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]

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}]]

Seventh: show everything together;
Show[G2,G3,G1]

Correlation bug
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}]]

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]

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]

Table
produce a set of numbers? If not, then it probably won't plot... $\endgroup$ – cormullion Jun 15 '13 at 16:30c
is not defined ? Can you try substitutingc
withExp
in yourtestprocess5
? $\endgroup$ – b.gates.you.know.what Jun 16 '13 at 6:37c
in your case; typically, you define the process forx[t]
whereas your price isExp[x[t]]
. $\endgroup$ – b.gates.you.know.what Jun 16 '13 at 8:55