Potentially Ambiguous Stochastic Calculus Question In the Lecture Notes for my Stochastic Calculus module, my lecturer provides the following question as an essential exercise in preparation for the exam. It reads as follows:

Let  $(W_t)_{t≥0}$ be a one-dimensional Wiener process. In the solution that you
found in the previous exercise, replace $G_t$ by $W_t$, that is, set $X_t = xe^{at + bW_t}$. Write an equation for X.

Now, the previous exercise asked us to find a solution to the equation: $$d X_t = aX_t\, dt + bX_t \, dG_t, \,\,\,\,\,\,\,\,\,\,\,\, X_0 = x,$$ where $G \in C^1$, i.e. $G$ is a continuously differentiable function and $G(0) = 0$, and  $a,b \in \mathbb{R}$.
My solution, as suggested in the original exercise was $$X_t = x e^{at + bG_t}$$
I'm not too sure what the process would be to answer the original question. It seems fairly ambiguous - am I essentially just reverse-engineering the 'previous exercise' and trying to find some sort of differential equation? If so, how would I do this?
 A: The question wants you to find a stochastic differential equation of the form $$dX_t = \mu (t,X_t) dt + \sigma (t, X_t) dW_t$$ for which $X_t = X_0 e^{at + b W_t}$ is a solution. You would approach this by applying Itô's formula on $X_t$. In particular, you should take $f(x,t) = e^{at + bx}$, find $f_x, f_t, f_{xx}$ and use Itô's formula to compute $df(t,W_t)$.
The reason why this is compared to the previous exercise is because when $G \in C^1$ you can formally substitute $dG = G^\prime (t) dt$. This is not something you can do for Brownian motion as its trajectories have unbounded variation.
A: The difference is that while $dG_t$ is of the scale $dt$, $dW_t\sim N(0,dt)$ is in $99\%$ of the cases of scale $\sqrt{dt}$. Thus while the higher powers of $dG_t$ are too small to contribute to the total, the square of $dW_t$ still is of size $dt$ and thus contributes.
$\newcommand{\D}{{\it\Delta}}\newcommand{\d}{\delta}$
Or to make that a little more exact, consider a time step $\D t$ with the increments $\D W_t=W_{t+\D t}-W_t$ etc. Then
$$
\D X_t=xe^{at+bW_t}(e^{a\D t+b\D W_t}-1)
$$
Using the exponential series this gives
$$
\D X_t=X_t\Bigl((a\D t+b\D W_t)+\tfrac12(a\D t+b\D W_t)^2+...\Bigr)
\\
=X_t\Bigl((a\D t+b\D W_t)+\tfrac12b^2(\D W_t)^2+O((\D t)^{3/2})\Bigr)
$$
To get a handle on how to treat the term $(\D W_t)^2$ consider an even smaller time step $\d t$ with $N\d t=\D t$. Then
$$
(\D W_t)^2=\left(\sum_{i=0}^{N-1}\d W_{t+i\d t}\right)^2
=\sum_{i=0}^{N-1}(\d W_{t+i\d t})^2+2\sum_{0\le i<j<N}\d W_{t+i\d t}\d W_{t+j\d t}
$$
Now by construction of the Wiener process, the $\d W_{t+i\d t}$ are independent and identically distributed $\sim N(0,\d t)$. By the law of large numbers,
$$
\sum_{i=0}^{N-1}(\d W_{t+i\d t})^2\approx N·\Bbb E[(\d W_{t})^2]=N·\d t=\D t
$$
and
$$
2\sum_{0\le i<j<N}\d W_{t+i\d t}\d W_{t+j\d t}\approx N(N-1)·\Bbb E[\d W_{t}\d W_{t+\d t}]=0
$$
so $(\D W_t)^2$ can be replaces with $\D t$, the deviations from that cancel out over larger time intervals.
