## 5 Example Where PPP Occurs without Approximation  
Thus far we have not demonstrated any error model which exactly (without approximation) produces a \(\Sigma\) that leads to PPP. A situation that leads to PPP without the approximations in Equation 1 is expressed in Theorem 3. _Theorem 3_. Suppose \(m_{1}=\mu+\epsilon_{R1}\) and \(m_{2}=\mu+\epsilon_{R2}\) where \(\epsilon_{R1}\) is random error in \(m_{1}\) with variance \(\sigma_{R1}^{2}\) and similarly for \(\epsilon_{R2}\). Then if \(y_{1}=m_{1}+\epsilon_{S}\) and \(y_{2}=m_{2}+\alpha\epsilon_{S}\), where \(\epsilon_{S}\sim N(0,\sigma_{S})\) and \(\alpha\) is any positive scale factor other than 1, the covariance matrix \(\Sigma\) of \((y_{1},y_{2})\) can lie in the PPP region. Proof. The proof is by demonstration. Specify any values \(\sigma_{1}\), \(\sigma_{2}\), and \(\sigma_{12}\) that satisfy the PPP condition \((\rho>\min\left(\frac{\sigma_{1}}{\sigma_{2}},\frac{\sigma_{2}}{\sigma_{1}} \right))\). Choose any \(\alpha>0\), then \(\sigma_{S}^{2}=\sigma_{12}^{2}/\alpha\), \(\sigma_{R1}^{2}=\sigma_{1}^{2}-\sigma_{S}^{2}\), and \(\sigma_{R1}^{2}=\sigma_{2}^{2}-\alpha^{2}\sigma_{S}^{2}\). As an example, let \(\sigma_{1}^{2}=0.1134\), \(\sigma_{2}^{2}=0.0504\), and \(\sigma_{12}^{2}=0.06\) as in Equation 3. Then if \(\alpha=0.7\), we have \(\sigma_{S}^{2}=\sigma_{12}^{2}/\alpha=0.06/0.7=0.0857\), \(\sigma_{R1}^{2}=\sigma_{1}^{2}-\sigma_{S}^{2}=0.1134-0.0857=0.0277\), and \(\sigma_{R2}^{2}=\sigma_{2}^{2}-\alpha^{2}\sigma_{S}^{2}=0.06-0.7^{2}\times 0.06/0.7=0.018\). Note:  
* If \(\alpha=1\), then \(\Sigma\) has the form of Equation 11, so PPP cannot occur ([2]). * If \(\alpha<0\) then PPP cannot occur. However, our applications have \(\alpha>0\). [PERSON] _et al._[10] showed that if \(\rho\rightarrow\pm 1\) and \(\sigma_{1}\
eq\sigma_{2}\), then \(\mu\) can be estimated with surprisingly small variance. That fact plus the known BLUE property of GLS could convince us to just \"live with\" the PPP because it can make sense for the GLS estimate \(\hat{\mu}\) to lie outside the range of \((y_{1},y_{2})\) in the case \(\alpha>0\), or \(\sigma_{12}>0\). One example in which the assumptions of Theorem 3 hold involves subtracting a background measurement from a region of interest (ROI) measurement to get a net result. Because the background measurement often involves a different number of channels than the ROI measurement, a scale factor \(k\) is introduced to estimate the net counts as \(\mathrm{net}=\mathrm{peak}-k\times\mathrm{background}\). Figure 2 illustrates a hypothetical example where each of three peak ROIs have a corresponding background measurement in the plot of the square root of detected neutron counts _versus_ neutron energy in arbitrary units (au). Suppose each ROI and corresponding background are analyzed separately, and consider the first ROI in Figure 2. The count times could vary between the two experiments, so \(\sigma_{R1}^{2}\
eq\sigma_{R2}^{2}\). Both experiment 1 and experiment 2 measure the ROI counts, but in many situations, the background measurement is made only by experiment 1. In that case, experiments 1 and 2 must use the background measurement made by experiment 1. Also, if the ROI is found by analyzing the shape of the curve (the \"spectrum\") that describes detected count rates _versus_ particle energy, then the ROI for experiments 1 and 2 could differ. We then have \(N_{1}=G_{1}-a_{1}\times B\) and \(N_{2}=G_{2}-a_{2}\times B\), where \(N\) is net counts, \(G\) is gross counts, \(B\) is background, \(a_{1}\) is the scale factor for experiment 1, and \(a_{2}\) is the scale factor for experiment 2. The scale factor \(a_{1}\) for experiment 1 is the ratio of the number of ROI channels to the number of background channels, and similarly for the scale factor \(a_{2}\) for experiment 2. The channel counts have variation from repeat to repeat so the detected counts will vary around the true counts with some error.