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3D graphs and projections


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1















I found this image on a presentation.
enter image description here
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 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}

%-------- Brownian motion in 2D only--------------------------
newcommand{BM}[5]{
% points, advance, rand factor, options, end label
draw[#4] (0,0)
foreach x in {1,...,#1}
{ -- ++(#2,rand*#3) coordinate (aux-x) % <- added coordinate names
}
node[right] {#5};
}

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)=100*sin(x*100)+0.2*cos(567*x); %% I'd like to generate a function brownian motion, starting at 100 with a sig standard deviation
}
]
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
defKK{100}
defTT{0.5}
defRR{0}
defSIG{0.15}

%addplot3[domain=50:150,samples y=1,samples=51,blue] (x,{g(x)},Call(0,1,KK,RR,SIG)}); %am doing something wrong here
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)};
addplot3[domain=50:150,samples y=1,samples=51] (x,{g(x)},
{Call(x,g(x),KK,RR,SIG)});
addplot3[domain=50:150,samples y=1,samples=51,red] (x,{1},
{Call(x,g(x),KK,RR,SIG)}); %am doing something wrong here too
end{axis}
end{tikzpicture}
end{document}









share|improve this question

























  • The projection of the red graph (yielding the cyan and blue graphs) is almost trivial: just set the y or z coordinate to zero. What's not trivial is to guess the red graph from your screen shot. So please add an MWE. An example, though in a slightly different context, can be found here. Yet this is unlikely the only example of this kind.

    – marmot
    Oct 24 '18 at 0:43


















1















I found this image on a presentation.
enter image description here
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 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}

%-------- Brownian motion in 2D only--------------------------
newcommand{BM}[5]{
% points, advance, rand factor, options, end label
draw[#4] (0,0)
foreach x in {1,...,#1}
{ -- ++(#2,rand*#3) coordinate (aux-x) % <- added coordinate names
}
node[right] {#5};
}

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)=100*sin(x*100)+0.2*cos(567*x); %% I'd like to generate a function brownian motion, starting at 100 with a sig standard deviation
}
]
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
defKK{100}
defTT{0.5}
defRR{0}
defSIG{0.15}

%addplot3[domain=50:150,samples y=1,samples=51,blue] (x,{g(x)},Call(0,1,KK,RR,SIG)}); %am doing something wrong here
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)};
addplot3[domain=50:150,samples y=1,samples=51] (x,{g(x)},
{Call(x,g(x),KK,RR,SIG)});
addplot3[domain=50:150,samples y=1,samples=51,red] (x,{1},
{Call(x,g(x),KK,RR,SIG)}); %am doing something wrong here too
end{axis}
end{tikzpicture}
end{document}









share|improve this question

























  • The projection of the red graph (yielding the cyan and blue graphs) is almost trivial: just set the y or z coordinate to zero. What's not trivial is to guess the red graph from your screen shot. So please add an MWE. An example, though in a slightly different context, can be found here. Yet this is unlikely the only example of this kind.

    – marmot
    Oct 24 '18 at 0:43
















1












1








1








I found this image on a presentation.
enter image description here
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 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}

%-------- Brownian motion in 2D only--------------------------
newcommand{BM}[5]{
% points, advance, rand factor, options, end label
draw[#4] (0,0)
foreach x in {1,...,#1}
{ -- ++(#2,rand*#3) coordinate (aux-x) % <- added coordinate names
}
node[right] {#5};
}

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)=100*sin(x*100)+0.2*cos(567*x); %% I'd like to generate a function brownian motion, starting at 100 with a sig standard deviation
}
]
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
defKK{100}
defTT{0.5}
defRR{0}
defSIG{0.15}

%addplot3[domain=50:150,samples y=1,samples=51,blue] (x,{g(x)},Call(0,1,KK,RR,SIG)}); %am doing something wrong here
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)};
addplot3[domain=50:150,samples y=1,samples=51] (x,{g(x)},
{Call(x,g(x),KK,RR,SIG)});
addplot3[domain=50:150,samples y=1,samples=51,red] (x,{1},
{Call(x,g(x),KK,RR,SIG)}); %am doing something wrong here too
end{axis}
end{tikzpicture}
end{document}









share|improve this question
















I found this image on a presentation.
enter image description here
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 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}

%-------- Brownian motion in 2D only--------------------------
newcommand{BM}[5]{
% points, advance, rand factor, options, end label
draw[#4] (0,0)
foreach x in {1,...,#1}
{ -- ++(#2,rand*#3) coordinate (aux-x) % <- added coordinate names
}
node[right] {#5};
}

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)=100*sin(x*100)+0.2*cos(567*x); %% I'd like to generate a function brownian motion, starting at 100 with a sig standard deviation
}
]
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
defKK{100}
defTT{0.5}
defRR{0}
defSIG{0.15}

