Deriving the Hull–White Bond Pricing Function A(t,T)

Summary

The response explains how to obtain the function A(t,T) in the one-factor Hull–White interest-rate model. Starting from the differential equation governing A, it divides by A to express the equation in logarithmic form. Integrating between the current time and maturity yields an expression for the logarithm of A as integrals involving the time-dependent drift parameter, the function B, and the volatility.

A key point is that the equation is solved backward from a terminal condition at maturity, rather than forward from an initial value. That direction determines the signs of the integrals, a detail the answer explicitly corrects. The remaining step is to substitute the already-derived formula for B(t,T) and evaluate the integral involving its square. The response gives a derivation path rather than carrying out all integrations or presenting the final closed form, so readers still need the model’s specific B expression and parameter conventions.

Key ideas

• Divide the differential equation for A by A to obtain an equation for the logarithm of A.
• Integrate from the current time to maturity, accounting for the backward terminal-value condition.
• The signs of the integral terms depend on solving the equation backward in time.
• Substitute the Hull–White expression for B and evaluate the remaining integrals to reach the closed form.
• The response outlines the derivation but does not complete the algebra or specify all parameter conventions.

Tags

# Function A(t,T) in one-factor Hull-White model


# Function A(t,T) in one-factor Hull-White model












I am struggling with Hull-White model now and have the following question: in the lecture notes under the link below I see how A(t,T) and B(t,T) are being derived. This requires the solution of ordinary differential equations. With B(t,T) it is more or less clear, still I don't understand how the author comes up with the formula (5) which expresses A(t,T).

Would be grateful for any hint.

http://www.math.nyu.edu/~benartzi/Slides10.3.pdf

## Answer by mbison (score 1, accepted)

https://quant.stackexchange.com/a/20776

looks like it comes directly from integration when I look at the slides. As per your slide we have that A(t,T) should satisfy:

$A_t − \theta(t)AB + 0.5 σ^2 AB^2 = 0$.

Simplifying above condition (by bringing all the A terms to the left side) we get:

$ \frac{1}{A}dA = \theta(t)B - 0.5 σ^2 B^2$

now we integrate both sides: $ln(A) = -\int_t^T \theta(s)B(s) ds + 0.5 \sigma^2 \int B^2 ds$

edited the signs of the integrals. Note the signs of the integral as we are dealing with a backward ODE instead of a regular forward ODE. This because we have terminal condition A(T,T) and B(T,T) instead of initial conditions.

You already were able to derive the formula for B. Stick it in and do the integration on the $B^2$ part. that should give you the answer as provided in the slides.

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