# Brownian Bridge general case

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?

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.