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I am working through the mathematics behind the Hull-White short rate model and am currently stuck on how to take the partial derivative of and evaluate a stochastic integral when looking at how bond prices change.

Here is my set-up:

$dr_t = \kappa(\theta_t - r_t) dt + \sigma_t dW_t^Q$

This is solved by

$r_s = r_t \exp(-\kappa(s - t)) + \int_t^s{\kappa \theta_u \exp(-\kappa(s - u))} du + \int_t^s{\sigma_u \exp(-\kappa(s - u))} dW_t^Q$

where $r_t$ is the short rate, $W_t$ is a Wiener process and $Q$ denotes the risk-neutral measure.

At time $t$, the price of a zero coupon bond with maturity $s$ is:

$ P(t, s) = \mathrm{E}^Q_t \bigg[\exp(-\int_t^s{r_u} du)\bigg]$

To explore the dynamics of the bond, I want $dP(t,s)$. For this, I apply Ito's lemma and expect to be evaluating terms for:

$dP(t, s) = \frac{\partial P}{\partial t}dt + \frac{\partial P}{\partial r_t}dr_t + \frac{\sigma_t^2}{2}\frac{\partial P}{\partial r_t}dt$.

To do this, I apply chain rule to the exponential term in $P(t, s)$. If I apply Leibniz integration rule wrt $r_t$, I get

$$ \begin{aligned} \frac{\partial P}{\partial r_t} &= \mathrm{E}_t^Q\bigg[\frac{\partial}{\partial r_t}\bigg(-\int_t^s{r_u}du\exp(-\int_t^s {r_u} du)\bigg)\bigg] \newline &= \mathrm{E}^Q_t\bigg[\bigg(-\int_t^s {\exp(-\kappa(u - t)} du\bigg)\exp\bigg(-\int_t^s{r_u} du\bigg)\bigg] \newline &= \frac{1}{\kappa}\bigg(\exp(-\kappa(s - t)) - 1 \bigg) P(t, s) \end{aligned} $$

And using this, I can find

$$ \begin{aligned} \frac{\partial^2 P}{\partial r_t^2} &= \bigg(\frac{1}{\kappa}\bigg(\exp(-\kappa(s - t)) - 1 \bigg)\bigg)^2 P(t, s) \end{aligned} $$

What I'm uncertain about is with the $\partial P / \partial t$ term:

$$ \begin{aligned} \frac{\partial P}{\partial t} &= \mathrm{E}_t^Q\bigg[\frac{\partial}{\partial t}\bigg(-\int_t^s{r_u}du\bigg) \exp\bigg(-\int_t^s {r_u} du)\bigg)\bigg] \newline &= \mathrm{E}_t^Q\bigg[ \bigg(-\int_t^s {\frac{\partial r_u}{\partial t} du + \frac{\partial t}{\partial t} r_t} \bigg) \exp\bigg(-\int_t^s {r_u} du)\bigg) \bigg] \newline &= \mathrm{E}_t^Q\bigg[ \bigg(r_t + (r_t - \theta_t) \bigg(\exp(-\kappa(s - t) - 1\bigg) - \int_t^s \frac{\partial}{\partial t} \bigg(\int_t^u {\sigma_v \exp(-\kappa(u - v))dW_v^Q} \bigg) du \bigg) \exp\bigg(-\int_t^s {r_u} du)\bigg) \bigg] \newline &= r_t P(t,s) + (r_t - \theta_t)(\exp(-\kappa (s - t) - 1))P(t,s) - \mathrm{E}_t^Q \bigg[ \int_t^s \frac{\partial}{\partial t} \bigg(\int_t^u {\sigma_v \exp(-\kappa (u - v))dW_v^Q} \bigg) du \exp\bigg(-\int_t^s {r_u} du)\bigg) \bigg] \newline \end{aligned} $$

Initially I wanted to use Leibniz integral rule again to get $\mathrm{E}_t^Q \bigg[ \int_t^s \frac{\partial}{\partial t} \bigg(-\frac{\partial t}{\partial t}{\sigma_t \exp(-\kappa (u - t))} \bigg) du \exp\bigg(-\int_t^s {r_u} du)\bigg) \bigg]$ but having read around a bit more, I'm unsure this would be correct.

