# A Model of Information Visualization Interpretation
## 3 Results
### Parametric Behavior of the Mean
#### 3.2.4 Analysis of the Correlation Parameter  
In order to do a parametric analysis of \(\mu(\rho;h,\tau)\), it is sufficient to use material in the proof of Theorem 1 and do a chain rule argument. The following corollary summarizes the results. **Corollary 1**. : \(\mu(\rho;h,\tau),\;\) _given in Equation (37), is a decreasing function of \(\rho\) over the interval \((-1,1)\)._  
Proof. : The proof proceeds by transforming \(\mu(\rho;h,\tau)\) into the intermediate form using  
\[u=\frac{h^{2}}{2\psi^{2}}\qquad\qquad\qquad\qquad\qquad\psi=\sqrt{2}h\tau\sqrt {1-\rho}.\]  
then taking the derivative with respect to \(\rho\) and showing it is negative. We rewrite Equation (37) as  
\[\mu(\rho;h,\tau)=\bigg{(}\sqrt{\frac{\pi}{2}}\sqrt{2}h\tau\sqrt{1-\rho}\exp \bigg{(}-\frac{h^{2}}{4h^{2}\tau^{2}(1-\rho)}\bigg{)}\,{}_{1}\mathrm{F}_{1} \bigg{(}3/2;1;\frac{h^{2}}{4h^{2}\tau^{2}(1-\rho)}\bigg{)}\bigg{)}.\]  
Noting  
\[\sqrt{2}h\tau\sqrt{1-\rho}=\frac{h}{\sqrt{2}u} \tag{56}\]  
and substituting, we obtain  
\[\mu(u;h)=h\frac{\sqrt{\pi}}{2}\frac{\mathrm{e}^{-u}}{\sqrt{u}}\,{}_{1} \mathrm{F}_{1}(3/2;1;u). \tag{57}\]  
By the chain rule,  
\[\frac{d\mu}{d\rho}=\frac{d\mu}{du}\frac{du}{d\psi}\frac{d\psi}{d\rho}. \tag{58}\]  
We have  
\[\frac{du}{d\psi}=-\frac{2}{\psi}u \tag{59}\]and  
\[\frac{d\psi}{d\rho}=-\frac{h\tau}{\sqrt{2}}\frac{1}{\sqrt{1-\rho}}. \tag{60}\]  
Using Equation (59), Equation (60), and Equation (50) in Equation (58) gives  
\[\frac{d\mu}{d\rho}=-h\frac{\sqrt{\pi}}{4}\frac{\mathrm{e}^{-u}}{u^{3/2}}\,_{1} \mathrm{F}_{1}(1/2;1;u)\biggl{(}-\frac{2}{\psi}u\biggr{)}\biggl{(}-\frac{h\tau }{\sqrt{2}}\frac{1}{\sqrt{1-\rho}}\biggr{)}. \tag{61}\]  
After canceling cross-terms, we apply  
\[\frac{1}{\sqrt{u}}=\frac{\sqrt{2}\psi}{h}\]  
and  
\[u=\frac{h^{2}}{2\psi^{2}}=\frac{1}{4\tau^{2}(1-\rho)}\]  
to return to the original variables, obtaining  
\[\mu(\rho;h,\tau)=\bigg{(}\sqrt{\frac{\pi}{2}}\sqrt{2}h\tau\sqrt{1-\rho}\exp \bigg{(}-\frac{h^{2}}{4h^{2}\tau^{2}(1-\rho)}\bigg{)}\,{}_{1}\mathrm{F}_{1} \bigg{(}3/2;1;\frac{h^{2}}{4h^{2}\tau^{2}(1-\rho)}\bigg{)}\bigg{)}.\]0  
From this, it is seen that \(d\mu/d\rho<0\) by part 2 of Lemma 1 over the interval \(-1<\rho<1\). Therefore, \(\mu(\rho;h,\tau)\) is decreasing as \(\rho\) approaches \(1\) from below. We now know that the expected length of a side of the triangle actually decreases as the fixation points become more strongly correlated in a positive sense. (We are discussing that the particular side connecting two points labeled one and two impacts the model only through the value of the noncentrality parameter. Ultimately, the value we labeled as \(h^{2}\) could become \(b^{2}\) or \(h^{2}+b^{2}\). We have chosen the stochastic edge \(D_{12}\) because of its association with the eye attempting to measure the height of the bar.) **Corollary 2**. : _For \(\mu(\rho;h,\tau)\), given in Equation (37),_  
\[\mu(\rho;h,\tau)=\bigg{(}\sqrt{\frac{\pi}{2}}\sqrt{2}h\tau\sqrt{1-\rho}\exp \bigg{(}-\frac{h^{2}}{4h^{2}\tau^{2}(1-\rho)}\bigg{)}\,{}_{1}\mathrm{F}_{1} \bigg{(}3/2;1;\frac{h^{2}}{4h^{2}\tau^{2}(1-\rho)}\bigg{)}\bigg{)}.\]1  
_as \(\rho\to 1^{-}\)._  
Proof. : It was shown in the proof of Corollary 1 that \(\mu(\rho;h,\tau)\) can be transformed into the intermediate form using  
\[\mu(\rho;h,\tau)=\bigg{(}\sqrt{\frac{\pi}{2}}\sqrt{2}h\tau\sqrt{1-\rho}\exp \bigg{(}-\frac{h^{2}}{4h^{2}\tau^{2}(1-\rho)}\bigg{)}\,{}_{1}\mathrm{F}_{1} \bigg{(}3/2;1;\frac{h^{2}}{4h^{2}\tau^{2}(1-\rho)}\bigg{)}\bigg{)}.\]2  
which can be written as  
\[\mu(\rho;h,\tau)=\bigg{(}\sqrt{\frac{\pi}{2}}\sqrt{2}h\tau\sqrt{1-\rho}\exp \bigg{(}-\frac{h^{2}}{4h^{2}\tau^{2}(1-\rho)}\bigg{)}\,{}_{1}\mathrm{F}_{1} \bigg{(}3/2;1;\frac{h^{2}}{4h^{2}\tau^{2}(1-\rho)}\bigg{)}\bigg{)}.\]3  
It is easily seen that \(u\to\infty\) as \(\rho\to 1^{-}\) and vice versa as \(\tau\) is fixed. In this case, Theorem 2 is applied to show us \(\Phi(u)\sim 1\) and, therefore, \(\mu(\rho;h,\tau)\sim h\).