1
$\begingroup$

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:

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

$Cov(X_s,X_t) = min(s,t)-st$

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

For the expectation and covariance I get

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

$Cov(X_s,X_1) = min(s,1)-s$

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 $\lim_{t \rightarrow 1} (1-t)\int_{0}^t \dfrac{dB_s}{1-s} = 0$ a.s.

Thanks in advance!

$\endgroup$
  • 1
    $\begingroup$ A first approach is to compute the second moment, since $$\mathrm{var}(X_t)=(1-t)^2\int_0^t\frac{ds}{(1-s)^2}=t(1-t),$$ one sees that $X_t\to1$ in $L^2$ when $t\to1$. $\endgroup$ – Did Jul 22 '15 at 10:07
  • $\begingroup$ Thank you! The second moment is $\mathbb{E}(X_t^2) = [a(1-t)+bt]^2 + t (1-t)$, as calculated here:math.stackexchange.com/questions/408620/brownian-bridge?rq=1, but I don't see how the estimate in math.stackexchange.com/questions/115727/… follows from that...how can I continue? $\endgroup$ – Max93 Jul 22 '15 at 10:24
  • $\begingroup$ The process is "bridge" between $a$ and $b$, hence $X_1=b$ so is $X_0=a$. $\endgroup$ – Math-fun Jul 22 '15 at 10:52
  • $\begingroup$ @Did $EX_1=b$ with variance vanishing at $1$ we obtain $X_1 \to b$. $\endgroup$ – Math-fun Jul 22 '15 at 10:55
  • $\begingroup$ @Math-fun Yeah, actually, the correct statement this yields is that $X_t\to b$ in $L^2$ when $t\to1$. $\endgroup$ – Did Jul 22 '15 at 11:38
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Obviously it might just be simpler to note that the covariance structure is the same as $B_t-tB_1$, but this method is more general. $\endgroup$ – Shalop Feb 14 at 14:51

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.