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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!

Michael Hardy
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Max93
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    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 $t\to1$. – Did Jul 22 '15 at 10:07
  • Thank you! The second moment is $\mathbb{E}(X_t^2) = [a(1-t)+bt]^2 + t (1-t)$, as calculated here:http://math.stackexchange.com/questions/408620/brownian-bridge?rq=1, but I don't see how the estimate in http://math.stackexchange.com/questions/115727/brownian-bridge-expression-for-a-brownian-motion?rq=1 follows from that...how can I continue? – Max93 Jul 22 '15 at 10:24
  • The process is "bridge" between $a$ and $b$, hence $X_1=b$ so is $X_0=a$. – Math-fun Jul 22 '15 at 10:52
  • @Did $EX_1=b$ with variance vanishing at $1$ we obtain $X_1 \to b$. – Math-fun Jul 22 '15 at 10:55
  • @Math-fun Yeah, actually, the correct statement this yields is that $X_t\to b$ in $L^2$ when $t\to1$. – Did Jul 22 '15 at 11:38
  • @Susan In general, $\displaystyle E\left(\left(\int_0^tu(s)dB_s\right)^2\right)=E\left(\int_0^tu(s)^2ds\right),$ this is often called Itô's isometry. – Did Jul 22 '15 at 11:40

1 Answers1

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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.

shalop
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    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. – shalop Feb 14 '19 at 14:51