I am analyzing the transition of the bond prices in the affine models in the form of $P(t,T)=e^{-A(t,T)-B(t,T)r_t}$

using the property that the diffusion and the drift of an affine model can be presented as $$b(t,r)=b(t)+\beta(t)r_t \quad \quad (1) $$ $$\sigma^2(t,r)=a(t)+\alpha(t)r_t \quad \quad (2) $$

where the A and B satisfy the system of the below ordinary differential equations $$\partial_t A(t,T) = \frac{1}{2} \ a(t) \ B^2(t,T) - b(t) \ B(t,T), \quad A(T,T)=0 \quad \quad (3)$$ $$\partial_t B(t,T) = \frac{1}{2} \ \alpha(t) \ B^2(t,T) - \beta(t) \ B(t,T) -1 , \quad B(T,T)=0 \quad \quad (4)$$

I am trying to derive bond price P(t,T) for the Ho-Lee model. $dr_t=\mu t dt + \sigma dW_t^*$
My understanding is that $b=\mu t, \ \beta=0, \ a=\sigma^2, \alpha=0$ will be the parameters for the eq (3) and (4), which gives

$$\partial A(t,T) = \frac{\sigma^2}{2} B^2(t,T) - \mu t \ B(t,T) \quad \quad (5)$$ $$\partial B(t,T) = -1 \quad \quad (6)$$ and further it should yield $$B(t,T)=T-t \quad \quad (7)$$

The last step is unclear to me.
My questions
1) why T-t is is substituted for, not t or T
2) why the integration $\partial B=-1$ results in (T-t) and not -(T-t)? integrating a function f'(x)=-1 should give f(x)=-x, shouldn't it?

$$A(t,T)=\frac{-\sigma^2}{6}(T-t)^3 + \int_t^T b(s)(T-s)ds \quad \quad (8)$$ giving the final equation $$P(t,T)=exp \ \big{(}\frac{-\sigma}{6}(T-t)^3 -\frac{\mu}{6} T^3 + \frac{1}{2} \mu T t^2 -\frac{1}{3} \mu t^3 - (T-t)r_t \big{)} \quad \quad (9)$$

Question 3)
How the integral $\int_t^T b(s)(T-s)ds$ gives $\frac{\mu}{6} T^3 - \frac{1}{2} \mu T t^2 + \frac{1}{3} \mu t^3$. I can't get the algebra here. Is this an ordinary integral?


1 Answer 1


From Equation (6), $B(t,T)=-t+c(T)$ for some function $c(T)$. $1=P(t,t)=e^{-A(t,t)-(c(t)-t)r_t}$ or $A(t,t)+(c(t)-t)r_t=0,\,\forall (r_t,t)$. So $c(t)=t, A(t,t)=0,\forall t$.

For Equation (8) you have missed the square on $\sigma$ and a factor of $\frac13$. Then you just need to substitute in the function for $b(s)$ and integrate the following to get the correct result. $$A(t,T) = -\frac{\sigma^2}{2}\int_t^T(T-s)^2ds +\mu\int_t^T s(T-s)ds.$$

  • $\begingroup$ did you mean P(T,T), A(T,T)? and than c(T)=T ? the notation is confusing $\endgroup$
    – Michal
    May 4, 2016 at 17:22
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    $\begingroup$ @Michal: Does it make any difference? The variables are place holders. You may as well write $P(x,x)$ and $c(x)=x$ or use any other symbols, Latin, Greek, Chinese, Arabic or Hobbitian. $\endgroup$
    – Hans
    May 4, 2016 at 17:57
  • $\begingroup$ ok, understood. One more question to that. Is the integral in the (8) an ordinary one. I can't get the right form of the P(t,T). Could you clarify that part please? $\endgroup$
    – Michal
    May 4, 2016 at 19:21
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    $\begingroup$ The solution manual solution is correct. Do the second integral in my answer. $\endgroup$
    – Hans
    May 4, 2016 at 23:45
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    $\begingroup$ Let us continue this discussion in chat. $\endgroup$
    – Hans
    May 5, 2016 at 0:59

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