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Consider the Vasicek model for the spot model:

$$ π‘‘π‘Ÿ = (𝛼 βˆ’ π›Ύπ‘Ÿ)𝑑𝑑 + βˆšπ›½π‘‘π‘Š $$

Suppose $𝛾 = 0.1, 𝛾 = 0.1$, and the volatility of the process is 0.02. The spot rate is 10%.

Assume the form of solution to the BPE (Bond Pricing Equation) is $Z(t, T; r) = e^{A(t, T) - rB(t,T)}$ and derive equations for $A$ and $B$.
Solve the equation and obtain the form of $Z(t, T; r)$, hence price the zero coupon bond.

How do I go about this problem? Do I need to substitute Z into the BPE and get the partial derivatives from there?

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  • $\begingroup$ Yes. They are giving you a hint (or Ansatz) for the BPE solution, so you calculate its derivatives $\frac{\partial Z}{\partial t},\frac{\partial Z}{\partial r},\frac{\partial Z}{\partial r^2}$ and plug the Z and its 3 derivatives into the BPE to check that it works and find out what A(t,T) and B(t,T) actually are. $\endgroup$
    – nbbo2
    Commented May 6 at 11:36
  • $\begingroup$ That makes sense. Do you know of any resources that would help me navigate this problem? I did find this paper, but most of it went over my head. $\endgroup$
    – user72282
    Commented May 6 at 11:53
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    $\begingroup$ The only skill required is taking derivatives and manipulating the resulting expressions carefully. Only tools needed are plenty of blank paper and a pencil with eraser. $\endgroup$
    – nbbo2
    Commented May 6 at 12:00
  • $\begingroup$ Thanks. I'll give it a try. $\endgroup$
    – user72282
    Commented May 6 at 12:03

1 Answer 1

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It is very easy to do it by direct calculation; it is a bit easier for the dynamics

$dr = a(b-r)dt + \sigma dW_t$

  1. Integrate it using the integrating multiple $e^{at}$ and moving $-are^{at}dt$ to the left side
  2. Compute $\int_0^T {r dt}$, for which you will need to change order of integration, to end up with $\int_0^T{ ... dW_t}$. By Ito isometry, this will be a normal variable with known mean and variance.
  3. Take expectation of the exponential of that normal variable

Everyone has to do this calculation once :) Good luck!

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  • $\begingroup$ Thanks for the clarification. Helps out a lot! $\endgroup$
    – user72282
    Commented May 6 at 12:19

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