# Fixed Income Mathematics

1. Oct 6, 2011

### bdw1386

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:

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

2. Relevant equations

N/A

3. The attempt at a solution

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

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

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:

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

What am I missing?

2. Oct 7, 2011

### process91

Your work here appears to be correct.

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