4
$\begingroup$

If I have a one-dimensional random walk with position $X(\text{t})$ with $X(0)=0$ and $\text{Var}(X(1)) = 1$, and I observe $X(1) = \text{c}$, what are the distributions of the minimum and maximum values of x between times 0 and 1 conditional on c, and what is the distribution of the range over this time interval?

Here is why I ask. I want to jointly simulate (min,max,range,final) for a random walk. To simulate the final value alone a normal random number generator can be used.

$\endgroup$
  • $\begingroup$ You must define your notation better! Random walk usually refers to discrete time, but your notation seems to indicate continuous time, so that you really are referring to a brownian motion! $\endgroup$ – kjetil b halvorsen Jun 15 '13 at 19:53
  • 1
    $\begingroup$ My question is about the maximum and minimum of $X(t)$ between times 0 and 1 in continuous time, conditional on $X(t) = c$. $\endgroup$ – Fortranner Jun 17 '13 at 11:41
  • 1
    $\begingroup$ @DJohnson That's not the case for random walks. This question asks a purely theoretical question concerning the joint distribution of $(X(1), M, m)$ where $M$ and $m$ are the maximum and minimum of the walk on the interval $[0,1]$. They are all almost surely finite, even when the increments are non-normal. $\endgroup$ – whuber Jul 27 '17 at 23:43
  • 2
    $\begingroup$ @DJohnson Since a random walk is defined to be the sum (or, for continuous processes, the integral) of its increments, the very fact that $X(1)$ is finite implies all intermediate values were finite, too, regardless of how the increments might have been distributed. $\endgroup$ – whuber Jul 28 '17 at 19:43
  • 1
    $\begingroup$ @DJohnson Yes, my comment is fully general. According to the rules of arithmetic of extended real numbers, if a sum or integral should ever reach infinity, it must always stay there. Thus, if by time $1$ it has the finite value $c$, it never could have been infinite at any previous moment. $\endgroup$ – whuber Jul 29 '17 at 13:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.