Fixed Income Mathematics

  • Thread starter bdw1386
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  • #1
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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.

Homework Statement



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]

Homework Equations



N/A

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?
 

Answers and Replies

  • #2
106
0
Your work here appears to be correct.
 

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