Skip to main content

finance - ItoProcess function


While looking in the help manual of Mathematica concerning the ItoProcess function I found the following:


ItoProcess[{a,b,c},x,t]: represents an Ito Process y(t)=c(t,x(t)), where dx(t)=a(t,x(t))dt+b(t,x(t)).dw(t)


In order to replicate and Plot this, I entered the following code:


(*Defining a process y(t)=c((xt)), where \[DifferentialD]x(t)=μ\
\[DifferentialD]t+σ\[DifferentialD]w(t)*)


ItoProcess[\[DifferentialD]x[
    t] == μ \[DifferentialD]t + σ \[DifferentialD]w[t], 
 c[x[t]], {x, 0}, t, w \[Distributed] WienerProcess[]]

Then I wanted to Plot one process by implementing drift and volatility.


(*Simulation of one Ito Process with μ=0.1 and σ=0.2, \
starting value x=0*)

testprocess5 = 

 ItoProcess[\[DifferentialD]x[t] == 
   0.1 *\[DifferentialD]t + 0.2 *\[DifferentialD]w[t],
c[x[t]], {x, 0}, t, w \[Distributed] WienerProcess[]]

Concerning this code I got the same output like in the Mathematica help manual just the drift and the volatility were substituted by 0.1 and 0.2 respectively.


However, when I tried to plot the process it did not work out.


ListLinePlot[
 Table[RandomFunction[
    testprocess5, {0 (*startis from t=0*), 5 (*ends at t=5*), 
     0.01 (*Δt*)}] ["Path"], {1(*number of paths*)}], 

 Joined -> True, AxesLabel -> {"time", "value"}, ImageSize -> 400, 
 PlotRange -> All]

I am not sure, why it is not workling, maybe it could be due to y (t) = c ((xt)). Does anyone of you have a solution for this problem or a suggestion how to change the code?


thanks



Answer



I'm not sure if I correctly understood what you want... However, I could read in the question that you want to simulate Ito Processes and, at the same time, to be able to change its parameters, especially the processes drifts and volatilities. In the comments I've read about the processes being correlated, so, let me try to put everything together in this answer...


First: set all the parameters you want to simulate;


iv1 = 1; (* initial value for process 1 *)
iv2 = -1; (* initial value for process 2 *)

drift1 = 0; (* drift of process 1 *)
drift2 = 0; (* drift of process 2 *)
diffusion1 = .2; (* diffusion of process 1 *)
diffusion2 = .2; (* diffusion of process 2 *)
correl = 0; (* Set the correlation value here *)
covar = correl*Sqrt[diffusion1]*Sqrt[diffusion2]; (* don't touch here! *)

Second: define the 2D-process;


proc = ItoProcess[{{drift1, drift2}, {{diffusion1, covar}, {covar, diffusion2}}}, {{w1,w2}, {iv1, iv2}}, t];


Third: compute the 2D-process means and variances;


processmean[x_] = Mean[proc[t]]; // Quiet
processvariance[x_] = Variance[proc[t]]; // Simplify // Quiet

Fourth: show the theoretical path intervals


G1 = Show[Plot[{processmean[t] - 2 Sqrt[processvariance[t]], processmean[t] + 2 Sqrt[processvariance[t]], processmean[t]}, {t, 0, 20}, Filling -> {1 -> {2}}], PlotRange -> All]

enter image description here


Fifth: generate the k desired amount of random paths;


k = 10; (* amount of paths to be generated for each individual process *)

path = RandomFunction[proc, {0, 20}, k]

Sixth: see the paths you've generated;


G2 = ListLinePlot[path["PathComponent", 1], PlotRange -> All, PlotStyle -> Directive[{Thin, Lighter@Red}]]
G3 = ListLinePlot[path["PathComponent", 2], PlotRange -> All, PlotStyle -> Directive[{Thin, Lighter@Green}]]

enter image description here enter image description here


Seventh: show everything together;


Show[G2,G3,G1]


enter image description here



According to this post there is a known bug affecting 2D-correlated Ito Processes. However, in my simulations I couldn't find any problem when generating correlated processes.


Consider, for instance, the 2D-process. In order to visualize the correlated processes I'll use the extrem case of a high negative correlation between the processes ($\rho=-0.95$). I'll also generate only two paths for better visualization/understanding.


ClearAll["Global`*"]
iv1 = 1;
iv2 = 1;
drift1 = 0;
drift2 = 0;
diffusion1 = .2;

diffusion2 = .2;
correl = -.95; (* Set correlation here *)
covar = correl*Sqrt[diffusion1]*Sqrt[diffusion2];
proc = ItoProcess[{{drift1, drift2}, {{diffusion1, covar}, {covar, diffusion2}}}, {{w1, w2}, {iv1, iv2}}, t];
k = 2;
path = RandomFunction[proc, {0, 20}, k]

Now you can visually observe the negative relationship between the generated processes:


G5 = ListLinePlot[path["PathComponent", 1], PlotRange -> All, PlotStyle -> Directive[{Thick, Lighter@Red}]]
G6 = ListLinePlot[path["PathComponent", 2], PlotRange -> All, PlotStyle -> Directive[{Thick, Lighter@Green}]]


enter image description here enter image description here


You might also want to see them all together with theoretical paths:


processmean[x_] = Mean[proc[t]]; // Quiet
processvariance[x_] = Variance[proc[t]]; // Simplify // Quiet
G7 = Show[Plot[{processmean[t] - 2 Sqrt[processvariance[t]], processmean[t] + 2 Sqrt[processvariance[t]], processmean[t]}, {t, 0, 20}, Filling -> {1 -> {2}}], PlotRange -> All];
Show[G5, G6, G7]

enter image description here


I hope this will help you.



