From 54cac035f77c8c617ae666790f6efca8c61d047a Mon Sep 17 00:00:00 2001
From: Jean Pouget-Abadie <jean.pougetabadie@gmail.com>
Date: Tue, 14 Apr 2015 18:47:29 -0400
Subject: fixing typo

---
 brainstorm/results.tex | 9 +++++----
 1 file changed, 5 insertions(+), 4 deletions(-)

(limited to 'brainstorm/results.tex')

diff --git a/brainstorm/results.tex b/brainstorm/results.tex
index 0b37049..6f2dc6d 100644
--- a/brainstorm/results.tex
+++ b/brainstorm/results.tex
@@ -662,14 +662,15 @@ $$\mathbb{P}_{S_1, \dots S_m \sim {\cal D}}\left(\mathbb{P}_{\hat S}
 The previous proposition can be reformulated in the slightly weaker formulation
 which is more intuitive:
 
-$$\mathbb{P}_{S_1, \dots S_m \sim {\cal D}}\left(\mathbb{E}(f(\hat S)) \geq
+$$\mathbb{P}_{S_1, \dots S_m \sim {\cal D}}\left(\mathbb{E}(f(\hat S)) \leq
 \frac{1}{n}OPT \right) \geq 1 - \frac{n^C}{2f(n)} \rightarrow 1 \text{ when } n
 \rightarrow +\infty$$
 
 Note that the restriction on ${\cal D}$ is verified by many distributions. In
-the case of the uniform distribution $ = \frac{1}{2^{n-1}}$ and in the case of
-observing all sets of size $k>1$ uniformly at random: $=0$ This example is not
-fully satisfying for two reasons:
+the case of the uniform distribution $ \mathbb{P}(\{A_1\}) + \mathbb{P}(\{A_2\})
+= \frac{1}{2^{n-1}}$ and in the case of observing all sets of a certain fixed
+size $k>1$ uniformly at random: $\mathbb{P}(\{A_1\}) + \mathbb{P}(\{A_2\})= 0$
+This example is not fully satisfying for two reasons:
 
 \begin{itemize}
   \item the coverage function is not monotone.
-- 
cgit v1.2.3-70-g09d2

