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bdw1386
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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.
Differentiate the following expression with respect to t:
<br /> 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)]<br />
N/A
Using the product rule and the FTC on both terms:
<br /> exp[\int_{0}^{t}d\tau \lambda(\tau)]\frac{dP(t)}{dt}+P(t)exp[\int_{0}^{t}d\tau \lambda(\tau)]\lambda(t)<br /> + <br /> R\int_0^tds(-\frac{dP(s)}{ds})exp[\int_0^sd\tau\lambda(\tau)](0)<br /> +<br /> R(-\frac{dP(t)}{dt})exp[\int_0^sd\tau\lambda(\tau)]<br />
The third term falls out, so we get:
<br /> exp[\int_{0}^{t}d\tau \lambda(\tau)]\frac{dP(t)}{dt}+P(t)exp[\int_{0}^{t}d\tau \lambda(\tau)]\lambda(t)<br /> +<br /> R(-\frac{dP(t)}{dt})exp[\int_0^sd\tau\lambda(\tau)]<br />
The textbook 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:
<br /> x=R\int_0^tds(-\frac{dP(s)}{ds})<br />
<br /> y=exp[\int_0^sd\tau\lambda(\tau)]<br />
<br /> \frac{d}{dt}[xy] = x\frac{dy}{dt}+y\frac{dx}{dt} = y\frac{dx}{dt}<br />
because \frac{dy}{dt} = 0.
What am I missing?
Homework Statement
Differentiate the following expression with respect to t:
<br /> 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)]<br />
Homework Equations
N/A
The Attempt at a Solution
Using the product rule and the FTC on both terms:
<br /> exp[\int_{0}^{t}d\tau \lambda(\tau)]\frac{dP(t)}{dt}+P(t)exp[\int_{0}^{t}d\tau \lambda(\tau)]\lambda(t)<br /> + <br /> R\int_0^tds(-\frac{dP(s)}{ds})exp[\int_0^sd\tau\lambda(\tau)](0)<br /> +<br /> R(-\frac{dP(t)}{dt})exp[\int_0^sd\tau\lambda(\tau)]<br />
The third term falls out, so we get:
<br /> exp[\int_{0}^{t}d\tau \lambda(\tau)]\frac{dP(t)}{dt}+P(t)exp[\int_{0}^{t}d\tau \lambda(\tau)]\lambda(t)<br /> +<br /> R(-\frac{dP(t)}{dt})exp[\int_0^sd\tau\lambda(\tau)]<br />
The textbook 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:
<br /> x=R\int_0^tds(-\frac{dP(s)}{ds})<br />
<br /> y=exp[\int_0^sd\tau\lambda(\tau)]<br />
<br /> \frac{d}{dt}[xy] = x\frac{dy}{dt}+y\frac{dx}{dt} = y\frac{dx}{dt}<br />
because \frac{dy}{dt} = 0.
What am I missing?