# Why do game theorists use a discounted payoff of this form?

Excuse the click-baity title. I notice the discounted payoff in the game theory literature usually takes the form

$$\sum_{t=1}^\infty\lambda(1-\lambda)^{t-1}R_t$$

This differs from the discounted payoff in the other dynamic optimization settings, e.g., see the Bellman equation in control theory.

Why the difference?

## 2 Answers

In my experience, it's mainly just for cleanliness for results.

Consider an infinite horizon repeated game, with discounted payoff representation (where I use $$\delta = (1-\lambda)$$ in your notation) $$(1-\delta)\sum_{t=0}^{\infty}\delta^t R_t$$ where $$0 < \delta < 1$$.

Suppose I play a strategy that gives me the same payoff, say $$a$$, for each period $$t$$. Then, $$(1-\delta)\sum_{t=0}^{\infty}\delta^t a = (1-\delta)\frac{a}{1-\delta} = a$$ which is cleaner than $$\frac{a}{1-\delta}$$ As a side note, multiplying the utility by a constant does not change preferences, so we keep the inherent preferences the same.

• In fact, arguably utility should be considered an equivalence class, and a particular formula simply an representative of that class. Two formulas in the same equivalence class are essentially the same utility. – Acccumulation Feb 28 '20 at 4:04

Are you referring to repeated games, in which player $$i$$ obtains a payoff of the form $$(1-\delta)\sum_{t=0}^{\infty}\delta^{t}u_{i}(x_{i}^{t})$$

from the sequence of payoffs $$\left\{x_{i}^{t}\right\}$$? The $$(1-\delta)$$ is appended to the front to give the average per period payoff.