Would someone be able to help outline the correct process for evaluating this integral/integrals in this manner?

I'm also confused as there appears to be a stochastic Leibniz rule where $dW_u^Q$ does not disappear, remarks about the expectation of an Ito integral being zero if it is a bounded martingale, and remarks about partial derivatives of Ito integrals not making sense as $dW_t^Q$ is stochastic and cannot be evaluated normally.

Any additional clarification on these points would also be greatly appreciated.

Thanks in advance for your help!

Learner248079
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    You brought the derivative wrt $t$ inside the conditional expectation, which depends on $t$. Are you sure about that? – Snoop Aug 02 '23 at 23:05
  • @Snoop thank you for pointing this out - I wasn't paying attention to that. Given I have $\mathrm{E}^Q[ | \mathrm{F}_t]$ and I'm trying to differentiate wrt $t$ when my expectation is conditional on knowing previous values up to $t$ I guess actually $\partial P/ \partial t = 0$ – Learner248079 Aug 04 '23 at 09:20
  • @KurtG. Following the link, is it correct to state that for $g(t) = \int_{\beta(t)}^{\alpha(t)} {f(t, u) dW_u}$, we can use Leibniz rule to find that $dg(t) = \frac{\partial \alpha}{\partial t} f(t, \alpha) dW_\alpha + \bigg(\int_{\beta(t)}^{\alpha(t)} {\frac{\partial f}{\partial t} dW_u}\bigg) dt -\frac{\partial \beta}{\partial t} f(t, \beta) dW_\beta$? Based off the post, is it correct to make the observation that you treat $dW_u$ almost as if it was part of the integrand function in the sense when you look at the upper/lower limits with Leibniz, evaluate the function at that value? – Learner248079 Aug 04 '23 at 10:08
  • We can focus on the simpler question if it is correct that for $g(t)=\int_0^{\alpha(t)}f(u),dW_u$ we have $dg(t)=\frac{\partial\alpha}{\partial t}f(\alpha),dW_\alpha,.$ The answer is clearly *no*. Even if we attempt to think that $dg(t)=\frac{\partial\alpha}{\partial t}f(\alpha(t)),dW_{\color{red}{t}}$ might be correct the answer is no. The simple reason is that Ito integrals are not differentiable w.r.t. their upper limit. There is no fundamental theorem of Ito calculus and no chain rule. – Kurt G. Aug 04 '23 at 17:29
  • The differential notation $dg(t)=f(t),dW_t$ is just a short cut for the integral notation $g(t)=g(0)+\int_0^tf(s),dW_s,.$ – Kurt G. Aug 04 '23 at 17:29
  • @KurtG. thanks for your reply. The first thing I'd like to clarify is that it is incorrect to try and apply non-stochastic Leibniz rule to any stochastic integral (due to the reasons you mentioned) and then include some stochastic factor - it is more an observation that the form in the link looks like it could be linked to a general rule in that way but it actually has no relation. – Learner248079 Aug 05 '23 at 11:13
  • The second thing I'd like to ask is, given $dg(t)$ is short hand for integral notation, is the point of the computation going from $g(t)$ to $dg(t)$ in the post shared more about expressing the separation of $g(t)$ into deterministic and stochastic parts as opposed to where Leibniz integral rule seeks to give the derivative of an integral? – Learner248079 Aug 05 '23 at 11:21
  • Again; stochastic integrals have no derivative. The point of $$ \boxed{\text{d}g(t)=\left(\int_0^t\frac{\partial f}{\partial t}(s,t)\text{d}W_s\right)\text{d}t+f(t,t)\text{d}W_t}$$ is to be a short hand notation for another stochastic integral that equals $$ g(t) = \int_0^tf(s,t)\text{d}W_s,.$$ Can you write that other integral out? I believe it could be a fruitful exercise. – Kurt G. Aug 06 '23 at 05:49
  • I have written it out! Thanks for clarifying - I think that's mainly what I've been a bit confused as I'm trying to draw links between non-stochastic and stochastic calculus (e.g. I've read that Ito's lemma is the stochastic dialogue of chain rule) but I think after this thread, it is easier for me to separate them as their own concepts rather than try to think of and categorise things as stochastic extensions of existing calculus concepts. Thanks again for your help and patience! – Learner248079 Aug 07 '23 at 20:19

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