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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.

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    $\begingroup$ 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
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    $\begingroup$ 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

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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 *)
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  • $\begingroup$ 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

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