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An example of the terminology that surrounds discounting that confuses me can be found in "Investment and Hysteresis" by Dixit:

Let future revenues be discounted at a positive rate $\rho > 0 \ldots$ Then, given a current level $R$ of revenues, the expected present value of the discounted future streams of revenues is $R / \rho$.

Hence, the net present value of the investment is quantified as $R/\rho - K$, where $K$ is the sunk cost.

What I would like to understand is what this formulation has to do with exponential discounting; in particular, where is the sum over the time periods (alluded to in the phrase "future revenue streams", but also made clear in the context of the paper)? Is this some kind of economist shorthand?

This matters because, as far as I can see, he uses the exact expression above for the NPV in the key derivation in the paper, which suggests to me that I've misunderstood his concept of "discounting".

(This notation is not unique to this paper: I've encountered this in several places; cf. Allcott and Greenstone "Is there an Energy Efficiency Gap".)

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Well this is 'exponential discounting'. An infinite sum of geometric series:

$$\sum R+ R\delta + R\delta^2.... R\delta^t = \frac{R}{1-\delta}, \text{ if } |\delta|<1$$

Now lets call the denominator rho $1-\delta= \rho$.

The exponential discounting is there since its an infinite sum of geometric series.

Edit: In response to Giskard's +1 comment I tried to look deeper into the paper and while I think that the Dixit is actually implying the above and just not being clear on terminology it is also possible that he means simply net present value of infinite constant stream as:

$$\sum \frac{R}{(1+\rho)^t}= \frac{R}{\rho}$$

However, I think the former interpretation is the right one because its more closer to how discount rates are treated in some other papers I have seen and I think he was just sloppy with terminology.

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    $\begingroup$ This might be the correct answer, but it seems strange to me that Dixit specifies "Let future revenues be discounted at a positive rate $\rho$". I.e. Dixit specifies the discount rate as $\rho$, not $\delta$. Is this normal? $\endgroup$ – Giskard Sep 1 at 11:27
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    $\begingroup$ @Giskard This is precisely the thing that confused me, and as I indicated in the question, it seems as though a lot of economics papers use this terminology. I guess it saves them from having to introduce a superfluous variable. Nevertheless, it would be helpful to call it the "lifetime" or "effective" discount rate, or something similar... $\endgroup$ – Anthony Sep 1 at 11:36
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    $\begingroup$ @Giskard I tried to address that in my answer. I see that while I was editing you actually already made an answer addressing that issue yourself I gave you +1. I think thats possible but I also dont think he meant to treat discount rate as interest rate $\endgroup$ – 1muflon1 Sep 1 at 11:50
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An answer (I am not sure this is the right one) is if the "positive rate $\rho$" refers to an interest rate. Sometimes interest rates are referred to as discount rates. In this case we would have a discount factor of $1/(1+ \rho)$, and the usual present value formula for a perpetual annuity yields

$$\frac{R}{1+ \rho} + \frac{R}{(1+ \rho)^2} + \frac{R}{(1+ \rho)^3} \dots = \frac{1}{1+ \rho}\frac{R}{1 - \frac{1}{1+ \rho}} = \frac{R}{\rho}.$$

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  • $\begingroup$ I assume you make the assumption here that $(1 + \rho) / \rho \approx 1/ \rho$? Then I believe that this answer is the right one since it would explain papers where two rates appear like so: $R / (\delta - \gamma)$ where $\delta$ is the discount rate and $\gamma$ is the rate of increase of the stream $R$, say (cf. Hassett and Metcalf, "Energy conservation investment"). The formulation in the above answer does not work here without a renaming of variables. The distinction is important I find, even if academic. $\endgroup$ – Anthony Sep 1 at 13:53
  • $\begingroup$ @Anthony I do not make that approximation. Why do you need it? $\endgroup$ – Giskard Sep 1 at 14:32
  • $\begingroup$ How do you from the second-last to the final equality then? wolframalpha.com/input/… $\endgroup$ – Anthony Sep 2 at 8:30
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    $\begingroup$ @Anthony My bad! The first equation did not hold in that form either, as I forgot to divide by $1 + \rho$, since we are starting the sequence from $R/(1 + \rho)$, not $R$. Edited, all equations should be correct now. $\endgroup$ – Giskard Sep 2 at 8:34
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I now have the full answer so I post it here for future reference (it is a generalisation of @Giskard's answer).

What Dixit (and others) mean is the following: $$ \int_0^\infty R e^{-\rho t}dt = \frac{R}{\rho}. $$ This is also why when the return is assumed to follow some kind of geomteric trend $\mu > 0$, one can write $$ \int_0^\infty (R_0 e^{\mu t}) e^{-\rho t} dt = \frac{R_0}{\rho - \mu}, $$ where the economically-natural condition $\rho > \mu$ is needed for convergence of the integral.

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