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I found this image on a presentation.

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
add a comment |
I found this image on a presentation.

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
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
add a comment |
I found this image on a presentation.

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
I found this image on a presentation.

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
pgfplots 3d
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
add a comment |
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
add a comment |
1 Answer
1
active
oldest
votes
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}

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
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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active
oldest
votes
active
oldest
votes
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}

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
add a comment |
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}

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
add a comment |
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}

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}

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
add a comment |
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
add a comment |
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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