# A Model of Information Visualization Interpretation
## 3 Results
### Parametric Behavior of the Mean
#### 3.2.5 Analysis of the Noncentrality Parameter  
Now, we consider the mean in Equation (27) as a function of the noncentrality parameter \(\lambda\).  
**Corollary 3**.: \(\mu(\lambda;h)\)_, given in Equation (39), is a decreasing function of \(\lambda\) over the interval \((0,\infty)\)._  
Proof.: For \(\mu\) given in Equation (39) and with \(u\) and \(\psi\) defined in terms of \(\lambda\) as  
\[u=\frac{1}{2}\frac{h^{2}}{\psi^{2}}\hskip 56.905512 pt\psi=\frac{h}{\lambda}\]  
we can transform Equation (39) into the intermediate form. Combining these transformations, we see that \(\lambda=\sqrt{2}\sqrt{u}\) and obtain  
\[\mu(u;h)=h\frac{\sqrt{\pi}}{2}\frac{\mathrm{e}^{-u}}{\sqrt{u}}\,_{1}\mathrm{F }_{1}(3/2;1;u). \tag{67}\]Now, we proceed as before and use the properties of the intermediate form. By the chain rule,  
\[\frac{d\mu}{d\lambda}=\frac{d\mu}{du}\frac{du}{d\psi}\frac{d\psi}{d\lambda}. \tag{68}\]  
We note  
\[\frac{d\psi}{d\lambda}=-h\psi^{-2}.\]  
Using this with Equation (59) and Equation (50) in Equation (68) gives  
\[\frac{d\mu}{d\lambda}=-h\frac{\sqrt{\pi}}{4}\frac{\mathrm{e}^{-u}}{u^{3/2}}\,_ {1}\mathrm{F}_{1}(1/2;1;u)\biggl{(}-\frac{2u}{\psi}\biggr{)}\biggl{(}-\frac{h} {\lambda^{2}}\biggr{)}. \tag{69}\]  
Now, we use  
\[\frac{1}{\sqrt{u}}=\frac{\sqrt{2}}{\lambda}\]  
and  
\[\frac{1}{\psi}=\frac{\lambda}{h}\]  
to obtain  
\[\frac{d\mu}{d\lambda}(\lambda;h)=-h\sqrt{\frac{\pi}{2}}\frac{\mathrm{e}^{- \frac{\lambda^{2}}{2}}}{\lambda^{2}}\,_{1}\mathrm{F}_{1}\biggl{(}1/2;1;\frac{ \lambda^{2}}{2}\biggr{)}. \tag{70}\]  
From this, it is seen that \(d\mu/d\lambda<0\) by Lemma 1 over the interval \((0,\infty)\). Therefore, \(\mu(\lambda;h)\) is decreasing.  
For \(\mu(\lambda;h)\), given in Equation (39),  
\[\mu(\lambda;h)\sim h \tag{71}\]  
as \(\lambda\) approaches \(\infty\)  
Proof.: It was just shown that under appropriate relationships among the parameters that \(\mu(\lambda;h)\) can be transformed into the intermediate form \(\mu(u;h)=h\Phi(u)\), where  
\[u=\frac{1}{2}\lambda^{2}. \tag{72}\]  
Undoubtedly \(u\to\infty\) as \(\lambda\to\infty\) and vice versa. Hence, we apply Theorem 2 and conclude that \(\Phi(u)\sim 1\); therefore, \(\mu(\lambda;h)\sim h\) as \(\lambda\to\infty\).  
Now, we consider the derivative of the mean with respect to the noncentrality parameter. For \(\frac{du}{d\lambda}(\lambda;h)\), given in Equation (70),  
\[\frac{d\mu}{d\lambda}(\lambda;h)\sim-\frac{h}{\lambda^{3}}\]  
as \(\lambda\to\infty\).  
Proof.: We change the variable to  
\[t=\frac{\lambda^{2}}{4}\]  
resulting in  
\[\frac{d\mu}{d\lambda}(t;h)=-h\sqrt{\frac{\pi}{2}}\frac{\mathrm{e}^{-2t}}{4t} \,_{1}\mathrm{F}_{1}(1/2;1;2t). \tag{73}\]As previously noted, the following relationship holds:  
\[{}_{1}\mathrm{F}_{1}(1/2;1;2t)=\mathrm{e}^{t}\,\mathrm{I}_{0}(t). \tag{74}\]  
Using this in Equation (73) gives  
\[\frac{d\mu}{d\lambda}(t;h)=-\frac{h}{4}\sqrt{\frac{\pi}{2}}\frac{\mathrm{e}^{- t}}{t}\,\mathrm{I}_{0}(t).\]  
As seen in Equation (66), for \(t\to\infty\),  
\[\frac{d\mu}{d\lambda}(t;h) \sim-\frac{h}{4}\sqrt{\frac{\pi}{2}}\frac{\mathrm{e}^{-t}}{t}\frac {\mathrm{e}^{t}}{\sqrt{2\pi t}}\] \[\sim-\frac{h}{8}\frac{1}{t^{3/2}}.\]  
Obviously, \(t\to\infty\) if and only if \(\lambda\to\infty\). Since \(1/t^{3/2}=8/\lambda^{3}\), we conclude that  
\[\frac{d\mu}{d\lambda}(\lambda;h)\sim-\frac{h}{8}\frac{8}{\lambda^{3}}=-\frac{ h}{\lambda^{3}} \tag{75}\]  
as \(\lambda\to\infty\).