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: $u(CE) = \mathbb E(U(X)) = \int_{-\infty}^{\infty} f(x)u(x)\,dx$


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


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

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.


1 Answer 1


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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