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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

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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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  • $\begingroup$ Dear mbison! Thanks for your very useful answer. Still I have a question to it. Everything is logical, but if we take exponential of both sides in ln(A)=∫Ttθ(s)B(s)ds−0.5σ2∫B2ds, the first term on the RHS will still have no minus... That is confusing... Am I missing something? Do you have any idea how the author comes to the solution in the lecture notes? $\endgroup$
    – QuackQuack
    Sep 19, 2015 at 12:00
  • $\begingroup$ @QuackQuack , There is a Different approach in this context. you can use $P(t,T)=\mathbb{E}^Q[exp(-\int_{t}^{T}r_u du|\mathcal{F}_t]$ .But you should calculate $-\int_{t}^{T}r_u du$ at first. $\endgroup$
    – user16651
    Sep 19, 2015 at 18:04
  • $\begingroup$ @QuackQuack: I see where the confusion with the + and - sign comes from. The reason is because normally ODEs go forward in time with a initial condition. However equation for bond is stated with a terminal condition $P(T,T) =1$, therefore you solve a backward ODE. So actually the author of your paper is absolutely correct with their signs and I actually screwed up the + and - sign in my answer. so it should be $ln(A) = - \int_{t}^T \theta(s)B(s)ds + 0.5\sigma^2 \int B^2 ds$ (due to the fact that you are solving the ODE backwards in time instead of forward). $\endgroup$
    – mbison
    Sep 19, 2015 at 22:02
  • $\begingroup$ @QuakQuack: might be helpful to look at page 9 of kurims.kyoto-u.ac.jp/EMIS/journals/HOA/JAMDS/8/11.pdf on page 9 of this paper they solve the same backward ode but for the regular Vasicek model. Concepts and maths are very similar to what you are looking for. $\endgroup$
    – mbison
    Sep 19, 2015 at 22:04
  • $\begingroup$ @mbison: thank you very much, now it works $\endgroup$
    – QuackQuack
    Sep 20, 2015 at 21:55

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