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I was wondering if there is a easy way (some native function) for interpolation of points, i.e. to connect some few given points by a curve (approximation of a function). I have some concrete formulas and I am able to calculate some coordinates. Additional problem is that the entire curve isn't injective (one X projects to more than one Y, here the axis contains mu and sigma respective return and risk).

In a nutshell: How to connect the first three points (0.9; 1.25), (1.12;1.5) and (2;2)? (pictures red line).

Note: functions are given for expected value and variance, diagram and table depict standard deviation (square root).

enter image description here

Update: As I've not found any alternative it must be the case that one has to calculate two functions mu in terms of sigma, solve the variance-equation for a yields to expressions (the lower and upper arm). TeX is able to ignore the injectivity-problem, so one can set the lower bound plot-value rather low.

\begin{center} 
\begin{tikzpicture}[scale=1.5]
  \begin{axis}[axis lines=middle,xmin=-0.3,xmax=3.2,ymin=-0.3,ymax=3.2,
    xlabel=$\scriptstyle \sigma$,
    ylabel=$\scriptstyle \mu$,
    tick label style={font=\tiny}, 
        ]
        \addplot+[no marks,blue,domain=0.6:2.5,samples=200, thick] {2 - 1/5* (sqrt(5 * x^2 - 4) + 4)} ;
        \addplot+[no marks,blue,domain=0.6:2.5,samples=200, thick] {2 - 1/5* (4 - sqrt(5 * x^2 - 4))} ;
  \end{axis}
\end{tikzpicture}
\end{center}

I was not able to construct the curve in the way that it looks continuous though. Maybe someone can resolve this last issue.

enter image description here

  • 1
    If you have a TI graphics calculator, you can perform quadratic regression. – John Kormylo Aug 21 '15 at 13:34
  • 1
    You could increase the number of samples, or just set the lower limit for x to be 0.894427191 (sqrt(4/5)). – John Kormylo Aug 22 '15 at 13:51

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