From f7112eca03ae59346b7f7c83836297868ad66176 Mon Sep 17 00:00:00 2001
From: jeanpouget-abadie <jean.pougetabadie@gmail.com>
Date: Tue, 13 Oct 2015 18:05:55 -0400
Subject: starting project proposal finale + added small paragraph to notes

---
 notes/extensions.tex | 27 +++++++++++++++++++++++++++
 1 file changed, 27 insertions(+)

(limited to 'notes/extensions.tex')

diff --git a/notes/extensions.tex b/notes/extensions.tex
index cb78e35..344555d 100644
--- a/notes/extensions.tex
+++ b/notes/extensions.tex
@@ -44,4 +44,31 @@ Interestingly however is that if we let the scale parameter of the sigmoid go to
 infinity, it is harder to violate the inequality. (TH) note that the LT model is
 NOT submodular for every fixed value of thresholds, but only in expectation over
 the random draw of these thresholds.
+
+\section{Logistic regression}
+
+Let's try to fit a logit model to the cascades. The variance of the parameters
+is approximated by $\hat V (\hat \beta) = (\sum p_i(1-p_i)X_i X_i^T)^{-1}$
+
+Let's have a look at the matrix we are taking the inverse of. If none of the
+parents of node $i$ are active at step t, then the $t^{th}$ term is $0$.
+Similarly, if the probability that node $i$ becomes active is $1$, then the term
+cancels out. We can therefore write:$$A = \sum_{g_i \notin \{0,1\}} g_i (1-g_i)
+X_i X_i^T$$
+
+Furthermore, we notice that $x^TAx = \sum_{g_i \notin\{0,1\}} g_i(1-g_i)x^TX_i
+X_i^Tx = \sum_{g_i \notin \{0,1\}} g_i(1-g_i)\|X_i^Tx\|_2^2$. This matrix-vector
+product is zero as soon as I can find one node $a$ which can never be active
+when the parents of my node are active (think line graph or circle with single
+source law).
+
+Suppose now that with high probability, each parent has been active at least
+once. If I consider the union of all nodes which were active just before $X_i$
+was, then by considering only this sub-space of parameters, I can avoid the
+previous pitfall. But is it invertible nonetheless?
+
+Let's consider a line and node $i \notin \{2, n-2\}$ along that line. The
+parents cannot be infected at the same time, depending on whether the source $s
+\geq i$ or $s \leq i$. The matrix $A$ is diagonal-by-block.
+
 \end{document}
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