%addplot3[domain=50:150,samples y=1,samples=51,blue] (x,{g(x)},Call(0,1,KK,RR,SIG)}); %am doing something wrong here
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)};
addplot3[domain=50:150,samples y=1,samples=51] (x,{g(x)},
{Call(x,g(x),KK,RR,SIG)});
addplot3[domain=50:150,samples y=1,samples=51,red] (x,{1},
{Call(x,g(x),KK,RR,SIG)}); %am doing something wrong here too
end{axis}
end{tikzpicture}
end{document}






pgfplots 3d






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 5 mins ago







Julien-Elie Taieb

















asked Oct 24 '18 at 0:39









Julien-Elie TaiebJulien-Elie Taieb

17319




17319













  • The projection of the red graph (yielding the cyan and blue graphs) is almost trivial: just set the y or z coordinate to zero. What's not trivial is to guess the red graph from your screen shot. So please add an MWE. An example, though in a slightly different context, can be found here. Yet this is unlikely the only example of this kind.

    – marmot
    Oct 24 '18 at 0:43





















  • The projection of the red graph (yielding the cyan and blue graphs) is almost trivial: just set the y or z coordinate to zero. What's not trivial is to guess the red graph from your screen shot. So please add an MWE. An example, though in a slightly different context, can be found here. Yet this is unlikely the only example of this kind.

    – marmot
    Oct 24 '18 at 0:43



















The projection of the red graph (yielding the cyan and blue graphs) is almost trivial: just set the y or z coordinate to zero. What's not trivial is to guess the red graph from your screen shot. So please add an MWE. An example, though in a slightly different context, can be found here. Yet this is unlikely the only example of this kind.

– marmot
Oct 24 '18 at 0:43







The projection of the red graph (yielding the cyan and blue graphs) is almost trivial: just set the y or z coordinate to zero. What's not trivial is to guess the red graph from your screen shot. So please add an MWE. An example, though in a slightly different context, can be found here. Yet this is unlikely the only example of this kind.

– marmot
Oct 24 '18 at 0:43












1 Answer
1






active

oldest

votes


















4














If you have a function, you can do the projections by, well, projecting the result.



documentclass[tikz,border=3.14mm]{standalone}
usepackage{pgfplots}
pgfplotsset{compat=1.16}
begin{document}
begin{tikzpicture}[scale=1.8,declare function={f(x,y)=exp(0.1*y);
g(x)=sin(x*100)+0.2*cos(567*x);}]
begin{axis}[view={45}{40},axis on top,
xlabel=$x$,ylabel=$y$,
mesh/interior colormap name=hot,
colormap/hot]
addplot3[domain=0:5,samples y=1,samples=51,blue] (x,{g(x)},{f(0,-2.5)});
addplot3[domain=0:5,domain y=-2.5:2.5,surf,shader =faceted interp,opacity=0.5]
{f(x,y)};
addplot3[domain=0:5,samples y=1,samples=51] (x,{g(x)},{f(x,g(x))});
addplot3[domain=0:5,samples y=1,samples=51,red] (x,{-2.5},{f(x,g(x))});
end{axis}
end{tikzpicture}
end{document}


enter image description here






share|improve this answer
























  • I finally found a way to input the correct functions in MWE. I am no far from a solution. Instead of the g function of cosinus, i'd rather generate a brownian motion. It would be projected on the Call function, and projected again in the frame (time, option value). I feel the magic will happen !

    – Julien-Elie Taieb
    14 mins ago











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1 Answer
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active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









4














If you have a function, you can do the projections by, well, projecting the result.



documentclass[tikz,border=3.14mm]{standalone}
usepackage{pgfplots}
pgfplotsset{compat=1.16}
begin{document}
begin{tikzpicture}[scale=1.8,declare function={f(x,y)=exp(0.1*y);
g(x)=sin(x*100)+0.2*cos(567*x);}]
begin{axis}[view={45}{40},axis on top,
xlabel=$x$,ylabel=$y$,
mesh/interior colormap name=hot,
colormap/hot]
addplot3[domain=0:5,samples y=1,samples=51,blue] (x,{g(x)},{f(0,-2.5)});
addplot3[domain=0:5,domain y=-2.5:2.5,surf,shader =faceted interp,opacity=0.5]
{f(x,y)};
addplot3[domain=0:5,samples y=1,samples=51] (x,{g(x)},{f(x,g(x))});
addplot3[domain=0:5,samples y=1,samples=51,red] (x,{-2.5},{f(x,g(x))});
end{axis}
end{tikzpicture}
end{document}


enter image description here






share|improve this answer
























  • I finally found a way to input the correct functions in MWE. I am no far from a solution. Instead of the g function of cosinus, i'd rather generate a brownian motion. It would be projected on the Call function, and projected again in the frame (time, option value). I feel the magic will happen !

