Let $(B_t)$ be a one-dimensional Brownian motion and $y \in \mathbb{R}$. Show that the solution to the SDE $$dX_t^y=dB_t + \frac{y-X_t^y}{1-t}dt$$ with initial value $X_0^y = 0$ on $[0,1)$ is given by $$X_t^y = yt+(1-t)\int_0^t \frac{1}{1-s}dB_s.$$
Ok, so I am new to the SDE theory and I do not know yet, how I need to proceed here. I need the Ito formula for sure, but applied to which function?!
I would appreciate any help!
Thanks in advance!
Ito formula describes the evolution of $F(Y_t)$ using integrals and derivatives that describe the evolution of $Y_t$ right?
Here you have an equation for the evolution of $X_t$ in terms of the evolution of $B_t$!
In other words, you have the equation: $$ X_t = f(B_t) $$
Can you finish from here?
$$dX_t = df(B_t)...$$
Edit: In general, using Ito's lemma for time-dependent functions is also referred as Ito-Doeblin formula, to derive it apply ITO's lemma to the $(n+1)$-dimensional semi-martingale $(t,X_t)$ (this will give you a formula you can use).
Nevertheless, for this example you don't need that much. Let me show you an easier approach. For simplicity, I'll do the case $y=0$ so you can do it for general $y$.
We wish to show that $$ X_t = (1-t) \int_0^t \frac{1}{1-s} dB_s $$ solves the SDE $$ dX_t = dB_t - \frac{X_t}{1-t} dt $$
What if we use things you already know? (Instead of having to derive Ito-Doeblin)
The product rule for ito integration says
$$ d(Y_tZ_t) = Y_tdZ_t + Z_t dY_t +d[Z,Y]_t $$
If we write $Y_t = 1-t $ and $ Z_t = \int_0^t \frac{1}{1-s} dB_s$ we have $$ X_t = Y_t Z_t $$ And we know that $ Y_t = 1-t $ is of finite variation, hence $d[Y,Z]_t = 0$, so by the product rule:
$$ dX_t = Y_t dZ_t + Z_t dY_t + 0 $$
Let us just substitute $Y$ and $Z$:
$$ dX_t = (1-t) \frac{1}{1-t} dB_t - \left( \int_0^t \frac{1}{1-s} dB_s \right) dt $$
But by definition of $X_t$ we can substitute the term in parenthesis for $ \frac{X_t}{1-t}$ and get
$$ dX_t = dB_t - \frac{X_t}{1-t} dt $$
This is the conclusion where looking for.