Brownian bridge sde

The SDE for the Brownian bridge is the following:

$$dX_t = \dfrac{b-X_t}{1-t} \, dt+dB_t$$

with the solution

$$X_t = a(1-t)+bt+(1-t)\int_{0}^t \dfrac{dB_s}{1-s}.$$

The expectation and covariance are:

$$\mathbb{E}(X_t) = a+(b-a)t$$

$$\operatorname{Cov}(X_s,X_t) = \min(s,t)-st$$

Now I want to have a look at what happens as $t\rightarrow 1$.

For the expectation and covariance I get

$$\mathbb{E}(X_1) = b,$$

$$\operatorname{Cov}(X_s,X_1) = \min(s,1)-s$$

But I'm having trouble to see what happens with $X_t$. The first two summands clearly go to b, and the last summand should go to 0 as Brownian bridge expression for a Brownian motion suggests. The prove in the last comment using Doob's maximal inequality and Borel-Cantelli is quite short and I don't understand, what's exactly happening there, especially not, where the last equation comes from. Would be great if someone could explain it more exact how I get $$\lim_{t \rightarrow 1} (1-t)\int_0^t \frac{dB_s}{1-s} = 0 \text{ a.s.} $$

Thanks in advance!

Solution 1:

We prove a.s. convergence to zero.

First notice that $\int_0^t f(s) dB_s$ has the same distribution as $B_{\int_0^t f(s)^2ds}$. This equality of distributions is true as processes in $t$ (not just for a single value of $t$). The way to prove this is to note that both are Gaussian processes with the same covariance kernel.

Using this with $f(s) = \frac{1}{1-s}$, one obtains that $\int_0^t \frac{dB_s}{1-s}$ is the same process (in law) as $B_{\frac{t}{1-t}}$. So we just need to show that $\lim_{t \to 1} (1-t)B_{\frac{t}{1-t}} = 0$ a.s. This is equivalent to showing $\frac{B_u}{u} \to 0$ as $ u \to \infty$. By time inversion, this is in turn equivalent to showing that $B_s \to 0$ as $s \to 0$, which is obvious from continuity of paths.