In footnote 3 of their paper "Patent Breadth, Patent Life, and the Pace of Technological Progress", O'Donoghue Scotchmer and Thisse take a discount rate of $r$ and write

If the flow of profit $\Delta$ lasts for length $t$, the discounted profit is $\Delta(1 − e^{−rt})/r$.

Can someone explain where the expression $\Delta(1 − e^{−rt})/r$ comes from?

Whilst I am familiar with continuous time discounting, I have never seen it formulated in this way before. I would usually expect the expression for discounted profit to be $\Delta e^{-rt}$.


From time $0$ to time $t$ the profit flow is constant $\Delta$. The discounted value of the profit flow of any instance $s$ is $\Delta \cdot e^{-r s}$. To get the total discounted profit we will need the integral of this from $0$ to $t$: $$ \int_0^t \Delta \cdot e^{-r s} \ ds = \left. \frac{\Delta \cdot e^{-r s}}{-r} \right]_0^t = \frac{\Delta \cdot e^{-r t}}{-r} - \frac{\Delta \cdot e^{-r \cdot 0}}{-r} $$ And as $e^{-r \cdot 0} = e^0 = 1$, this is $$ - \frac{\Delta \cdot e^{-r t}}{r} + \frac{\Delta \cdot 1}{r} = \Delta \cdot \frac{1 -e^{-r t}}{r}. $$

  • $\begingroup$ Yes, I had confused flows with stocks. Schoolboy error! thanks for for clearing it up. $\endgroup$ – Ubiquitous Jul 30 '15 at 15:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.