# Estimating Tail Probabilities of Random Sums of Phase-Type Scale Mixture Random Variables +
## 5 Numerical Experiments
Then, the likelihood ratio is given by_ \[\frac{f(X_{N})}{f_{\theta}(X_{N})}=\frac{\beta^{k}}{\beta_{\theta}}\frac{X_{N} ^{(k-1)}\mathrm{e}^{-(\beta-\beta_{\theta})X_{N}}}{(k-1)!}\,.\] _We solve_ \(\zeta^{\prime}(\theta(N))=\frac{Z_{N-1}^{\star}\vee(u-S_{N-1})}{\mathbb{E}[W_ {N}]}\) _using_ \(\zeta(\theta(N))=\ln\frac{\beta}{\beta-\theta(N)}\)_, conditioning on_ \(N\)_, resulting in_ \(\theta(N)=\beta-\frac{\mathbb{E}[W_{N}]}{Z_{N-1}^{\star}\vee(u-S_{N-1})}\)_, and_ \(\beta_{\theta}=\frac{\mathbb{E}[W_{N}]}{Z_{N-1}^{\star}\vee(u-S_{N-1})}\)_. The likelihood ratio for (_10_), conditioning on_ \(N\)_, is given by_ \[f(x_{i})=\frac{\beta^{k}x_{i}^{k-1}\operatorname{e}^{-\beta x_{i}}}{(k-1)!},\ \ x_{i}\geq 0\,,\]0  
_We take a sample size of \(10^{5}\), \(W_{i}\overset{i.i.d}{\sim}\mathrm{Exp}(\mu)\), \(X_{i}\overset{i.i.d}{\sim}\mathrm{Erlang}(k,\beta)\), where \(i=1,\ldots,N\) and \(N\sim\mathrm{Geo}(p)\) with \(\mu=1\), \(k=1\), \(\beta=3\), and \(p=0.2\). The results are shown in \[f(x_{i})=\frac{\beta^{k}x_{i}^{k-1}\operatorname{e}^{-\beta x_{i}}}{(k-1)!},\ \ x_{i}\geq 0\,,\]1 Proposition 1 does not provide a theoretical efficiency guarantee for this example as the number of summands is random, although the scaling random variable has a Weibullian tail. From numerical example results, we can observe that estimator \(\widehat{\ell}_{\mathrm{ConAK2+CV}}\) is not providing reliable estimates; however, estimators \(\widehat{\ell}_{\mathrm{ConAK1+IS+CV}}\) and \(\widehat{\ell}_{\mathrm{ConAK2+IS+CV}}\) perform better._  
**Example 4** (Heavy-tailed Scaling: Weibull). : _Let the scaling random variable \(W_{i}\) be Weibull-distributed, with PDF_  
\[f(x_{i})=\frac{\beta^{k}x_{i}^{k-1}\operatorname{e}^{-\beta x_{i}}}{(k-1)!},\ \ x_{i}\geq 0\,,\]2  
_and consider \(X_{i}\overset{i.i.d}{\sim}Erlang(k,\beta)\), where \(i=1,\ldots,N\)._  
1. _When conditioning on the scaling random variable, the resulting hazard-rate twisted PDF of the scaling random variable is_ \[f(x_{i})=\frac{\beta^{k}x_{i}^{k-1}\operatorname{e}^{-\beta x_{i}}}{(k-1)!},\ \ x_{i}\geq 0\,,\]3 _Thus, conditioning on_ \(N\)_, and taking into account_ \(\mathbb{E}[X_{N}]\)_, the likelihood ratio of a single element associated with this change of measure is_ \[f(x_{i})=\frac{\beta^{k}x_{i}^{k-1}\operatorname{e}^{-\beta x_{i}}}{(k-1)!},\ \ x_{i}\geq 0\,,\]4 _The likelihood ratio for (_7_), conditioning on_ \(N\)_, is given by_ \[f(x_{i})=\frac{\beta^{k}x_{i}^{k-1}\operatorname{e}^{-\beta x_{i}}}{(k-1)!},\ \ x_{i}\geq 0\,,\]5 _where_ \(\theta(N)=1-1/\Lambda((Z_{N-1}^{\star}\vee(u-S_{N-1}))/\mathbb{E}[X_{N}])\)_. We note that from (_11_), we have_ \(\overline{H}_{\theta}(W_{N})=\big{[}\overline{H}(W_{N})\big{]}^{1-\theta}\)_, and so it is straightforward to generate the twisted scaling random variables by using the inverse transform method._
2. _When conditioning on the_ \(PH\) _random variable, the resulting PDF of the Erlang random variable is Erlang_\((k,\beta_{\theta})\)_; then, the likelihood ratio is given by_ \[f(x_{i})=\frac{\beta^{k}x_{i}^{k-1}\operatorname{e}^{-\beta x_{i}}}{(k-1)!},\ \ x_{i}\geq 0\,,\]6 _We solve_ \(\zeta^{\prime}(\theta(N))=\frac{Z_{N-1}^{\star}\vee(u-S_{N-1})}{\mathbb{E}[W_ {N}]}\) _using_ \(\zeta(\theta(N))=k\ln\frac{\beta}{\beta-\theta(N)}\)_, conditioning on_ \(N\)_, resulting in_ \(\theta(N)=\beta-\frac{\mathbb{E}[W_{N}]}{Z_{N-1}^{\star}\vee(u-S_{N-1})}\)_, and_ \(\beta_{\theta}=\frac{k\mathbb{E}[W_{N}]}{Z_{N-1}^{\star}\vee(u-S_{N-1})}\)_. The likelihood ratio for (_10_), conditioning on_ \(N\)_, is given by_ \[f(x_{i})=\frac{\beta^{k}x_{i}^{k-1}\operatorname{e}^{-\beta x_{i}}}{(k-1)!},\ \ x_{i}\geq 0\,,\]7  
\[f(x_{i})=\frac{\beta^{k}x_{i}^{k-1}\operatorname{e}^{-\beta x_{i}}}{(k-1)!},\ \ x_{i}\geq 0\,,\]8 The random variable \(N\) is geometrically distributed with success probability \(p=0.2\), \(W_{i}\overset{i.i.d}{\sim}\mathrm{Exp}(1)\), \(X_{i}\overset{i.i.d}{\sim}\mathrm{Erlang}(1,3)\), where \(i=1,\ldots,N\). Black squares: ConAK1+IS+CV (conditioning and IS on exponential). Red stars: ConAK2+CV.