I have come up with the following Leibniz stochastic rule and I want to check that:
- The result is correct;
- The proof is right.
Statement: let $f(\cdot,t):s \rightarrow f(s,t)$, $s \in \mathbb{R}^+$, be some function parameterised by a real number $t \in \mathbb{R}^+$ and $W_s$ a standard Brownian Motion. We define the function $g(t)$ as follows:
$$ g(t) = \int_0^tf(s,t)\text{d}W_s $$
Then:
$$ \boxed{\text{d}g(t)=\left(\int_0^t\frac{\partial f}{\partial t}(s,t)\text{d}W_s\right)\text{d}t+f(t,t)\text{d}W_t}$$
Proof: we consider an infenitesimal increment of $t$, $\text{d}t$:
\begin{align*}
\text{d}g(t) & = \text{d}\left(\int_0^tf(s,t)\text{d}W_s\right)
\\[12pt]
& = \int_0^{t+\text{d}t}f(s,t+\text{d}t)\text{d}W_s-\int_0^tf(s,t)\text{d}W_s
\\[12pt] & = \int_0^{t+\text{d}t}\left(f(s,t)+\frac{\partial f}{\partial t}(s,t)\text{d}t\right)\text{d}W_s-\int_0^tf(s,t)\text{d}W_s
\end{align*}
Separating the first integral in two parts:
\begin{align*}
\text{d}g(t) & = \int_t^{t+\text{d}t}\left(f(s,t)+\frac{\partial f}{\partial t}(s,t)\text{d}t\right)\text{d}W_s+\int_0^{t}\left(f(s,t)+\frac{\partial f}{\partial t}(s,t)\text{d}t\right)\text{d}W_s
\\ & \qquad -\int_0^tf(s,t)\text{d}W_s
\end{align*}
Integrals cancel and we get:
\begin{align*}
\text{d}g(t) & = \int_t^{t+\text{d}t}\left(f(s,t)+\frac{\partial f}{\partial t}(s,t)\text{d}t\right)\text{d}W_s+\left(\int_0^{t}\frac{\partial f}{\partial t}(s,t)\text{d}W_s\right)\text{d}t
\end{align*}
By definition of the Ito integral and properties of Brownian Motion:
\begin{align*}
\int_t^{t+\text{d}t}\left(f(s,t)+\frac{\partial f}{\partial t}(s,t)\text{d}t\right)\text{d}W_s & = \left(f(t,t)+\frac{\partial f}{\partial t}(t,t)\text{d}t\right)(W_{t+\text{d}t}-W_t)
\\[12pt]
& = \left(f(t,t)+\frac{\partial f}{\partial t}(t,t)\text{d}t\right)\text{d}W_t
\\[12pt]
& = f(t,t)\text{d}W_t
\end{align*}
Hence:
\begin{align*}
\text{d}g(t) & = \left(\int_0^t\frac{\partial f}{\partial t}(s,t)\text{d}W_s\right)\text{d}t+f(t,t)\text{d}W_t\end{align*}
which completes the proof.
