1
$\begingroup$

Let $\{W'_t\}_{t\in [0,1]}$ be the Brownian motion on the real line obtained by taking the standard Brownian motion $\{W_t\}_{t\ge 0}$ and conditioning on the events $W_1 = 1$ and $0\le W_t\le 1$ for $t\in[0,1]$. My first question is whether $W'$ has a name.

Let $T$ be chosen uniformly at random in $[0,1]$. Secondly, is it possible to calculate explicitly the p.d.f. of $W'_T$?. In words, I am wondering how much does $W'$ favor the middle parts of $[0,1]$ and whether this can be calculated explicitly.

Improve this question
$\endgroup$
4
  • 1
    $\begingroup$ Assuming $W_t$ starts at $W_0=0$, the event $W_t\geq 0$ for $t\in[0,1]$ has probability zero, so $W'$ is not defined. $\endgroup$
    – John Pardon
    Commented Nov 11, 2017 at 23:56
  • 1
    $\begingroup$ Check Brownian excursion $\endgroup$
    – Thomas Kojar
    Commented Nov 12, 2017 at 0:06
  • $\begingroup$ Thanks a lot. I think I have phrase this question as the limiting case of the discrete random walks of length tn by equipartitioning [0,1] into n pieces. $\endgroup$
    – boinkboink
    Commented Nov 12, 2017 at 0:06
  • 2
    $\begingroup$ This conditioning can be made rigorous using a space-time Doob $h$-transform. Aside from the usual references, you can find some nice examples at toywiki.xyz/doob_transform.html $\endgroup$
    – HMPanzo
    Commented Nov 12, 2017 at 0:42

1 Answer 1

Reset to default
4
$\begingroup$
  • Without the condition $0\le W_t\le1$ on $[0,1]$, it calls a Brownian bridge: $$ W'_T \sim B_T + T, $$ where $B=\left(B_t\right)_{t\ge0}$ is the standard Brownian motion.

  • And with the condition, it has probability $0$ to occur: the Brownian motion will take negative values around $t=0$ with probability $1$ (consequence of Proposition 9, p. 28).

    Also, from the Arcsin Law:

    Theorem : The probability that the Brownian motion $\left(B\right)_{t\ge0}$ has no zeros in the time interval $(a,b)$ is given by $\frac{2}{\pi}\arcsin\sqrt{\frac{a}{b}}$.

    So for $a=0$, it is zero, and by independent increments the probability to remain positive in a neighborhood of $0$ is null.

Improve this answer
$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .