From de7df38121cc8a39ee72899c9ed0628421dab7e1 Mon Sep 17 00:00:00 2001
From: jeanpouget-abadie <jean.pougetabadie@gmail.com>
Date: Wed, 28 Jan 2015 22:27:09 -0500
Subject: updated results section

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 paper/sections/appendix.tex | 36 ++++++++++++++++++++++++++++++++++++
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+\subsection*{Proof of support recovery lemma}
+
+
+\subsection*{Upper-bound for $\|\nabla f(\theta^*)\|_\infty$}
+
+We show an upper-bound for $ 2 \|\nabla f(\theta^*) \|_\infty$. If we only consider the first measurement of every cascade, this can be done easily. Let $N$ be the number of cascades (to distinguish from $n$ number of total measurements). Begin by noting that:
+
+\begin{align}
+\nabla_k f(\theta) & = \sum^n_{i=1} \frac{b^i x^i_k}{1 - e^{\langle x^i, \theta \rangle}} - \sum^n_{i=1} x^i_k \nonumber \\
+& = \sum_{i=1}^n x^k_i \left( \frac{b^i}{\mathbb{P}(\text{node } \alpha \text { infected})} - 1\right) \nonumber 
+\end{align}
+
+\begin{lemma}
+\label{lem:subgaussian_variable}
+$\nabla f(\theta^*)$ is a $N/p_{\min}$-subgaussian variable, where $p_{\min}$ is the smallest non-zero link to node $\alpha$.
+\end{lemma}
+
+\begin{proof}
+We show that its expectation is $0$ by conditioning on the seed set $S$:
+
+\begin{align}
+\mathbb{E} \left( \nabla_k f(\theta) \right) & = \sum_{i=1}^N \mathbb{E} \left[ x^i_k \left( \frac{b^i}{\mathbb{P}(\text{node } \alpha \text { infected})} - 1\right) \right] \nonumber \\
+& = \sum_{i=1}^N \mathbb{E}_S \left[ \mathbb{E}\left[x^i_k \left( \frac{b^i}{\mathbb{P}(\alpha \text { infected})} - 1\right) \middle| S \right] \right] \nonumber \\
+& = \sum_{i=1}^N  \mathbb{E}\left[x^i_k \left( \frac{ \mathbb{E}_S \left[ b^i \middle| S \right]}{\mathbb{P}(\alpha \text { infected})} - 1\right)  \right] \nonumber \\
+& = 0 \nonumber
+\end{align}
+Therefore, $\nabla f(\theta^*)$ is the sum of zero-mean variables, upper-bounded by $1/p_{\min}$. It follows that $\nabla f(\theta^*)$ is $N/p_{\min}$-subgaussian.
+\end{proof}
+
+By union bound and characterization of sub-gaussian variables:
+
+\begin{equation}
+\mathbb{P}(\| \nabla f(\theta) \|_{\infty} > \lambda) \leq 2 \exp \left( -\frac{\lambda^2 p_{\min}}{2n} + \log p \right)
+\end{equation}
+
+In conclusion, for $\delta>0$, $\lambda := 2 \sqrt{\frac{n^{\delta + 1} \log p}{p_{\min}}}$ is a valid regularizer with probability $1 - \exp(-n^\delta \log p )$
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