3
$\begingroup$

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?

$\endgroup$
3
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.