# Integration in the Hull-White SDE

I'm stuck in solving the SDE in Hull-White interest rate model. I do not have a thorough background in math (only Real Analysis during my blissful undergrad years), so I am having trouble understanding the integration process in explicitly solving the Hull-White SDE.

So, the Hull-White interest model follows the SDE $$dR(u) = (a(u) - b(u) R(u)) du + \sigma(u) d\tilde{W}(u)$$ It says the explicit solution can be obtained by applying Ito's Lemma to $$e^{\int_0^u b(v) dv} R(u)$$ and integrating both sides.

This is where I am having trouble understanding. $$\int_t^T d\left(e^{\int_0^u b(v) dv} R(u)\right) = e^{\int_0^T b(v) dv} R(T) - e^{\int_0^t b(v) dv} R(t)$$ It seems that we are naively replacing $u$ with $T$ in the first term and with $t$ in the second term. Could we simply do this due to the fundamental theorem of Calculus? Or is there some other working mechanism behind the scenes?

• What do you get when applying Ito's Lemma ? – Dark Apr 27 '16 at 8:26
• @Dark The exponential survives, while canceling out the $b(u)$ drift in the original SDE. – Astaboom Apr 27 '16 at 8:34
• I don't understand how you ran into such an integral by applying Ito's lemma and integrating both sides. Could you provide more details on your calculations? – Quantuple Apr 27 '16 at 8:38
• You should get: $dY_t = e^{\int_0^t b(v) dv} ( a(t) dt + \sigma(t) dW_t)$ with $Y_t = e^{\int_0^t b(v) dv} r_t$ after applying Itô to $Y_t = f(t,r_t)$, with $r_t$ unique solution of SDE $dr_t = (a(t)-b(t)r_t)dt + \sigma(t)dW_t$. From there you should integrate both sides go back to $r_t$ using the fact that $r_t = Y_t e^{-\int_0^t b(v) dv}$ and $r_0=Y_0$. – Quantuple Apr 27 '16 at 8:39
• @Quantuple Maybe I should've been more specific in asking the question. The integral above is the result of integrating the LHS, or $\int_t^T d(e^{\int_0^u b(v) dv} R(u))$. And my bad about the typo. – Astaboom Apr 27 '16 at 9:30

Applying Itô's lemma to $$Y_t := e^{\int_0^t b(v) dv} r_t$$

You get \begin{align} dY_t &= b(t) e^{\int_0^t b(v) dv} r_t dt + e^{\int_0^t b(v) dv} dr_t + 0\\ &= e^{\int_0^t b(v) dv} (b(t) r_t dt + dr_t) \\ &= e^{\int_0^t b(v) dv} (a(t) dt + \sigma(t) dW_t) \end{align} where the last line is obtained by using the fact that $$dr_t = (a(t)-b(t)r_t) dt + \sigma(t) dW_t$$

Your question regards the integration of the LHS, well it is simply given by \begin{align} \int_t^T dY_u &= Y_T - Y_t \\ &= e^{\int_0^T b(v) dv} r_T - e^{\int_0^t b(v) dv} r_t \end{align}

For the RHS, you get $$\int_t^T e^{\int_0^u b(v) dv} (a(u) du + \sigma(u) dW_u) = \int_t^T a(u) e^{\int_0^u b(v) dv} du + \int_t^T \sigma(u) e^{\int_0^u b(v) dv} dW_u$$ which you cannot really simplify much more.

To obtain the solution to the original equation, work from here using the fact that $$r_t := e^{-\int_0^t b(v) dv} Y_t$$

[Remark]

$$\int_0^t dY_u = (Y_t - Y_0)$$

is merely a consequence of how we define stochastic integrals to begin with (wheter Ito or Stratonovich). Assuming Itô formalism, for a sufficiently well-behaved integrand $\psi_t$ and a semi-martingale $X_t$, the stochastic integral writes $$I_t := \int_0^t \psi_u dX_u = \lim_{\Vert P \Vert \rightarrow 0} \sum_{i=1}^N \psi_{t_{i-1}} (X_{t_i}-X_{t_{i-1}})$$ where the limit (when it exists) is taken in the mean square sense, as the partition $P = \{ 0=t_0 < \dots < t_N=t \}$ is refined more and more (the maximum interval $t_{i}-t_{i-1}$ tends towards 0).

In your particular case, just replace $\psi_t$ by $1$ to see that: $$I_t = \int_0^t dX_u = \lim_{\Vert P \Vert \rightarrow 0} \sum_{i=1}^N X_{t_i}-X_{t_{i-1}} = X_t - X_0$$ due to the telescopic sum, as in ordinary calculus.

• @Astaboom, you're welcome. Actually I've added more details on the 'mechanism' since you've asked why. Hope this will help :) – Quantuple Apr 27 '16 at 10:35
• Your remark is actually very enlightening. I have never thought about it in that light. Thank you again! – Astaboom Apr 28 '16 at 23:31