# Proof of the Hull & White Model calibration

I have a question about the demonstration of the formula which states that: If we have an Hull & White Model for the short rate diffusion such that

Then the model is fully calibrated if and only if:

Where

f^M is the market instantaious forward rate.

Now to demonstrate this formula I managed to arrive at the fact that the Zero Coupon bond price under the Hull and White Model for maturity T is given by

Where:

And :

Now we only need to derive the bond prices in order to get the instantanious forward rate, so we get to the following:

My question is about the derivation of

Which is somehow equal to

How can we proove this? It seems to me that this derivation is incorrect because of the fact that the T variable is present both in the integral limits and in B(u,T).

• I think you are supposed to plug in $0$ for the $t$ value and that's why the integral limits don't match. Beyond that I think you may need to use leibniz integral rule or integration by parts to get the expressions to match. – Slade Jun 4 at 16:21
• Also it's fine for the variable in the integral limits to be in the integral as well. It's just like an 'external constant' provided to the integral – Slade Jun 4 at 17:10
• @Slade, thanks for your comment! I think that choosing t or 0 makes no difference because after all t is a variable that can be replaced by 0... I tried integration by parts and it leads nowhere because of the dependency of both u and T inside the integral... And for the fact of treating T in the integral limit as a constant, I don't think it's possible because during the whole proof of the Hull&White calibration the T inside the variable B(t,T) is the same as the one in the integral limits so either both of them are constant or both of them are variable at the same time. – Xman Jun 4 at 20:13

I found the answer to my question!

It consists of separating the terms $$1$$ and $$e^{-a(T-u)}$$ from $$B(u,T)$$ and isolating the $$e^{aT}$$ term from the integral and it's straight forward then!

All in all it's the form of $$B(u,T)$$ that makes it possible to state such a formula.

PS: In order to continue the demonstration, we need to derive $$-\frac{\partial^2}{\partial^2 T}\ln(P(t,T)) = \frac{\partial^2}{\partial^2 T}(\int_t^T b_u.B(u,T) du ) = b_T - a.\int_t^T b_u.e^{-a(T-u)} du$$ ...

This is also a straight forward formula...

Thanks all!