Probability – Existence of a Process on R^2 Like Two Independent Brownian Bridges

brownian motionpr.probabilityprobability distributionsstochastic-processes

Consider any probability density function $f(x)$ that has mean zero variance one and say all finite moments. You may assume standard normal density if you like.

Given $a_1,a_2>0$, I consider two copies of independent Random Walk bridges $(S_0^{(i)}=0,S_1^{(i)},S_2^{(i)},\ldots,S_n^{(i)}=a_p\sqrt{n})_{p=1,2}$ each of length $n$ started at $0$ and ending at $a_p$ with increments drawn from $f$. Thus each has a joint density given by

$$g_{a_p}(y_0,y_1,y_2,\ldots,y_n) \propto \delta_{y_0=0}\delta_{y_n=a_p}\prod_{i=1}^{n} f(y_{i}-y_{i-1}), \quad p=1,2$$

We define the event $\Lambda_n:=(S_1^{(1)}+S_1^{(2)}>0,\ldots,S_n^{(1)}+S_n^{(2)}>0)$.

Consider Donsker type scaling: $X_n(t;a_1,a_2)=(\frac{S_{nt}^{(1)}}{\sqrt{n}},\frac{S_{nt}^{(2)}}{\sqrt{n}})_{t\in [0,1]}$, where as usual the function is linearly interpolated in between.

My question is:

Conditioned on $\Lambda_n$, Does the process $X_n(t;a_1,a_2)=(\frac{S_{nt}^{(1)}}{\sqrt{n}},\frac{S_{nt}^{(2)}}{\sqrt{n}})$ converges weakly to some process $Y(t;a_1,a_2)$ taking values on $\mathbb{R}_{\ge 0}^2$?

Loosely speaking the limit should be like taking two independent Brownian bridges $B_1(x)$ and $B_2(x)$ on $[0,1]$ from $0$ to $a$ and then conditioning $B_1(x)+B_2(x)\ge 0$ on $[0,1]$ (of course one needs to make sense of what conditioning means in that case).

Note that for one random walk (not the bridge one, though bridge case can be done by Doob-$h$ transform from there I think) case, the limit is given by Brownian meander as shown in Iglehart paper. The proof is very computational in my opinion. It follows by showing finite-dimensional convergence and involves lot of density formulas using the maximum of Brownian motion.

I am not sure if this question has been addressed in literature before. At least I did not find any references. Also since my question is just about the weak convergence to some process, with no interest in the finite-dimensional distribution, I wonder if it can be proven by some soft techniques.

Best Answer

There is a paper by Durrett, Iglehart and Miller, which also sounds related to what you want for the sum, Weak convergence to Brownian meander and Brownian excursion, published in Ann Probab 1977. There they show that you can do the bridge case for one Brownian motion and it gives Brownian excursion. There is a Brownian excursion decomposition of Brownian motion. For example in the textbook of Revuz and Yor. In the textbooks of Rogers and Williams there is some historical context in Volume 2, Chapter VI.42.

There is also a well-developed theory of non-intersecting Brownian motions in 1d. If you take $S_2$ and reflect it (assuming $f$ is symmetric) then your condition becomes $S_1>S_2$ at all times. Then it seems like you would want non-intersecting Brownian bridges. A search turned up this paper by Gi Bao Nguyen Non-intersecting Brownian bridges and the Laguerre Orthogonal Ensemble in Ann I H Poincare-Pr 2015. That paper assumes $a_1=a_2=0$. But there are references therein that may treat the more general case you want.

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