## 2 Materials and Methods
### Tempered EM
#### 2.3.3 Sketch of Proof  
The detailed proof of Theorem 3.2 can be found in Appendix A.2, we propose here an abbreviated version where we highlight the key steps.  
Under the current assumptions, it is sufficient to verify the second part of condition (5) in order to apply Theorem 1. To that end, we must control the integral  
\[\int_{z}\biggl{(}\frac{\tilde{p}_{\theta,n}(z)}{p_{\theta}(z)}-1\biggr{)}^{2} p_{\theta}(z)dz\,,\]  
for all \(\theta\) in a compact \(K\subseteq\Theta\). First, with a [PERSON] development in the term \(\Bigl{(}\frac{1}{T_{n}}-1\Bigr{)}\), which converges toward \(0\) when \(n\rightarrow\infty\), we control the difference \((\tilde{p}_{\theta,n}(z)-p_{\theta}(z))^{2}\):  
\[\biggl{(}\frac{p_{\theta}(z)^{\frac{1}{T_{n}}}}{\int_{z^{\prime}}p_{\theta}(z^ {\prime})^{\frac{1}{T_{n}}}}-p_{\theta}(z)\biggr{)}^{2}\leq 2\biggl{(}\frac{1}{T_{n}}-1\biggr{)}^{2}p_{ \theta}(z)^{2}\biggl{(}\Bigl{(}\ln p_{\theta}(z)\,e^{\mu(z,\theta,T_{n})} \Bigr{)}^{2}\!A(\theta,T_{n})+B(\theta,T_{n})\biggr{)}\,.\]  
where the terms \(A(\theta,T_{n})\), \(B(\theta,T_{n})\) and \(a(z,\theta,T_{n})\) come from the [PERSON] development. Then, we can control the integral of interest:  
\[\begin{split}\int_{z}\frac{\biggl{(}\frac{p_{\theta}(z)^{\frac{1} {T_{n}}}}{\int_{z^{\prime}}p_{\theta}(z^{\prime})^{\frac{1}{T_{n}}}}-p_{\theta }(z)\biggr{)}^{2}}{p_{\theta}(z)}dz\leq& 2\biggl{(}\frac{1}{T_{n}}-1\biggr{)}^{2}A(\theta,T_{n})\int_{z}p_{ \theta}(z)e^{2a(z,\theta,T_{n})}\ln^{2}p_{\theta}(z)dz\\ &+2\biggl{(}\frac{1}{T_{n}}-1\biggr{)}^{2}B(\theta,T_{n})\,.\end{split} \tag{7}\]From the properties of the Taylor development, we prove that \(A(\theta,T_{n})\) and \(B(\theta,T_{n})\) both have upper bounds involving only \(\int_{z}p_{\theta}(z)\ln p_{\theta}(z)\) and \(\int_{z}p_{\theta}(z)^{\frac{1}{n}}\ln p_{\theta}(z)\). In a similar fashion, the term \(\int_{z}p_{\theta}(z)e^{2a(z,\theta,T_{n})}\ln^{2}p_{\theta}(z)\) has an upper bound involving only \(\int_{z}p_{\theta}(z)\ln^{2}p_{\theta}(z)dz\) and \(\int_{z}p_{\theta}(z)^{\frac{2}{n}-1}\ln^{2}p_{\theta}(z)dz\).  
Using the hypotheses of the Theorem, we prove that for any \(\alpha\in\overline{B}(1,e)\) and \(\theta\in K\) the terms \(\int_{z}p_{\theta}(z)^{\alpha}\ln p_{\theta}(z)\) and \(\int_{z}p_{\theta}(z)^{\alpha}\ln^{2}p_{\theta}(z)\) are both upper bounded by a constant \(C\) independent of \(\theta\) and \(\alpha\).  
Since \(T_{n}\underset{n\rightarrow\infty}{\longrightarrow}1\), then when \(n\) is large enough, \(\frac{1}{T_{n}}\in\overline{B}(1,e)\) and \(\frac{2}{T_{n}}-1\in\overline{B}(1,e)\). Hence, the previous result applies to the upper bounds of \(A(\theta,T_{n})\), \(B(\theta,T_{n})\) and \(\int_{z}p_{\theta}(z)e^{2a(z,\theta,T_{n})}\ln^{2}p_{\theta}(z)dz\). As a result, these three terms are respectively upper bounded by \(C_{1}\), \(C_{2}\) and \(C_{3}\), three constants independent of \(\theta\) and \(T_{n}\).  
The inequality (7) then becomes:  
\[\int_{z}\frac{1}{p_{\theta}(z)}\left(\frac{p_{\theta}(z)^{\frac{1}{n}}}{\int_ {z^{\prime}}p_{\theta}(z^{\prime})^{\frac{1}{n}}}-p_{\theta}(z)\right)^{2}dz \leq 2\bigg{(}\frac{1}{T_{n}}-1\bigg{)}^{2}C_{1}C_{2}+2\bigg{(}\frac{1}{T_ {n}}-1\bigg{)}^{2}C_{3}\,.\]  
By taking the supremum in \(\theta\in K\) and the limit when \(n\longrightarrow\infty\), we get the desired result:  
\[\sup_{\theta\in K}\,\int_{z}\frac{1}{p_{\theta}(z)}\left(\frac{p_{\theta}(z)^{ \frac{1}{n}}}{\int_{z^{\prime}}p_{\theta}(z^{\prime})^{\frac{1}{n}}}-p_{ \theta}(z)\right)^{2}dz\underset{n\rightarrow\infty}{\longrightarrow}0\,.\]  
With condition (5) verified, the conclusions of Theorem 1 are applicable, which concludes the proof.