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Let $B_t$ be a standard Brownian motion in $\mathbb R$, then the Brownian bridge on $[0,1]$ is defined as $$ Y_t = a(1-t)+bt+(1-t)\int\limits_0^t\frac{\mathrm dB_s}{1-s} $$ for $0\leq t<1$. Here $Y_0 = a$ and $\lim\limits_{t\to 1} Y_t = b$ a.s. The latter implies $$ \lim\limits_{t\to 1}\;(1-t)\int\limits_0^t\frac{\mathrm dB_s}{1-s} = 0\text{ a.s.} $$ and using integration by parts: $$ \lim\limits_{t\to 1}\;(1-t)\int\limits_0^t\frac{B_s}{(1-s)^2}\mathrm ds = B_1 \text{ a.s.} $$

I wonder if the latter formula has been shown to have a particular interesting meaning. Maybe there a known relationship with a Cauchy's integral formula.

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@ Ilya : If there any link between those 2 formulas maybe it can be revealed by considering Karhunen-Loeve decomposition of the Brownian Bridge and then by passing to an analytical prolongation of the integral expression might lead to something. Best regards – TheBridge Mar 6 '12 at 11:23
up vote 4 down vote accepted

If $f$ is any continuous function, then L'Hopital's rule gives \[ \lim_{t\to1}\; (1-t)\int_0^t\frac{f(s)}{(1-s)^2}\,ds = f(1). \] In terms of generalized functions, this says that if \[ \mu_t(s) = \frac{1-t}{(1-s)^2}1_{[0,t]}(s), \] then $\mu_t\to\delta_1$. The Brownian bridge can also be written in terms of generalized functions: \[ Y_t= a(1 - t) + bt + \langle\partial B,\nu_t\rangle, \] where \[ \nu_t(s) = \frac{1-t}{1-s}1_{[0,t]}(s), \] and $\partial B$ is the (random) distributional derivative of the Brownian sample path. Note that $\langle\partial B,\nu_t\rangle = -\langle\partial\nu_t,B\rangle$, and \[ \partial\nu_t = (1-t)\delta_0 + \mu_t - \delta_t. \] Hence, \begin{align*} \lim_{t\to1}\;\langle\partial B,\nu_t\rangle &= \lim_{t\to1}\;-\langle(1-t)\delta_0 + \mu_t - \delta_t,B\rangle\\ &= \lim_{t\to1}\;\langle\delta_t - \mu_t,B\rangle\\ &= \langle\delta_1-\delta_1,B\rangle = 0. \end{align*}

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+1 for the first sentence. This isn't a special property of Brownian motion. – Nate Eldredge Mar 13 '12 at 18:45
Thanks - so does it only use the continuity of Brownian motion? – Ilya Mar 20 '12 at 16:05
Correct. The final equation in your question can be proved using only the continuity of the Brownian sample paths. As I mentioned, it is just an elementary application of L'Hopital's rule (and the fundamental theorem of calculus). – Jason Swanson Mar 20 '12 at 19:30

Well, without generalized functions, the proof is based on Doob's inequality for martingales: If $\{X_t\}_{t \in [0,T]}$ is continuous quadratic integrable martingale, then for all $C>0$ we have: $$P\left(\sup\limits_{t \in [0,T]}|X_t|\geq C\right)\leq \dfrac{EX^2(T)}{C^2}$$ So if we denote $X_t=\int_{0}^t\dfrac{dB_s}{1-s}$, then $$P\left(\sup\limits_{1-2^{-n}\leq t\leq 1-2^{-n-1}}(1-t)|X_t|>C\right)\leq 2 C^{-2}2^{-n},$$ take $C=2^{-n/4}$ and apply the Borel-Cantelli lemma. The rest is almost evident ($X_t \to 0$ a.s.)

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