Differentiating Integral Exponents?

  • Thread starter rwinston
  • Start date
36
0

Main Question or Discussion Point

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!
 

Answers and Replies

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

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.
 
HallsofIvy
Science Advisor
Homework Helper
41,738
897
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]
 
36
0
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?
Thanks guys - yes, sorry, I had a typo in the integral - it should have been T as the upper limit of integration.
 
36
0
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:
exk
119
0
If I may ask, why is r(t)=0?
 
exk
119
0
nm got it.
 

Related Threads for: Differentiating Integral Exponents?

Replies
17
Views
4K
  • Last Post
Replies
4
Views
5K
Replies
1
Views
7K
  • Last Post
Replies
2
Views
9K
  • Last Post
Replies
8
Views
3K
  • Last Post
Replies
7
Views
11K
  • Last Post
Replies
7
Views
3K
  • Last Post
5
Replies
100
Views
3K
Top