# Hull White Extended Vasicek model

I am trying to understand a formula in the landmark paper by Hull&White "Pricing Interest Rate Derivative Securities" (1990). I cannot see how rearranging (11) and applying boundary conditions results in (13) on page 6 (as below).

Please could you give me a pointer and apologies if its my rudimentary calculus that is at fault. • Please at least explain your reason for downvote whoever you are. Aug 28, 2020 at 13:10

You are right, equation (11) is derived mechanically from (7) (by taking the derivative wrt to $$T$$ and then combining is with (7)), and somehow they think that (13) can be obtained from (11) without remembering (7). Maybe by smartly integrating (note for example that $$B_tB_T - BB_{tT}$$ is the numerator of the derivative wrt to $$t$$ of fraction $$B/B_T$$) and using the boundary condition (I couldn't figure it out).

Of course, what we can do is solve the first-order linear equation in $$t$$ (7)

$$B_t = a(t)B-1.$$

With the usual primitive functions:

$$\alpha'(t) = a(t), \; \; \beta'(t) = -{\rm e}^{-\alpha(t)},$$

the general solution to equation (7) is

$$B(t,T) = c(T){\rm e}^{\alpha(t)} + {\rm e}^{\alpha(t)}\beta(t),$$

with $$c(T)$$ arbitrary function of $$T$$.

As $$B(T,T)=0$$, we must have:

$$c (T)= -\beta(T).$$

So:

$$B(t,T) = -{\rm e}^{\alpha(t)} \left(\beta(T) - \beta(t)\right).$$

We can then easily check that this solution respects (13):

$$B(0,T) = -{\rm e}^{\alpha(0)} \left(\beta(T) - \beta(0)\right)$$

$$B(0,t) = -{\rm e}^{\alpha(0)} \left(\beta(t) - \beta(0)\right)$$

$$\partial B(0,t)/\partial t = -{\rm e}^{\alpha(0)}\beta'(t) = {\rm e}^{\alpha(0)} {\rm e}^{-\alpha(t)}$$

Edit: Note that (11) can be written as:

$$(B_T)_t =\frac{1-B_t}{B}B_T$$ which is equivalent to $$(\ln B_T)_t = \frac{1-B_t}{B}.$$

At this point we need to remember from (7) that the right hand side is a function of $$t$$ only, $$a(t)$$, otherwise it's getting cumbersome to progress from here. The solution is $$B_T = {\rm e}^{\alpha (t) + \gamma (T)}$$ for $$\gamma (T)$$ an arbitrary function of $$T$$. Integrating wrt to $$T$$, we get:

$$B(t,T) = {\rm e}^{\alpha (t)} (\Gamma (T) + \eta (t))$$ for $$\eta (t)$$ an arbitrary function of $$t$$ and $$\Gamma^\prime = {\rm e}^{\gamma}$$.

Boundary condition $$B(T,T)=0$$ then forces:

$$\Gamma(T) = -\eta(T).$$

So,

$$B(t,T) = -{\rm e}^{\alpha(t)} \left(\eta(T) - \eta(t)\right).$$

One more time, noting that

$$B_t = -{\rm e}^{\alpha(t)}a(t)\eta(T) + {\rm e}^{\alpha(t)}a(t) \eta(t) + {\rm e}^{\alpha(t)} \eta^\prime (t),$$

(7) then implies:

$$\eta(t) = \beta(t).$$

• Thank you very much. I can see the boundary conditions fit from your explanation. Readers can google Hull&White's paper for more context if interested. Would be nice to know how they thought formula 13 follows 11 but at least I now feel I'm not alone in being puzzled by that step. I emailed John Hull last week but doubt he'll respond. Aug 29, 2020 at 21:59
• I added a note where I started from (13) and obtained the same solution. I did use (7) in the process, in particular that it is stating that $(1-B_t)/B_T$ is a function of $t$ only.
– ir7
Aug 30, 2020 at 0:45
• I meant starting from (11), not (13).
– ir7
Aug 30, 2020 at 0:52
• Many thanks. I think there's typo ... arbitrary function of t. A lot of effort much appreciated and I wish I could vote you up for it but I'm only a minion. Aug 30, 2020 at 6:29
• Fixed it. Wish you all the best.
– ir7
Aug 30, 2020 at 15:35