Average Discounted Payoffs in Continuous Time
While reading some theory papers that model a continuous time problem, I have noticed the way they represent the expression for average discounted payoff is of this form, where $ r> 0 $ is the discount rate,$$ r\int_0^\infty e^{-rs}u\left(x_s\right)ds $$ Can someone explain to me where does the $ r $ in this expression come from? I understand the use of $ e^{-rs} $ since it is the discount factor, but can’t understand why we also multiply by $ r $
For example see this paper https://elischolar.library.yale.edu/cgi/viewcontent.cgi?article=3011&context=cowles-discussion-paper-series#page=12. On page 10 (PDF page 12).
The $ r $ , or rather $ \frac{1}{1/r} $ , is the “average” part of the average discounted payoff. As you note, $ \mathrm e^{-rs} $ is the discount factor. Observe that $$ \begin{equation} \int_0^\infty \mathrm e^{rs},\mathrm ds = \frac1r. \end{equation} $$ If the payoff is $ p $ in every instant of time, the discounted payoff (with no averaging) is $ p/r $ . But if we want to get $ p $ as the average discounted payoff, we would weight the discount factor by $ \frac{1}{1/r} $ so that $$ \begin{equation} \int_0^\infty \frac{\mathrm e^{rs}}{1/r}p,\mathrm ds = r\int_0^\infty \mathrm e^{rs}p,\mathrm ds = p. \end{equation} $$
In the paper you referred to, the average discounted payoff on page 10 comes from: $$ \begin{equation} \int_t^\infty (1)\frac{\mathrm e^{-rs}}{1/r},\mathrm ds - \int_0^t \frac{\mathrm e^{-rs}}{1/r} c_i(u_{i,s}),\mathrm ds \end{equation} $$ where the discount factor is weighted by $ \frac{1}{1/r} $ .