Integrating factor change of terminals

Mar 2008
31
2
USA
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.

Capture.PNG

TIA
 

HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
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.
 
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Mar 2008
31
2
USA
Thank you for your reply. My follow up question is

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:

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

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

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

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

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

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

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

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

Comments welcome. Thanks again:)