Appendix stuff

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+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 $\alpha$: