## 2 Materials and Methods
### Tempered EM
#### 2.3.4 Examples of Models That Verify the Conditions
The complete likelihood of the model, not accounting for irrelevant constants, is:  
\[h(z;\theta)=e^{\theta\sum_{i=1}^{N}x^{(i)}}.exp\left(-e^{\theta}\sum_{i=1}^{N} e^{z^{(i)}}\right). \tag{10}\]\[g(\theta)=\int_{z}h(z;\theta)dz\] can be computed analytically up to a constant: \[g(\theta) =\int_{z\in\mathbb{R}^{N}}h(z;\theta)dz \tag{11}\] \[=e^{\theta\sum_{i=1}^{N}x^{(i)}}\int_{z\in\mathbb{R}^{N}}exp\left( -e^{\theta}\sum_{i=1}^{N}e^{z^{(i)}}\right)dz\] \[=e^{\theta\sum_{i=1}^{N}x^{(i)}}\prod_{i=1}^{N}\int_{z^{(i)}\in \mathbb{R}}exp\left(-e^{\theta}e^{z^{(i)}}\right)dz^{(i)}\] \[=e^{\theta\sum_{i=1}^{N}x^{(i)}}\left(\int_{u\in\mathbb{R}_{+}} \frac{exp(-u)}{u}du\right)^{N}\] \[=e^{\theta\sum_{i=1}^{N}x^{(i)}}E_{1}(0)^{N}\,,\]  
where \(E_{1}(0)\) is a finite, non zero, constant, called \"exponential integral\", in particular independent of \(\alpha\) and \(\theta\). Condition on \(\int_{z}p_{\theta}^{\alpha}(z)dz\). : Let \(K\) be a compact in \(\Theta\). We have \(p_{\theta}(z)=\frac{h(z;\theta)}{g(\theta)}\). Let us compute \(\int_{z}h(z;\theta)^{\alpha}\) for any positive \(\alpha\). The calculations work as in Equation (11):  
\[\int_{z}p_{\theta}^{\alpha}(z)dz=\sum_{k}p_{\theta}^{\alpha}(z=k)\,.\]0 \[\int_{z}p_{\theta}^{\alpha}(z)dz=\sum_{k}p_{\theta}^{\alpha}(z=k)\,.\]1  
Hence:  
\[\int_{z}p_{\theta}^{\alpha}(z)dz=\sum_{k}p_{\theta}^{\alpha}(z=k)\,.\]2  
Since \(E_{1}(0)\) is finite, non zero, and independent of \(\theta\), we easily have:  
\[\int_{z}p_{\theta}^{\alpha}(z)dz=\sum_{k}p_{\theta}^{\alpha}(z=k)\,.\]3  
\(\theta\) does not even have to be restricted to a compact. Condition on \(\int_{z}S_{u}^{2}(z)p_{\theta}^{\alpha}(z)dz\). : Let \(K\) be a compact in \(\Theta\) and \(\alpha\) a positive real number. In this Poisson count model, \(S(z)=\sum_{i=1}^{N}e^{z^{(i)}}\in\mathbb{R}\). We have:  
\[\int_{z}p_{\theta}^{\alpha}(z)dz=\sum_{k}p_{\theta}^{\alpha}(z=k)\,.\]4  
First, let us prove that the integral is finite for any \(\theta\). We introduce the variables \(u_{k}:=\sum_{l=1}^{k}e^{z_{l}}\). The Jacobi matrix is triangular and its determinant is \(\prod_{k}e^{z^{(i)}}=\prod_{k}u_{k}\). \[\int_{z}p_{\theta}^{\alpha}(z)dz=\sum_{k}p_{\theta}^{\alpha}(z=k)\,.\]5  
Which is proportional to:  
\[\int_{z}p_{\theta}^{\alpha}(z)dz=\sum_{k}p_{\theta}^{\alpha}(z=k)\,.\]6where we removed the finite constant \(\frac{1}{E_{1}(0)^{\alpha N}}\) for clarity. This integral is finite for any \(\theta\) because the exponential is the dominant term around \(+\infty\). Let us now prove that \(\theta\mapsto\int_{z}S^{2}(z)p_{\theta}^{\alpha}(z)dz\) is continuous. From Equation (12), we have that  
* \(z\mapsto S^{2}(z)p_{\theta}^{\alpha}(z)\) is measurable on \(\mathbb{R}^{N}\). * \(\theta\mapsto S^{2}(z)p_{\theta}^{\alpha}(z)\) is continuous on \(K\) (and on \(\Theta=\mathbb{R}\)). * With \(\theta_{M}:=\underset{\theta\in K}{min}\,\theta\), then \(\forall\theta\in K,\;0\leq S^{2}(z)p_{\theta}^{\alpha}(z)\leq S^{2}(z)p_{ \theta_{M}}^{\alpha}(z)\)  
Since we have proven that \(S^{2}(z)p_{\theta_{M}}^{\alpha}(z)<\infty\), then we can apply the intervertion Theorem and state that \(\theta\mapsto\int_{z}S^{2}(z)p_{\theta}^{\alpha}(z)dz\) is continuous. It directly follows that:  
\[\int_{z}p_{\theta}^{\alpha}(z)dz=\sum_{k}p_{\theta}^{\alpha}(z=k)\,.\]7  
Note that after the change of variable, the integral could be computed explicitly, but involves \(N\) successive integration of polynomial \(\times\) exponential function products of the form \(P(x)e^{-\alpha\theta^{x}x}\). This would get tedious, especially since after each successful integration, the product with the next integration variable \(u_{k-1}\) increases by one the degree of the polynomial, i.e., starting from 3, the degree ends up being \(N+2\). We chose a faster path.