Is there a way to adjust the estimated value of the WienerProcess after n steps?

Eg. could we evaluate the value of

data = RandomFunction[WienerProcess[.3, .5], {0, 1, 0.01}]

after n steps?

The sample code is from the documentation and I could not find the example that would answer my question.

  • 2
    $\begingroup$ From the documentation: The state at time $t$ follows NormalDistribution[μ*t,σ*√t]. $\endgroup$
    – Roman
    Commented Feb 4, 2022 at 10:49
  • $\begingroup$ Thanks @Roman! So t is a step, yes? $\endgroup$
    – matzar
    Commented Feb 4, 2022 at 11:40
  • 2
    $\begingroup$ There are no "steps" in the Wiener process. It's a continuous process. $t$ is the variable describing the continuous process; usually $t$ is time. $\endgroup$
    – Roman
    Commented Feb 4, 2022 at 12:38

1 Answer 1


The parameter you specify: 0.3 and 0.5 are actually the mean and the standard deviation (STD) at t==1.

You can get the mean and STD directly from the definition. E.g. at t==1:

Mean[WienerProcess[.3, .5][1]]
(* 0.3 *)
StandardDeviation[WienerProcess[.3, .5][1]]
(* 0.5 *)

And at t==2:

Mean[WienerProcess[.3, .5][2]]
(* 0.6 *)
StandardDeviation[WienerProcess[.3, .5][2]]
(* 0.707107 *)
  • $\begingroup$ So would this RandomFunction[ WienerProcess[ Mean[WienerProcess[.3, .5][ StandardDeviation[WienerProcess[.3, .5][2500]]]], sd], {0, 1, 0.01}] mean I get a random Weiner walk adjusted for 2500steps? Would t == 1000 mean 1000steps? What is a step really? This is a new area to me so sorry for rudimentay questions. $\endgroup$
    – matzar
    Commented Feb 4, 2022 at 11:39
  • $\begingroup$ Your syntax for WienerProcess is wrong. Look it up in the help. I think the help answers your questions. $\endgroup$ Commented Feb 4, 2022 at 13:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.