I could really do with some help on this question, have no idea where to start. Any advice would be much appreciated, thank u in advance.
I am given
$$\begin{align}dx(t)&=(1+x(t))dt + x(t) dw(t)\\x(0)&=1\end{align}$$
and
$$\begin{align}dy(t)&=y(t)dt + y(t) dw(t)\\y(o)&=1\end{align}$$
we define another stochastic process $$z(t)=\int_0^t \frac{1}{y(s)} ds$$ It can be shown via the multidimensional ito formula that the stochastic process $(y(t)z(t))$ satisfies:
$$d(y(t)z(t))=z(t)dY(t)+dt.$$
Using this show that the solution $x(t)$ of the above given SDE is:
$$x(t)=y(t) (1+z(t))$$
Where $z(t)$ is defined as above, i.e. $$z(t)=\int_0^t \frac{1}{y(s)} ds$$
The second part of the problem is to show that x(t)-y(t) is equal in distribution to z(t)!!
Recall that an SDE of the form
$$dX_t = b(X_t) \, dt + \sigma(X_t) \, dW_t$$
has a unique solution whenenver the coefficients $b$, $\sigma$ are globally Lipschitz continuous. In particular, we see that the SDE
$$dX_t = (1+X_t) \, dt+ X_t \, dW_t$$
has a unique solution. Now let $(Y_t)_t$ the (unique) solution of the SDE $$dY_t = Y_t \, dt + Y_t \, dW_t \tag{1}$$ and set $$Z_t := \int_0^t \frac{1}{Y_s} \, ds$$ i.e. $$dZ_t = \frac{1}{Y_t} \, dt.$$
From Itô's formula we get
$$d(Y_t \cdot Z_t) = Z_t \, dY_t + dt \tag{2}$$
Consequently, we see that
$$\begin{align*} d(Y_t(1+Z_t)) &= dY_t + d(Y_t \cdot Z_t) \stackrel{(2)}{=} (1+Z_t) \, dY_t + dt \\ &= (1+Z_t) Y_t \, dW_t + (1+Z_t) Y_t \, dt + dt \\ &\stackrel{(1)}{=} (1+Z_t) Y_t \, dW_t + ((1+Z_t)Y_t+1) \, dt. \end{align*}$$
This means that the process $A_t := (1+Z_t) Y_t$ solves the SDE
$$dA_t = A_t \, dW_t + (A_t+1) \, dt.$$
Now the claim follows from the uniqueness of the solution.
The second part works analogously.