Brownian motion on $[0,1]$

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.

• 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. – John Pardon Nov 11 '17 at 23:56
• Check Brownian excursion – OOESCoupling Nov 12 '17 at 0:06
• 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. – boinkboink Nov 12 '17 at 0:06
• 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 – HMPanzo Nov 12 '17 at 0:42

• 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).
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.