Determining dynamics w.r.t. convolution

Question

For a random variable $X_t$ I'm trying to better understand the dynamics.

Consider the SDE \begin{align*} &dX_t=\mu(t,X_t)dt+\sigma(t,X_t)dB_t+\alpha(t,X_t)d\mathfrak{L}_t\\ &\mathfrak{L}_t=(\mathfrak{K}\ast L)_t,\qquad L_t=\mathbb{P}(t\leq \tau)\\ \end{align*} where $\tau$ is a stopping time and $\mathfrak{K}\in\mathcal{W}^{1,1}_0(\mathbb{R}_+)$ (the Sobolev space with zero trace).

My comprehension of stochastic dynamics is not good enough to get the idea behind $d\mathfrak{L}_t$. Thus I'm asking you for help in understanding this, maybe by finding another form, something like: \begin{align*} d\mathfrak{L}_t = \mathfrak{K}(0)L_tdt \end{align*} which I got by reshaping $d(\int_0^t\mathfrak{K}(t-s)L_sds)$, but it seems a bit pointless, as $\mathfrak{K}(0)=0$.

Thanks for your answers, please let me know, if I forgot any crucial information.

Answer 1

I "solved" my problem doing the following steps:

1. $f(t):=\int_0^t \mathfrak{K}(t-s)L_sds=(\mathfrak{K}\ast L)_t$
2. $df(t)=f'(t)dt \big( =\int_0^t f'(s)ds\big)$
3. $\big( (\mathfrak{K}\ast L)_t\big) '=(\mathfrak{K}'\ast L)_t$, where I used the properties of the Sobolev space to get the derivative.
4. $d(\mathfrak{K}\ast L)_t=(\mathfrak{K}'\ast L)_tdt$