<p>No, <span class="math-container">$u_1(x_1-E[x_2])$</span> is agent <span class="math-container">$1$</span>'s <em>utility of the expected value</em>, not the <em>expected utility of the value</em>.</p>
<p>Generally <span class="math-container">$u(E(x)) \neq E(u(x))$</span>. A simple example:</p>
<p>Let <span class="math-container">$x$</span> take value <span class="math-container">$-1$</span> with probability <span class="math-container">$50\%$</span> and value <span class="math-container">$1$</span> with probability <span class="math-container">$50\%$</span>. Let <span class="math-container">$u(x) = x^2$</span>. Then
<span class="math-container">$$E(x) = 50\% \cdot (-1) + 50\% \cdot 1 = 0$$</span>
<span class="math-container">$$u(E(x)) = 0^2 = 0$$</span>
<span class="math-container">$$E(u(x)) = 50\% \cdot (-1)^2 + 50\% \cdot 1 = 1$$</span></p>
<p>There are special cases where <span class="math-container">$u(E(x)) = E(u(x))$</span> does hold, e.g., this is always true when <span class="math-container">$u$</span> is an affine function. The equality never holds when <span class="math-container">$u$</span> is strictly convex, see <a href="https://en.wikipedia.org/wiki/Jensen%27s_inequality" rel="noreferrer">Jensen's inequality</a>.</p>

CC BY-SA 4.0
2022-08-26T06:59:45.263Z
2022-08-26T06:59:45.263Z
Giskard
<p>Lets say Agent 1 has a utility function that depends on the other person, i.e.,  <span class="math-container">$u_1(x_1-x_2)$</span>, where <span class="math-container">$x_i$</span> is the choice of Agent <span class="math-container">$i$</span>. Suppose the expected value of <span class="math-container">$x_2$</span> is denoted <span class="math-container">$E[x_2]$</span>.</p>
<p>Can <span class="math-container">$u_1(x_1-E[x_2])$</span> be defined as Agent 1's expected utility from choosing some <span class="math-container">$x_1$</span>?</p>

<econometrics><expected-utility>
Expected value inside a utility function
Eli J
# Expected value inside a utility function

(Asked by: username_0 on 2022-08-26)

Lets say Agent 1 has a utility function that depends on the other person, i.e., $u\_1(x\_1-x\_2)$, where $x\_i$ is the choice of Agent $i$. Suppose the expected value of $x\_2$ is denoted $E[x\_2]$.

Can $u\_1(x\_1-E[x\_2])$ be defined as Agent 1's expected utility from choosing some $x\_1$?

**Tags:** econometrics, expected-utility


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## 1 answers

### Answer by username_1 on 2022-08-26. Score: 6 | Accepted answer

No, $u\_1(x\_1-E[x\_2])$ is agent $1$'s *utility of the expected value*, not the *expected utility of the value*.

Generally $u(E(x)) \neq E(u(x))$. A simple example:

Let $x$ take value $-1$ with probability $50\%$ and value $1$ with probability $50\%$. Let $u(x) = x^2$. Then
$$E(x) = 50\% \cdot (-1) + 50\% \cdot 1 = 0$$
$$u(E(x)) = 0^2 = 0$$
$$E(u(x)) = 50\% \cdot (-1)^2 + 50\% \cdot 1 = 1$$

There are special cases where $u(E(x)) = E(u(x))$ does hold, e.g., this is always true when $u$ is an affine function. The equality never holds when $u$ is strictly convex, see [Jensen's inequality](https://en.wikipedia.org/wiki/Jensen%27s_inequality).


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