Let us assume that $T$ is an absolutely continuous random variable such that $T>0$ almost surely and $\mathbb{E} e^T < \infty$. Let us define
$$ X_t = \mathbb{I}_{ [ t <T ] }, \ t \geq 0 .$$
I would like to write $X$ in the form of a stochastic differential equation, i.e., I want find functions $A$ and $\sigma$ such that
$$ \text{d} X_t = A(t) X_t \text{d} t + \sigma (t) \text{d} W_t . $$
Could somebody give me some advice or a reference to a similar problem.
Update
We see that
$$ \mu_t = \mathbb{E} [ X_t ] = \mathbb{P} ( t <T) . $$
If we define $p_t = \mathbb{P} ( t<T)$, then
$$ C(s,t) = \text{cov} (X_s, X_t ) = p_{ t \vee s } ( 1 - p_{s \wedge t } ) .$$
We also want
$$ X_t = \int_{0}^t A(s) X_s \text{d} s + \int_{0}^t \sigma (s) \text{d} W_s .$$
We can compute the expected value function and the covariance function based on the last expression for $X_t$ and then compare it to $\mu_t$ and $C(s,t)$ to infer some properties of $A$ and $\sigma $.
We see that
$$ \mathbb{E} [X_t] = \mathbb{E} \bigg[ \int_0^t A(s) X_s \text{d} s \bigg] = \int_0^t A(s) \mathbb{E}_s \text{d}s .$$
Let us treat $p_t$ as a function $p(t)$. Then the last expression gives us
$$ p(t) = \int_0^t A(s) p(s) \text{d}s \Rightarrow p' (t) = A(t) p(t) \Rightarrow A(t) = (\ln p(t))' . $$
We also assume that $T$ is idependent of $W$. Thus, we also see that
$$ \int_0^t A(s) X_s \text{d} s $$
is independent of
$$ \int_0^t \sigma (s) \text{d} W_s .$$
Hence, we get
$$ \mathbb{E} X_t^2 = \int_0^t \sigma^2 (s) \text{d} s + 2p(t) \ln (p(t)), $$
ergo
$$ \sigma^2 (t) = - p' (t) [ 1+ 2 p(t) + \ln (p (t)) ]. $$
Does my approach seem to be correct, or are there any mistakes?
