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Thread: Integrating factor change of terminals

  1. #1
    Newbie mkmath's Avatar
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    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.

    Integrating factor change of terminals-capture.png

    TIA
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  2. #2
    MHF Contributor

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    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.
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  3. #3
    Newbie mkmath's Avatar
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    Re: Integrating factor change of terminals

    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:

     \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}.

    Comments welcome. Thanks again
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