It is easy to compute the gradient and Hessian matrix of $\mathcal{L}$:
\begin{multline*}
    \nabla \mathcal{L}(\theta) =
    \frac{1}{|\mathcal{T}|}\sum_{t\in \mathcal{T}}x^t\bigg[
        x_i^{t+1}\frac{f'}{f}(\inprod{\theta_i}{x^t})\\
    - (1-x_i^{t+1})\frac{f'}{1-f}(\inprod{\theta_i}{x^t})\bigg]
\end{multline*}
\begin{multline*}
    \nabla^2 \mathcal{L}(\theta) =
    \frac{1}{|\mathcal{T}|}\sum_{t\in \mathcal{T}}x^t(x^t)^T\bigg[
        x_i^{t+1}\frac{f''f - f'^2}{f^2}(\inprod{\theta_i}{x^t})\\
    - (1-x_i^{t+1})\frac{f''(1-f) + f'^2}{(1-f)^2}(\inprod{\theta_i}{x^t})\bigg]
\end{multline*}

Since $\E[x_i^{t+1}|x^t] = f(\inprod{\theta_i}{x^t})$,
$\nabla\mathcal{L}(\theta)$ can be written
$\frac{1}{|\mathcal{T}|}\sum_{t\in\mathcal{T}} Y_t$ where $\E[Y_t|Y_{t-1}]
= 0$. It follows from the law of expected iterations that
$\E[\nabla\mathcal{L}(\theta)] = 0$.

\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 for a node $\alpha$:

\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}(\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 m \right)
\end{equation}

In conclusion, for $\delta>0$, $\lambda := 2 \sqrt{\frac{n^{\delta + 1} \log m}{p_{\min}}}$ is a valid regularizer with probability $1 - \exp(-n^\delta \log m )$

\subsection{Proposition~\ref{prop:irrepresentability}}
In the words and notation of Theorem 9.1 in \cite{vandegeer:2009}: 
\begin{lemma}
\label{lemm:irrepresentability_proof}
Let $\phi^2_{\text{compatible}}(L,S) \defeq \min \{ \frac{s \|f_\beta\|^2_2}{\|\beta_S\|^2_1} \ : \ \beta \in {\cal R}(L, S) \}$, where $\|f_\beta\|^2_2 \defeq \{ \beta^T \Sigma \beta \}$ and ${\cal R}(L,S) \defeq \{\beta : \|\beta_{S^c}\|_1 \leq L \|\beta_S\|_1 \neq 0\}$. If $\nu_{\text{irrepresentable}(S,s)} < 1/L$, then $\phi^2_{\text{compatible}}(L,S) \geq (1 - L \nu_{\text{irrepresentable}(S,s)})^2 \lambda_{\min}^2$. 
\end{lemma}

Since ${\cal R}(3, S) = {\cal C}$, $\|\beta_S\|_1 \geq \|\beta_S\|_2$, and $\|\beta_S\|_1 \geq \frac{1}{3} \|\beta_{S^c}\|_1$ it is easy to see that $\|\beta_S\|_1 \geq \frac{1}{4} \|\beta\|_2$ and therefore that: $\gamma_n \geq \frac{n}{4s}\phi^2_{\text{compatible}}(3,S)$

Consequently, if $\epsilon > \frac{2}{3}$, then $\nu_{\text{irrepresentable}(S,s)} < 1/3$ and the conditions of Lemma~\ref{lemm:irrepresentability_proof} hold.