Please consider the
$$
X_t = X_0 + \int^t_of(X_s,s)ds+\int^t_og(X_s,s)dB_s
$$ where $B_t$ is Brownian Motion. This can also be expressed as:
$$dX_t=f()dt+g()dB_t$$
What does it mean that $X_T$ is a solution to the above stochastic differential equation?
I can't seem to intuitive explain this. The way I will put it is:
For a given Brownian motion realization it $X_T$ is a solution to the SDE above.. Is this correct or is there more it?
Best Answer
There's actually two definitions of solution to SDE. Strong and weak.
There is also
Note that strong solution implies weak solution.
As an example of something that has weak but not strong solution, consider Tanaka's equation
$$dX(t)=\text{sign}(X(t))dB(t)$$
It can be shown that this has no strong solution by considering local times. However let $X(t)$ be a Brownian motion and then $X(t)$ is a weak solution. Note that $\int_0^t\text{sign}^2(s)ds=t<\infty$ so the Ito integral $\int_0^t\frac{1}{\text{sign}(s)}dX(s)=\int_0^t \text{sign}(s) dX(s)$ exists. Noting that the quadratic variation of this integral is $t$, and Ito integrals are continuous martingales, so we know $\int_0^t\frac{1}{\text{sign}(s)}dX(s)=\tilde{B}(t)$ is a Brownian motion.
Thus, $dX(t)=\text{sign}(t)d\tilde{B}(t)$.
The difference here is that the Brownian motion is constructed in terms of our solution not the other way around.