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Fixed Income Mathematics

  1. Oct 6, 2011 #1
    This problem comes from a book on fixed income finance. The solution is provided, so I gave it a shot and had a slight discrepancy. Probably just due to my rustiness, so hopefully it's an easy one for you guys.

    1. The problem statement, all variables and given/known data

    Differentiate the following expression with respect to t:

    [tex]
    exp[\int_{0}^{t}d\tau \lambda(\tau)]P(t)+R\int_0^tds(-\frac{dP(s)}{ds})exp[\int_0^sd\tau\lambda(\tau)]
    [/tex]

    2. Relevant equations

    N/A

    3. The attempt at a solution

    Using the product rule and the FTC on both terms:
    [tex]
    exp[\int_{0}^{t}d\tau \lambda(\tau)]\frac{dP(t)}{dt}+P(t)exp[\int_{0}^{t}d\tau \lambda(\tau)]\lambda(t)
    +
    R\int_0^tds(-\frac{dP(s)}{ds})exp[\int_0^sd\tau\lambda(\tau)](0)
    +
    R(-\frac{dP(t)}{dt})exp[\int_0^sd\tau\lambda(\tau)]
    [/tex]

    The third term falls out, so we get:
    [tex]
    exp[\int_{0}^{t}d\tau \lambda(\tau)]\frac{dP(t)}{dt}+P(t)exp[\int_{0}^{t}d\tau \lambda(\tau)]\lambda(t)
    +
    R(-\frac{dP(t)}{dt})exp[\int_0^sd\tau\lambda(\tau)]
    [/tex]

    The text book matches my solution exactly EXCEPT that the integral in the last term goes from 0 to t, rather than 0 to s. I couldn't figure out why the s changed to a t. The final exp[.] expression comes from the product rule and is simply copied from the original equation:

    [tex]
    x=R\int_0^tds(-\frac{dP(s)}{ds})
    [/tex]
    [tex]
    y=exp[\int_0^sd\tau\lambda(\tau)]
    [/tex]
    [tex]
    \frac{d}{dt}[xy] = x\frac{dy}{dt}+y\frac{dx}{dt} = y\frac{dx}{dt}
    [/tex]
    because [itex]\frac{dy}{dt}[/itex] = 0.

    What am I missing?
     
  2. jcsd
  3. Oct 7, 2011 #2
    Your work here appears to be correct.
     
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