<p>You basically have the proof, just remember that for any function <span class="math-container">$f(x)$</span> strictly increasing, then <span class="math-container">$f(a)\leq f(b)$</span> if and only if <span class="math-container">$a\leq b$</span>. It is also true that for any strictly increasing function its inverse exists, but there is probably no need for that. </p>

CC BY-SA 4.0
2019-07-01T19:42:48.610Z
2019-07-01T19:42:48.610Z
Regio
<p>I would like to show for a randomly distributed variable <span class="math-container">$x$</span> with CDF <span class="math-container">$F(\cdot)$</span> , given a Bernoulli utility function <span class="math-container">$u(x)$</span> the following property holds:</p>

<p>The certainty equivalent, <span class="math-container">$CE(\cdot)$</span>, is smaller than the expected value, <span class="math-container">$\mathbb E(\cdot)$</span>, if and only if the decision maker is risk-averse. </p>

<p><span class="math-container">$$CE(F,u) \leq \int_{-\infty}^{\infty} x dF(x) \quad  \forall F(\cdot)$$</span> </p>

<p>By Jensens Inequality for risk-averse agents we have: 
<span class="math-container">$$\int_{-\infty}^{\infty} u(x) f(x)dx \leq u(\int_{-\infty}^{\infty} x f(x) dx)$$</span></p>

<p>I know that the CE is defined as: 
<span class="math-container">$u(CE) = \mathbb E(U(X)) = \int_{-\infty}^{\infty} f(x)u(x)\,dx$</span> </p>

<p>Hence, </p>

<p><span class="math-container">$$u(CE)  \leq u(\int_{-\infty}^{\infty} x f(x) dx)$$</span> </p>

<p>and</p>

<p><span class="math-container">$$CE \leq \int_{-\infty}^{\infty} x f(x) dx$$</span></p>

<p>Is it possible to kind of invert the u() function is we assume it is strictly increasing?</p>

<p>I also thought about rearranging it to: </p>

<p><span class="math-container">$$u(CE) = \int_{-\infty}^{\infty} f(x) dx * \int_{-\infty}^{\infty} u(x) dx$$</span></p>

<p>but I fail to see how that helps.</p>

<p>I thought perhaps showing that the risk premium is positive for risk-averse agents is equivalent but I could not get started either. </p>

<p>I would be glad for a hint to work out the solution myself.</p>

<microeconomics><utility><expected-utility><risk><risk-aversion>
Proof: Risk averse; Certainty Equivalent smaller than expected value
user333444
# Proof: Risk averse; Certainty Equivalent smaller than expected value

(Asked by: username_0 on 2019-07-01)

I would like to show for a randomly distributed variable $x$ with CDF $F(\cdot)$ , given a Bernoulli utility function $u(x)$ the following property holds:

The certainty equivalent, $CE(\cdot)$, is smaller than the expected value, $\mathbb E(\cdot)$, if and only if the decision maker is risk-averse.

$$CE(F,u) \leq \int\_{-\infty}^{\infty} x dF(x) \quad \forall F(\cdot)$$

By Jensens Inequality for risk-averse agents we have:
$$\int\_{-\infty}^{\infty} u(x) f(x)dx \leq u(\int\_{-\infty}^{\infty} x f(x) dx)$$

I know that the CE is defined as:
$u(CE) = \mathbb E(U(X)) = \int\_{-\infty}^{\infty} f(x)u(x)\,dx$

Hence,

$$u(CE) \leq u(\int\_{-\infty}^{\infty} x f(x) dx)$$

and

$$CE \leq \int\_{-\infty}^{\infty} x f(x) dx$$

Is it possible to kind of invert the u() function is we assume it is strictly increasing?

I also thought about rearranging it to:

$$u(CE) = \int\_{-\infty}^{\infty} f(x) dx \* \int\_{-\infty}^{\infty} u(x) dx$$

but I fail to see how that helps.

I thought perhaps showing that the risk premium is positive for risk-averse agents is equivalent but I could not get started either.

I would be glad for a hint to work out the solution myself.

**Tags:** microeconomics, utility, expected-utility, risk, risk-aversion

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## 1 answers

### Answer by username_1 on 2019-07-01. Score: 0 | Accepted answer

You basically have the proof, just remember that for any function $f(x)$ strictly increasing, then $f(a)\leq f(b)$ if and only if $a\leq b$. It is also true that for any strictly increasing function its inverse exists, but there is probably no need for that.


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