<p>In my experience, it's mainly just for cleanliness for results. </p>

<p>Consider an infinite horizon repeated game, with discounted payoff representation (where I use <span class="math-container">$\delta = (1-\lambda)$</span> in your notation)
<span class="math-container">$$
(1-\delta)\sum_{t=0}^{\infty}\delta^t R_t
$$</span>
where <span class="math-container">$0 &lt; \delta &lt; 1$</span>. </p>

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

CC BY-SA 4.0
2020-02-28T04:04:53.440Z
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
CC BY-SA 4.0
2020-02-27T22:50:30.070Z
2020-02-27T22:50:30.070Z
Walrasian Auctioneer
<p>Excuse the click-baity title. I notice the discounted payoff in the game theory literature usually takes the form</p>

<p><span class="math-container">$$\sum_{t=1}^\infty\lambda(1-\lambda)^{t-1}R_t$$</span></p>

<p>This differs from the discounted payoff in the other dynamic optimization settings, e.g., see the <a href="https://en.wikipedia.org/wiki/Bellman_equation" rel="nofollow noreferrer">Bellman equation</a> in control theory.</p>

<p>Why the difference?</p>

<game-theory><dynamic-programming><optimal-control><bellman-equations><discount>
Why do game theorists use a discounted payoff of this form?
jonem
# Why do game theorists use a discounted payoff of this form?

(Asked by: username_0 on 2020-02-27)

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](https://en.wikipedia.org/wiki/Bellman_equation) in control theory.

Why the difference?

**Tags:** game-theory, dynamic-programming, optimal-control, bellman-equations, discount

---

## 1 answers

### Answer by username_1 on 2020-02-27. Score: 2 | Accepted answer

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.

> **Comments:**
> **username_2**: 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.  

---

