I'm having some issues with solving a Stochastic Differential Equation.
Here there is the problem
"Consider the stochastic process $X(t)$ which is the strong solution of this stochastic differential equation: $$dX_t= \frac{1}{2} b(X_t) b'(X_t)dt+ b(X_t)dW_t$$ $$X_0=c_0 \in \mathbb{R}$$ $$b \in C^1(\mathbb{R})$$ Consider the function $F(X_t)$ defined in this way: $$F(X_t)= \int_a^{X_t} \frac{1}{b(u)} du$$ Assume as a fact that $F$ is invertible. Deduce the solution of: $$dX_t= – \frac{1}{2}e^{-2X_t}dt+ e^{-X_t}dW_t \hspace{5mm}(*)$$ Hint: Compute the stochastic differential of $F$ and express $X_t$ as an expression of the inverse of $F$."
In order to do so, I computed the stochastic differential of $F(X_t)$ using Ito-Doeblin formula, and I got:
$$dF(X_t)= dW_t$$
From which I found out that:
$$X_t= F^{-1} (W_t+ F(c_0))$$
I'm sure that the last two equations are correct.
However, I have no idea how to find the solution of $(*)$.
Can someone help me please? Thank you
Best Answer
To complete the solution you need to find what $F^{-1}(x)$ is.
We start by finding the form of $b(X_t)$.
We see from comparing the form of $dX_t=\frac{1}{2}b(X_t)b'(X_t)dt + b(X_t) dW_t$ with $dX_t=-\frac{1}{2}\exp(-2X_t) + \exp(-X_t) dW_t$ that our function $b(X_t) = \exp(-X_t)$.
We can calculate $F(X_t)=\int_a^{X_t}\frac{1}{b(u)}~du=\int_a^{X_t}\exp(u)~du=\exp(X_t)-\exp(a)$ from using what we have found for $b(X_t)$.
Now we want to find $F^{-1}(x)$ such that $F^{-1}(F(X_t))=X_t$. Thus, from the form of $F(X_t)$ above it is hopefully clear that this is of the form $F^{-1}(x)=\ln(x+\exp(a))$.
We now calculate $X_t = F^{-1}(W_t + F(c_0))=\ln(W_t+\exp(c_o))$ and we have our solution.