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How would you show that whenever additively represents an agent’s preferences, then so does any function ′ that differs from only by the choice of zero and unit.

If we assume that additively represents an agent’s preferences, so that for some subvalue functions 1, 2, ... , , (⟨1, 2, ... , ⟩) = 1(1) + 2(2) + ... + ().

and assume ′ differs from only by a different choice of unit and zero, which means that there are numbers > 0 and such that ′(⟨1, 2, ... , ⟩) = ⋅ (⟨1, 2, ... , ⟩)+.

How do we show that that there are subvalue functions1′,2′,...,′ such that ′(⟨1, 2, ... , ⟩) = 1′(1) + 2′(2) + ... + ′().

jayil
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1 Answers1

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Let $V(a_1,...,a_n)$ be an additive separable utility.

$$V(a_1,...,a_n)=\sum_iv_i(a_i)$$

Let $\alpha>0$ and $\beta\in \mathbb{R}$ be given. Let $U(a_1,..,a_n):=\alpha V(a_1,..,a_n)+\beta$. $U$ also represents the preferences. I claim that $U(a_1,..,a_n)$ also has an additive representation

$$U(a_1,...,a_n)=\sum_i\left(\alpha v_i(a_i)+\frac{\beta}{n}\right)$$

Define $u_i(a_i)=\alpha v_i(a_i)+\frac{\beta}{n}$

$$U(a_1,...,a_n)=\sum_{i}u_i(a_i)$$

Therefore, $U$ also is additively separable

VARulle
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  • are we able to show 1′,2′,...,′ such that ′(⟨1, 2, ... , ⟩) = 1′(1) + 2′(2) + ... + ′() – jayil Oct 27 '22 at 17:39
  • Yes. I simply used U instead of V'. In your original question, take $V_i'=xV_i+\frac{y}{n}$ –  Oct 27 '22 at 17:54