Consider a Hull-White model

$dr(t)=\left(\theta(t)-a(t) r(t)\right) dt + \sigma dW(t)$

with parameters

  • $a=0.1$
  • $\sigma=0.3$
  • $\theta(t)$ was calibrated to match
    • $P(0,t)=\exp(-\mu t)$
    • with: $\mu=0.2$

At time $t=0$ the forward discount factor is:

$P(0,t,T) = \exp (-\mu (T-t))$

For example $P(0,5,10)=\exp(-0.2\times (10-5))\approx 0.37$

I can calculate this discount factor also in HW model (the discount factor will depend on the observed rate $r(t)$) at time $t=5$ with the formula:

$P(t,T)=\exp\left(A(t,T)-B(t,T) r(t)\right)$

If I do so (with $t=5$ and $T=10$ below), I get:

  • $\mathbb{E}\ r(5)=0.9$
  • $A(5,10)=-2.4$
  • $B(5,10)=3.9$
  • $P(t,T)=\exp(-2.4 - 3.9 \times 0.9)=0.003$

Now I ask myself: is this result correct? The discount factor between $t=5$ and $T=10$ was diminished from $0.36$ to $0.003$! Of course I do not have very realistic numbers (if $t$ is in years, then at $t=5$ the interest rate is 90% per year), but maybe I misunderstand how to use the formulas?!

Thanks a lot in advance!


2 Answers 2


A good advice when it comes to the Hull-White model is to never work with the short rate $r(t)$ directly. It will typically be quite unstable and depend on interpolation on the yield curve. Instead introduce the new variable $$ x(t) = r(t) - f(0,t), $$ where $f(0,t)$ is the instantaneous forward rate at time $t$ seen from time $0$. By rewriting the Hull-White model in this variable, the only quantities you will need from the yield curve to price the bond $P(t,T)$ are the market discount factor to $t$ and $T$.

Here, the book "Interest Rate Modeling" volume II by L. Andersen and V. Piterbarg is an excellent reference.


When you use $E[r(t)]$, you need to include an additional term to ensure the no-arbitrage condition.

Try to use simulated $r(t)$ and calculate the expectation of $P(t,T)$ because $P(t,T)$ is a random variable. ($N$ = number of simulation)

$$ P^1 (t,T) = \exp(A(t,T) - B(t,T)r^1(t)) $$ $$ P^2 (t,T) = \exp(A(t,T) - B(t,T)r^2(t)) $$ $$ ... $$ $$ P^N (t,T) = \exp(A(t,T) - B(t,T)r^N (t)) $$

$$ E[P (t,T)] = \frac{1}{N} \sum_{i=1}^{N} P^i (t,T) $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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