sde for brownian bridge

Question

Let $B(t)$ be a standard Brownian motion and \begin{align*} Y(t)=B(t)-tB(1) && Z(t) = \left\{ \begin{array}{ll} Z(t)=(1-t)B\left(\frac{t}{1-t}\right)& t\in [0,1)\\ 0 & t=1 \\ \end{array} \right. \end{align*} with $Y(0)=Z(0)=Y(1)=Z(1)=0$ two Brownian bridges. Its easy to see, that $Z(t)$ satisfies the sde $dZ(t)=-\frac{Z(t)}{1-t}dt+dB_t$. \ Is there also a sde satisfied by $Y(t)$?

Answer 1

First to be clear the representation

$$b_{t}=B_{t}-tB_{1}$$

is not Markov with respect to the usual Brownian filtration (since it requires evaluating in the future at $t=1$) and so it will not satisfy an SDE process with strong solution.

The way around this is to use a different filtration.

We observe that $b_{t}=B_{t}-tB_{1}$ is independent of $B_{1}$ because indeed $$E[B_{1}b_{t}]=t-t=0,$$

for $t\leq 1$. In particular, the conditioned process $(B_{t})_{t\in [0,1]}|B_{1}=0$ is equal in law to $b_{t}=B_{t}-tB_{1}$. So one candidate filtration is

$$\mathcal{F}_{t}^{brig}:=\bigcap_{\epsilon>0}\mathcal{F}_{t+\epsilon}^{B}\vee \sigma(B_{1}), t\in [0,1].$$

In Brownian Bridges they describe this situation in detail.

Thm: A continuous process $\{B_t\}_{t\in[0,T]}$ is a Brownian bridge if and only if $B_0=0$ (almost surely) and it is Markov (with respect to its natural filtration $\mathcal{F}_{\cdot} $) such that for all $0\le s < t\le T$ then, conditional on $\mathcal{F}_s, B_t$ is normally distributed with mean and variance given by, \begin{aligned} &{\mathbb E}[B_t\;\vert \mathcal F_s]=\frac{T-t}{T-s}B_s,\\ &{\rm Var}(B_t\;\vert \mathcal F_s)=\frac{(t-s)(T-t)}{T-s}. \end{aligned}

And indeed in Filtration enlargement and Brownian bridge SDE, as you mentioned they develop an SDE with strong solution.