Stochastic Leibniz Rule for HJM Yield Dynamics

# Ito's lemma for special case


# Ito's lemma for special case












Assume a HJM framework with the same Brownian motion driving the dynamics for every tenor. $$ df(t,T) = \alpha(t, T)dt + \sigma(t,T) dw_t \,, $$ with $\alpha(t, T) = \sigma(t,T)\int_t^T \sigma(t,s)ds$.

It can be proved that: $$ -\ln(P(t, T)) = \int_0^T f(0, u) du + \int_0^t \int_s^T \alpha(s, u) du ds + \int_0^t \int_s^T \sigma(s, u) du dw_s - \int_0^t r(u) du . $$ Define the yield for a fixed maturity $$ Y_\tau(t) := Y(t, t+\tau) = -\frac{\ln(P(t, t+\tau))}{\tau} . $$ I would like to write the SDE of this process.

This reduces to a problem having a stochastic process defined by: $$ X_t = \int_0^t h(s,t) dw_s $$ For $h$ a "well behaved" function.

Is there a way to apply Ito's lemma or any other similar method to get the corresponding SDE $dX_t$?

I am also happy to take any other approach to solve the main problem, i.e. find $dY_\tau(t)$.

I am aware that the solution is not simply as for a function depending only of $s$: $$ dX_t = h(t,t) dw_t. $$ Also I realize that if the function $h(s, t)$ is separable then I can simply take out the part depending on $t$ and apply Ito. For example for $X_t=tw_t$ were one can easily get $dX_t = w_tdt + tdw_t$, which can also be verified to be the correct answer by integrating.

However this is not the case and the function is far from separable.

I have seen this and this slightly related questions, and this proof but the lack of bibliography does not give me enough confidence to apply the result. Is there any reference I could use?

## Answer by Alper Hekimoglu (score -1)

https://quant.stackexchange.com/a/66153

This seems to be solved direct application of differentiation under Integral Sign, a stochastic version of Leibniz rule since $h(s,t)$ is a well-behaving deterministic function: \begin{equation} dX(t)=\left[\int_{0}^{f(t)}\frac{\partial h(s,t)}{\partial t}dW(s)\right]dt+\frac{\partial f(t)}{\partial t}h(t,t)dW(t). \end{equation} Here $f(t)=t$ therefore we have, \begin{equation} dX(t)=\left[\int_{0}^{t}\frac{\partial h(s,t)}{\partial t}dW(s)\right]dt+h(t,t)dW(t). \end{equation} We can cross-check this using OU process where we can write the SDE and solution: \begin{equation} dZ(t)=-\theta Z(t)dt+\sigma dW(t). \end{equation} \begin{equation} Z(t)=Z(0)e^{-\theta t}+\sigma \int _{0}^{t}e^{-\theta (t-s)}dW(s). \end{equation} After we apply the Leibniz rule: \begin{equation} dZ(t)=-\theta\left[Z(0)+\sigma \int _{0}^{t}e^{-\theta (t-s)}dW(s)\right]dt+\sigma e^{-\theta(t-t)}dW(t)\\ =-\theta Z(t)dt+\sigma dW(t). \end{equation} now we are back into the original OU process. Then for $dY_{\tau}(t)$ using the expression for $-\ln(P(t,T))$ given by @Diego, we can write, \begin{equation} dY_{\tau}(t)=-\frac{d\ln(P(t,t+\tau)}{\tau} =\frac{f(0,t+\tau)+\left[\int_{t}^{t+\tau}\alpha(t+\tau,u)du+\int_{0}^{t}\alpha(s,t+\tau)ds+\left(\int_{t}^{t+\tau}\sigma(t,u)du\right)dW(t)+\left[\int_{0}^{t}\sigma(s,t+\tau)dW(s)\right]dt-r(t)dt\right]}{\tau} \end{equation}