I am a little bit stucked with the following integral process, using Fubini's method, this is an intermediate step of short rate Merton Model.

$\int_{t}^{T} W(s)ds=\int_{0}^{\hat {T}}ds\int_{0}^{s}dW(u)\\=\int_{0}^{\hat {T}}dW(u)\int_{u}^{\hat {T}}ds\\=\int_{0}^{\hat {T}}(\hat{T}-u)dW(u)$

My more specific question is how did the change of integration variables proceed, as the process described by above integration is not very intuitive to me.

Many thanks!


\begin{align*} \int_0^T W(t)\, dt &{}= \int_0^T\!\!\int_0^t dW(u)\,dt \\ &{}= \int_0^T\!\!\int_u^T dt\, dW(u) \\&{}= \int_0^T (T - u)\,dW(u) \\&{}= TW(T) - \int_0^T u\, dW(u) \end{align*}

however i am not sure if it is what you are asking for

  • $\begingroup$ Thanks for your reply. However the second step is not clear for me, how did you change the integral variables from 0 - t with u to T? :D $\endgroup$ – Donkey_JOHN Jul 4 '16 at 15:08
  • 1
    $\begingroup$ @Donkey_JOHN It's easily seen on a graph with 2 axes: one for the integration variable $t $, say the vertical one, and one of the integration variable $u$, say the horizontal. You then see that the surface covered by all $t \in [0,T] $ and for each fixed $t$, $u \in [0,t] $ (i.e. the upper triangle on the graph) is the same as the one covered by all $u \in [0,T] $ and for each fixed $u $, $t \in [u, T] $. The first view corresponds to the first integral on the RHS, while the second reflects the second integral on the RHS. Hope this makes sense to you. $\endgroup$ – Quantuple Jul 4 '16 at 17:08
  • $\begingroup$ @Quantuple Thanks for your explanation. I think I got what you mean, perfectly. Appreciated. $\endgroup$ – Donkey_JOHN Jul 4 '16 at 22:33
  • $\begingroup$ @Donkey_JOHN no problem. If the answer has helped you please consider accepting it. It might help others :) $\endgroup$ – Quantuple Jul 6 '16 at 7:37

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