For fixed \(\alpha\) and \(\beta\) consider the stochastic differential equation

\[

dY(t)=\frac{\beta-Y(t)}{1-t} dt + dB(t) ~,~~ 0\leq t < 1 ~,~~Y(0)=\alpha.

\]

Verify that \(\lim_{t\to 1}Y(t)=\beta\) with probability one. ( This is called the Brownian bridge from \(\alpha\) to \(\beta\).)

Hint: In the problem “Solving a class of SDEs“, you found that this equation had the solution

\begin{equation*}

Y_t = a(1-t) + bt + (1-t)\int_0^t \frac{dB_s}{1-s} \quad 0 \leq t <1\; .

\end{equation*}

To answer the question show that

\begin{equation*}

\lim_{t \rightarrow 1^-} (1-t) \int_0^t\frac{dB_s}{1-s} =0 \quad \text{a.s.}

\end{equation*}

# Category Archives: Brownian Bridge

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