I found this image on a presentation.
[![enter image description here](https://i.stack.imgur.com/TZV1X.png)](https://i.stack.imgur.com/TZV1X.png)
I am working on MWE but I was wondering if you had ever come through that type of representation with a projection **on** the 3d graph ?
It could look quite like [TeXexample](http://pgfplots.net/tikz/examples/contour-and-surface/) but impossible to adapt to real data so far. MWE to follow.
The green graph is projected on the 3D graph (transformation) and projected on the axis below.

Following @marmot answer, I adapted the code with the correct 3D functions (Call).

```
\documentclass[tikz,border=3.14mm]{standalone}
\usepackage{pgfplots}
\pgfplotsset{compat=1.12}
   \begin{document}
   \begin{tikzpicture}[scale=1.8,   declare function={ 
    Nprime(\x) = 1/(sqrt(2*pi))*exp(-0.5*(pow(\x,2))); 
    normcdf(\x,\m,\SIG) = 1/(1 + exp(-0.07056*((\x-\m)/\SIG)^3 - 1.5976*(\x-\m)/\SIG));
    d2(\x,\y,\KK,\RR,\SIG) = (ln(\x/\KK)+(\RR-(pow(\SIG,2)/2)*\y))/(\SIG*(sqrt(\y)));
    d1(\x,\y,\KK,\RR,\SIG) = d2(\x,\y,\KK,\RR,\SIG) + (\SIG*(sqrt(\y)));
    Call(\x,\y,\KK,\RR,\SIG) = \x*normcdf(d1(\x,\y,\KK,\RR,\SIG),0,1)-\KK*exp(-\RR*\y)*normcdf(d2(\x,\y,\KK,\RR,\SIG),0,1); 
       Brownian(\x)= ; %% I'd like to generate a  function brownian motion, starting at 100 with a \sig standard deviation over time
    }
    ]
        \begin{axis}[view={20}{20},axis on top,xlabel=$S$,ylabel=Time,zlabel=Option 
   price,mesh/interior colormap name=hot,colormap/hot,3d box=complete,grid,grid 
   style={thin,gray!40},axis line style={gray!40}]

    % I fix the following parameters of the Call function
    \def\KK{100}
    \def\TT{0.5}
    \def\RR{0}
    \def\SIG{0.15}

    \addplot3[line width=0.5pt,surf, opacity=0.25, shader=flat,y 
    domain=0.1:1,domain=50:150] {Call(\x,\y,\KK,\RR,\SIG)};
    \end{axis}
    \end{tikzpicture} 
    \end{document}

```

[![What I try to reach ](https://i.stack.imgur.com/1TCry.jpg)](https://i.stack.imgur.com/1TCry.jpg)