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Assuming an Affine term structure model, where bond prices arebe defined as: $$P(t,T)=\exp({A(t,T)-B(t,T)r_t)}$$ and describing the Q-dynamics of the short rate according to the model: $$dr_t=ar_tdt+\sigma dW_t$$hence having: $$ \partial_t{A(t,T)}=-\frac{\sigma^2}{2}B^2(t,T) \\\partial_tB(t,T)=-aB(t,T)-1$$ What is the Q-dynamics of the bond prices $dp(t,T)$?

Would it be correct to start from the P-dyanimics: $$ dp(t,T)=((B(t,T)+1)a+1)r_tp(t,T)dt-\sigma p(t,T)dW_t$$and perform the change of measure by defining the new Brownian motion as $$dW_t^{\mathcal{Q}}=dW_t-\frac{(B(t,T)+1)ar_t}{\sigma}dt$$

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2 Answers 2

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You can simply use Ito's lemma under the risk neutral measure $Q$.For the log-bond price $p(t,T)$ this gives

$$dp(t,T)=(A_t(t,T)-B_t(t,T)r_t)dt-B(t,T)dr_t$$

$$=[A_t(t,T)-(B_t(t,T)+B(t,T)a)r_t]dt-B(t,T)\sigma dW_t$$

Here $A_t(t,T)$ and $B_t(t,T)$ are partial derivatives wrt $t$ and $W_t$ is Wiener process under $Q$.

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  • $\begingroup$ I think I got your point that we already are in the Q-measure and no change of mesure is required. However it seems that you are forgetting the quadratic variation term from Ito $\endgroup$
    – Mr Frog
    Apr 11, 2021 at 7:27
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    $\begingroup$ @MrFrog I am not forgetting it. That term is zero because the second derivative of the log-price wrt $r$ is zero. $\endgroup$
    – fes
    Apr 11, 2021 at 8:21
  • $\begingroup$ And why exactly are we taking the log-price? $\endgroup$
    – Mr Frog
    Apr 11, 2021 at 8:22
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    $\begingroup$ @MrFrog For convenience; the expression looks cleaner, this is a common trick. You can always back out the actual price simply with exp. $\endgroup$
    – fes
    Apr 11, 2021 at 8:35
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Just adding my two cents. Without taking the logarithm of the price, the Ito's Lemma should result in:

$d p(t,T) = \left( \partial_t A(t,T) - \partial_t B(t,T) r + \frac{1}{2}\sigma^2B(t,T)^2 \right)p(t,T) dt - B(t,T) p(t,T) dr_t$

substituting now the partial derivatives and the differential $dr_t$, and simplifying the identical terms:

$d p(t,T) = r_t p(t,T) d t - \sigma B(t,T) p(t,T) d W_t$

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