Brownian Bridge general case

Question

The SDE for the Brownian bridge is the following:

$dY_t=\frac{b-Y(t)}{1-t}dt+dW(t)$

with solution:

$Y(t)=Y(0)(1-t)+bt+(1-t)\int_0^t \dfrac{dW(s)}{1-s}$

Can someone help me on proving that $$\lim_{t\rightarrow 1^-} Y(t)=b$$ using the Dambis-Dubins-Schwarz theorem and the law of large numbers?

Answer 1

We need to show that $$\lim_{t\to1^-} (1-t)\int_0^t\frac1{1-s}dW_s \stackrel{\text{a.s.}}= 0.$$

$(M_t)_{t<1}=\Big(\int_0^t\frac1{1-s}dW_s\Big)_{t<1}$ is a martingale, so we can use Dambis-Dubins-Schwarz and say that $M_t = B_{\langle M\rangle_t}$ for a Brownian motion $B$ (with a different filtration of course).

However, $\langle M\rangle_t = \int_0^t \frac1{(1-s)^2}ds = \int_{1-t}^1\frac1{s^2}ds = \frac1{1-t}-1 = \frac t{1-t}.$

This means that we are left to show that $$\lim_{t\to1^-}(1-t)B_{\frac t{1-t}} \stackrel{\text{a.s.}}=0.$$ If we denote $u:=\frac t{1-t}$, we obtain $t = \frac u{1+u}$ and $1-t = \frac1{1+u}$, thus we must show that $$\lim_{u\to\infty} \frac{B_u}{u+1} \stackrel{\text{a.s.}}=0,$$ which is well-known.