# A Model of Information Visualization Interpretation
## 3 Results
### Parametric Behavior of the Mean
#### 3.2.6 Analysis of the Scale Parameter  
Now, we consider the influence of the \(\tau\) parameter. Recall \(\tau=\sigma/h\), which is the ratio of the spread of the fixation points to the height of the bar. It gives the scale of the visualization problem. Let us consider the sensitivity of the mean to it.  
**Corollary 5**.: \(\mu(\tau;h,\rho)\)_, given in Equation (38), is an increasing function of \(\tau\) over the interval \((0,\infty)\)._  
Proof.: Using \(u\) and \(\psi\) satisfying  
\[u =\frac{1}{2}\frac{h^{2}}{\psi^{2}} \psi^{2} =2h^{2}\tau^{2}(1-\rho)\]  
then \(\mu(\tau;h,\rho)\) can be transformed into the intermediate form in a fashion identical to what has already be done for \(\mu(\rho;h,\tau)\).  
By the chain rule,  
\[\frac{d\mu}{d\tau}=\frac{d\mu}{du}\frac{du}{d\psi}\frac{d\psi}{d\tau}. \tag{76}\]  
We need \(d\psi/d\tau\):  
\[\frac{d\psi}{d\tau}=\frac{d}{d\tau}(\sqrt{2}h\tau\sqrt{1-\rho})=\sqrt{2}h \sqrt{1-\rho}. \tag{77}\]  
Using Equation (77), Equation (59), and Equation (50) in Equation (76) gives  
\[\frac{d\mu}{d\tau}=-h\frac{\sqrt{\pi}}{4}\frac{\mathrm{e}^{-u}}{u^{3/2}}\,_{ 1}\mathrm{F}_{1}(1/2;1;u)\biggl{(}-\frac{2u}{\psi}\biggr{)}(\sqrt{2}h\sqrt{1 -\rho}).\]  
We must eliminate \(u\) and \(\psi\) from the expression, leaving the parameters \(h\) and \(\rho\). To this end, we transform it according to  
\[\frac{1}{\psi}=\frac{1}{\sqrt{2}h\tau\sqrt{1-\rho}} \frac{1}{\sqrt{u}} =2\tau\sqrt{1-\rho}.\]After making the appropriate substitutions followed by basic manipulations, we obtain  
\[\frac{d\mu}{d\tau}(\tau;h,\rho)=h\sqrt{\pi}\sqrt{1-\rho}\,\mathrm{e}^{-\frac{1}{4 \tau^{2}(1-\rho)}}\,_{1}\mathrm{F}_{1}\bigg{(}1/2;1;\frac{1}{4\tau^{2}(1-\rho)} \bigg{)}. \tag{78}\]  
From this, it is seen that \(d\mu/d\tau>0\) by Lemma 1 over the interval \((0,\infty)\). Therefore, \(\mu(\tau;h,\rho)\) is increasing.  
For \(\mu(\tau;h,\rho)\), given in Equation (38),  
\[\mu(\tau;h,\rho)\sim h \tag{79}\]  
as \(\tau\to 0^{+}\).  
Proof.: As has already been demonstrated for \(\mu(\rho;h,\tau)\), it is possible to transform this expression into the intermediate form. The details are seen in Corollary 1 and shall be omitted here.  
Once \(\mu(\tau;h,\rho)\) is transformed into \(\mu(u;h)=h\Phi(u)\), we note that, unlike the other key parameters, as it becomes small, \(u\) becomes large, as can be seen from the definition  
\[u=\frac{1}{4}\tau^{-2}\frac{1}{1-\rho}. \tag{80}\]  
That is, \(u\to\infty\) as \(\tau\to 0^{+}\) and vice versa as \(\rho\) is a parameter that is held constant. Hence, Theorem 2 yields \(\Phi(u)\sim 1\), and therefore, \(\mu(\tau;h,\rho)\sim h\).  
Finally, we calculate the derivative with respect to \(\tau\) and look at its asymptotic properties, as has been done similarly before.  
For \(\frac{d\mu}{d\tau}(\tau;h,\rho)\), given in Equation (78),  
\[\frac{d\mu}{d\tau}(\tau;h,\rho)\sim 2h(1-\rho)\tau \tag{81}\]  
as \(\tau\to 0^{+}\).  
Proof.: We change the variables to  
\[v=\frac{1}{8\tau^{2}(1-\rho)}. \tag{82}\]  
Substituting Equation (82) into Equation (78), we obtain  
\[\frac{d\mu}{d\tau}(v;h,\rho)=h\sqrt{\pi}\sqrt{1-\rho}\,\mathrm{e}^{-2v}\,_{1} \mathrm{F}_{1}(1/2;1;2v). \tag{83}\]  
Again, we have  
\[{}_{1}\mathrm{F}_{1}(1/2;1;2v)=\mathrm{e}^{v}\,\mathrm{I}_{0}(v)\]  
which we use in Equation (83):  
\[\frac{d\mu}{d\tau}(v;h,\rho)=h\sqrt{\pi}\sqrt{1-\rho}\,\mathrm{e}^{-v}\, \mathrm{I}_{0}(v).\]Recalling the asymptotic behavior of the modified Bessel function given in Equation (66) and substituting into the previous equation, we obtain  
\[\frac{d\mu}{d\tau}(v;h,\rho) \sim h\sqrt{\pi}\sqrt{1-\rho}\,\mathrm{e}^{-v}\,\frac{\mathrm{e}^{v}}{ \sqrt{2\pi v}}\] \[\sim \frac{h}{\sqrt{2}}\frac{1}{\sqrt{v}}\sqrt{1-\rho}\]  
as \(v\to\infty\). We invert the mapping from \(v\) to \(\tau\) using  
\[\frac{1}{\sqrt{v}}=2\sqrt{2}\tau\sqrt{1-\rho}.\]  
Note that as \(v\to\infty\), we have \(\tau\to 0^{+}\)  
\[\frac{d\mu}{d\tau}(\tau;h,\rho) \sim \frac{h}{\sqrt{2}}2\sqrt{2}\tau(1-\rho)\] \[\sim 2h(1-\rho)\tau.\]