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Let $W_t$ be Wiener process and $B_t$ a Brownian Bridge pinned at $0$ between $t$ and $T$. That is: $B_t = B_T = 0$. It is well known that $B_t$ has (among many) the following two equivalent (in terms of moments) solutions/representations:

  1. Non-anticipatory version: $$ B_t = (T-t) \int^t_0 \frac{1}{T-s} dW_s $$

  2. Anticipatory version (need to know $W_T$ at $t < T$): $$ B_t = W_t - \frac{t}{T}W_T $$

It's relatively simple to show that the non-anticipatory representation satisfies the SDE (see this answer here for example - mind that I interchange $W$ and $B$):

$$ dB_t = \left( \frac{-B_t}{T-t}\right) dt + dW_t $$

Question: is it possible to find a SDE satisfied by the anticipatory version? Note that this SDE shall necessarily be not Wiener-driven (see @Nate comment)

Gabriele Pompa
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    I don't see how it can be, since the solution of any SDE driven by $W_t$ is necessarily adapted to the filtration of $W_t$. However you might be able to get it to satisfy an SDE driven by some Brownian motion other than $W_t$. – Nate Eldredge Mar 26 '21 at 16:20
  • @NateEldredge you are totally right and I was completely wrong. Didn't realized such a simple implication of SDEs. I eventually also found that the non-anticipatory representation is indeed the path-wise unique solution of the B.B. SDE that I was mentioning above [see Karatzas Shreve (2ed) Corollary 6.26]. What is true as well is that the anticipatory solution is a B.B. too as it is a gaussian process from $0$ to $0$ in $[0,T]$ with appropriate B.B. moments [see Karatzas Shreve (2ed) Problem 6.14]. Afaik, not much more can be said on the anticipatory representation. – Gabriele Pompa Mar 26 '21 at 17:04
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    Yes there is a natural SDE for that process which is driven by a different Brownian Motion. Basically, you just need to find the semi-martingale decomposition of $W_t$ in the (initially) enlarged filtration $\mathcal{G}_t = \mathcal{F}_t^W \vee \sigma(W_T).$ In that filtration, by construction, you can treat $W_T$ as a constant. Brownian Bridge is a standard example of the process. Have a look at Section 2.2 of https://www.minet.uni-jena.de/Marie-Curie-ITN/EoF/talks/jeanblanc_introduction.pdf This is also essentially what is going on in your non-anticipatory version as well. – Chris Janjigian Mar 26 '21 at 17:47
  • @ChrisJanjigian thanks for the clarification it helped... but, is this decomposition of the Wiener on the enlarged filtration actually known explicitly? Also, what is its relation with the anticipatory B.B. representation? Please, feel free to post an answer. Thanks! – Gabriele Pompa Mar 29 '21 at 11:14
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    Look at Proposition 2.2.2 in the notes I linked. The answer is that your the SDE for your "non-anticipatory" representation of Brownian Bridge is the semi-martingale decomposition of a Brownian Motion in its natural filtration enlarged by its value at time $T$ (just set $B_1 = 0$ in Prop 2.2.2). Note that there is a typo where the - should be a + in front of the integral, by the way. – Chris Janjigian Mar 29 '21 at 13:20
  • I'm trying to express my non-anticipatory repr along the lines of the decomposition in Prop 2.2.2. I used integration by parts (no quadratic covariation contribution) but this is not what I expected to find... $$ B_t = (T-t) \int^t_0 \frac{1}{T-s} dW_s = (T-t) \left[ \frac{W_t}{T-t} - \int^t_0 \frac{W_s}{(T-s)^2}ds\right] = W_t - (T-t) \int^t_0 \frac{W_s}{(T-s)^2}ds $$ – Gabriele Pompa Mar 29 '21 at 15:54
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    It is useful to remember that all stochastic differential equations are just shorthand for stochastic integral equations. You already saw that Ito's lemma implies $dB_t = -\frac{B_t}{T-t}dt + dW_t$. This really means that $B_t = - \int_0^t \frac{B_s}{T-s}ds + W_t = B_t = \int_0^t \frac{0 - B_s}{T-s}ds + W_t$ on the event where you condition on $B_T = 0$, in the notation of your original post. This is exactly what is in Prop 2.2.2 (up to the typo I mentioned) – Chris Janjigian Mar 29 '21 at 16:38
  • Thanks @Chris. Sorry for raising this futher, but I feel that my expression with $\frac{1}{(T-s)^2}$ as integrating factor is actually equivalent with the one from Prop. 2.2.2, but I don't see how to prove (or disprove) such a conjecture. – Gabriele Pompa Apr 19 '21 at 12:47
  • Or, in other words, I don't see where I might be wrong in deriving it, if it is the case. Thanks in advance for the help – Gabriele Pompa Apr 19 '21 at 12:53
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    I don't understand the question. The expression you get by integrating by parts is equal to the original process as a process, but it isn't the semi-martingale decomposition. The semi-martingale decomposition writes a process as a local martingale plus a finite variation process. – Chris Janjigian Apr 19 '21 at 14:35
  • Thanks Chris. Well, you actually answered my question. Was just curious whether - using identities - one could go from mine expression to the semi-martingale decomposition one. – Gabriele Pompa Apr 19 '21 at 14:46

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