Is the distribution of the maximum of 3D Bessel Bridge same as the distribution of the maximum of 1D Brownian Excursion ?
Brownian excursion process: a Brownian motion that is conditioned always be positive when 0<t<1 to 0 at t=1.
three-dimensional Bessel bridge with duration 1: starts from the origin at time t = 0 and return to the origin at time t = 1.
My impression is that the distribution of the maximum of 3D Bessel Bridge is same as the distribution of the maximum of 1D Brownian Excursion, is THIS CORRECT ?
Any intuitive reason for this (or a simple derivation of this) ?
Here is the intuitive reason, which can be formalized in the same way that Pitman proved his $2M-B$ Theorem (see [1] or [2], Sec. 5.5 page 140).
The 3D Bessel process $\{Z_t\}$ is a Doob $h$-transform of Brownian motion on $[0,\infty)$, with respect to the harmonic function $h(x)=x$ on $(0,\infty)$. Thus $Z_t$ can be interpreted as Brownian motion on $[0,\infty)$, conditioned to stay positive forever. Consider the graph $(t,Z_t)$ of that process and condition it to reach $\{1\}\times [0,\epsilon)$, then let $\epsilon \downarrow0$. Then the conditioned process will converge both to the Bessel bridge and to a Brownian excursion.
Edit: The question asked by the OP was considered by Willians [3] who showed that the Bessel(3) bridge has the same law as a Brownian excursion (and not just their maxima). See also the discussion after figure 2 in [4].
[1] J.W. PITMAN. One-dimensional Brownian motion and the three-dimensional Bessel process. Advances in Appl. Probability. 7, 511–526 (1975).
[2] Mörters, Peter, and Yuval Peres. Brownian motion. Vol. 30. Cambridge University Press, 2010. https://www.yuval-peres-books.com/brownian-motion/
[3] D. Williams. Decomposing the Brownian path. Bull. Amer. Math. Soc., 76:871–873, 1970.
[4] Pitman, Jim, and Marc Yor. "The law of the maximum of a Bessel bridge." Electronic Journal of Probability 4 (1999): 1-35.