0
$\begingroup$

I have been trying to understand the H&W model expression for zero coupon bond price volatilities:

$\nu_B(t_0,t_M)=-\frac{\nu_r}{m}(1-e^{-m\tau_{0,M}})$,

where $\nu_B(t_0,t_M)$ is zero coupon bond price volatility, $\nu_r$ is the short rate volatility, $m$ is the mean-reversion level (or speed?) and $\tau_{0,M}$ is the time to maturity.

I have looked in all the associated papers but found no exact match for this expression. What is the intuition and how exactly do you get this expression?

Edit: made notation clearer

$\endgroup$
  • $\begingroup$ Your notations are not clear: what are $m$, $\tau_0$, $\nu_r$ etc.? $\endgroup$ – Gordon Jun 22 '17 at 14:38
  • $\begingroup$ Thanks Gordon for pointing that out, I have added what the terms are. $\endgroup$ – Vivek Patel Jun 22 '17 at 15:38
2
$\begingroup$

Based on this question, for the Hull-White model of the form \begin{align*} dr_t = (\theta(t)-a r_t) dt + \sigma dW_t, \end{align*} where $a$ and $\sigma$ are constants, $a(t)$ is a deterministic function, and $W_t$ is a standard Brownian motion, the price at time $t$ of a zero-coupon bond with maturity $T$ and unit face value is given by \begin{align*} P(t, T) &= A(t, T) e^{-B(t, T) r_t}, \end{align*} where \begin{align*} B(t, T) = \frac{1}{a}\Big(1-e^{-a(T-t)} \Big), \end{align*} and \begin{align*} A(t, T) &= \exp\left(- \int_t^T \theta(u) B(u, T) du -\frac{\sigma^2}{2a^2}\big(B(t, T) -T+t\big)-\frac{\sigma^2}{4a}B(t, T)^2\right). \end{align*} Note that \begin{align*} \ln P(t, T) = \ln A(t, T) -B(t, T) r_t. \end{align*} Therefore, \begin{align*} d\ln P(t, T) &=\frac{\partial \ln A(t, T)}{\partial t}dt - r_t \frac{\partial B(t, T)}{\partial t} dt- B(t, T) dr_t\\ &=\frac{\partial \ln A(t, T)}{\partial t}dt - r_t \frac{\partial B(t, T)}{\partial t} dt- B(t, T) (\theta(t)-a r_t) dt - \sigma B(t, T) dW_t. \end{align*} That is, the zero-coupon bond price volatility is of the form \begin{align*} \sigma B(t, T) = \frac{\sigma}{a}\Big(1-e^{-a(T-t)} \Big). \end{align*}

$\endgroup$
  • $\begingroup$ Hi! I'm having some problems understanding your procedure, so it would be great if you could clarify, or just point me to some resources. Appreciate your efforts in any case. First thing - I don't quite understand the reasoning behind calculating the differential of the bond price once we already have it expressed quite explicitly through $\ln P(t,T)=\ln A(t,T)-B(t,T)r_t$. Couldn't we just use the fact that $r_t$ is normally distributed with known variance, which implies that the bond price $P(T,t)$ is log-normally distributed and we $\endgroup$ – Milan Jul 7 '17 at 22:13
  • $\begingroup$ can again easily calculate its variance using the known expression. for the variance of the Hull-White process. $\endgroup$ – Milan Jul 7 '17 at 22:13
  • $\begingroup$ Another thing is how you derive the volatility itself. I might be overlooking something, but it seems to me like you go from the fact that $d \ln P(t,T)=\text{Y}dt-\sigma B(t,T)dW_t$ to conclude that $\text{Std}(P(t,T))=-\sigma B(t,T)$. Here I am using $Y$ to shorten the notation. But if we consider the simplest case $d \left( \ln X_t \right)=\sigma dW_t$, we have $\ln X_t = \ln X_0+\int_0^t \sigma dW_u$, implying that $X_t=X_0e^{\sigma Z_t}$, where $Z_t \sim \mathcal{N}\left( 0, t \right)$. We end up with $\text{Std}(X_t)=X_0\sqrt{e^{\sigma^2t}\left( e^{\sigma^2 t-1} \right)} \neq \sigma$. $\endgroup$ – Milan Jul 7 '17 at 22:21
  • $\begingroup$ Again, I might be overlooking something - especially because in this case $Y$ is actually stochastic itself, and it contains a double integral term, where one of the integrals is an Ito integral, which is too advanced for me to handle. So, maybe there is actually some other reasoning behind your derivation of the final volatility, but in that case I think that maybe a couple of steps more could be included. Many thanks for any insight on all of this. $\endgroup$ – Milan Jul 7 '17 at 22:27
  • $\begingroup$ The SDE for $P(t, T)$ is based on the SDE for $r_t$. the volatility function is the coefficient of the $dW_t$ term of the SDE for $d\ln P(t, T)$. $\endgroup$ – Gordon Jul 7 '17 at 23:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.