    – Julien-Elie Taieb
    14 mins ago
















4














If you have a function, you can do the projections by, well, projecting the result.



documentclass[tikz,border=3.14mm]{standalone}
usepackage{pgfplots}
pgfplotsset{compat=1.16}
begin{document}
begin{tikzpicture}[scale=1.8,declare function={f(x,y)=exp(0.1*y);
g(x)=sin(x*100)+0.2*cos(567*x);}]
begin{axis}[view={45}{40},axis on top,
xlabel=$x$,ylabel=$y$,
mesh/interior colormap name=hot,
colormap/hot]
addplot3[domain=0:5,samples y=1,samples=51,blue] (x,{g(x)},{f(0,-2.5)});
addplot3[domain=0:5,domain y=-2.5:2.5,surf,shader =faceted interp,opacity=0.5]
{f(x,y)};
addplot3[domain=0:5,samples y=1,samples=51] (x,{g(x)},{f(x,g(x))});
addplot3[domain=0:5,samples y=1,samples=51,red] (x,{-2.5},{f(x,g(x))});
end{axis}
end{tikzpicture}
end{document}


enter image description here






share|improve this answer
























  • I finally found a way to input the correct functions in MWE. I am no far from a solution. Instead of the g function of cosinus, i'd rather generate a brownian motion. It would be projected on the Call function, and projected again in the frame (time, option value). I feel the magic will happen !

    – Julien-Elie Taieb
    14 mins ago














4












4








4







If you have a function, you can do the projections by, well, projecting the result.



documentclass[tikz,border=3.14mm]{standalone}
usepackage{pgfplots}
pgfplotsset{compat=1.16}
begin{document}
begin{tikzpicture}[scale=1.8,declare function={f(x,y)=exp(0.1*y);
g(x)=sin(x*100)+0.2*cos(567*x);}]
begin{axis}[view={45}{40},axis on top,
xlabel=$x$,ylabel=$y$,
mesh/interior colormap name=hot,
colormap/hot]
addplot3[domain=0:5,samples y=1,samples=51,blue] (x,{g(x)},{f(0,-2.5)});
addplot3[domain=0:5,domain y=-2.5:2.5,surf,shader =faceted interp,opacity=0.5]
{f(x,y)};
addplot3[domain=0:5,samples y=1,samples=51] (x,{g(x)},{f(x,g(x))});
addplot3[domain=0:5,samples y=1,samples=51,red] (x,{-2.5},{f(x,g(x))});
end{axis}
end{tikzpicture}
end{document}


enter image description here






share|improve this answer













If you have a function, you can do the projections by, well, projecting the result.



documentclass[tikz,border=3.14mm]{standalone}
usepackage{pgfplots}
pgfplotsset{compat=1.16}
begin{document}
begin{tikzpicture}[scale=1.8,declare function={f(x,y)=exp(0.1*y);
g(x)=sin(x*100)+0.2*cos(567*x);}]
begin{axis}[view={45}{40},axis on top,
xlabel=$x$,ylabel=$y$,
mesh/interior colormap name=hot,
colormap/hot]
addplot3[domain=0:5,samples y=1,samples=51,blue] (x,{g(x)},{f(0,-2.5)});
addplot3[domain=0:5,domain y=-2.5:2.5,surf,shader =faceted interp,opacity=0.5]
{f(x,y)};
addplot3[domain=0:5,samples y=1,samples=51] (x,{g(x)},{f(x,g(x))});
addplot3[domain=0:5,samples y=1,samples=51,red] (x,{-2.5},{f(x,g(x))});
end{axis}
end{tikzpicture}
end{document}


enter image description here







share|improve this answer












share|improve this answer



share|improve this answer










answered Dec 30 '18 at 1:51









marmotmarmot

105k4126241




105k4126241













  • I finally found a way to input the correct functions in MWE. I am no far from a solution. Instead of the g function of cosinus, i'd rather generate a brownian motion. It would be projected on the Call function, and projected again in the frame (time, option value). I feel the magic will happen !

    – Julien-Elie Taieb
    14 mins ago



















  • I finally found a way to input the correct functions in MWE. I am no far from a solution. Instead of the g function of cosinus, i'd rather generate a brownian motion. It would be projected on the Call function, and projected again in the frame (time, option value). I feel the magic will happen !

    – Julien-Elie Taieb
    14 mins ago

















I finally found a way to input the correct functions in MWE. I am no far from a solution. Instead of the g function of cosinus, i'd rather generate a brownian motion. It would be projected on the Call function, and projected again in the frame (time, option value). I feel the magic will happen !

– Julien-Elie Taieb
14 mins ago





I finally found a way to input the correct functions in MWE. I am no far from a solution. Instead of the g function of cosinus, i'd rather generate a brownian motion. It would be projected on the Call function, and projected again in the frame (time, option value). I feel the magic will happen !

– Julien-Elie Taieb
14 mins ago


















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