# Solve the SDE $dX_t=dB_t+\frac{c-X_t}{T-t}dt$

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

$$X_t=B_t+\frac{t(c-B_T)}{T}$$

Tips for solving the equation are welcome!

• 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 Jan 30, 2020 at 17:27

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.
• 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?
• The SDE may only have solution for $t\in[0,T)$ (but I have not examined that issue in details). Jan 31, 2020 at 14:25
• $\int_{0}^{t}\frac{T-t}{T-s}dB_{s}=(T-t)\int_{0}^{t}\frac{dB_{s}}{T-s}$. 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 $(T-t)^{2}\{\frac{1}{T-t}-\frac{1}{T}\}.$ Jan 31, 2020 at 15:29
• $B_{t}-\frac{t}{T}B_{T}=B_{t}(1-\frac{t}{T})-\frac{t}{T}(B_{T}-B_{t})$. 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 $(1-\frac{t}{T})^{2}t+(\frac{t}{T})^{2}(T-t)$. Jan 31, 2020 at 15:37