Hull–White Bond Discounts: Short Rates, Forward Curves, and Expectations

# Discount factor in Hull-White model


# Discount factor in Hull-White model












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!

## Answer by Jesper Tidblom (score 2)

https://quant.stackexchange.com/a/55117

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.

## Answer by sh lee (score 1)

https://quant.stackexchange.com/a/54079

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)

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

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