How to find the solution to stochastic differential equation (SDE):
$dY = YdW$,
where $W(t)$ is Brownian motion and $Y(0)=1$? The solution to the corresponding ordinary differential equation is easy to find:
$Y=e^{W(t)}$
However the solution to the SDE, according to Ito's formula, is supposed to be:
$Y=e^{W(t)-t/2}$,
and I don't understand how to find this solution. I tried to compare Ito's formula (left-hand side) with the SDE (right-hand side):
$(y'b+\frac{y''}{2})dt+y'dW = YdW$,
where $b$ is some function of $t$, but wasn't able to make progress. Also I tried to take the derivative of $Y$:
$Y'=(W'(t)-1/2)e^{W(t)-t/2}$,
but could't see how this relates to Ito's formula (e.g. there is no $y''$ term).
Best Answer
If you know what the solution should look like then you can just apply Ito's formula to the thing you expect the solution to be and check it works. To apply Ito's formula you want to write $Y$ as a function of $W(t)$ and $t$. We have $Y_t = f(t,W(t))$ where $f(t,x) = e^{x - t/2}$. Then applying Ito's formula gives $$dY_t = \frac{\partial f}{\partial t} dt + \frac{\partial f}{\partial x} dW_t + \frac12 \frac{\partial^2 f}{\partial x^2} dt$$ since $W(t)$ has quadratic variation $t$. Computing the partial derivatives gives \begin{align} dY_t &= - \frac12 f(t,W(t)) dt + f(t,W(t)) dW(t) + \frac12 f(t,W(t))dt \\& = f(t,W(t))dW(t) = Y_t dW(t) \end{align} as desired.
If you didn't know the right solution beforehand, one way you might proceed is as follows. Notice that right hand side of Ito's formula is the same as you get in the usual calculus except for the second order Ito correction term. This means you might hope that you can start with a solution to the corresponding ODE and guess the right change to cancel out the second order correction. Here this method works.
We have $$de^{W(t)} = e^{W(t)} dW(t) + \frac12 e^{W(t)} dt$$ Now a natural guess is to assume that you can separate the variable and write $Y(t) = f(t) e^{W(t)}$ for some $f$. By the product rule for Ito calculus (or an application of Ito's formula) $$d(f(t)e^{W(t)}) = e^{W(t)}f'(t) dt + f(t)e^{W(t)} dW(t) + \frac12 f(t) e^{W(t)} dt$$ and so $Y(t) = f(t) e^{W(t)}$ is the desired solution if and only if $f'(t) = -\frac12 f(t)$ which is if and only if $f(t) = e^{-t/2}$