# Stochastic Leibniz rule

We have the following single-factor HJM model $$d_tf(t,T)=\sigma(t,T)dW_t+\alpha(t,T)dt$$ $$f(t,T)=f(0,T)+\int_0^t\sigma(s,T)dW_s+\int_0^t\alpha(s,T)ds$$ The discounted T bond is then \begin{align} Z(t,T)&=\exp-\bigg(\int_0^Tf(0,u)du+\int_0^t\int_s^T\sigma(s,u)dudW_s+\int_0^t\int_s^T\alpha(s,u)duds\bigg)\\ &=\exp\bigg(-\int_0^Tf(0,u)du+\int_0^t\underbrace{-\int_s^T\sigma(s,u)du}_{\Sigma(s,T)}dW_s-\int_0^t\int_s^T\alpha(s,u)duds\bigg) \end{align} By Ito's Lemma $$d_tZ(t,T)=Z(t,T)\Bigg(\bigg(\frac{1}{2}\Sigma^2(t,T)-\int_t^T\alpha(t,u)du\bigg)dt+\Sigma(t,T)dW_t\Bigg)$$ My question is, how was Ito's Lemma applied to the term $\int_0^t\int_s^T\sigma(s,u)dudW_s$, which contains an integral of Brownian motion?

Claus Munk's Fixed Income Modelling proves the following stochastic Leibniz rule on page 57-58 $$Y_t=\int_t^Tf(0,u)du+\int_t^T\int_0^t\alpha(s,u)dsdu+\int_t^T\int_0^t\sigma(s,u)dW_sdu$$ $$dY_t=\bigg(\int_t^T\alpha(t,u)du-f(t,t)\bigg)dt+\bigg(\int_t^T\sigma(t,u)du\bigg)dW_t$$ however due to the different limits in the integrals, I am unable to extrapolate from Munk's example to solve my question.

Any help is appreciated.

• As you correctly wrote down, we have $\int_0^t \int_s^T \sigma(s,u) du dW_s = \int_0^t \Sigma(s,T) dW_s$. Under some mild regularity conditions for the function $\Sigma$, this represents an Itô integral. As such taking the differential should be straightforward: there is no use of Leibniz rule here. It's basic stochastic integration/differentiation. See an answer to a question you've already asked here: quant.stackexchange.com/questions/30655/… specifically the treatment of $d(X_t)$. – Quantuple Oct 24 '16 at 8:14