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Differentiating Integral Exponents?

  1. Apr 11, 2008 #1
    Hi

    I have a question about rearranging the following equation (I saw this in a finance book):

    If we rearrange and differentiate

    [tex]
    Z(t;T) = e^{-\int_{t}^{\tau}r(\tau)d\tau}
    [/tex]

    We get

    [tex]
    r(T) = -\frac{\partial}{\partial{T}}(\log{Z(t;T)})
    [/tex]

    My question is: how do we differentiate the exp(-int()) portion? How can we simplify the integral as an exponent?

    Thanks!
     
  2. jcsd
  3. Apr 11, 2008 #2

    By the fundamental theorem of calculus you have
    [tex]
    \frac{\partial}{\partial T}\int_t^T{d\tau\,r(\tau)} = r(T)
    [/tex]

    If it is in the exponent you must use the chain rule. As for the second question: I don't understand it.:smile:

    You might also try to first solve your equation Z(t;T) for the integral and then taking the derivative.
     
  4. Apr 11, 2008 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    The answer to your second question is: we can't simplify the exponent.

    But I have to say I don't understand your notation. Your exponent is
    [tex]\int_t^\tau r(\tau) d\tau[/tex]
    so you have the variable of integration in the limits of integration.

    Since you only have t and T in Z, did you intend
    [tex]\int_t^T r(\tau) d\tau[/tex]?
    And, if so, is T a constant?

    If that is the case you need to use the chain rule, to get
    [tex]\frac{dZ}{dt}= (\int_t^T r(t,\tau)d\tau)e^{\int_t^T r(t,\tau)\dtau}[/tex]
    and Laplace's rule:
    [tex]\frac{d}{dt}\int_{a(t)}^{b(t)} F(t, \tau)d\tau= \int_{a(t)}^{b(t)} \frac{\partial F}{\partial t} d\tau+ \frac{db}{dt}F(t, b(t))- \frac{da}{dt}F(t,a(t))[/itex]
    to differentiate the integral.
    In this case, r is not a function of t so the first term is 0. The upper limit of integration is T, so the second term is 0. a= t so da/dt= 1. The derivative of the integral is just -r(t) which is basically what Pere Callahan was saying.
    You derivative is
    [tex]\frac{dZ}{dt}= -r(t)e^{\int_t^T r(\tau)d\tau[/itex]
     
  5. Apr 11, 2008 #4
    Thanks guys - yes, sorry, I had a typo in the integral - it should have been T as the upper limit of integration.
     
  6. Apr 11, 2008 #5
    Actually I think I was making this more complicated than it needed to be.

    [tex]
    Z(t;T) = exp^{-\int_{t}^{T}{r(\tau) d\tau}}
    [/tex]
    [tex]
    -\log{Z(t;T)}=\int_{t}^{T}{r(\tau) d\tau}
    [/tex]
    [tex]
    -\frac{\partial}{\partial{T}}=r(T)-r(t)
    [/tex]

    r(t)=0, so:

    [tex]
    r(T) = -\frac{\partial}{\partial{T}}\log{Z(t;T)}
    [/tex]

    Im still not sure why we are differentiating wrt T - I would have expected t. I suspect that this is due to a misunderstanding of the application of the FTOC in this case.
     
    Last edited: Apr 11, 2008
  7. Apr 11, 2008 #6

    exk

    User Avatar

    If I may ask, why is r(t)=0?
     
  8. Apr 11, 2008 #7

    exk

    User Avatar

    nm got it.
     
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