From b89c767264b3ea4d16c3a0e660fcc25013d1e195 Mon Sep 17 00:00:00 2001
From: Thibaut Horel <thibaut.horel@gmail.com>
Date: Thu, 25 Sep 2014 00:01:56 -0400
Subject: [ps2] end. Just in time!

---
 ps2/main.tex | 42 +++++++++++++++++++++++++++++++++++++++++-
 1 file changed, 41 insertions(+), 1 deletion(-)

(limited to 'ps2/main.tex')

diff --git a/ps2/main.tex b/ps2/main.tex
index 45638ab..c4a4ad3 100644
--- a/ps2/main.tex
+++ b/ps2/main.tex
@@ -39,6 +39,7 @@
 \begin{document}
 
 \maketitle
+Collaborators: Eric Balkanski and Jean Pouget-Abadie
 \section*{Problem 1}
 
 \paragraph{(a)} Let us consider the function $f$ defined for $x>-1$ by:
@@ -141,6 +142,14 @@ a global Chernoff bound.}:
 where $C=\frac{1}{5e}$. The right-hand side can be bounded from above by
 $2e^{-C'\mu\eps^2}$ where $C' = C\log(5/4)>0$. This concludes the proof.
 
+\paragraph{(c)} The case were the $p_i$ are not all the same can be reduced to
+the previous case by considering the random variables $X_i' = X - p_i$. These
+random variables have mean zero and variance $p_i(1-p_i)\leq p_i$. The previous
+analysis applies almost verbatim. The only difference being that
+$\E[X_{i_1}\ldots X_{i_{2l}}]$ can now be bounded from above by $p_{j_1}\ldots
+p_{j_k}$ where $j_1,\ldots j_k$ are the unique indices (they are repeated an
+even number of times so the moments of even orders are bounded by the
+variance).
 
 \section*{Problem 2}
 
@@ -302,7 +311,38 @@ choosing $k=\left\lceil\frac{3}{4\eps^2}\right\rceil$ is enough to obtain:
 
 \paragraph{(b)}
 \begin{itemize*}
-    \item In case we want a probability of success of $\delta$,
+    \item In case we want a probability of success of $\delta$, the previous
+        analysis shows that we can choose $k=\frac{1}{4\eps^2\delta}$.
+    \item If the $s_i$ were fully independent, then we would be able to apply
+        the Chernoff bound and obtain:
+        \begin{displaymath}
+            \Pr[|Z-J(A,B)|\geq \eps]\leq 2e^{-C\eps^2\frac{k}{p}}\leq2e^{-C\eps^2k}
+        \end{displaymath}
+        Hence, choosing
+        $k=\left\lceil\frac{1}{C\eps^2}\log\frac{2}{\delta}\right\rceil$ will
+        yield a probability of failure of at most $\delta$.
+    \item We note that if the median of $Z_1\ldots Z_t$ (denoted by $Z$) is at
+        distance more than $\eps$ of $J(A,B)$, this implies that at least half
+        of the $Z_i$'s are at distance at least $\eps$ of $J(A,B)$. This
+        implies that there is a subset of size $\frac{t}{2}$ of the $Z_i$'s at
+        distance at most $\eps$ of $J(A,B)$. Since there are $\binom{t}{t/2}$
+        such subsets, applying a union bound:
+        \begin{displaymath}
+            \Pr[|Z-J(A,B)|\geq \eps]\leq\binom{t}{\frac{t}{2}}\frac{1}{3^t}
+        \end{displaymath}
+        where $\frac{1}{3^t}$ is the probability that all the $Z_i$'s in a set
+        of size $\frac{t}{2}$ are at distance at least $\eps$ of $J(A,B)$ (the
+        $Z_i's$ are all independent). Using the standard bound on binomial
+        coefficients:
+        \begin{displaymath}
+            \Pr[|Z-J(A,B)|\geq \eps]\leq\left(\frac{\sqrt{2e}}{3}\right)^t
+        \end{displaymath}
+        This implies that choosing:
+        \begin{displaymath}
+            t = \left\lceil\frac{\log\frac{1}{\delta}}{\log\frac{\sqrt{2e}}{3}}\right\rceil
+            = O\left(\log\frac{1}{\delta}\right)
+        \end{displaymath}
+        will give us a probability of failure of at most $\delta$.
 \end{itemize*}
 
 \section*{Problem 4}
-- 
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