How to solve the following SDE characterizing Brownian motion with fixed end point $c$ at time $T$?

$$\mathrm dX_t=\mathrm dB_t+\frac{c-X_t}{T-t}\,\mathrm dt$$

I do not believe a strong solution exists, by ito lemma it would seem contradictory. Although, intuitively the solution should be something like


Tips for solving the equation are welcome!

  • 1
    $\begingroup$ This is a Brownian bridge you can check Protter's book where there is a chapter dedicated "augmented" filtration which is the underlying subject but this a very studied matter so there is plenty of literature on this on hte web for example here : randomservices.org/random/brown/Bridge.html $\endgroup$
    – TheBridge
    Jan 30, 2020 at 17:27

1 Answer 1


Your SDE is a special case for Hull-White SDE, which can be solved analytically. Recall that $X$ is a Hull-White process if $X$ satisfies a SDE of the form: $$ dX_{t}=\beta_{t}(\alpha_{t}-X_{t})dt+\sigma_{t}dB_{t}, $$ where $\alpha,\beta,\sigma$ are deterministic processes (and are usually identified as functions with domain $[0,\infty)$ and codomain $\mathbb{R}$). For your case, $\alpha_{t}=c$, $\beta_{t}=\frac{1}{T-t}$, $\sigma_{t}=1$.

The trick to solve such SDE is to consider a suitable integrating factor $\mu$ (a deterministic process)

Apply Ito's lemma, then \begin{eqnarray*} & & d(X_{t}\mu_{t})\\ & = & \mu_{t}'X_{t}dt+\mu_{t}dX_{t}. \end{eqnarray*} Choose $\mu_{t}$ such that $\mu_{t}'X_{t}+\mu_{t}\beta_{t}(-X_{t})=0$. That is, $\mu_{t}'=\beta_{t}\mu_{t}$. Note that, there is more than one choice for $\mu$ but any choice is OK. Then we observe that $$ d(\mu_{t}X_{t})=\gamma_{1}(t)dt+\gamma_{2}(t)dB_{t} $$ for some deterministic functions $\gamma_{1}$ and $\gamma_{2}$. The above SDE can be integrated directly.

  • $\begingroup$ Thank you, very nice answer. Just a quick question, how do I deal with the discontinuity at $T$, since $\mu_t=\frac{1}{T-t}$? Is it OK to still have the same solution? Further, is $\int_0^t \frac{(T-t)}{T-s}dB_s$ in law the same as $B(t)-\frac{t}{T}B(T)$ then? $\endgroup$
    – Dole
    Jan 31, 2020 at 13:02
  • $\begingroup$ The SDE may only have solution for $t\in[0,T)$ (but I have not examined that issue in details). $\endgroup$ Jan 31, 2020 at 14:25
  • $\begingroup$ $\int_{0}^{t}\frac{T-t}{T-s}dB_{s}=(T-t)\int_{0}^{t}\frac{dB_{s}}{T-s}$. Now, it is well-known that if the integrand $f$ is deterministic ( and square-integrable), $\int_{0}^{t}f(s)dB_{s}$ is normally distributed with zero mean and variance $\int_{0}^{t}f^{2}(s)ds$. For our case, let $t\in[0,T)$ be fixed. then $\int_{0}^{t}\frac{1}{(T-s)^{2}}ds=\frac{1}{T-t}-\frac{1}{T}$. Hence, $\int_{0}^{t}\frac{T-t}{T-s}dB_{s}$ is normally distributed with zero mean and variance $(T-t)^{2}\{\frac{1}{T-t}-\frac{1}{T}\}.$ $\endgroup$ Jan 31, 2020 at 15:29
  • $\begingroup$ $B_{t}-\frac{t}{T}B_{T}=B_{t}(1-\frac{t}{T})-\frac{t}{T}(B_{T}-B_{t})$. Note that $B_{t}$, $B_{T}-B_{t}$ are independent, normally distributed with zero mean and variance $t$ and $T-t$ respectively, so $B_{t}(1-\frac{t}{T})-\frac{t}{T}(B_{T}-B_{t})$ is normally distributed with zero mean and variance $(1-\frac{t}{T})^{2}t+(\frac{t}{T})^{2}(T-t)$. $\endgroup$ Jan 31, 2020 at 15:37

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