EDITED


Another example of negative-correlated processes, this time with positive drifts.


iv1 = 1;
iv2 = 1;
drift1 = .1;
drift2 = .3;
diffusion1 = .15;
diffusion2 = .25;
correl = -.95; (* Set correlation here *)
covar = correl*Sqrt[diffusion1]*Sqrt[diffusion2];

proc = ItoProcess[{{drift1, drift2}, {{diffusion1, covar}, {covar, diffusion2}}}, {{w1, w2}, {iv1, iv2}}, t];
processmean[x_] = Mean[proc[t]]; // Quiet
processvariance[x_] = Variance[proc[t]]; // Simplify // Quiet
G8 = Show[Plot[{processmean[t] - 2 Sqrt[processvariance[t]], processmean[t] + 2 Sqrt[processvariance[t]], processmean[t]}, {t, 0, 20}, Filling -> {1 -> {2}}], PlotRange -> All]
k = 1;
path = RandomFunction[proc, {0, 20}, k]
G9 = ListLinePlot[path["PathComponent", 1], PlotRange -> All, PlotStyle -> Directive[{Thick, Lighter@Red}]]
G10 = ListLinePlot[path["PathComponent", 2], PlotRange -> All, PlotStyle -> Directive[{Thick, Lighter@Green}]]
Show[G8, G9, G10]


enter image description here enter image description here enter image description here enter image description here


Comments

Post a Comment

Popular posts from this blog

plotting - Filling between two spheres in SphericalPlot3D

Manipulate[ SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, Mesh -> None, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], {n, 0, 1}] I cant' seem to be able to make a filling between two spheres. I've already tried the obvious Filling -> {1 -> {2}} but Mathematica doesn't seem to like that option. Is there any easy way around this or ... Answer There is no built-in filling in SphericalPlot3D . One option is to use ParametricPlot3D to draw the surfaces between the two shells: Manipulate[ Show[SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], ParametricPlot3D[{ r {Sin[t] Cos[1.5 Pi], Sin[t] Sin[1.5 Pi], Cos[t]}, r {Sin[t] Cos[0 Pi], Sin[t] Sin[0 Pi], Cos[t]}}, {r, 1, 2 - n}, {t, 0, Pi}, PlotStyle -> Yellow, Mesh -> {2, 15}]], {n, 0, 1}]
Post a Comment
Read more

plotting - Plot 4D data with color as 4th dimension

I have a list of 4D data (x position, y position, amplitude, wavelength). I want to plot x, y, and amplitude on a 3D plot and have the color of the points correspond to the wavelength. I have seen many examples using functions to define color but my wavelength cannot be expressed by an analytic function. Is there a simple way to do this? Answer Here a another possible way to visualize 4D data: data = Flatten[Table[{x, y, x^2 + y^2, Sin[x - y]}, {x, -Pi, Pi,Pi/10}, {y,-Pi,Pi, Pi/10}], 1]; You can use the function Point along with VertexColors . Now the points are places using the first three elements and the color is determined by the fourth. In this case I used Hue, but you can use whatever you prefer. Graphics3D[ Point[data[[All, 1 ;; 3]], VertexColors -> Hue /@ data[[All, 4]]], Axes -> True, BoxRatios -> {1, 1, 1/GoldenRatio}]
Post a Comment
Read more

plotting - Mathematica: 3D plot based on combined 2D graphs

I have several sigmoidal fits to 3 different datasets, with mean fit predictions plus the 95% confidence limits (not symmetrical around the mean) and the actual data. I would now like to show these different 2D plots projected in 3D as in but then using proper perspective. In the link here they give some solutions to combine the plots using isometric perspective, but I would like to use proper 3 point perspective. Any thoughts? Also any way to show the mean points per time point for each series plus or minus the standard error on the mean would be cool too, either using points+vertical bars, or using spheres plus tubes. Below are some test data and the fit function I am using. Note that I am working on a logit(proportion) scale and that the final vertical scale is Log10(percentage). (* some test data *) data = Table[Null, {i, 4}]; data[[1]] = {{1, -5.8}, {2, -5.4}, {3, -0.8}, {4, -0.2}, {5, 4.6}, {1, -6.4}, {2, -5.6}, {3, -0.7}, {4, 0.04}, {5, 1.0}, {1, -6.8}, {2, -4.7}, {3, -1.
Post a Comment
Read more