[Math] Stochastic Leibniz Rule

proof-verificationself-learningstochastic-analysisstochastic-calculusstochastic-integrals

I have come up with the following Leibniz stochastic rule and I want to check that:

  1. The result is correct;
  2. 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.

Best Answer

I have not been able to find bibliography on this, but I think the result is correct. I also think your proof is fine, you use properties of the Brownian motion very nicely.

Here is just an alternative proof which I personally find clearer. Using Fubini (and assuming $f$ is nice enough): \begin{align} g_T - g_0 = & \int_0^T \big(f(s, T) - f(s, s) \big) dW_s + \int_0^T f(t, t) dW_t, \\ = & \int_0^T\int_s^T \frac{\partial f}{\partial t}(s, t) dt dW_s + \int_0^T f(t, t) dW_t, \\ = & \int_0^T \int_0^t \frac{\partial f}{\partial t}(s, t) dW_s dt + \int_0^T f(t, t) dW_t, \end{align} which is just the integral form of your SDE.

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