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@@ -38,6 +38,42 @@ the proof.
\subsubsection{Approximate sparsity proof}
\subsubsection{RE with high probability}
+\begin{proof}Writing $H\defeq \nabla^2\mathcal{L}(\theta^*)$, if
+ $ \forall\Delta\in C(S),\;
+ \|\E[H] - H]\|_\infty\leq \lambda $
+ and $\E[H]$ verifies the $(S,\gamma)$-(RE)
+ condition then:
+ \begin{equation}
+ \label{eq:foo}
+ \forall \Delta\in C(S),\;
+ \Delta H\Delta \geq
+ \Delta \E[H]\Delta(1-32s\lambda/\gamma)
+ \end{equation}
+ Indeed, $
+ |\Delta(H-E[H])\Delta| \leq 2\lambda \|\Delta\|_1^2\leq
+ 2\lambda(4\sqrt{s}\|\Delta_s\|_2)^2
+ $.
+ Writing
+ $\partial^2_{i,j}\mathcal{L}(\theta^*)=\frac{1}{|\mathcal{T}|}\sum_{t\in
+ T}Y_t$ and using $(LF)$ and $(LF2)$ we have $\big|Y_t - \E[Y_t]\big|\leq
+ \frac{4(M+2)}{\alpha}$.
+ Applying Azuma's inequality as in the proof of Lemma~\ref{lem:ub}, this
+ implies:
+ \begin{displaymath}
+ \P\big[\|\E[H]-H\|_{\infty}\geq\lambda\big] \leq
+ 2\exp\left(-\frac{n\alpha\lambda^2}{4(M+2)} + 2\log m\right)
+ \end{displaymath}
+ Thus, if we take $\lambda=\sqrt{\frac{12(M+2)\log m}{\alpha
+ n^{1-\delta}}}$, $\|E[H]-H\|_{\infty}\leq\lambda$ w.p at least
+ $1-e^{-n^{\delta}\log m}$. When $n^{1-\delta}\geq
+ \frac{M+2}{21\gamma\alpha}s^2\log m$, \eqref{eq:foo} implies
+ $
+ \forall \Delta\in C(S),\;
+ \Delta H\Delta \geq \frac{1}{2} \Delta \E[H]\Delta,
+ $ w.p. at least $1-e^{-n^{\delta}\log m}$ and the conclusion of
+ Proposition~\ref{prop:fi} follows.
+\end{proof}
+
\subsection{Other continuous time processes binned to ours: proportional
hazards model}