeled as Bernoulli distributed
-random variables so that
+Next, probabilities for negative results $\outcome = 0$ were calculated as
 \[
 \prob{\outcome = 0| \features = \featuresValue} = \dfrac{1}{1+\exp\{-\featuresValue\}}.
 \]
+and then the result variable $\outcome$ was sampled from Bernoulli distribution with parameter $1-\frac{1}{1+\exp\{-\featuresValue\}}$.
 
-The decision variable $\decision$ was set to 0 if the probability $\prob{\outcome = 0| \features = \featuresValue}$ resided in the top $(1-\leniencyValue)\cdot 100 \%$ of the subjects appointed for that judge. \antti{How was the final Y determined? I assume $Y=1$ if $T=0$, if $T=1$ $Y$ was randomly sampled from $\prob{\outcome| \features = \featuresValue}$ above? Delete this comment when handled.}
+The decision variable $\decision$ was set to 0 if the probability $\prob{\outcome = 0| \features = \featuresValue}$ resided in the top $(1-\leniencyValue)\cdot 100 \%$ of the subjects appointed for that judge.
 Results for estimating the causal quantity $\prob{\outcome = 0 | \doop{\leniency = \leniencyValue}}$ with various levels of leniency $\leniencyValue$ under this model are presented in Figure \ref{fig:without_unobservables}.
 
 \begin{figure}
 \begin{center}
 \includegraphics[width=\columnwidth]{img/without_unobservables.png}
 \end{center}
-\caption{$\prob{\outcome = 0 | \doop{\leniency = \leniencyValue}}$ with varying levels of acceptance rate. Error bars denote standard error of the mean.}
+\caption{$\prob{\outcome = 0 | \doop{\leniency = \leniencyValue}}$ with varying levels of acceptance rate without unobservables. Error bars denote standard error of the mean.}
 \label{fig:without_unobservables}
 \end{figure}