Consider \(dY_t= \frac{b-Y_t}{1-t}dt + dB_t\) with \(0 \leq t < 1 \ ,Y_0 = a\)
Verify that \(Y_t = a(1-t) + bt + (1-t)\int_{0}^{t} \frac{dB_s}{1-s};\ \ 0\leq t <1\)
solves the equations and prove that \(\displaystyle \lim_{t \to 1} Y_t = b\) a.s.
\[dg/dt(t,Y_t)dt =(-a+b- \int_{0}^{t}\frac{dB_s}{1-s}) dt\],
\[dg/dx(t,Y_t)dY_t = \frac{1-t}{1-t}dB_t = dB_t\],
\[1/2*d^2g/dx^2(t,Y_t) \cdot (dY_t)^2 = 1/2* 0 \cdot dB_t\cdot dB_t = 0dt\]
From this, we get
\[(-a+b- \int_{0}^{t}\frac{dB_s}{1-s}) dt + dB_t\]
which can be transformed into
\[\frac{-a(1-t) + (1-t)b - (1-t)\int_{0}^{t}\frac{dB_s}{1-s})}{(1-t)} +dB_t\]
\[\frac{b-a(1-t) -bt - (1-t)\int_{0}^{t}\frac{dB_s}{1-s})}{(1-t)} +dB_t\]
\[\frac{b-Y_t}{(1-t)} +dB_t\]
To prove that \(\displaystyle \lim_{t \to 1} Y_t = b\), we can use the epsilon delta approach. This means that for every number \(\epsilon >0\) there is a corresponding number \(\delta>0\) such that
\[0<|x-a|<\delta \to |f(x) - L| < \epsilon\]
where our limit is defined as \(\displaystyle \lim_{x \to a}f(x) = L\)
In this case, our \(\alpha=1\) and \(L=b\). Thus, our
\[|Y_t - L|< \epsilon \to |a(1-t)+bt+(1-t)\int_{0}^{t}\frac{dB_s}{1-s} - b |< \epsilon\] \[|a(1-t)+b(t-1)+(1-t)\int_{0}^{t}\frac{dB_s}{1-s}| < \epsilon\] \[|(t-1)(-a+b-\int_{0}^{t}\frac{dB_s}{1-s}) < \epsilon\]
\[|(t-1)| < \frac{\epsilon}{(-a+b-\int_{0}^{t}\frac{dB_s}{1-s})}\]
and we are trying to show that
\[|t-1| < \delta\]
If we let \(\delta = \frac{\epsilon}{(-a+b-\int_{0}^{t}\frac{dB_s}{1-s})}\), we can see that for any value of |(t-1)|>0, we can find an epsilon so that the inequality holds true. Thus completing our proof