Thread: Integrating factor change of terminals

1. Integrating factor change of terminals

Hi,

Can anyone help me with the explicit steps to get from the first equation [f(t,T)=...] to the second equation [P(t,T)=...]?

In particular how would I know to integrate from t to T by changing the integration variable to u?

This is taken from a Coursera lecture slide on Interest rates.

TIA

2. Re: Integrating factor change of terminals

You know you can't use "T" as variable because it is used as one of the bounds on the integral. If it is specifically, "u" that you are worrying about, any letter other that "t" or "T" could be used.

3. Re: Integrating factor change of terminals

How did I know that I can't use "T" as the variable given that in the first equation I am taking a derivative with respect to "T" on the right hand side?

But I think I can see it if I work backwards:

$\log{P(t,T)} = -\int_{t}^{T}f(t,u)du$

$\log{P(t,T)} = -g(t,t) + g(t,T) + C$

where $f(t,u) = \frac{\partial g(t,u)}{\partial u}$.

I know that at $T=t, \log{P(t,t)} = 0$ so $C=0$

Next take the partial derivative of both sides with respect to "T" such that

$\frac{\partial \log{P(t,T)}}{\partial T} = -\frac{\partial g(t,t)}{\partial T} + \frac{\partial g(t,T)}{\partial T}$

$\frac{\partial \log{P(t,T)}}{\partial T} = f(t,T)$

by our earlier definition of $f(t,u) = \frac{\partial g(t,u)}{\partial u}$.