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.

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  • $\begingroup$ Please at least explain your reason for downvote whoever you are. $\endgroup$
    – rupert
    Aug 28, 2020 at 13:10

1 Answer 1


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:

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


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

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:

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


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

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:

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

  • 1
    $\begingroup$ 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. $\endgroup$
    – rupert
    Aug 29, 2020 at 21:59
  • $\begingroup$ 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. $\endgroup$
    – ir7
    Aug 30, 2020 at 0:45
  • $\begingroup$ I meant starting from (11), not (13). $\endgroup$
    – ir7
    Aug 30, 2020 at 0:52
  • $\begingroup$ 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. $\endgroup$
    – rupert
    Aug 30, 2020 at 6:29
  • $\begingroup$ Fixed it. Wish you all the best. $\endgroup$
    – ir7
    Aug 30, 2020 at 15:35

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