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I need to show that the Hull-White model $$dr=(\theta(t)-ar)dt+\sigma dW^Q$$ corresponds to the Heath-Jarrow-Morton formulation $$df(t,T)=\alpha(t,T)dt+\sigma e^{-a(T-t)}dW^Q.$$ I obtained the drift by with drift condition $$\alpha(t,T)=\sigma(t,T)\int_t^T\sigma(t,s)ds,$$ where $\sigma(t,T)=\sigma e^{-a(T-t)}.$ Then, I integrated the resulting $df(t,T)$ and set $T=t\rightarrow f(t,t)=r(t)$. Finally, I looked for the differential $dr(t)$, but the resulting expression looks completely diffeent form the Hull-White formulation.

Could you show me how to perform needed calculations?

Thank you, Giulio

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  • $\begingroup$ what is your $f(0, T)$? $\endgroup$
    – Gordon
    Apr 30 '20 at 12:19
  • $\begingroup$ I set the initial condition as $f^\star(0,T)$, i.e. market forward rate. I am following Björk (2009) p.393. $\endgroup$ Apr 30 '20 at 14:15
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Note that \begin{align*} f(t, T) = f(0, T) + \int_0^t\alpha(u,T)du+\int_0^t\sigma e^{-a(T-u)}dW_u, \end{align*} where, based on this question, \begin{align*} f(0, T) = \int_0^T \theta(u) e^{-a(T-u)} du - \frac{\sigma^2}{2a^2}\big(e^{-a T} -1\big)^2 + e^{-a T} r_0. \end{align*} Note also that \begin{align*} \int_0^t\alpha(u,T)du &= \int_0^t\sigma(u,T)\int_u^T\sigma(u,s)dsdu\\ &=\int_0^t\sigma e^{-a(T-u)}\int_u^T\sigma e^{-a(s-u)}dsdu\\ &=-\frac{\sigma^2}{2a^2}\Big[\big(e^{-a(T-t)}-1\big)^2 - \big(e^{-aT}-1 \big)^2 \Big]. \end{align*} Therefore, \begin{align*} r_t &= f(t, t)\\ &=e^{-at}\int_0^t \theta(u) e^{au} du + e^{-a t} r_0+e^{-at}\int_0^t\sigma e^{au}dW_u. \end{align*} Then, it is clear that \begin{align*} dr_t &= -ar_t dt + \theta(t) dt + \sigma dW_t\\ &= (\theta(t) - a r) dt + \sigma dW_t. \end{align*}

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  • $\begingroup$ Thank you very much, Gordon! :) $\endgroup$ May 6 '20 at 18:00
  • $\begingroup$ You are welcome. $\endgroup$
    – Gordon
    May 6 '20 at 18:33

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