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?

  • $\begingroup$ Do you mean the SDE of Y - i.e., $dY_{\tau}(t)$? $\endgroup$ Commented Jul 14, 2019 at 11:28
  • $\begingroup$ Yes, please also note this is not the usual yield to maturity as there is some time dependence on the second component. $\endgroup$ Commented Jul 15, 2019 at 12:58
  • $\begingroup$ Could you clarify the dependence on the second parameter pls? HJM models fixed maturity as opposed to fixed remaining maturity. $\endgroup$ Commented Jul 15, 2019 at 18:26
  • $\begingroup$ Sure sorry, I should have mentioned that I am considering there is only one Brownian motion driving the whole IR structure (or equivalently all with correlation 1). This is the case in which separability of the volatility functions (something my case does not assume) implies the model to be equivalent to a short rate one with $r(t)$ Markov, as seen in Chapter 5.2 of Brigo. Let me add more details to the question. $\endgroup$ Commented Jul 16, 2019 at 16:38

1 Answer 1


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}

  • $\begingroup$ This answer has mistakes. Given $X_t=\int_0^t h(s,t)dW_s$ one gets $dX_t=h(t,t)dW_t+\{\int_0^t\partial_2h(s,t)dW_s\}\color{red}{dt}$. $\endgroup$
    – Kurt G.
    Commented Aug 17, 2022 at 6:00
  • $\begingroup$ Thanks for editing the question. OU wanted an SDE for $Y_\tau(t)$ . I don't see it . $\endgroup$
    – Kurt G.
    Commented Aug 19, 2022 at 7:58

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