Let $X_t$ be a one-dimensional process satisfying the Itô SDE $$ dX_t = f(X_t)dt + \sigma dW_t, $$ where $\sigma \in \mathbb R$ and with initial condition $X_0 \in \mathbb R$ given. Now let $\Delta > 0$ and $$ Y_t := \begin{cases} \frac1t \int_0^t X_s \, ds, &0<t<\Delta,\\ \frac1\Delta \int_{t-\Delta}^t X_s \, ds, &t\geq \Delta,\end{cases} $$ i.e. the moving average of $X_t$. Clearly, $Y_0 = X_0$.
I found that one can express (for $t \geq \Delta$) the process as $$ Y_t = X_{t-\Delta} + \frac1\Delta \int_{t-\Delta}^t (t-s)f(X_s) \, ds + \frac1\Delta \int_{t-\Delta}^t (t-s)\sigma \, dW_s. $$ Is there a nice way to write the differential $dY_s$? I suspect that this process satisfies a stochastic delay differential equation (SDDE), but I do not know how to proceed.
For example, for $0 < t \leq \Delta$, we have $$ tY_t = \int_0^t X_s \, ds, $$ which implies, as $d(tY_t) = t\,dY_t + Y_t \,dt$, $$ d(tY_t) = X_t \, dt \implies dY_t = \frac{X_t - Y_t}{t}\,dt. $$