Applying Itô’s Lemma to Zero-Coupon Bond Prices Under Stochastic Rates

# Differential bond price stochastic rates


# Differential bond price stochastic rates












Suppose that the short rate follows the process $$dr(t) = a(t, r(t))dt + \sigma(t, r(t))dW(t)$$

If $B(t) = exp(-\int_0^t r(u) d u)$, can one still write the differential $dB(t)$ a-la-Ito? Thanks.

## Answer by Kevin (score 2)

https://quant.stackexchange.com/a/53744

I assume you are interested in the bond price. Let $B_t=B(t,r_t)=\mathbb{E}^\mathbb{Q}[\exp\left(-\int_t^T r_u\mathrm{d}u\right)\mid\mathcal{F}_t]$ be the time $t$ price of a default-free zero-coupon bond maturing at $T$.

That is, by the way, very different to $\exp\left(-\int_0^t r(u)\mathrm{d}u\right)$ which relates to a money-market account (bank account, savings account) $M_t=\exp\left(\int_0^t r(u)\mathrm{d}u\right)$, where $\mathrm{d}M_t=r_tM_t\mathrm{d}t$. The bond price has to have an expectation around the exponential! Remember that the bond price, at time $t$, is a real number, the value of your bank account at time $t$ is a random variable!

A priori, all you can do with $B_t=B(t,r_t)$ is to write \begin{align*} \mathrm{d}B_t = \underbrace{\left(\frac{\partial B_t}{\partial t} + a(t,r_t)\frac{\partial B_t}{\partial r_t}+\frac{1}{2}\sigma^2(t,r_t)\frac{\partial^2B_t}{\partial r_t^2}\right)}_{\mu_B(t,r_t)}\mathrm{d}t+\underbrace{\sigma(t,r_t)\frac{\partial B_t}{\partial r_t}}_{\sigma_B(t,r_t)}\mathrm{d}W_t \end{align*} That is what Itô's Lemma gives you.

Allow me to make two points:

- Using a dynamic hedge and the Black-Scholes line of argument, you can find a second order PDE for the bond price. You need to be careful though because the ``underlying'', the short rate, is not a traded asset.

- Many popular short rate models (e.g. Vasicek, Hull-White, CIR) are affine term-structure (ATS) models meaning that $B_t=e^{A_t+C_tr_t}$, where $A_t$, $C_t$ are deterministic functions of time. Then, of course, \begin{align*} \frac{\partial B_t}{\partial t} &= \left(\frac{\partial A_t}{\partial t}+\frac{\partial C_t}{\partial t}r_t\right)B_t \\ \frac{\partial B_t}{\partial r_t} &= C_tB_t,\\ \frac{\partial^2 B_t}{\partial r_t^2} &= C_t^2B_t. \end{align*} This allows you to simplify the equation for $\mathrm{d}B_t$, \begin{align*} \mathrm{d}B_t = \left(\frac{\partial A_t}{\partial t}+\frac{\partial C_t}{\partial t}r_t + a(t,r_t)C_t+\frac{1}{2}\sigma^2(t,r_t)C_t^2\right)B_t\mathrm{d}t+\Big(\sigma(t,r_t)C_t\Big)B_t\mathrm{d}W_t \end{align*}